QCD at low energies

QCD at low energies
K.Goeke, WS 1999/2000
January 10, 2006
Contents
I Basic concepts of field theory 5
1 Lagrangeans and action 5
1.1 Hamilton principle . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Free massive boson field with spin S=0 (Klein-Gordon): . . . . . 6
1.3 Free massive fermion field with spin S=1/2 (Dirac): . . . . . . . 7
1.4 Free massles bosonic field of Spin=1 (Maxwell) . . . . . . . . . . 8
1.5 Fermion-Boson-coupling . . . . . . . . . . . . . . . . . . . . . . . 9
1.6 Bilinear covariants: . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.7 Parity, time reversal, charge conjugation . . . . . . . . . . . . . . 10
2 Canonical field quantization 11
2.1 General remarks: . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Quantized Klein-Gordon-Field: . . . . . . . . . . . . . . . . . . . 12
2.3 Quantized Dirac-field: . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4 Quantized Maxwell-field: . . . . . . . . . . . . . . . . . . . . . . . 16
3 Symmetries and currents 18
3.1 Elements of Lie-Group theory . . . . . . . . . . . . . . . . . . . . 18
3.1.1 Lie-Groups SU(N) . . . . . . . . . . . . . . . . . . . . . . 18
3.1.2 Representations . . . . . . . . . . . . . . . . . . . . . . . . 19
3.1.3 Gell-Mann matrices . . . . . . . . . . . . . . . . . . . . . 20
3.2 Noether Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2.1 General structure . . . . . . . . . . . . . . . . . . . . . . . 21
3.2.2 Noether-Theorem and Lie-groups . . . . . . . . . . . . . . 23
3.2.3 Alternative expression for several fields, Energy momentum
tensor . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.3 Current algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.4.1 Free massless fermions, chirality, helicity . . . . . . . . . . 28
3.4.2 Free massive Dirac field, Fermion-number, U(1)-transformation 31
1

3.4.3 Free massless Dirac field, Axial U A (1)-transformation . . 32
3.5 Isospin symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.6 Linear (chiral) Sigma-model . . . . . . . . . . . . . . . . . . . . . 35
4 Symmetry breaking 39
4.1 Realization of symmetries . . . . . . . . . . . . . . . . . . . . . . 39
4.2 Principle of Spontaneous symmetry breaking . . . . . . . . . . . 40
4.3 Spontaneous symmetry breaking in the linear sigma model . . . . 43
4.3.1 The Goldstone Mode . . . . . . . . . . . . . . . . . . . . . 45
4.3.2 Broken chiral symmetry and PCAC . . . . . . . . . . . . 47
5 The gauge principle 50
5.1 Abelian gauge theory: . . . . . . . . . . . . . . . . . . . . . . . . 50
5.2 Non-abelian gauge theory . . . . . . . . . . . . . . . . . . . . . . 55
6 The QCD-Lagrangean 61
6.1 The gauge transformation . . . . . . . . . . . . . . . . . . . . . . 61
6.2 Strong coupling constant . . . . . . . . . . . . . . . . . . . . . . . 63
6.3 Mandelstam variables . . . . . . . . . . . . . . . . . . . . . . . . 68
7 Symmetries and anomalies 70
7.1 Mass terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
7.2 Vector symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . 72
7.2.1 Global fermion symmetry . . . . . . . . . . . . . . . . . . 72
7.2.2 Global (iso-)vectorial symmetry . . . . . . . . . . . . . . . 73
7.3 Axial (non-)symmetries: . . . . . . . . . . . . . . . . . . . . . . . 75
7.3.1 Flavour symmetry and U A (1) anomaly . . . . . . . . . . . 75
7.3.2 Axial Vector symmetry and Axial Anomaly . . . . . . . . 76
7.4 Other symmetries, Theta-vacuum . . . . . . . . . . . . . . . . . . 78
7.4.1 Discrete symmetries . . . . . . . . . . . . . . . . . . . . . 78
7.4.2 Scale invariance and trace-anomaly . . . . . . . . . . . . . 78
7.4.3 Theta-Vacuum . . . . . . . . . . . . . . . . . . . . . . . . 82
8 Anomalies 85
9 Electroweak Interactions in the Standard Model 85
9.1 Electromagnetic Quark Currents . . . . . . . . . . . . . . . . . . 86
9.2 Weak Quark Currents . . . . . . . . . . . . . . . . . . . . . . . . 88
9.3 Leptonic currents . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
2

10 Chiral symmetry breaking 90
10.1 Chiral Symmetry and Current algebras . . . . . . . . . . . . . . . 90
10.2 Spontaneous symmetry breaking . . . . . . . . . . . . . . . . . . 97
10.3 Pion decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
10.4 PCAC: partial conservation of axial current . . . . . . . . . . . . 103
10.5 The chiral condensate . . . . . . . . . . . . . . . . . . . . . . . . 105
10.6 LSZ-Reduction formulae . . . . . . . . . . . . . . . . . . . . . . . 107
10.7 Pion-Nucleon coupling constant . . . . . . . . . . . . . . . . . . . 108
10.8 Goldberger-Treiman relation and pion pole . . . . . . . . . . . . 110
10.9 Vacuum Sigma Term and Gell-Mann–Okubo . . . . . . . . . . . 112
10.10Pion-Nucleon Sigma Term Σ πN . . . . . . . . . . . . . . . . . . . 115
10.11Ward-Identities and Low energy pion-nucleon theorems . . . . . 123
10.12Isospin-Violation . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
11 Group theory, Flavour structure and Quark model 128
11.1 Elements of Lie-group theory . . . . . . . . . . . . . . . . . . . . 129
11.2 SU(2)-algebra: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
11.3 SU(2)-multiplets in nature . . . . . . . . . . . . . . . . . . . . . . 133
11.4 SU(3)-algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
11.5 SU(3)-multiplets in nature . . . . . . . . . . . . . . . . . . . . . . 136
11.6 Quarks and SU(3) . . . . . . . . . . . . . . . . . . . . . . . . . . 139
11.6.1 Quarks and fractional charges . . . . . . . . . . . . . . . . 139
11.6.2 Construction of higher multiplets by triplets, Confinement
(phenomenological) . . . . . . . . . . . . . . . . . . . . . . 141
11.7 Fock states and non-relativistic quark model . . . . . . . . . . . . 147
11.7.1 Mass splittings . . . . . . . . . . . . . . . . . . . . . . . . 152
11.7.2 Quark model: Calculations . . . . . . . . . . . . . . . . . 153
12 Effective bosonic Lagrangean 155
12.1 Review: Spontaneous breakdown of chiral symmetry . . . . . . . 156
12.1.1 Current algebra and transformation properties . . . . . . 156
12.1.2 Chiral Quark Condensate . . . . . . . . . . . . . . . . . . 156
12.2 Representations of the chiral boson fields . . . . . . . . . . . . . 157
12.2.1 Linear Sigma Model . . . . . . . . . . . . . . . . . . . . . 157
12.2.2 Non-linear Sigma Model . . . . . . . . . . . . . . . . . . . 158
12.2.3 Generalization to finite Goldstone Boson mass . . . . . . 159
12.3 QCD and chiral fields . . . . . . . . . . . . . . . . . . . . . . . . 161
12.4 The minimal effective bosonic Lagrange-density . . . . . . . . . . 163
12.4.1 Massless case . . . . . . . . . . . . . . . . . . . . . . . . . 163
12.4.2 Massive case . . . . . . . . . . . . . . . . . . . . . . . . . 166
12.4.3 Higher order terms . . . . . . . . . . . . . . . . . . . . . . 168
12.5 Application: pion-pion scattering . . . . . . . . . . . . . . . . . . 174
12.5.1 Calculation: . . . . . . . . . . . . . . . . . . . . . . . . . . 174
12.5.2 Comparison with experiment . . . . . . . . . . . . . . . . 177
12.6 Chiral perturbation theory and QCD: Functional integrals . . . . 180
12.6.1 Internal symmetries and Ward-Identities . . . . . . . . . . 180
3

12.6.2 External fields and Greens functions . . . . . . . . . . . . 183
12.6.3 Local chiral symmetry . . . . . . . . . . . . . . . . . . . . 186
12.7 Application: Pion decay . . . . . . . . . . . . . . . . . . . . . . . 189
12.8 The chiral Lagrange-Density in ord(p 4 ) or ord(E 4 ) and renormalization.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
12.8.1 The Lagrangean . . . . . . . . . . . . . . . . . . . . . . . 191
12.8.2 Chiral perturbation theory, renormalization program . . . 192
12.9 Application in order(p 4 ): Masses of Goldstone bosons . . . . . . 194
12.9.1 Aim of this section . . . . . . . . . . . . . . . . . . . . . . 194
12.9.2 Calculation of the pole mass . . . . . . . . . . . . . . . . 195
12.9.3 Calculation of the self energy . . . . . . . . . . . . . . . . 198
13 Lattice Gauge Theory 202
13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
13.2 Quantum Mechanics: Transition amplitudes and path integrals . 203
13.2.1 Free Motion (Propagator) . . . . . . . . . . . . . . . . . . 205
13.2.2 Propagators ( Harmonic oscillator) . . . . . . . . . . . . . 209
13.2.3 Extraction of information, Euklidean time . . . . . . . . . 212
13.2.4 Correlators . . . . . . . . . . . . . . . . . . . . . . . . . . 214
13.2.5 Partition function . . . . . . . . . . . . . . . . . . . . . . 217
13.2.6 Metropolis Formalism . . . . . . . . . . . . . . . . . . . . 218
13.3 Boson Quantum Field on the lattice . . . . . . . . . . . . . . . . 222
13.3.1 Quantum Field Theory with functional integrals . . . . . 222
13.3.2 Euklidean Field Theory . . . . . . . . . . . . . . . . . . . 225
13.3.3 Scalar boson field: Discretization of space-time . . . . . . 228
13.4 Fermion Quantum fields on the lattice . . . . . . . . . . . . . . . 233
13.4.1 Grassmann-algebra . . . . . . . . . . . . . . . . . . . . . . 234
13.4.2 Fermionic path integral . . . . . . . . . . . . . . . . . . . 235
13.4.3 Wilson fermions, quenched lattice. etc. . . . . . . . . . . . 237
13.5 Gauge Fields on the Lattice . . . . . . . . . . . . . . . . . . . . . 238
13.5.1 Abelian gauge fields: QED . . . . . . . . . . . . . . . . . 238
13.5.2 Non-abelian gauge fields on the lattice: QCD . . . . . . . 243
13.5.3 Continuum limit . . . . . . . . . . . . . . . . . . . . . . . 244
13.6 Practical applications of lattice QCD . . . . . . . . . . . . . . . . 247
13.6.1 Metropolis algorithm in pure gauge theory . . . . . . . . . 247
13.6.2 Pure gauge calculations . . . . . . . . . . . . . . . . . . . 248
13.6.3 Problems with Fermions . . . . . . . . . . . . . . . . . . . 254
14 Linear chiral Sigma model (Gell-Mann–Levy) 255
14.0.4 Fock states and variational principle . . . . . . . . . . . . 255
14.0.5 Projection techniques . . . . . . . . . . . . . . . . . . . . 257
4

15 Restbestände 259
Part I
Basic concepts of field theory
1 Lagrangeans and action
1.1 Hamilton principle
The dynamics of classical interacting fields
φ i (x) :::::::: i = 1,.....,N
with x = (t,x) is determined often by the Lagrangean density L:
L = L(φ i (x),∂ µ φ i (x)) (1)
This sort of Lagrangean is very common in connection with elementary particle
theories, i.e. QED, QCD, Glashow-Salam-Weinberg theories and simple (linear)
theories of strong interaction. Effective theories, of baryon-meson interaction
etc. and field theories for many body systems are often more complicated and
show higher derivatives. We will not consider them at the moment but later.
The action of the fields is defined as:
∫ ∫
S = dtL = d 4 xL(φ i (x),∂ µ φ i (x))
The Lagrange density (we refer to it simply as lagrangean) is an expression,
where φ i (x) and ∂ µ φ i (x) are considered as independent variables. It is just a
mathematical expression, which has to fulfil certain general conditions. The
actual motion of fields, i.e. evolution in time, is determined by the equations
of motion and by initial conditions. By Hamiltons principle the equations of
motion are fixed by the Euler-Lagrange equations of the variation principle
without subsidiary conditions:
δS = 0
The explicit variation of φ i (x) and ∂ µ φ i (x) is performed at a given but arbitrary
space-time point x by varying the value of the field φ keeping the x fixed. The
equations of motion describe how a field , given at the time t 1 in the whole
space x evolves to a later time t 2 in the whole space. The theory of variation
tells that one has to perform the variation (for simplicity in case of only one)
field in the following way
∫
δS = d 4 x[L(φ + δφ,∂ µ φ + δ∂ µ φ) − L(φ,∂ µ φ)]
5

∫
=
∫
=
∫
=
[ ∂L
d 4 x
∂φ δφ + ∂L ]
∂(∂ µ φ) δ(∂ µφ)
[ ∂L
d 4 x
∂φ δφ + ∂L ]
∂(∂ µ φ) ∂ µ(δφ)
[ ( ∂L ∂L
d 4 x
∂φ δφ − ∂ µ
∂(∂ µ φ)
)
δφ + ∂ µ
( ∂L
∂(∂ µ φ) δφ )]
The last term can be turned into a surface integral over the boundary of the
four-dimensional space-time region of integrateion. Since the initial and final
field configuration φ(t 1 ,x) and φ(t 1 ,x) are assumed given the δφ vanishes at
the temporal beginning t 1 and temporal end t 2 in the whole integration region.
This lets the surface integral vanish: We obtain
∫ [ ]
∂L
δS = d 4 x
∂φ − ∂ ∂L
µ δφ
∂(∂ µ φ)
Since the δφ is besides its boundary conditions at t 1 ,t 2 an arbitrary variation,
the contents of the bracket must be zero. This yields the equation of motion:
∂ µ
∂L
∂(∂ µ φ i ) − ∂L = 0 i = 1,...,N (2)
∂φ i
If one does a snapshot of the field at the time t in the whole space, then the
equations of motion (2) describe how the field propagates in space and time.
For the quantization of the fields one needs always the conjugate momentum
field
π i (x) = ∂L
∂ 0 φ i
(x) (3)
There are several basic sorts of classical Lagrangeans, which are used in
elementary particle physics. These Lagrangeans describe fields, which we meet
in nature.We will generally use = c = 1
1.2 Free massive boson field with spin S=0 (Klein-Gordon):
This is the simplest field. The famous higgs-field, or the scalar σ-field of the
nucleon-nucleon interaction, or a neutral pion field is e.g. a free massive scalar
boson field. After quantization (see later) the corresponding field particles have
a finite mass and the spin zero and move freely without interaction. (In reality
one is of course interested in the interaction of the particles with others. This
will be discussed later): The Lagrangean of such a field, if it is not interacting
with anything else, is given by
L = 1 (
∂λ φ∂ λ φ − µ 2 φ 2) (4)
2
The corresponding equation of motion is the famous Klein-Gordon equation:
(
∂ 2 + µ 2) φ = 0 (5)
with ∂ 2 = ∂ λ ∂ λ = ∂ 2 0 − ∇ 2 .
6

1.3 Free massive fermion field with spin S=1/2 (Dirac):
The corresponding field quanta are Fermions with a mass m and Spin= 1 2 .The
massive fermion field is described in the simplest form by the Lagrangean
L = ψ [iγ µ ∂ µ − m]ψ (6)
We have by definition
¯ψ(x) = ψ † (x)γ 0
The corresponding equation of motion is the famous Dirac equation:
[iγ µ ∂ µ − m]ψ = 0 (7)
Fermions have Spin=1/2, in a non-relativistic theory they have to be described
by a two-component spinor. Since we are dealing in particle physics with relativistic
systems the description with 2 components is not sufficient, but one
needs at least a 4-component spinor (Dirac spinor)
ψ(x) =
⎛
⎜
⎝
ψ 1 (x)
ψ 2 (x)
ψ 3 (x)
ψ 4 (x)
The γ µ are 4x4-matrices. There are various equaivalent ways, one can express
them. Usually the γ-matrices are expressed as generalizations of the Pauli-
Spin-matrices. Explicit forms are given in books on Dirac-theory (convention:
µ,ν = 0,1,2,3 and i,j = 1,2,3:
( ) ( )
γ 0 I 0
=
γ i 0 σ
i
=
0 −I
−σ i (8)
0
One shoud note the commutation rules
and the hermiticity properties:
⎞
⎟
⎠
{γ µ ,γ ν } = 2g µν (9)
(
γ
0 ) †
= γ
0
(
γ
i ) †
= −γ
i
(10)
The Pauli spin matrices are given as
( ) (
σ 1 0 1
=
σ 2 0 −i
=
1 0
i 0
One often needs
γ 5 = iγ 0 γ 1 γ 2 γ 3 =
γ 0 γ µ γ 0 = (γ µ ) † (11)
( 0 I
I 0
)
)
σ 3 =
(
1 0
0 −1
(
γ
5 ) †
= γ
5
)
(12)
(13)
7

with
{
γ 5 ,γ ν} = 0 (14)
and furthermore
σ µν = i 2 [γµ ,γ ν ] (15)
Trγ µ = Trγ 5 = 0
One often needs the following formulae: From the Dirac equation we obtain
γ µ ∂ µ ψ = −imψ (16)
We get for the hermitian conjugate with eq.(11) and multiplying from left with
γ 0 :
(iγ µ ∂ µ ψ) † = −i∂ µ ψ † (γ µ ) † = −i∂ µ ψ † γ 0 γ 0 (γ µ ) † = −i∂ µ ¯ψγ0 (γ µ ) †
or
−i∂ µ ¯ψγ0 (γ µ ) † γ 0 = mψ † γ 0
∂ µ ¯ψγ µ = im ¯ψ (17)
and
and similarly we obtain from iγ µ ∂ µ ψ = mψ+ γ µ A µ
Here we must put in final 12
γ µ ∂ µ ψ = −imψ − iγ µ A µ ψ (18)
∂ µ ¯ψγ µ = im ¯ψ + i ¯ψγ µ A µ (19)
1.4 Free massles bosonic field of Spin=1 (Maxwell)
The typical example of a free massless boson field of spin=1 is the electromagnetic
field. The corresponding field quanta are the photons. The Lagrangean of
the free field is
L = − 1 4 F µνF µν (20)
and it is expressed by the antisymmetric field tensor, which is directly related
to the classical electric and magnetic fields:
⎛
⎞
0 E x E y E z
F µν (x) = ⎜ −E x 0 B z −B y
⎟
⎝ −E y −B z 0 B x
⎠ (21)
−E x B y −B x 0
1 put here the free Dirac states consistent with the quantized Dirac field
2 Here we must display plane wave states consistently with quantization of Dirac field
8

The F µν is related to the 4-potential A µ by:
F µν (x) = ∂ µ A ν (x) − ∂ ν A µ (x) (22)
The electric and magnetic fields E and B have as sources charges and currents.
We write them as
j µ (x) = (ρ(x),j(x))
and then we can note the Lagrangean of the classical electromagnetic field coupled
to the charge current as
L = − 1 4 F µνF µν − j µ A µ (23)
The equations of motion an be derived using (2) and they are the famous
Maxwell equations in covariant form:
∂ ν F µν (x) = j µ (x) inhom.eq. (24)
∂ λ F µν (x) + ∂ µ F νλ (x) + ∂ ν F λµ (x) = 0 hom.eq. (25)
If one writes these equations explicitely in terms of E and B one obtains the
usual Maxwell equations of classical electrodynamics in the non-covariant form.
1.5 Fermion-Boson-coupling
The free field lagrangeans are not very interesting because they are interacting
with nothing. The most popular interaction, which occurs e.g. in all elementary
gauge fields, is the interaction of a fermion field with a boson field. If we take
the simplest case of scalar boson field coupled minimally to the fermion field we
have
L = ψ [iγ µ ∂ µ − m]ψ + 1 (
∂λ σ∂ λ σ − µ 2 σ 2) − gψσψ
2
One can write down the equations of motion immediately:
∂ µ
∂L
∂(∂ µ ψ) − ∂L
∂ψ = 0 ⇒ [iγµ ∂ µ − m]ψ = gσψ
∂ µ
∂L
∂(∂ µ σ) − ∂L
∂σ = 0 ⇒ (
∂ 2 + µ 2) σ = gψψ
Apparently the bosonic field σ modifies the equation of motion of the fermion,
such that this is no longer a free fermion. Vice versa the fermion field in the
form of a scalar current gψψ serves as a source for the boson scalar field σ. We
will meet this simple structure very often.
9

1.6 Bilinear covariants:
In the previous subsection we had a very simple coupling of the fermion field
to a boson field. A realistic coupling of a fermion field with charge q to the
electromagnetic field is given by the Lagrangean:
with the electromagnetic current
L = ψ [iγ µ ∂ µ − m]ψ − 1 4 F µνF µν − j µ A µ (26)
j µ = −ieψγ µ ψ
Apparently here the source for the A-field is of the form gψγ µ ψ, which is a
vector-current. Actually there are more complicated currents known, which
have other 4-tensorial characters. The structure of the field and the structure
of the source must be consistent, since the source generates the bosonic field
and the total Lagrangean incorporating the coupling must be a scalar function.
We have the following covariant bilinears, which form a complete set and which
could serve as sources for the corresponding bosonic fields:
Bilinear Covariant (27)
field
ψψ
ψγ µ ψ
ψγ 5 ψ
ψγ µ γ 5 ψ
ψσ µν ψ
name
S(x) scalar
J µ (x) vector
P(x) pseudoscalar
(x) pseudo-vector
T µν (x) tensor
J µ 5
Covariance means, that these bilinears have in all inertial systems the same
structure. I.e. if you perform a Lorentz-transformation from IS to IS’
x ′ µ = Λ ν µx ν (28)
then e.g. the 4-vector (ψγ µ ψ) ρσ in IS gets transformed into (ψγ µ ψ) ′ ρσ =
(Λ ν µψγ ν ψ) ρσ in IS’. The ρσ are indices of the matrixelements in the Gammamatrices.
1.7 Parity, time reversal, charge conjugation
There are some fundamental properties of space time, which play an important
role in physics and field theory:
Parity:
x = (x 0 ,x) → x P = (x 0 , −x)
Time reversal: x = (x 0 ,x) → x T = (−x 0 ,x)
The effect of these transformation on a fermion field ψ(x) will be implemented
by a unitary operator P for parity and an antiunitary operator T for
time reversal. We have
Pψ(x)P −1 = γ 0 ψ(x P ) ::::::::::::::::: Tψ(x)T −1 = iγ 1 γ 3 ψ(x T )
10

There is a third transformation, which maps matter into antimatter, this is
Charge conjugation:
Cψ(x)C −1 = iγ 2 γ 0 ψ(x) T
The effects on a scalar boson field are
Pφ(x)P −1 = φ(x P )
Tφ(x)T −1 = φ(x T )
The effects on a photon field are
Cφ(x)C −1 = φ ∗ (x)
PA µ (x)P −1 = A µ (x P )
TA µ (x)T −1 = A µ (x T )
CA µ (x)C −1 = −A µ (x)
The response of the fermion bilinears to these transformations can be easily
calculated and is given in the table
Discrete Transformations (29)
C P T
S(x) S(x P ) S(x T )
P(x) −P(x P ) −P(x T )
−j µ (x) J µ (x P ) j µ (x T )
j µ 5 (x) −j 5µ(x P ) j 5µ (x T )
−T µν (x) T µν (x P ) −T µν (x T )
(Attention: moving of indices µ,ν because {γ µ ,γ ν } = 2g µν )
The Standard Model involving electromagnetic, weak and colour-gauge fields,
is invariant under CPT = 1. For the following it will be important to note the
behaviour of those covariants.
2 Canonical field quantization
2.1 General remarks:
The fundamental degrees of freedom are the interacting fields. In the above section
the fields are assumed to be classical. In analogy to simple non-relativistic
quantum mechanics we proceed now to quantize the fields. This step, which is
11

here only sketched shortly, is necessary if one wants to describe modern experiments
in the field of particle and radiation physics. We review her the canonical
quantization process. It consists in quantizing the free massive fermion and
boson fields and in treating interaction terms by means of perturbation theory
lieading to Feynman-diagrams. There is also the quantization procedure of interacting
fields by means of path integrals, which will primarily be used later in
quantizing interacting non-abelian fields.
The canonically quantized fields give rise to the definition of physical particles.
These particles can be encountered in detectors by “click” or a flash of
of light. The particles have angular momentum, parity etc. which are characteristics
corresponding to the ones of the original fields and the corresponding
Lagrangean and equation of motion. As we will see, the quantization process
takes in a natural way into account, that the particles are identical and that we
have fermions (antisymmetrized) and bosons (symmetrized). The mass appearing
in the Lagrangean is (after renormalization etc.) closely related to the mass
of the freely propagating particle.
The quantization of non-relativistic quantum mechanics consisted in the
following recipe: Consider a Lagrangean L(q i (t), ˙q(t)). Calculate the canonical
momentum p i (t) canonical to q i (t) by
p i (t) =
∂L
∂ ˙q i (t)
and promote the canonical momentum to an operator ̂p i (t) with the demand
[
q i (t), ̂p
]
j (t) = iδ ij
The quantization takes place for both quantities at the same time t.
The quantization of fields is similar, but more complicated. In particular
we have the situation, that the different sort of fields (Klein-Gordon, Dirac,
Maxwell) require different quantization presriptions, because their field quanta
are bosons and fermions and photons, repectively..
2.2 Quantized Klein-Gordon-Field:
Take the above Klein-Gordon Lagrangean density eq.(4)
L = 1 2
(
∂λ φ∂ λ φ − µ 2 φ 2)
and calculate the conjugate momentum
π(x) =
∂L
∂(∂ 0 φ)
⇒
π(x) = ∂ 0 φ(x)
Then demand the equal time commutation rules
[
]
φ i (t,x),π j (t,x ′ ) = iδ (3)
ij (x − ) (30)
x′
12

[
] [
]
φ i (t,x),φ j (t,x ′ ) = π i (t,x),π j (t,x ′ ) = 0
Actually this demand is sufficient for a complete quantization. Mostly one
formulates this in another way using the plane wave solutions of the free Klein-
Gordon equation as a basis to expand the fields. (This is convenient, but by
no means necessary, one also could use an expansion into partial waves.). Then
one has for the classical field
∫
d 3 k
]
φ(t,x) = √
[a(k)exp(i(kx − ω k t) + a(k) ∗ exp(−i(kx − ω k t)
(2π)3 2ω k
∫
π(t,x) = −i
√
d 3 ωk
[
]
k
(2π) 3 a(k)exp(i(kx − ω k t) − a(k) ∗ exp(−i(kx − ω k t)
2
with ω k = + √ k 2 + µ 2 , such that the φ(t,x) fulfills the Klein Gordon equation
(5) The demand to the above expressions, equivalent in both directions, is
realized by promoting the coefficients a(k)and a ∗ (k) to operators, i.e.to creation
and annihilation operators of the field quanta. Thus we demand the well known
commutator rules of bosonic single particles
[ ]
a(k),a † (k ′ ) = δ (3) (k − k ′ ) (31)
[ ] [ ]
a(k),a(k ′ ) = a † (k),a † (k ′ ) = 0
By this the fields φ and π become operators as well (we do not indicate this by
a ”head”):
∫
φ(t,x) =
d 3 k
]
√
[a(k)exp(−ikx + a(k) † exp(+ikx)
(2π)3 2ω k
(32)
∫
π(t,x) = −i
√
d 3 ωk
[
]
k
(2π) 3 a(k)exp(−ikx) − a(k) † exp(+ikx)
2
(33)
with k = (k 0 = ω k ,k). By construction the quantized fields (32) and (33)
fulfill the commutation rules of the scalar boson fields (30). It is now clear
how one proceeds in general. Any quantity, depending on the classical fields
φ, π becomes now an operator by promoting the φ, π to operator fields. Hence
in principle the full theory is now quantized. Calculations become much more
complicated.
There are several important further developements in the quantized field
theory, which all have their analogies in classical field theory where commutators
are replaced by Poisson-brackets. We define the operator of the hamiltonian
density
H(x) = π(x,t)∂ 0 φ(x,t) − L
13

and the hamiltonian operator
∫
H(t) =
d 3 xH(x)
One obtains by easy calculations the following features
i[H(t),π(x,t)] = ∂ 0 π(x,t)
i[H(t),φ(x,t)] = ∂ 0 φ(x,t)
One defines also the momentum operator
∫
P k (t) = d 3 x [ π(x,t)∂ k φ(x,t) + φ(x,t)∂ 0 π(x,t) ]
with the properties
i [ P k (t),π(x,t) ] = ∂ k π(x,t)
i [ P k (t),φ(x,t) ] = ∂ k φ(x,t)
The proofs of these equations are simple: One inserts the quantized fields, uses
the commutator rules (30), gets zeros and delta-funtions, performs the integrals
over space and gets the above formulae. If one combines H and P k to one
4-Vector operator P µ and considers a general operator F which is a function of
the fields and depends through them on (x,t) then one can write
An often used feature is
i[P µ (t),F(x,t)] = ∂ µ F(x,t)
F(x) = exp(iPx)F(0)exp(−iPx) (34)
F(0) = exp(−iPx)F(x)exp(iPx)
If one takes matrix elements of such an operation between many body states,
generated by the a(k) † , of given 4-momentum p and p ′ one obtains the often
used formula
〈p ′ |F(x) |p〉 = exp(i(p ′ − p)x)) 〈p ′ |F(0) |p〉 (35)
One needs in the following the (bare) Feynman-Propagator of the Klein-
Gordon field. It is defined as
with the time ordering operator given by
i∆ F (x) = 〈0| T {φ(x)φ † (0) |0〉 (36)
T {φ(x)φ † (y)} = θ(x 0 − y 0 )φ(x)φ † (y) + θ(y 0 − x 0 )φ † (y)φ(x) (37)
The expression for the Feynman propagator is
∫
d 4 k exp(−ikx)
∆ F (x) =
(2π) 4 k 2 − m 2 + iε
(38)
14

2.3 Quantized Dirac-field:
The quantization of the Dirac-field is conceptually similar to the one of the
Klein-Gordon field. We again calculate the spinor field π(t,x) conjugate to the
spinor ψ(t,x) using (3) applied to the Lagrangean of the free Dirac field (6).
The conjugate fermion momentum appears to be
π(t,x) = iψ † (t,x)
Technically the quantization prescription of the fermion fields (Dirac fields) is
different from the one of the boson fields (Klein Gordon fields) because the
corresponding field quanta are fermions rather than bosons. We demand
{
}
ψ i (t,x),π j (t,x ′ )
= iδ (3)
ij (x − x′ )
{
} {
}
ψ i (t,x),ψ j (t,x ′ ) = π i (t,x),π j (t,x ′ ) = 0
where the first equation can also be replaced by
{
ψi (t,x),ψ † (t,x) } = δ (3)
ij (x − x′ )
The indices i,j are Dirac indices of the spinor commponents. The expansion
of the quantized fields in terms of the plane wave solutions u(k) and v(k) of
the free Dirac equation (7) with mass m reads with k = (k 0 = ω k ,k) and
ω k = + √ k 2 + m 2 :
ψ(t,x) = ∑ ∫
d 3 √
k m
[
]
√ c r (k)u r (k) exp(−ikx) + d †
(2π)
3 ω
r(k)v r (k)exp(+ikx)
k
r=1,2
(39)
with the commutaion rules for the fermionic creation and annihilation operators
for particles c and anti-particles d:
{ }
c r (k),c † s(k ′ ) = δ rs δ (3) (k − k ′ ) (40)
{ } { }
c r (k),c s (k ′ ) = c † r(k),c † s(k ′ ) = 0
The analogous expressions hold for the operators d,d † of the antifermions. { In }
addition any anticommutator between a c,c † and a d,d † vanishes, e.g. c r (k),d † s(k ′ ) =
0
One needs in the following the (bare) Feynman-Propagator of the Dirac field.
It is defined as
iS F (x) = 〈0|T {ψ(x) ¯ψ(0) |0〉
with the time ordering operator given [with a different sign compared to (37)]
by
T {ψ(x) ¯ψ(y)} = θ(x 0 − y 0 )ψ(x) ¯ψ(y) − θ(y 0 − x 0 ) ¯ψ(y)ψ(x) (41)
15

and it is given by
or
S F (x) = (iγ µ ∂ µ + m) ∆ F (x)
∫
S F (x) = d 4 k
exp(−ikx)
γ µ k µ − m + iε
(42)
2.4 Quantized Maxwell-field:
In principle the quantization of the Maxwell-field proceeds similarly to the one
of the Klein-Gordon- and the Dirac-field. There is, however, a fundamental
difference.
If we take the Lagrangean (20) of the free Maxwell field and calculate the
conjugate momentum field (3), we obtain
π µ (x) =
∂L
∂(∂ 0 A µ ) = −F µ0 (x)
This yields π 0 (x) = −F 00 (x) = 0, which does not allow to get a quantization
presription for µ = 0. On the other hand one likes to preserve Lorentz-covariance
so that one cannot distinguish a particular Lorentz-indices µ. Thus one must
proceed differently. A Lagrangean, which is suitable for the quantization procedure
has been suggested by Fermi and is
Here we get
L = − 1 2 (∂ µA ν (x))(∂ ν A µ (x)) − j µ (x)A ν (x)
π µ (x) =
∂L
∂(∂ 0 A µ ) = −∂ 0A µ (x)
This can be quantized, demanding
[
A µ (t,x), A ˙
]
ν (t,x ′ ) = −ig µν δ (3) (x − x ′ )
. The problem, however, is now, that the equations of motion of this Lagrangean
do not give the Maxwell equations. There appear additional terms which are
only vanishing in the Laurentz-Gauge:
∂ µ A µ (x) = 0 (43)
This means, the equations of motion reduce to Maxwell equations only in a
special gauge, which contradicts gauge invariance.
All these problems have been solved in a quantization formalism invented by
Gupta and Bleuler. In this theory one quantizes the photon field (i.e. Maxwell
field) analogous to eq.(30) leading then to the quantized photon field operator
A µ (x) =
3∑
∫
r=0
d 3 k [
√ ε
µ
r (k)a r (k)exp(−ikx + ε µ r(k)a(k) † exp(+ikx) ]
(2π)3 2ω k
16

with the polarization vectors given by
ε µ 0 (k) = nµ = (1,0,0,0) ε µ r(k) = (0, ε r (k) for r = 1,2,3
and ε r (k) being a orthogonal dreibein. In covarianter form one can write
and
One can write in a covariant way
ε µ r(k)ε µr (k) = −ζ r δ rs
∑
ζ r ε µ r(k)ε ν r(k) = −g µν
r
ζ 0 = −1 ζ 1 = ζ 2 = ζ 3 = +1
ε µ 3 (k) = kµ − (kn)n µ
[(kn) 2 − k 2 ] 1/2
and we call ε µ 1 (k),εµ 2 (k) transversal, εµ 3 (k) longitudial and εµ 0 (k) skalar or timelike.edvia
plane waves to creation and annihilation operators for photons with
the following commutator rules:
[ ]
a r (k),a † s(k ′ ) = δ rs δ (3) (k − k ′ ) (44)
[ ]
a r (k),a s (k ′ ) = [ a † r(k),a † s(k ′ ) ] = 0
for the four spinor components of A µ µ = 0,1,2,3. and the four polarization
components of the phon field r,s = 0,1,2,3. However the Lorentz-condition (43
leads to a additional constraint for physical (measurable) photon states |Ψ〉:
[a 3 (k) − a 0 (k)] |Ψ〉 = 0 for all k
Das hat zur Folge, daß reale physikalische Photonen weder Longitudinalkomponente
noch eine Zeitkomponente besitzen und transversal sind.
The Gupta-Bleuler formalim has been formulated for abelian gauge theories.
In principle one can generalize it to non-abelian ones, however, nowadays one
prefers path-integral quantization procedures which can equally well be applied
to abelian and non-abelian gauge fields like the gluon-field or the fields of the
elctro-weak bosons.
One needs in the following the (bare) Feynman-Propagator of the Maxwell
field. It is defined analogous to eqs.(37,38) as
iD µν
F (x) = 〈0|T {Aµ (x)A ν (0) |0〉
and it is given by (in the limit of boson mass in the bosonic propagator going
to zero)
D µν
F
(x) = − lim
m→0 gµν ∆ F (x,m)
or
∫
D µν
F (x) = −gµν d 4 k exp(−ikx)
k 2 (45)
+ iε
17

3 Symmetries and currents
3.1 Elements of Lie-Group theory
3.1.1 Lie-Groups SU(N)
Symmetries play a tremendous role in elementary particle field theory and
hadronic physics, particularly those associated with Lie-Groups . Thus we discuss
in this section some basic properties of Lie-groups. Details can be found in
group theory books (e.g. F. Stancu). We consider the following transformation
applied to an abstract Hilbert vector |Ψ〉:
with
|Ψ〉 → |Ψ ′ 〉 = U(θ a ) |Ψ〉
U(θ a ) = exp(−iθ a F a ) (46)
The θ a are real numbers, a = 1,2,...,N 2 − 1. The operators U(θ a ) are
members of a group, i.e. the Lie-group SU(N), if the operators F a are hermitian
and are the generators of the Lie-group SU(N). By definition the F a fulfill a
closed algebra obbeying the following commutation rules
[
F a ,F b] = if abc F c (47)
with the normalization
Tr(F a F b ) = 1 2 δab
The numbers f abc are called structure functions of the Lie-group. They are
totally antisymmetric ( i.e. f abc = −f bac = −f acb = −f cba etc.) and determine
the group properties uniquely. For the SU(N) group one has
f abc f abd = Nδ cd
In SU(2) we jave f abc = ε abc . The anticommutator of the generators defines the
symmetric coefficients d abc with d abc = d bac = d acb = d cba etc:
{
F a ,F b } = 1 3 Iδab + d abc F c
The Lie-Group SU(N) can be shown to be of rank r = N − 1. This means
that among the generators F a there exist r = N −1 generators, which commute
with each other. Each Lie-group SU(N) has N −1 Casimir-operators C i i =
1,2,...N − 1, which are defined by the property that they commute with each
of the generators and hence with each of the group members:
[C i ,F a ] = 0 i = 1,2,...N − 1 a = 1,....,N
18

The Casimir operators and the commuting generators have common eigenvectors.
They are grouped in multiplets |c 1 ,...c N−1 ,α 1 ,....,α r 〉 .The multiplets are
characterized by the eigenvalues c i of the Casimir operators C i , and all eigenvectors
within a multiplet are characterized by r = N − 1 quantum numbers of
the commuting generators. The multiplets can be explicitely constructed from
the commutator rules of SU(N).
3.1.2 Representations
In the following we need representations of a Lie-group. One needs this representation
if one represents the Hilbert vector |Ψ〉 by a n-tupel with n components
in some basis (n-spinor). A representation of a group is a mapping of the group
element U(θ a ) on a unitariy n · n matrix U(θ a ):
U(θ a ) ↦→ U(θ a )
where the mapping is such that the group properties are fulfilled. That means
the representation of a product of two operators equals the product of the two
representation. Furthermore we have
UU † = 1 and detU = 1
Usually the representation of a generator F a is called 1 2 λa . The matrices 1 2 λa
are traceless and fulfill the following commutator rules
[ λ
a
2 , λb
2
]
abc λc
= if
2
(48)
with normalization
The anticommutator of the λ-matrices fulfills:
{
λa
2 , λ }
b
= 1 2 3 δab + d abc λ c
2
Tr(λ a λ b ) = 2δ ab (49)
(50)
The representation of the operator (46) is given by the N · N matrix U(θ a )
with
U(θ) = exp(−i λa θ a
2 ) (51)
For SU(N) with N > 1 we have a non-abelian group with
U(θ 1 )U(θ 2 ) ≠ U(θ 2 )U(θ 1 )
There exists a fundamental representation of SU(N). It is the smallest nontrivial
representation and is formed by N-spinors, which transform by the N ·Nmatrix
ψ α −→ Uβ α ψ β α,β = 1,..,N
19

Complex conjugate N-spinors (which should be presented by a row) transform
by the hermitian-conjugate matrix
ψ † α −→ ψ † β U †β
α
α,β = 1,..,N
where the following conventions hold: ψ α † = (ψ α ) † and U α
†β
clear that the combination ψ αψ † α is an SU(N) invariant
= ( U †) β
. It is
α
ψ † αψ α −→ ψ † β U †β
α U α γ ψ γ = ψ † αψ α = ψ † ψ
There are also quantities transforming according to the (N 2 − 1)-dimensional
adjoint represntation:
A a −→ O ab A b
with
O ab = 1 2 Tr ( U † λ a Uλ b) and O ab O ac = δ bc
To verify the lawst eq. you need the Fiertz identity
Fiertz identity
( ) λ
a α ( λ
a
β
2 2
) γ
δ
= − 1
2N δα βδ γ δ + 1 2 δα δ δ γ β
One can show that the following two comintions are invariants of the SU(N)
transformations
A a A a = inv A a ψ † λ a ψ=inv
3.1.3 Gell-Mann matrices
The representations of SU(N) are not unique. The most used ones for SU(2)
are the Pauli-matrices (12). For SU(3) we have the Gell-Mann matrices:
The Gell-Mann-matrices are given
⎛
λ 1 = ⎝
⎛
λ 4 = ⎝
⎛
λ 7 = ⎝
0 1 0
1 0 0
0 0 0
0 0 1
0 0 0
1 0 0
0 0 0
0 0 −i
0 i 0
⎞
Gell-Mann-Matrices (52)
⎛
⎠ λ 2 = ⎝
⎞
⎛
⎠ λ 5 = ⎝
⎞
⎠
λ 8 = 1 √
3
⎛
⎝
0 −i 0
i 0 0
0 0 0
0 0 −i
0 0 0
i 0 0
⎞
⎛
⎠ λ 3 = ⎝
⎞
1 0 0
0 1 0
0 0 −2
⎛
⎠ λ 6 = ⎝
1 0 0
0 −1 0
0 0 0
0 0 0
0 0 1
0 1 0
The structure coefficients f abc are totally antisymmetric, i.e if one interchanges
two indices one obtains a minus-sign:
⎞
⎠
Antisymmetric Coefficients (53)
⎞
⎠
⎞
⎠
20

f 123 = 1
f 147 = −f 156 = f 246 = f 257 = f 345 = −f 367 = 1 2
f 458 = f 678 = 1 2√
3
All other f abc either obtained by a transposition of two arbitrary indices ( for
each transpposition you have a minus-sign) pre they are Zero. The symmetric
coefficients d abc are given by
Symmetric coefficients (54)
d 118 = 1 √
3
,d 146 = 1 2 ,d 157 = 1 2 ,d 228 = 1 √
3
d 247 = − 1 2 ,d 256 = 1 2 ,d 338 = √ 1 ,d 344 = 1 3 2
d 355 = 1 2 ,d 366 = − 1 2 ,d 377 = − 1 2 ,d 448 = − 1
2 √ 3
d 558 = − 1
2 √ 3 ,d 668 = − 1
2 √ 3 ,d 778 = − 1
2 √ 3 ,d 888 = −√ 1 3
All other d abc are obtained by a tranpositon of two arbitrary indices (for each
transposition you have a plus-sign) or are Zero. The SU(3)-Lie-group has the
rank 2, because you find maximally 2 generators, which commute with each
other, as one can see at the Gell-Mann Matrices where λ 3 and λ 8 are diagonal.
This corresponds to the operator relations:
[F 3 ,F 8 ] = 0 ⇒ F 3 , F 8 : can be diagonalized simultaneously
3.2 Noether Theorem
3.2.1 General structure
For any field theory we have an important theorem: Noether Theorem:
Noether Theorem: To every continuous symmetry of a Langrangean there
exists a conserved current
More precisely stated: Consider the
Lagrangean (1), i.e. L = L(φ i (x),∂ µ φ i (x)) We assume that L depends only
on φ and ∂ µ φ, d.h. NOT on ∂ µ ∂ ν φ :::::: ∂ 2 φ ::::etc. We can have basically any
power of ∂ µ φ, however, if we want a renormalizable theory we are restricted
to the second power. There are also non-renormalizable theories and they also
have conserved currents and they can also be derived in the way shown right
now:
21

In detail the Noether-theorem reads in the simplest case of one field as
follows: The L be invariant under an infinitesimal change δφ i (x), i.e.
L(φ + δφ) = L(φ)
In fact one needs less, namely that only the action S = ∫ d 3 xL is invariant.
If the symmetry is continuous this implies the existence of a conserved current
J µ (x)
∂ µ j µ (x) = 0
with the charge defined by
which is a constant of motion:
Q(t) = const
∫
Q(t) = d 3 xj 0 (x)
or
dQ(t)
dt
Actually the reasoning for the last equation is trivial, because the surface term
at infinity is negligeably small. On always assumes that at infinity the fields are
vanishing such that any current vanishes. This is not correct for a simple plane
wave, which is of course an idealization, but it is always correct for a physically
generated field with a source of finite size.:
dQ(t)
dt
= 0
∫
∫ ∫ ∫
= ∂ 0 d 3 xj 0 (x) = −∂ i d 3 xj i (x) = d 3 x∇j(x) = dfj(x) = 0
since the integral goes along the surface at infinity, where the fields composing
the current j are vanishing.
The proof of the Noether theorem is simple: Consider the Lagrangean under
the transformation
φ → φ + δφ
this means
or
δL = L(φ + δφ) − L(φ)
δL = ∂L
∂φ δφ + ∂L
∂(∂ µ φ) δ(∂ µφ)
we can rewrite the first term by the equation of motion eq.(2) such that
( ) ∂L
δL = ∂ µ
∂(∂ µ φ) δφ
or
with the current
j µ (x) =
∂ µ j µ (x) = δL (55)
∂L
δφ(x) (56)
∂(∂ µ φ(x))
22

There is often factor −i or +i, which is purely by convention, it is often omitted.
This means, if the Lagrangean is invariant under the transformation, we have
∂ µ j µ (x) = 0 (57)
This is the general structure of the Noether-Theorem. The current gives rise
to the definition of the charge, which is time independent if the lagragean is
invariant under the transformation. If not, the charge is time dependent.
Since we have
∫
Q(t) =
d 3 xj 0 (x) :::::::::: dQ(t)
dt
π(x) =
∂L
∂(∂ 0 φ)
the charge can be rewritten as
∫
Q(t) = d 3 xπ(x,t)δφ(x,t)
= 0 (58)
3.2.2 Noether-Theorem and Lie-groups
One mostly uses the Noether Theorem if several fields are mixed by the transformation
and the Lagrangean remains unchanged. This we will discuss now:
Assume the Lagrangean invariant under the transformation
φ i → φ i + δφ i
i = 1,...,n
with
δφ i (x) = iɛ a t a ijφ j (x)
and t a with a = 1,2,....,N are n-dimensional representations of the generators of
the SU(N) Lie-group.This is the most often used case because the Lagrangeans
of the elementary forces whe have in nature show several of those symmetries and
the interaction of elementary particles with external fields can also be formulated
in terms of those Noether currents.The fact that we work with a Lie-group means
that the t a fulfill the commutation rules (48) We can derive easily the structure
of the current and also its conservation if the Lagrangean is invariant under the
above transformation: We have the variation
δL = ∂L δφ i +
∂L
∂φ i ∂(∂ µ φ i ) δ(∂ µφ i )
We can use now the equation of motion eq(2) to modify the first term on RHS
∂L
∂φ i
yielding
∂L
δL =∂ µ
∂(∂ µ φ i ) δφ + ∂L
∂(∂ µ φ i ) ∂ µ(δφ i )
23

Since the ɛ a are arbitrary we have
( ) ∂L
δL = ɛ a ∂ µ
∂(∂ µ φ i ) ita ijφ j
Defining the Noether-current for each a as
jµ(x) a δL
= −i
δ(∂ µ φ i (x)) ta ijφ j (x) (59)
one can write analogously to eq(55)
δL = ɛ a ∂ µ j a µ(x)
or, if the Lagrangean is invariant, we have analogously to eq.(57):
∂ µ j a µ(x) = 0
a = 1,2,....,N
On can define the charges, which are in general time dependent. If the Lagrangean
is invariant under the considered Lie-group the charges are time independent:
∫
Q a (t) = d 3 xj0(x)
a
Since we have
dQ a (t)
dt
π k (x) =
= 0 a = 1,2,....,N (60)
∂L
∂(∂ 0 φ k )
the charge operator can be rewritten as
∫
Q q (t) = −i d 3 xπ k (x,t)t a kjφ j (x,t) (61)
3.2.3 Alternative expression for several fields, Energy momentum
tensor
Actually there is an equivalent and sometimes more convenient way to calculate
the current of a Lagrangean under a symmetry transformation (without proof):
This consists in performing an infinitesimal field transformation, and assuming
the ɛ artificially and only for the moment to be a function of x, this beeing only
a technical trick in order to get another and often more convenient derivation
of the current. Hence one considers the infinitesimal transformation
̂φ i (x) = φ i (x) + ɛ a (x)F a i (φ j (x))
24

The F is a function of the set φ ⃗ i . In the restiriction to constant ɛ the lagrangean
becomes invariant. For an internal symmetry the Noether current is then obtained
by
j a(x) µ ∂
=
∂(∂ µ ɛ a (x)) L( ̂φ i ,∂ µ ̂φi )
One also obtains immediately
∂ µ j µ a(x) =
∂L
∂ɛ a (x)
In practice one simply expands the Lagrangean
L( ̂φ i (x), ̂ ∂µ φ i (x)) = L(φ i (x),∂ µ φ i (x)) + ∂ µ ɛ a (x)F µ a (x)
performs the derivative and can read off the current.
Proof: Under the above transformation we can use the following formula
δL = ∂L δφ i +
∂L
∂φ i ∂(∂ µ φ i ) δ(∂ µφ i )
= − ∂L ɛ a Fi a −
∂L
∂φ i ∂ (∂ µ φ i ) [(∂ µɛ a ) Fi a + ɛ a (∂ µ Fi a )]
Using the Euler Eqs. of motion we have
( ) ∂L
δL = −∂ µ ɛ a Fi a −
∂L
∂(∂ µ φ i ) ∂(∂ µ φ i ) [(∂ µɛ a ) Fi a + ɛ a (∂ µ Fi a )]
yielding
( ∂L
δL = −ɛ a ∂ µ
∂(∂ µ φ i ) F i
a
Defining the current as
j µa (x) = −
∂L
∂(∂ µ φ i ) F i
a
) ( ) ∂L
− (∂ µ ɛ a )
∂(∂ µ φ i ) F i
a
associated with the above infinitesimal transformation with the x-dependent
epsilon we have
δL = = ɛ a ∂ µ j µa + (∂ µ ɛ a )j µa
From this follows that the current may also be defined as
j µa =
which is equivalent to eq.(59). and furthermore
∂
∂ (∂ µ ε a δL (62)
)
∂ µ j µa = ∂ δL (63)
∂εa 25

Without proof, which one finds in any textbook, we quote the concept of
energy momentum tensor. This is defined as
T µν =
∂L
∂(∂ µ φ i ) ∂ν φ i − g µν L (64)
For a system, which is invariant under a shift in space and time, i.e. timeindependent
and translationally invariant, we haave the following conservation:
∂ µ T µν = 0 (65)
The various components have a clear meaning: The 00-component equals to the
hamiltonian of the system: ∫
H = d 3 xT 00 (66)
And the other 0-components set up the operators of the momentum, altogether
a 4-momentum:
∫
P µ = d 3 xT µ0 (67)
Actually we know an important property of P µ which was given alrady in
eq.(34). There it was derived in the context of a free boson field. Here we
see,however, that it is also valid for a general Lagrangean:
3.3 Current algebras
F(x) = exp(iPx)F(0)exp(−iPx)
We consider a = 1,....N generators t a ij of the Lie-group G, which means that
the t a fulfill the commutation rules of eq.(48). Following the previous section,
we have also a = 1,...,N charges Q a (t), see eq.(60). By now the Noether
theorem was completely on the classical level. If we quantize the fields φ i (x) in
a canonical way then the charges get quantized as well and become operators.
From nowon we treat them as operators, omitting, however, any indication of
such a promotion. The assertion is that the quantized charge operators follow
the algebra of the Lie-group considered and hence may be used themselves as
generators of the group fulfilling of course eq.(48):
[
Q a (t) ,Q b (t) ] = if abc Q c (t) (68)
The proof is simple, and we demonstate it for a boson field: The quantization
is well defined and requires only the knowledge of the field π(x) conjugate to φ(x)
and the usual quantization procedure. We have (now all fields are operators)
eq.(61) from which we can write
[
Q a (t),Q b (t) ] ∫
=
d 3 xd 3 y [ π i (x)t a ijφ j (x),π k (y)t b kmφ m (y) ] x 0 =y 0
26

with x 0 = y 0 .With the operator identity [AB,CD] = A[B,C]D − C [D,A] B
(valid if[A,C] = [D,B] = 0) we have now
[
Q a (t),Q b (t) ] ∫
= − d 3 xd 3 yπ i (x,t)t a [
ij φj (x,t),π k (y,t) ] t b kmφ m (y,t)
∫
+
d 3 xd 3 yπ k (x,t)t b [
km πj (x,t),φ m (y,t) ] t a ijφ j (y,t)
∫
= −
∫
= −
d 3 xπ k (x,t)i [ t a ,t b] kj φ j(x,t)
d 3 xπ k (x,t)φ j (x,t)f abc t c kj
Witlh the definition Q q (t) given in eq. (61) we get
[
Q a (t),Q b (t) ] = if abc Q c (t) qed
Thus the charge operators generate the changes of the fields like the generators
do: A simple proof using [AB,C] = A[B,C] + [A,C] B yields with the
same techniques as above:
[Q a (t),φ k (x,t)] = −t a kjφ j (x,t)
[Q a (t),π k (x,t)] = −t a kjπ j (x,t)
This means, if there is only one field, one immediately determines the charge of
a field by calculating the commutator of the field with the charge operator.
He above algebra independent whether the Lagrangean is invariant under
the considered Lie-group transformation or not. However, if the Lagrangean
is invariant the charges commute with the Hamiltonian of the system.This is
obvious, however it can be shown explicitely with the above techniques.
The Hamiltonian is defined
H = ∑ i
π i (x)∂ 0 φ(x) − L(φ i (x),∂ µ φ i (x))
giving (without proof) for all a
[H,Q a (t)] = 0
One can generalize the charge algebras with the same and simple operator
techniques to current algebras. One obtains using the above techniques (without
explicit but rather simple proof)
[
Q a (t),j b 0(x,t) ] = if abc j c 0(x,t)
or more generally [
Q a (t),j b µ(x,t) ] = if abc j c µ(x,t) (69)
27

and [
j
a
0 (x,t),j b 0(y,t) ] = if abc j c 0(x,t)δ (3) (x − y) (70)
However be careful: The analogy for space like componends is not valid due to
the so called “Schwinger term”:
[
j
a
0 (x,t),j b i (y,t) ] = if abc j c i (x,t)δ (3) (x − y) + S ab
ij (x) ∂
∂y j
δ (3) (x − y)
The Schwinger term vanishes upon integration over the whole space. By this it
does not modify the charge algebra.
3.4 Examples
3.4.1 Free massless fermions, chirality, helicity
We a consider a system of free massles fermions. Let ψ be the solution of this
massles field
[iγ µ ∂ µ ]ψ = 0
We multiply this euation from the left with γ 5 and use the {γ 5 ,γ µ } = 0 anticommutator
rules to obtain another solution:
[iγ µ ∂ µ ]γ 5 ψ = 0
We superimpose these solutions to form by construction the combinations of
definite “chirality” (watch in books the definition of the γ 5 , they often differ in
sign and hence the definitions of right- and left-handed look different. Here we
use the definition of Bjorken-Drell, Peskin-Schroeder, Izykson-Zuber)
ψ R = 1 2 (1 + γ 5)ψ ψ L = 1 2 (1 − γ 5)ψ (71)
In the massles case the chirality of a free particle is a Lorentz-invariant
concept. For example, particle which is left handed to one observer will appear
left handed to all observers.
On the Lagrangean level we can write
or
L = ψ R [iγ µ ∂ µ ]ψ R + ψ L [iγ µ ∂ µ ]ψ L (72)
with
L = L R + L L
L R = ψ R [iγ µ ∂ µ ]ψ R
L L = ψ L [iγ µ ∂ µ ]ψ L
28

These Lagrange densities are both invariant under the global chiral phase transformations
ψ R (x) → ψ ′ R(x) = exp(−iα R )ψ R (x)
¯ψR (x) → ¯ψ ′ R(x) = exp(+iα R )ψ R (x)
ψ L (x) → ψ ′ L(x) = exp(−iα L )ψ L (x)
¯ψR (x) → ¯ψ ′ R(x) = exp(+iα R )ψ R (x)
where the phase is real-valued and arbitrary. The associated Noether currents
for the present massless free lagrangean are
j µ R (x) = ¯ψ R (x)γ µ ψ R (x) ∂ µ j µ R (x) = 0
j µ L (x) = ¯ψ L (x)γ µ ψ L (x) ∂ µ j µ L (x) = 0
For massless free fermions these chiral currents are conserved. We can construct
chiral charges
∫
Q R,L (t) = d 3 xjR,L(x)
0
From the chiral currents we can construct the vector and axial vector currents
by linear combinations:
j µ (x) = j µ R + jµ L
vector
j µ 5 (x) = jµ R − jµ L
axial vector
. We also can define vector and axial vector charges
∫
Q(t) = d 3 j 0 (x)
∫
Q 5 (t) =
d 3 j 0 5(x)
The above properties hold if the considered fermions are massless or perhaps
coupled to a field which does not mix the left-handed with the right handed
fermion fields. Actually if we have a mass term in the Lagrangean represented
by a real mass or a scalar field σ(x) then immediately right- and left-handed
components are mixed and the separate right- and left-handed currents are no
longer conserved:
L = ψ R [iγ µ ∂ µ ]ψ R + ψ L [iγ µ ∂ µ ]ψ L − m(ψ L ψ R + ψ R ψ L ) (73)
In this lagrangean we have no more invariance separately under SU(2)-L and
SU(2)-R and the above charge operators are time dependent. One might think
that the latter case is the realistic one and the one with massless fermions only
29

an academic construction since in nature there are no massless fermions. In fact
the massless case is important since in e.g. high energy lepton scattering one
has the situation that the momentum transfer Q 2 of the exchanged photon to
the quarks is very much larger than the mass of the quarks, i.e. Q 2 >> m 2 .In
such a case the mass of the quark is negligible and the quarks can be treated
massless.
There is another interesting difference between the massive and the massless
case. This concerns the concept of both “helicity” and “chirality”. Generally
a particle (with or without mass) with definite helicity is a particle, whose
spin and momentum are aligned. This can always be achieved by chosing the
momentum direction as axis of quantization. Positive helicity means then →⇒
and negative helicity means →⇐. In general a massive particle with positive
helicity does not have definite chirality and vice versa. For a massive particle
one can always find a Lorentz-transformation into a inertial frame where the
helicity is positive and into another frame where the helicity is negative. If we
have a particle with mass, then the construction of ψ R does not yield a state
with definite helicity, since helicity depends on the fram whereas chirality does
not Only in the case of a massless particle, which has no rest frame and which
always moves with speed of light the helicity is a lorentz invariant quantity.
For a massless particle positive helicity implies right handedness and vice versa:
This means, if we have a massles particle with definite momentum →then for
ψ R the spin and momentum are aligned and for ψ L they are anti-aligned.
massless: ψ R right-handed →⇒ positive helicity
massless: ψ L left-handed →⇐ negative helicity
In the massless limit the concept of chirality is a Lorentz invariant concept and
hence it is a natural label to be used for massless fermions and a collection of
such particles may be characterized by the number of left-handed and righthanded
particles.
The proof, that in the massless limit helicity and chirality are identical is
simple: Consider the Dirac eqation for a massless quark and a plane wave
solution in 3-direction
Plane wave in 3-direction yields
p 2 = m 2 = 0 = (p 0 ) 2 − (p 1 ) 2 − (p 2 ) 2 − (p 3 ) 2
p 1 = p 2 = 0 :::::⇒:::: p 0 = p 3 = p
ψ = exp (−ipx) ψ 0 ⇒ γ µ p µ ψ = 0 ⇒ p(γ 0 − γ 3 )ψ = 0 ::::⇒:::: γ 0 ψ = γ 3 ψ
hence with
J 3 = 1 2 σ12 = i 4
[
γ 1 ,γ 2] = i 2 γ1 γ 2
30

J 3 ψ L = i 2 γ1 γ 2 ψ L = i 2 γ0 γ 0 γ 1 γ 2 ψ L = i 2 γ0 γ 1 γ 2 γ 0 ψ L = i 2 γ0 γ 1 γ 2 γ 3 ψ L
where we have used
γ 0 ψ = γ 3 ψ,γ 5 = i γ 1 γ 2 γ 3 , ( γ 5) 2
= I, {γ µ ,γ ν } = 2g µν , {γ 1 ,γ 2 } = 0
Reshuffling gives
J 3 ψ L = + 1 2 γ5 ψ L = − 1 1
2 2 (1 − γ 5)ψ = − 1 2 ψ L
J 3 ψ R = + 1 2 ψ R :::::::::: qed
3.4.2 Free massive Dirac field, Fermion-number, U(1)-transformation
Consider the lagrangean for the free massive fermion field (6). Apparently this
Lagrangean is invariant under the “global U(1)-transformation” with:
ψ(x) → ψ ′ (x) = exp(−iα)ψ(x) (74)
ψ(x) → ψ ′ (x) = ψ(x) exp(+iα)
Here α is a real constant. The proof is trivial. We can immediately calculate
the corresponding Noether-current (since δψ = −iαψ) using eq.(56):
yielding
j µ = ∂L 0
∂(∂ µ ψ) δψ
j µ = ψγ µ ψ (75)
The α is a free number, hence it is omitted in the expression for the current j.
The Noether theorem says
∂ µ j µ (x) = 0
This current agrees with the vector current V µ = j µ R + jµ L
. Apparently the
divergence of the current vanishes for massless as well as massive fermions.
Apparently we have
j 0 (x) = ψγ 0 ψ = ψ † ψ
This quantity is positive definite. If one interpretes ψ as a particle, then we
have here the particle density and hence can use the probability interpretation.
31

3.4.3 Free massless Dirac field, Axial U A (1)-transformation
We consider the free massles fermion lagrangean eq.(6) with m = 0. We consider
the axial U A (1)-transformation
ψ(x) → ψ ′ (x) = exp(−iαγ 5 )ψ(x) (76)
corresponding to (attention: sign in the exponent does not change due to anticommutation
rules (9) of the γ-matrices)
ψ(x) → ψ ′ (x) = ψ(x) exp(−iαγ 5 ) (77)
Wit the anti-commutation rules of the γ-matrices (9) one can show by direct
calculation, that the free massles fermion lagrangean is invariant under the
above transformation:
L ′ 0 = ψ ′ [iγ µ ∂ µ ]ψ ′ = L 0 = ψ [iγ µ ∂ µ ]ψ
and the corresponding Noether current is the axial current (with :
with
j µ 5 = ψγµ γ 5 ψ (78)
∂ µ j µ 5 (x) = 0
This current agrees with the axial vector current j µ 5 = jµ R − jµ L
. It is interesting
to consider axial transformations in the case of a massive fermion field eq.(6):
L 0 = ψ [iγ µ ∂ µ − m]ψ
If we now perform the transformation eq.(76,77) we see that the Lagrangean is
no longer invariant:
L ′ = ψ [iγ µ ∂ µ ]ψ − mψ exp(−2iαγ 5 )ψ
. When we calculate the current, using eq.(56) we see of course that it is identical
to the one calculated with m = 0, since the definition of the current (56) uses
only terms coming from the kinetic energy of the Lagrangean. Howvever if we
calculate the divergence of the current we see from eq.(55) that it is of course
no longer zero:
∂ µ j µ A (x) = ∂ µψγ µ γ 5 ψ = i2mψγ 5 ψ (79)
Hence: A mass term destroys the invariance of the free fermion field under the
U A (1)-transformation. Remember the mass term did not destroy the invariance
of the free fermion field under the U(1)-transformation.
32

3.5 Isospin symmetry
There is a simple example of an effective model, which shows SU(2) isospin
symmetry, which we are going to meet often and particulaly in QCD. We will
practice some useful techniques here. The example considered is the nucleonpion
system. Consider a doublet of nucleon fields each being a 4-spinor field
( )
p(x)
ψ(x) =
n(x)
and a triplet of pion fields π a .......a = 1,2,3 with the lagrangean
L = ψ [iγ µ ∂ µ − m]ψ + 1 (
∂λ π a ∂ λ π a − m 2
2
ππ a π a) − gψiτ a π a γ 5 ψ − λ 4 (πa π a ) 2
(80)
with τ a being the Pauli-isospin matrices and the mass matrix m is given by the
nucleon masses M N : ( )
MN 0
m =
0 M N
This Lagrangean describes a bare nucleon (i.e. proton and neutron described
by the Dirac fields p(x) and n(x)) interacting with a pion field in the most
simple way. The M prot = 938 MeV and the neutron is about two MeV heavier.
Thus it is a good approximation to neglect the mass difference between proton
and neutron, which would be an isospin violating effect. The Lagrangean is
to be understood purely classical. This means, that ψ describes nucleons with
positive energy (i.e. valence nucleons), there does not exist a Dirac-sea, the
vacuum is hence described by ψ = 0, the boson fields do not show any quantum
fluctuation. For any value of M N the Lagrangean is invariant under the global
SU(2) transformation (rotation in isospin space) of the fields
ψ → ψ ′ = Uψ ::::::::::: U(α) = exp(−i τa α a
2 ) (81)
ψ → ψ ′ = ψU †
::::::::::: U † (α) = exp(+i τa α a
2 )
where the are the SU(2) isospin matrices or the Pauli matrices (12). We have
UU †
= U † U = I. This unitarity can be shown explicitely by performing a
Taylor expansion. We show it to second order:
UU † =
(
1 − i τa α a
2
= 1−i τa α a
2 +iτc α c
2 −iτa α a
2 iτc α c
+ 1 ( 2 i
τ
2! 2) a π a τ b π
)(1 b + i τc α c
+ 1 ( )
2 i
τ
2 2! 2) c π c τ d π d
2 ++ 1 2!
( i
2) 2
τ a π a τ b π b + 1 2!
( i
2
) 2
τ c π c τ d π d = 1
The higher orders go similarly. The invariance of the mass tern and of
the kinetic energy term of the fermions under the above transformation is obvious.
The nucleon-pion interaction is invariant provided the pion fields are
33

transformed as
τ a π a → τ a π ′a = Uτ a π a U †
In this case the action of U on the fermion- and on the boson-field compensate
each other. Thus we demand this transformation properties for the pionic field.
In order to prove then that also π a π a is invariant one should use the identity
π a π a = 1 2 Tr(τa π a τ b π b ) (82)
from which we see immediately that π ′a π ′a = π a π a ,because we have under the
action of the transformation U.
π ′a π ′a = 1 2 Tr(τa π ′a τ b π ′b ) = 1 2 Tr(Uτa π a U † Uτ b π b U † ) = 1 2 Tr(τa π a τ b π b U † U) = 1 2 Tr(τa π a τ b π b ) = π a π a
Thus the total Lagrangean eq.(80)is invariant under (81). The transformation
properties of the pion field are defined in a slightly indirect way, which is not
that simple as for the fermion field. However, the response of the individual
pion components to the isospin transformation can be found from multiplying
the transformation equation for the pion field by τ b and taking the trace using
a well known trace-property of all Lie-groups:
We obtain explicitely:
Tr(τ a τ b ) = 2δ ab (83)
Tr(τ b τ a )π a → Tr(τ b τ a )π ′a = Tr(τ b Uτ a π a U † )
and finally
2δ ab π a → 2δ ab π ′a = Tr(τ b Uτ a π a U † )
π ′a = R ab (α)π b with R ab (α) = 1 2 Tr(τa Uτ b U † )
To determine the isospin current, one can directly apply the formulae for the
conserved current. One also can consider the spacetime-dependent transformation
with α a (x) now infinitesimal (the ɛ abc enters because of the commutator
rules of the isospin Pauli matrices=:
̂ψ = (1 − iτ a α a (x)/2)ψ ::::::::::: ̂π a = π a − ɛ abc π b α c (x)
Performing this transformation on the lagrangean gives
L( ̂ψ, ̂π a ) = L(ψ,π a ) + 1 2 ψγµ τ a ∂ µ α a (x)ψ − ɛ abc (∂ µ π a )π b ∂ µ α c (x)
Aplying the general expression for a current eq.(62)
j µ (x) =
∂
∂(∂ µ ɛ(x)) L(̂φ,∂ µ̂φ)
34

in the present situation yields
V a µ =
∂
∂(∂ µ α a (x)) L(̂φ,∂ µ̂φ)
Explicit and straight forward calculation shows that we obtain the triplet of
conserved iso-vector currents:
V a µ = ψγ µ
τ a
2 ψ + ɛabc π b ∂ µ π c (84)
This current is conserved for each a, as one can explicitely show by direct
calculation using the equations of motion for ψ and π a . We see that the current
is composed of a fermion part and a pion part, which equally contribute. The
mesonic contribution is called ”meson exchange current”. One can calculate
the charges60 and one can show (without proof) that they satisfy the charge
commutation rules eq.(68):
[
Q a (t),Q b (t) ] = iε abc Q c (t)
as the Isospin-Pauli-Matrices do, eq.(). This is expected from the general
rules. Actually this current is not the charge current. The charge current
involves only the upper component of ψ (because only the proton is charged)
and only the charged pion fields π + and π − . The isospin current involves upper
and lower component and all pion fields.
Actually in case of the nucleon, i.e. the fermion field representing proton and
neutron, the Lagrangean has been and still is very popular. The Lagrangean
implies (without proof) features, which are well corresponding to experiment
for the masses:
m 0 = m p = m n m π + = m π − = m π 0
and for the pion-nucleon-nucleon coupling constants
g ppπ 0 = −g nnπ 0 = g pnπ + / √ 2 = g npπ − / √ 2 π ± = (π 1 ± iπ 2 )/ √ 2
yielding a common value of g2
4π = 14.3. In real life, howevere, we have m n −
m p = 2MeV with m p = 938MeV , and with m π + = 139.5MeV and m π 0 =
134.9MeV MeV, which means that the isospin symmetry, described below, is
slightly broken.
3.6 Linear (chiral) Sigma-model
With a few modifications the above Lagrangean (80) becomes one of the most
instructive of all field theoretical models for the nucleon and for the general
concept of spontaneously broken symmetry. We note the Gell-Mann–Levy lagrangean,
where we add a scalar field σ and remove the bare mass
L = ψ [iγ µ ∂ µ ]ψ + 1 2
(
∂λ π a ∂ λ π a + ∂ λ σ∂ λ σ ) (85)
35

−gψ(σ + iτ a π a γ 5 )ψ − µ2
2 (σ2 + π a π a ) − λ 4 (σ2 + π a π a ) 2
We shall see later that this model exhibits the famous phenomenon of spontaneous
symmetry breaking which is extremely important and interesting, since
it is shown by the QCD as well. When it was invented it was first used to describe
a nucleon, interpreting the ψ as a doublet of proton and neutron field as
we did in the previous section. The nucleon was in the end a sysstem of interacting
fields, where the bare nucleon field given by ψ was ”dressed” by the sigma
and pion field. The result was moderate since the mass of the nucleon was too
small by a factor of two. Around 1985 the Gell-Man-Levy eq.(85) lagrangean
was used by interpreting the ψ as a quark field doublet
( ) u(¯r,t)
ψ(¯r,t) =
(86)
d(¯r,t)
In this form it was very successfull in the description of nucleon properties.
Today the Lagrangean in appreciated since it has basic properties of QCD, i.e.
it is completely field theoretical, it shows spontaneous chiral symmetry breaking
(see later) and isovector symmetry (see right now).
In treating this Lagrangean and in investigating its symmetry properties it
is useful to rewrite the mesons in terms of a mtrix field
Σ(x) = σ(x) + iτ a π a (x)
such that
σ 2 + π a π a = 1 2 Tr(Σ† Σ)
where we have used the identity (82).
Then we obtain by simple and direct calculation after separating into rightand
left-hand fermions for the Gelll-Man-Levy lagrangean (85) :
L = ψ R [iγ µ ∂ µ ]ψ R + ψ L [iγ µ ∂ µ ]ψ L + 1 4 Tr(∂ µΣ∂ µ Σ † )
− 1 4 µ2 Tr(ΣΣ † ) − λ [
Tr(ΣΣ † ) ] 2 (
− g ψL Σψ R + ψ R Σ † )
ψ L
16
The lagrangean (85), written in this way, shows directly its invariance, as we
see in the following: The left-handed and right-handed fermions are coupled to
each other only in the interaction with the Σ-field, i.e the last tern on the rhs.
The full lagrangean
⊗
has separate left- and right-invariances, i.e. it is invariant
under SU(2) L SU(2)R :
ψ R,L (x) → ψ ′ R,L(x) = U R,L ψ R,L (x)
Σ ′ = U L ΣU † R
with U R,L being two arbitrary SU(2) matrices analogous to (81).
U R,L (α) = exp(−i τa αR,L
a )
2
36

For the fermions the SU(2) L
⊗ SU(2)R transformation involves just a separate
rotation of the left- and right-handed fields. For the mesons we have a
combination of a pure isospin rotation among the pionic fields together with a
transformation between the σ and π fields. These invariances have the following
conserved Noether-currents:
J a Lµ = ψ L γ µ
τ a
2 ψ L − i 8 Tr ( τ a ( Σ∂ µ Σ † − ∂ µ ΣΣ †))
JRµ a τ a
= ψ R γ µ
2 ψ R − i 8 Tr ( τ a ( Σ † ∂ µ Σ − ∂ µ Σ † Σ ))
These can be rewritten into a conserved vector current and axial vector current:
V a µ = J a Rµ + J a Lµ = ψγ µ
τ a
2 ψ + ɛabc π b ∂ µ π c (87)
A a µ = JRµ a − JLµ a τ a
= ψγ µ γ 5
2 ψ + πa ∂ µ σ − σ∂ µ π a (88)
In the infinitesimal form the corresponding symmetry transformations are:
Isovector with β a = αR a + αa L :
ψ → ψ ′ = (1 − iβ a τa
)ψ (89)
2
σ → σ ′ = σ
and axial with β a = α a R − αa L :
π a → π ′a = π a + ɛ abc β b π c
ψ → ψ ′ = (1 − iβ a τa
2 γ 5)ψ (90)
σ → σ ′ = σ + β a π a
π a → π ′a = π a − β a σ
As one sees, the vector current is identical to the isospin current (84) in the
simple effective model (80), it contains only the fermion- and pion-fields. The
axial current, however, is composed of the fermion-, and pion- and the sigmafields.
One sees that the axial transformation mixes σ- and π-field, whereas the
(iso-)vector transformation does not mix thes two different types of fields.
There are (without proof) closed charge algebras and current algebras for
the left-hand and right-hand current analogous to eqs.(68) separately (with
f abc = ε abc ):
37

[
Q
a
R (t),Q b R(t) ] = if abc Q c R(t) (91)
[
Q
a
L (t),Q b L(t) ] = if abc Q c L(t)
[
Q
a
L (t),Q b R(t) ] = 0
One can now write down the algebras of the vector and axial currents, and
these expressions remind us of the expressions we had in the general chapter on
Noether currents69,70:
[
Q
V
a (t),V µ
b (xt)] = if abc V c µ (x) (92)
[
Q
V
a (t),A µ b (xt)] = if abc A µ c (x)
[
Q
A
a (x),A µ b (y)] = if abc V µ
c (x)
and
[
Q
A
a (x),V µ
b (y)] = −if abc A µ c (x)
δ(x 0 − y 0 ) [ V 0
a (x),V 0
b (y) ] = iδ(x − y)f abc V 0
c (x) (93)
δ(x 0 − y 0 ) [ V 0
a (x),A 0 b(y) ] = iδ(x − y)f abc A 0 c(x)
δ(x 0 − y 0 ) [ A 0 a(x),A 0 b(y) ] = iδ(x − y)f abc V 0
c (x)
These relations are obtained directly by explicit calculations (without proof).
They hold, even if there are finite quark masses, because the definition of the
currents involves only the kinetic energy of the quarks and the transformation
done on the system. Of course, in such a case the charges are time dependent.
The left hand and right hand currents form each a closed algebra, but the linear
combination does not: In fact, the vector currents provide a closed algebra,
whereas the axial currents do not.One should note that these relations hold
even if there is no pion field in the lagrangean, i.e. for a lagrangen where several
fermion fields are ordered in a multiplet and being free or coupled to a gauge
field. This simple feature will be the case in QCD, where the quark fields are
coupled to a flavour blind gluon field.
38

4 Symmetry breaking
4.1 Realization of symmetries
If the Lagrangean shows a symmetry, then there are various ways this symmetry
manifests itself. This is by no means trivial. We have actually the following
possibilities:
1. The symmetry is exactly and always fulfilled, on the classical as well as
on the quantum level. That means, any solution of the Lagrangean will
show it. This is e.g the case for electromagnetic U(1)-symmetry in QED
and QCD, and also charge symmetry, flavour symmetry, baryon number
symmetry in QCD.
2. The symmetry may have an anomaly. This means, that it is not a true
symmetry, because it is only a symmetry of the classical lagrangean but
not of the quantized lagrangean. It means that the quantized symmetry
current is not conserved, or better to say, the symmetry is not a symmetry
of the quantized theory. This is actually not unsusual, so the name
“anomaly” is misleading. However, as a matter of fact and based on
fundamental field theoretical properties (and without proof), symmetry
currents directly associated to the gauge transformation must not have an
anomaly.
3. The symmetry may be explicitely broken. Example is the chiral symmetry
of a massless Dirac-equation, which is explicitely broken by small
massterm in the lagrangean. Isospin symmetry is broken by mass differences
of the upper and lower component of the field e.g. for proton-neutron
doublets, or triplets of up-, down- and strange quarks, or by electromagnetic
forces (charges). In real life this explicit breaking very often happens,
but - depending on the process or structure considered - can often can be
ignored if it is small.
4. The symmetry may be ”hidden”, as one names this phenomenon. This
means, the symmetry is an invariance of the quantized Lagrangean but
not of the ground state of the system. Thus one does not recognize it
by investigating the properties of the ground state or some of the excited
states of the system. Such a feature is also called “spontaneously broken
symmetry”. Such a phenenon is only possible in infinite systems. Actually
there are different physical mechanisms, which produce the spontaneous
symmetry breaking
(a) If in the lagrangean a scalar field occurs and if it acquires a nonvanishing
vacuum expectation value (i.e. it is not dependent on
space and time), then we have a spontaneously broken symmetry.
An example is the Higgs-field in the electro-weak theory, whose nonvanishing
vacuum value gives rise to finite values of the fermion and
39

gauge boson masses due to the meson-fermion coupling term. Another
example is the sigma-field in the linear chiral soliton model
(85), which gives the fermions a dynamical mass..
(b) Even in the absence of scalar fields, quantum effects can lead to the
dynamical breaking of a symmetry. This is the case of the spontaneously
broken chiral symmetry in QCD, one of the most important
symmetries in the universe.
4.2 Principle of Spontaneous symmetry breaking
We consider a system with the Lagrangean (1) involving one field φ. Suppose
the Lagrangean is invariant under a simple continuous symmetry which has a
conserved Noether current as consequence. This means, if Q is the charge of
the symmetry as obtained from the Noether-current, the Q can be used as the
generator of the symmetry group.
Nonbroken Symmetry: In the usual case of a non-broken field theory
the ground state of the system |0〉 shows the symmetries of the Lagrangean and
hence the ground state is invariant under
|0〉 → exp(−iαQ) |0〉
where α is a continuous parameter. Thus for such a (normal) sytem we have
or
exp(−iαQ) |0〉 = |0〉
Q |0〉 = 0 (94)
Obvioulsly we have only one ground state, and application of the symmetry
operation does not change it.
Spontaneously broken symmetry: We have by definition spontaneous
symmetry breaking, if under the above transformation the vacuum changes and
or
exp(−iαQ) |0〉 ≠ |0〉
Q |0〉 ≠ 0 (95)
Such a phenomenon does not occur in finite systems. One needs infinite systems
in space-time and a field theory with its infinite number of particles and
antiparticles provides this. In this case we have
exp(−iαQ) |0〉 = |α〉 :::::::::::: |α〉 ≠ |0〉 (96)
Examples for this are the ferromagnetism in solid states. The hamiltonian of
the atom-atom interaction is rotationally smmetric and hence the hamiltonian
for the many body system as well. Nevertheless we have in ferromagnets an
orientation of the magnetic moments inside the Weiss-regions. If we rotate all
40

spins of a Weiss-region we get a set of degenerate states. Another example is
the gravitaional field of the earth. It is rotationally symmetric, nevertheless for
a pen in the gravitational field and on the spherical surface of the earth the
lowest energy state is not the pen pointing upright but lying on the ground.
And one obtains degenerate states, if one rotates the pen on the surface and
changes by this its orientation. Although the states |α〉 of eq.(96) for different
α are different, their energy is the same. This one can see easily: We have from
Noethers theorem
[H,Q] = −i dQ
dt
Since, however, Q is a charge operator of a symmetry current, and it is hence
independent on time. Therfore the RHS is zere and hence it commutes with the
Hamiltonian. Thus the energy of the state |α〉 is the one of the ground state
|0〉:
H |α〉 = H exp(−iαQ) |0〉 = exp(−iαQ)H |0〉 = exp(−iαQ)E 0 |0〉 = E 0 |α〉
Because the symmetry transformation is continuous there must occur a continuous
family of degenerate states |α〉.
Actually there is an important consequence of the above considerations.
Assertion: There are matrixelements of the current, which connect the vacuum
with one particular excited state of the system.This is a particular excitation of
the ground state (vacuum), namely the so called Goldstone Boson. The logic is:
dQ
dt = 0 ⇒ d
〈0|[Q(t),φ(0)] |0〉 = 0
dt
⇒ 〈0|[Q(t),φ(0)] |0〉 = η ≠ 0 (97)
If we have no spontaneous symmetry breaking the η = 0, since according to
eq.(94) we have Q |0〉 = 0. With spontaneous symmetry breaking we have η ≠ 0
sincewith eq.(95) we have Q |0〉 ≠ 0 and η is time independent since Q(t) is time
independent. This simple feature eq.(97) has the immediate consequence that
there exists a Goldstone boson. In order to see this, we represent the charge
by an integral over the current and use the translation operation and insert a
complete set of states |n〉, i.e. ground state of the system and excited ones. This
looks in detail as follows
η = 〈0|[Q(t),φ(0)] |0〉 =
=
∫
d 3 x 〈0|[j 0 (x,t),φ(0)] |0〉 =
= ∑ n
∫
d 3 x{〈0| j 0 (x,t) |n〉 〈n| φ(0) |0〉 − 〈0|φ(0) |n〉 〈n| j 0 (x,t) |0〉}
41

The |n〉 are physical states, which can in principle be observed. They have the
quantum numbers, which can be created by applying the charge operator to the
vacuum. Now we use that F(x) = exp(iPx)F(0)exp(−iPx) shown in eq.(34)
and obtain
η = ∑ n
∫
d 3 x {〈0|exp(iPx)j 0 (x = 0,t = 0)exp(−iPx) |n〉 〈n| φ(0) |0〉 −
− 〈0| φ(0) |n〉 〈n|exp(iPx)j 0 (x = 0,t = 0)exp(−iPx) |0〉 }
We have 〈0|exp(iPx) = 〈0|exp(iE 0 t − ip 0 x) with E 0 = 0 and p 0 = 0 because
the vacuum is time independent and homogeneous. We have analogue
expressions for the application of the shift operator on |n〉 .This yields then
after integration over d 3 x
η = ∑ n
(2π) 3 { δ (3) (p n ) 〈0|j 0 (0) |n〉 〈n| φ(0) |0〉 exp(−iE n t)−
−δ (3) (−p n ) 〈0|φ(0) |n〉 〈n|j 0 (0) |0〉 exp(iE n t) }
In order that the right hand side is non-vanishing and time-independent we
must have the following situation:
1) consider only states with p n = 0.
2) Consider the vacuum state |n〉 = |0〉 with E 0 = 0. This yields a term,
which does NOT contribute the the above sum for η, since this term is obviously
zero. Thus in order to get something unequal zero at least one of the other states
|n〉 must have a non-vanishing matrix element.
3) For an arbitrary state with E n ≠ 0 the time-exponentials are out of phase
and hence yield a nonvanishing contribution, which is time dependent in contrast
to η which is time independent. Thus all those “normal” states do not contribute
in the sum, that is their matrix elements must be zero: 〈0| j 0 (0) |n〉 〈n| φ(0) |0〉 =
0. There must be one particular state, call it |n = G〉, which must have the
property:
E G = 0
〈0| j 0 (0) |G〉 ≠ 0 ::: and :::: 〈0|φ(0) |G〉 ≠ 0 (98)
because then its contribution to η is not time dependent and unequal zero. This
particular state is the state of the Goldstone Boson. It is an excitation of the
vacuum and hence belongs to the set of states |n〉. Thus the current and the field
connects the vacuum with the Goldstone boson |G〉. This is a very prominent
feature, which will later on in QCD give rise to concepts like conserved axial
current, partially conserved axial current, pion decay constant, etc.
What does this mean physically In a quantum field theory the ground
state of the system is the lowest state. For e.g. baryon number zero it is the
vacuum, for baryon number one it is the proton. In a quantum field theory
42

any excitation of the ground state becomes quantized and describes in some
way a particle. The minimum excitation energy corresponds to the particles
mass. Thus the above excitations, having no excitation energy, correspond
to the creation of a particle with vanishing mass. These particles carry the
quantum number of the symmetry which generates them (via the field or via
the current), and they are called Goldstone Bosons. If the symmetry is not
exact but slightly broken, then the Goldstone bosons acquire a small mass. As
we will see, pions are the Goldstone bosons of the spontaneously broken chiral
SU(2)-symmetry in QCD. Their mass should hence be zero, but in reality it
is m π = 139 : MeV because m u = 6 : MeV and m d = 10 : MeV . For
the spontaneously broken SU(3)-symmetry, involving also strange quarks, the
kaons should also be Goldstone bosons, their mass is already m K = 493 : MeV
due to the fact that m s = 180 : MeV .
Spontaneous symmetry breaking is a very general phenomenon. We will
illustrate it at one particular model, namely the chiral linear sigma model (85).
4.3 Spontaneous symmetry breaking in the linear sigma
model
The lagrangean of the linear sigma model is given by eq.(85). It is an excellent
example to illustrate the spontaneous breakdown of thea chiral symmetry, as
it in fact happens in QCD. Its symmetries and invariances (vector and axial
vector) are known from previous sections an will play a prominent role. It
seems that there are mass terms for σ and π in the Lagrangean. However, the
quartic term also gives quadratic terms in the boson fields and hence the mass is
given as second derivative of the bosonic potential, which collects all the bosonic
terms which are neither coupled to a fermionic field nor have a derivative. In
the following we want to interpret the Lagrangean in a classical way. This is
not really possible because the ψis a Grassmann variable and only functional
integrals over it make sense. Nevertheless if we want to investigate the properties
of the vacuum one can first consider the properties of the bosonic part of the
lagrangean using ψ = 0 for the vacuum and after having investigated that, add
the fermion. This procedure has its reason in the fact that the bosonic field can
be as large as you want whereas the fermionic one is basically restricted to one
due to Pauli principle.
We can bring the Lagrangean into a slightly modified but equivalent form
L = ψ [iγ µ ∂ µ ]ψ + 1 2
(
∂λ π a ∂ λ π a + ∂ λ σ∂ λ σ ) ::::::::::::::::::::
:::::::::::::::::::: −gψ(σ + iτ a π a γ 5 )ψ − C 2 (σ 2 + π a π a − A) 2
To simplify the following considerations we consider classical limit of this lagrangean.
We want to determine the ground state of the system, for this we
need the hamilton function and its minimum. We obtain the hamilton function
by considering the energy momentum tensor of the mesonic part, see eq.(64) for
43

the definition. The fermionic part is ignored since we consider the vacuum:
T 00 = −g 00 +
= − 1 2
= 1 2
∂L
∂ (∂ 0 π a ) ∂ 0π a +
∂L
∂ (∂ 0 σ) ∂ 0σ
[
(∂ µ σ) 2 + (∂ µ
−→ π )
2 − C 2 (σ 2 + π a π a − A) 2] + (∂ 0 σ) 2 + (∂ 0
−→ π )
2
( ∂σ
∂t
) 2
+ 1 2
( ∂
−→ ) 2 π
+ 1 ( −→∇
)
−→ 2 1
( −→∇σ
) 2
π + + C 2 (σ 2 + −→ π 2 − A) 2
∂t 2 2
Determining the minimum of the energy density means determining in a the energy
per unit volume and minimize this. Thus we consider the bosonic potential
energy of the system. This is
V (σ, π a ) = C 2 (σ 2 + π a π a − A) 2
and we look for the minimum of V (σ,π a ) with respect to σ and π a . In this
simple picture masses of σ and π a are given by the second derivative of V at
the vacuum values of σ and π. Obviously we have for the vacuum two different
cases:
.
Case A < 0: Wigner Mode
The minimum is at σ V = 0 and πV a = 0. And the masses are given by
(
1 ∂ 2 V
2 m2 σ =
∂σ
)V
2 = −2AC 2 > 0
(
1 ∂ 2 V
2 m2 π =
∂π
)V
2
= −2AC 2 > 0
Apparently σ and π are both massive and degenerated. We have complete
symmetry with respect to sigma and pi. This vacuum state is invariant under
(iso-)vector transformation eq.(89):
σ → σ ′ = σ → σ ′ V = σ V = 0
π a → π ′a = π a + ɛ abc α b π c → (π a V ) ′ = π a V = 0
and axial vector transformation eq.():
σ → σ ′ = σ + α a π a → σ ′ V = σ V = 0
π a → π ′a = π a − α a σ → (π a V ) ′ = π a V = 0
Thus altogether the Wigner mode shows vanishing mesonic fields in the
vacuum and this vacuum is invariant under the vector and axial vector transformation
44

4.3.1 The Goldstone Mode
This mode is given for A > 0. In such a case there is not one definite minimum
but an infinite multitude of minima which all are only required to fulfill
σ 2 + π a π a = A > 0
All these minima are of equal right. If we demand positive parity from the
vacuum we chose π a V = 0 and σ V = √ A. This vacuum is invariant under the
isovector transformation eq.(89):
σ → σ ′ = σ → σ ′ V = σ V
π a → π ′a = π a + ɛ abc α b π c → (π a V ) ′ = π a V = 0
However it is not invariant under the infinitesimal axial transformation eq.().
In fact this consists in rotating along the chiral circle σ 2 + π a π a = A. For an
infinitesimal transformation we have:
σ → σ ′ = σ + α a π a → σ V → σ ′ V = σ V + α a π a V → σ ′ V = σ V
π a → π ′a = π a −α a σ → π a V → π ′a
V = π a V −α a σ V → (π a V ) ′ = −α a σ V ≠ π a V = 0
The axial transformation changes all the possible vacua into each other, however
for an infinitesimal transformation only the pion field gets an ifinitesimal
contribution proportional to the vacuum value of the sigma field..
We can now calculate the masses of the pion and sigma. They are given by
the second derivative of the potential, however not around the trivial vacuum
but around the new vacuum. In order to do so we rewrite the Lagrangean by
defining new fields, which describe the deviation from the vacuum field: Instead
of σ we use now σ = ˜σ + σ V , such that the vacuum value of ˜σ is zero. Doing
this we obtain the same Lagrangean with the new fields as:
L = ψ [iγ µ ∂ µ − gσ V ]ψ + 1 (
∂λ˜σ∂ λ˜σ − 8C 2 σV 2 ˜σ 2) + 1 2
2 ∂ λπ a ∂ λ π a
− gψ(˜σ + iτ a π a γ 5 )ψ + 4C 2 σ V ˜σ(˜σ 2 + π a π a ) − C 2 (˜σ 2 + π a π a ) 2
Again we can now calculate the masses of the pion and sigma:
1
2 m2 σ =
( ∂ 2 V
∂σ 2 )V
= 4AC 2 = 4σ 2 V C 2 > 0
(
1 ∂ 2 V
2 m2 π =
∂π
)V
2 = 0
Apparently there is no mass term for the pion. The pion appears as massless
Goldstone boson and the mass-squared of the sigma has increased twice.
45

There is also an interesting consequence for the fermions. In the original
Lagrangean the fermions were assumed to be massles. Thus their energy
spectrum is given by ɛ(k) = ± √ k 2 After transforming to the new vacuum the
fermions appear as massive particles with a mass of m = gσ V . This process is
called “dynamical mass generation”. Thus their energy spectrum is given by
ɛ(k) = ± √ k 2 + m 2 and hence there appears a gap of 2m between the occupied
and unoccupied single particle states.
ε (k)
single quark energies
Wigner mode
ε (k)
M
Single quark energies: Goldstone mode
The Lagrangean looks actually quite different compared to the original one.
However the smmetries are unchanged and the Lagrangean is still invariant
under SU(2) L
⊗ SU(2)R .On the other hand the σand πdo not form a degenerate
multiplet. Thus the symmetry is ”hidden”, or spontaneously broken. Also, the
full set of original symmetry currents remain conserved. The vector current 87is
unchanged. The axial current, however, looks different. It was given by eq.(88)
and is now is now
A a µ = ψγ µ γ 5
τ a
2 ψ + πa ∂ µ σ − ˜σ∂ µ π a − σ V ∂ µ π a
and, however, still has a vanishing divergence ∂ µ A a µ = 0. Thus the fact that the
fermion field got a mass due to canamical mass generation has not change the
divergence of the currents.
[A comment for the advanced reder: The above procedure is in principle
much more complex: If we want to solve this Lagrangean we have in
priciple to perform functional integrals to obtain the partition function, from
which one can derive basically everything. The partition function is given by
Z = ∫ DψDψDσDπ exp [ i ∫ d 4 xL(x) ] .First we perform the integral over the
grassmann variables. By this you integrate out the fermionic fields yielding a
preexponential factor, i.e. the fermion determinnt=Trlog. For the following
considerations it is helpful to consider the socalled small coupling limit. To perform
this limit we change the variables in the functional integral by introducing
σ → √ σ λ
and π a → √ πa
λ
. This does not affect the preexponential factor if we
assume that the fermion-boson coupling constant scales with √ λ.The stationary
phase approximtion is now given by the minimum of the bosonic action. After
46

having obtained the bosonic field in the minimum we insert this into the Trlog.
The stationary phase approximation is justified in the limit λ → 0 because then
the bosonic terms dominate.]
4.3.2 Broken chiral symmetry and PCAC
In the following section we move with the chiral linear sigma model a bit coloser
to reality, since in experiment the pions have a small but finite mass of 139
MeV. This will lead to the principle of partial conservation of the axial current,
which is very imprtant. In order to simulate this QCD-feature in the Gell-man
Levy model (85)it is sufficinet to add to the Lagrangean a term
L → L+L m = L + Dσ
With this term the vector current is still conserved, however the divergence of
the axial current does not vanish any more, as we see immediately. In order
to calculate the axial current we consider the infinitesimal axial transformation
(90)
σ → σ ′ = σ + ε a π a
π a → π ′a = π a − ε a σ
Therefore we have for the mesonic part δL m = Dπ a ε a (x) and hence with
eq.(62) and (63) one obtains immediately
∂ µ j µa
5 =
∂
∂(ε a (x)) L m =
∂
∂(ε a (x)) Dπb ε b (x) = Dπ a
This equation holds even if we consider the full Lagranean including the Ferionic
fields. Since the divergence of the axial current is not zero, the chiral symmetry
is ecplicitely broken. This has immediate effect on the classical ground state of
the theory as we can see at the bosonic potential
We obtain the minimum of U by
U = C 2 ( σ 2 + −→ π 2 − A ) 2
− Dσ
∂U
∂σ = 4C2 σ ( σ 2 + −→ π 2 − A ) − D = 0 (99)
∂U
∂π a = 4C2 π a ( σ 2 + −→ π 2 − A ) = 0
Again we denote the ground state values with the index V: σ V and π V . The
Wigner-mode is trivially changed. More interesting is the Goldstone mode with
A > 0. There we obtain from eq.(99)
σ 2 V − A =
47
D
4C 2 σ V
(100)

If we again expand the sigma field around the vacuum value σ V we obtain
again that the fermions get a mass of M = gσ V . The mesonic potenatial is now
changed into by eliminating A by eq.(100)
U = C 2 ( σ 2 + −→ π 2 − A ) 2
− Dσ
= C 2 ( (˜σ + σ V ) 2 + −→ π 2 − A ) 2
− D(˜σ + σV )
(
= C 2 ˜σ 2 + −→ π 2 +
D ) 2
4C 2 + 2˜σσ V − D(˜σ + σ V )
σ V
We find immediately by taking the derivative w.r. to π 2 that the mass or the
pion is now finite:
m 2 π = D σ V
Using this equation and M = gσ V we can express the yet unknown value of D
in the following way
D = m2 πM
g
This relation holds only in the Goldstone phase. Now we are able to rewrite the
divergence of the current:
∂ µ j µa
5 = Dπ a = m2 πM
π a (101)
g
Apparently the current is almost conserved because the pion mass is small and
in practice the M = 350MeV and g = 5 roughly. Still we have to get to know
the RHS better.
The objective of the following lines is, to express the right hand side of
eq.(101) in terms of the pion decay constant, since this is an important and
well known quantity f π = 93MeV . For this consider the matrixelement for the
pion decay (detailled reasoning is given in sect on pion decay), i.e. in the view
of strong interaction from a 1-pion state to the vacuum. The relevant process
for this is for instance π − → µ − + ν µ .Since the pion has negative parity only
the axial current contributes and we have hence the definition of the pion decay
constant
< 0|j µa
5 (0)|πb (k) >= ik µ δ ab f π
This is equation is the concretization of the more abstract expression (98).
Using the shift operator eq.(34) one can write
< 0|exp(−iPx)j µa
5 (x)exp(+iPx)|πb (k) >= ik µ δ ab f π
or
< 0|j µa
5 (x)|πb (k) >= ik µ δ ab f π exp(−ikx)
and one gets for the divergence because of k 2 = m 2 π
< 0|∂ µ j µa
5 (x)|πb (k) >= m 2 πf π δ ab exp(−ikx)
48

Using the expression for the quantized boson field (32) and the norm of the
boson creation operators (31) one gets for the properly normalized one-pion
state
|π b (k) >= (2ω k ) 1 2
a † (k)|0 > (102)
so that
Thus we obtain
yielding
< π a (k)|π b (k ′ ) >= 2ω k (2π) 3 δ(k − k ′ ) (103)
< 0|π a (x) |π b (k) >= δ ab exp(−ikx)
< 0|∂ µ j µa
5 (x)|πb (k) >= m 2 πf π < 0|π a (x) |π b (k) >
If we go back to the classical level we obtain
∂ µ j µa
5 (x) = m2 πf π π a (x) (104)
This is an extremely important equation, which expresses the principle of
PCAC, i.e. partially conserved axial current.
One can apply this in the following way. We had the divergence of the
current as ∂ µ j µa
5 = m2 π M
g
π a and ∂ µ j µa
5 (x) = m2 πf π π a yielding M = gf π and
hence
σ V = f π
and hence the vacuum value of the sigma-field equals the pion decay constant.
If one takes M = gf π , as we have shown above, and interpretes the fermion
field as the nucleon matter field, then the coupling constant g equals to the
pion-nucleon coupling constant g πNN We have from experiment the numbers
M N = 938MeV
g πNN = 13.45
f π = 93MeV
which fits the equation M = gf π within 30%. The fact that such a simple effective
model yields such an agreement with experiment is really an achievement.
One should note: A small quantity as the pion decay constant, being related to
the weak decay of the pion, is related via the pion-decay constant to a large
quantity, i.e. the nucleon mass. This is an astonishing feature.
Actually one can improve this model by interpreting the fermion field as a
field for quarks interacting with the pion field. In this way (without proof) one
can incorporate the substructure of the nucleon. If one does it (see later) we
obtain a modified equation, the so called Goldberger-Treimann equation:
g A = 1.26
M N g A = g πNN f π
49

with the axial vector coupling constant of the nucleon with experimental value
g A = 1.26. The Goldberger Treiman relation is satisfied at the 5% level. In
this way one has a phenomenological description of the dynamic mass generation
caused by the spontaneous breaking or the chiral symmetry: The massless
quarks (so called current quarks, or QCD-quarks) get via this breaking a mass
of about 350 to 400 MeV, such that three of them yield basically the mass of the
nucleon. One sees that without the breaking we would not have an explanation
for the mass of the nucleon.
5 The gauge principle
The gauge principle has been established in the last 30 years. It is basically a
recipe how to construct field theories, which describe the forces of nature. The
principle has not been derived, but it has been proven extremely succesfull. We
first consider for this the Maxwell theory, which is an abelian gauge theory, and
then a Yang-Mills theory, which is a non-abelian gauge theory. The Quantum
Chromodynamics, which is the theory for strong interaction is a non-abelian
gauge theory, as well as the Glashow-Salam-Weinberg theory describing the
electro-weak interaction. The gauging of a theory and the derivation of a gauge
field will be exemplified at the Dirac equation.
5.1 Abelian gauge theory:
Global symmetry: Consider the lagrangean equation (6) for the free electron.
Apparently this Lagrangean is invariant under the “global U(1)-transformation”
(74) Here α is a real constant. The symmetry is a trivial symmetry of the free
Dirac Lagrangean, each free Dirac particle shows this. Actually the symmetry is
an abelian symmetry since two successive symmetry transformations commute.
Local symmetry: Now we want to generalize this symmetry into a “local
U(1) symmetry” by demanding, that the Lagrangean should be invariant under
this symmetry even if α(x) is an arbitrary differentiable function. We will later
see, that this seemingly arbitrary demand is highly relevant and will generate
in principle all present theories of elementary particles and their interaction:
ψ(x) → ψ ′ (x) = U(α(x))ψ(x) = exp(−iα(x))ψ(x) (105)
ψ(x) → ψ ′ (x) = ψ(x)U −1 (α(x)) = ψ(x) exp(+iα(x))
This is a very strong demand, it says that we can change at any space-time point
the phase of the Dirac field in an arbitrary way and the Lagrangean, i.e. the
dynamics of the Dirac field should be the same. This is a lot, in non-relativistic
quantum mechanics we can multiply the wave function only with a constant
phase, never with a space- and time-dependent one. However, the demand is
not stupid, in fact, this demand has appeared over the last 20 years as the basic
50

tool to construct theories of elementary particles and their interactions. Electrodynamics,
Quantum Chromodynamics and Elctroweak theory are constructed
in this way.
Immediately we see that the above Lagrangean does NOT fulfil the demand
of a local U(1) symmetry. The mass term is indeed invariant, however the
kinetic term is not since the derivative acts on α(x):
ψ(x)∂ µ ψ(x) → ψ ′ (x)∂ µ ψ ′ (x) = ψ(x) exp(+iα(x))∂ µ exp(−iα(x))ψ(x)
= ψ(x)∂ µ ψ(x) − iψ(x)(∂ µ α(x))ψ(x)
The second term on the RHS destroys the invariance of the Lagrangean. In
order to fulfill our demand, we have to modify the Lagrangean. We do this by
introducing a new field A µ (x) and the so-called gauge-covariant derivative:
D µ ψ(x) = (∂ µ + ieA µ (x))ψ(x) (106)
The covariant derivative contains the coupling constant e of the fermion field
ψ(x) to the gauge field A µ (x). In the present U(1) theory this coupling constant
e is the charge of the fermion. [The electron has the charge e = −e 0 , where e 0
is the elementary charge.] We consider the new Lagrangean, where we have
replaced ∂ µ by D µ :
L = ψ [iγ µ D µ − m]ψ (107)
and demand invariance of this Lagrangean (107) under the gauge transformation
() of the fermion field. However, this demand cannot be fulfilled without
demanding simultaneously the following transformation property of the A-field:
A µ (x) → A ′ µ(x) = A µ (x) + i e (∂ µU(α))U −1 (α) = A µ (x) + 1 e ∂ µα(x) (108)
Assertion: The Lagrangean (107) with the covariant derivative (106) is invariant
under simultaneous gauge transformation of the fermion field and the gauge field
( 105,108).
Proof: On can see immediately in a direct calculation that we have under
the new combined gauge transformation (108) the property
D µ ψ(x) → [D µ ψ(x)] ′ = exp (−iα(x)) D µ ψ(x) (109)
and hence the kinetic energy term is invariant:
ψ(x)D µ ψ(x) → ψ ′ (x)[D µ ψ(x)] ′
( [
= ψ(x) exp(+iα(x)) ∂ µ + ie A µ (x) + 1 ])
e ∂ µα(x) exp(−iα(x))ψ(x)
51

= ψ(x) (∂ µ + ieA µ (x)) ψ(x) = ψ(x)D µ ψ(x) q.e.d.
By now the A-field is an additional field with no dynamical properties. To
make it a true dynamical variable we have to complement the Lagrandian density
(107) by a term corresponding to the kinetic energy of the A-field. The
additional term should not destroy the gauge invariance (108) and should contain
derivatives in order to provide dynamics. A candidate is the Lagrangean
(20) of the free Maxwell field. We will show below that − 1 4 F µνF µν is indeed
gauge invariant. Therefore we can note the full Lagrangean, being gauge invariant
under U(1), for a fermion field of charge e coupled to a Maxwell field
:
L = ψ [iγ µ D µ − m]ψ − 1 4 F µνF µν (110)
with the field tensor known from classical electrodynamics (22)
F µν (x) = ∂ µ A ν (x) − ∂ ν A µ (x) (111)
The proof of the gauge invariance of the kinetic energy of A is simple: By a
direct calculation one can easily show (first) that
(D µ D ν − D ν D µ ) ψ = ieF µν ψ (112)
The proof of eq.(112) is simple. [One should note for the derivation that ∂ µ acts
on everything on the right, when it is situated in D µ ]:
(D µ D ν − D ν D µ ) ψ =
= [(∂ µ + ieA µ (x)) (∂ ν + ieA ν (x)) − (∂ ν + ieA ν (x)) (∂ µ + ieA µ (x))]ψ
= ie(∂ µ A ν (x) − ∂ ν A µ (x))ψ
. Direct calculation yields the above identity. Furthermore from eq.(109) we
derive immediately (second)
[(D µ D ν − D ν D µ ) ψ] ′ = exp(−iα(x))(D µ D ν − D ν D µ ) ψ
The proof of this equation is simple because
(D µ D ν ψ) ′ = D ′ µD ′ ν exp(−iα(x)ψ(x) = D ′ µ exp(−iα(x)D ν ψ(x)
You can consider D ν ψ as some particular spinor and hence you get:
D ′ µ exp(−iα(x)D ν ψ(x) = exp(−iα(x)D µ D ν ψ(x)
which proofs the assertion. In summary we have
F ′ µνψ ′ = exp(−iα(x))(F µν ψ)
52

Because ψ is an arbitrary spinor and exp(−iα(x)) and F µν commutes since F µν is
a function, we have finally the important gauge invariance of the field tensor
F ′ µν = F µν qed (113)
This feature, remember our section on classical Maxwell field, is necessary. The
F µν are directly related to the electric and magnetic fields E and B, which are
observables, since they appear directly in the Lorentz force acting on a charged
particle moving in these fields: K = q (vxB) .Du to the motion of the particle
the electric and magnetic fields are observable quantities. The follwing remarks
are in order:
1. The above Lagrangean can be rewritten as
with the current
L = ψ [iγ µ ∂ µ − m]ψ − 1 4 F µνF µν − ej µ A µ (114)
j µ = ψγ µ ψ (115)
A comparison shows that we have tacitly derived the Lagrangean of a
Dirac field and a Maxwell-field, where the Dirac field provides the source
(current) to generate the Maxwell field. If we put e = −e 0 we describe
electrons.
2. There is no mass term for the photon. It shoud have the form m phot A µ A µ ,
which is obviously NOT gauge invariant. One sees this directly looking at
the gauge transformation property of the A-field eq .().
3. In the covariant derivative (106) we have a rather simple coupling of the
gauge field to the fermion field. This coupling is called “minimal coupling”.
This coupling results from the wish to introduce a gauge field
in the utmost simple way into the original lagrangean in order to make
the lagrangean of the free fermion invariant under the U(1) gauge transformation.
The coupling is therefore completely independent on further
properties of the fermion field. Indeed any charged fermion field couples in
the same way to the gauge field ( i.e. to the scalar Φ- and vector A- field
of the Electrodynamics). This property is calles universality. Other
higher dimensional gauge-invariant couplings such as ψs µν ψF µν are ruled
out by the requirement of renormalizability of the resulting lagrangean.
There is, however, one degree of freedom in the coupling, that is the value
of the charge. We could simply add to the lagrangean with the fermion
field ψ another fermion field φ with a charge e ′ = be with an arbitrary
b. As one sees in the next formulae has in fact one electromagnetic field
coupled to the two different fermion fields
4.
L = ψ [iγ µ D µ − m]ψ + φ[iγ µ D µ − m]φ − 1 4 F µνF µν
53

D µ ψ(x) = (∂ µ + ieA µ (x)) ψ(x)
D µ φ(x) = (∂ µ + iebA µ (x))φ(x)
ψ(x) → ψ ′ (x) = exp(−iα(x))ψ(x)
φ(x) → φ ′ (x) = exp(−ibα(x))φ(x)
A µ (x) → A ′ µ(x) = A µ (x) + i e (∂ µU(α))U −1 (α)
= A µ (x) + i
be (∂ µU(bα))U −1 (bα)
= A µ (x) + 1 be ∂ µ[bα(x)] = A µ (x) + 1 e ∂ µα(x)
and there would be exactly one electromagnetic field being generated by
both fermion fields. This will be different for non-abelian gauge fields.
There the charges are not arbitrary.
5. From the above Lagrangean (114) one derives immediately the inhomogeneous
Maxwell equations (24). They are linear in the field tensor F µν or
in the potential A µ . In the Maxwell-equations the only coupling of A is
to the current (115) in the form j µ A µ . They do not contain terms of the
sort AA or AAA or AAAA. Those would correspond to a selfcoupling of
the gauge field. This selfcoupling does not appear here and this is consistent
with the fact that the A-field is not charged. Thus a photon-photon
interaction does not exist. It exists only in higher order quantum field
theory.
6. The features 1. to 3. occur in all field theories, abelian and non-abelian in
one or the other form. The selfcoupling, however, is fundamentally different
in non-abelian field theories: It really exists and makes the treatment
and phenomenology of these theories often quite different.
The coupling of the photon field A to the fermion field ψ can graphically be
depicted as
f
γ
54

5.2 Non-abelian gauge theory
We assume the ψ-field to be an n-tupel
⎛
ψ(x) =
⎜
⎝
ψ 1 (x)
ψ 2 (x)
...
ψ n (x)
The free field is described by the well known Lagrangean (6) and we assume the
m to be a diagonal n · n-matrix with identical masses as entries.
Global symmetry: We consider now a SU(N)-transformation (51 ) and
consider the representation with dimension n i.e. the representation of the
generators 1 2 τa are n · n-matrices:
⎞
⎟
⎠
ψ(x) → ψ ′ (x) = exp(−i τa θ a
)ψ(x) (116)
2
ψ(x) → ψ ′ (x) = ψ(x) exp(+i τa θ a
2 )
The 1 2 τa (a = 1,2,3...N 2 − 1) are n · n matrices of the n-dimensional
representation of the SU(N)-group. In fact they are the representations of
the generators of the SU(N). The θ a are a real constants, that is why the
transformation is called global, since the θ a are not dependent on space or time.
The τ a satisfy the commutation rules (48)
[ ] τ
a
2 , τb τ c
= if abc
2 2
As we know the structure coefficients f abc define uniquely the group SU(N).
Due to the fact that they do not commute with each other the SU(N)-Lie-group
is non-abelian. This means that two successive transformations give a different
result when one interchanges the order. We denote
and then we have
U(θ) = exp(−i τa θ a
2 )
U(θ 1 )U(θ 2 )ψ ≠ U(θ 2 )U(θ 1 )ψ
The invariance of the free Lagrangean under SU(N) is a remarkable one.
It says that one can start from the fields ψ k and is allowed to perform in a
well defined way linear combinations ot those fields to get new fields ψ ′ k and
these new fields, combined again to a n-plet, obbey the same Lagrangean. This
seemingly absurd thing is the SU(N)-symmetry of L 0 and it is really true.
Local symmetry: We demand now that the L 0 Lagrangean is locally
invariant under SU(N), which means that we demand invariance allowing the
55

θ a = θ a (x) to be differentiable functions depending on the space-time points
x. This is a very strong demand, it says that we can change at any space-time
point the phase of the Dirac field independently of the phase at any other point
and the Lagrangean, i.e. the dynamics of the Dirac field should be the same.
We have
U(θ) = exp(−i τa θ a (x)
) (117)
2
The assertion, which we have to proof is the following: The Lagrangean L 0 =
ψ [iγ µ ∂ µ − m]ψ as such is invariant under the global SU(N) transformation
but NOT invariant under the local one. Similar to the abelian case we have
to complement the Lagrangian by adding an additional field A µ (x). Since our
invariance is now more complicated one gauge field is not enough and instead we
have to consider A a µ(x) with a = 1,2,3...N 2 − 1. Apparently we have so many
A a -fields, as there are generators. For this we define the covariant derivative
analogously to eq.(106)
(
D µ ψ(x) = ∂ µ + ig τa A a )
µ(x)
ψ(x) (118)
2
and we demand for the field A a µ(x) something similar as in the non-abelian case
eq.(109):
D µ ψ(x) → [D µ ψ(x)] ′ = U(θ)[D µ ψ(x)] (119)
or explicitely
D µ ψ(x) → [D µ ψ(x)] ′ = exp(−i τa θ a (x)
)[D µ ψ(x)] (120)
2
We damand this because (proof below) then the more general Lagrangean
is gauge invariant:
L = ψ [iγ µ D µ − m]ψ
This lagrangean is exactly the like the one from the abelian theory eq.(107)
except that now the ψ is a multiplet and the covarinat derivative contains a
more complicated field A a µ(x) raather than just A µ (x). The above demand is
actually a demand for the behauviour of the gauge field A a µ(x) under gauge
transformations, and it implies analogously to eq.(108):
τ a A a µ(x)
2
→ τa A a µ(x) ′
2
( τ a A a µ(x)
= U(θ(x))
2
Proof: The above demand implies
(∂ µ + ig τa A a µ(x) ′
2
)
(U(θ)ψ) = U(θ)
or [
(∂ µ U(θ)) + ig τa (A a µ(x)) ′
2
]
U(θ)
56
)
U −1 (θ(x))+ i g [∂ µU(θ(x))]U −1 (θ(x))
(
∂ µ + ig τa A a )
µ(x)
ψ
2
ψ = igU(θ) τa A a µ(x)
ψ
2
(121)

or
τ a (A a µ(x)) ′
2
= U(θ) τa A a µ(x)
U −1 (θ) + i 2 g [∂ µU(θ(x))]U −1 (θ(x))
qed
Actually we have reached our goal somehow, since the Lagrangean L 0 = ψ [iγ µ D µ − m]ψ
is now SU(N)-gauge invariant.
We have now D ′ µψ ′ = UD µ ψ or D ′ µUψ = UD µ ψ from which follows, since
ψ is arbitrary, that
D ′ µ = UD µ U −1
Analogously to the abelian theory (112) we define now
(D µ D ν − D ν D µ ) ψ = ig τa
2 F a µνψ (122)
or explicitely and analogously to eq.(111)
F a µν(x) = ∂ µ A a ν(x) − ∂ ν A a µ(x) − gɛ abc A b µ(x)A c ν(x) (123)
The gauge transformation proterties of F µν can now easily be inferred: They
are slightly more complicated than in the abelian case of eq.(113)
Proof: We have from eq.(122)
τ a
2 F ′a
µν = U τa
2 F a µνU −1 (124)
or
and
[D µ ,D ν ] = ig τa
2 F a µν
[D ′ µ,D ′ ν] = ig τa
2 F ′a
µν
[D ′ µ,D ′ ν] = [UD µ U −1 ,UD ν U −1 ] = U[D µ ,D ν ]U −1 = igU τa
2 F a µνU −1 qed
Knowing this we can now write down the complete Lagrangean including kinetic
terms for the boson fields and being invariant under SU(2) (local) gauge
transformation. I looks to the QED lagrangean (110), however the meaning of
the quanities is different:
L = ψ [iγ µ D µ − m]ψ − 1 4 F a µνF µν
a (125)
[Note: Unlike the Lorentz indices, µ,ν there is no difference between color
indices, a,b raised or lowered. The choice is dictated by conveninece in writing.]
57

For this we have to proof, that the kinetic energy term is indeed gauge invariant:
Proof: We apply (49):
F ′a
µνF ′µν
a = 1 2
∑
ab
= 2 ∑ ab
= 2 ∑ ab
F ′a
µνF ′µν
b
2δ ab = 1 2
Tr(UFµν
a τ a
F a µνTr( τa
2
∑
ab
2 U −1 UF µν
b
τ b µν
)F
b
= ∑ 2
ab
F µνF ′a ′µν
b
Tr[τ a τ b ] = 2 ∑ ab
Tr(Fµν
a τ a
τ b
2 U −1 ) = 2 ∑ ab
Tr(F µν
′a τ a
2 F µν
b
F a µνF µν
b
δ ab qed
τ b
2 )
2 F ′µν
b
The Lagrangean looks nearly identical to the corresponding one in the abelian
case (110). There is, however, a fundamental difference. In the term FµνF a a
µν
there are factors contained, which are trilinear and quadrilinear in A:
F a µνF µν
a
∼ −gf abc ∂ µ A a νA bµ A cν − g2
4 fabc f ade A b µA c νA dµ A eν
These terms correspond to a selfcoupling of the gauge field with itself. This
means, that the gauge field is ”charged” and hence interacts with itself. This
property makes the theory fundamentally different from the abelian case, where
there is no selfcoupling. Actually the Lagrangean can be rewritten in the following
way
L = ψ [iγ µ ∂ µ − m]ψ − 1 4 F a µνF µν
a
− igj a µA µ a
with the current
jµ a = ψγ µ τa
2 ψ
apparently this current generates the gauge field
The interactions of this lagrangean can be depicted as follows:
τ b
2 )
Several remarks are in order:
1. The coupling of the non-abelian A-field to the current is universal as it is
in the case of an abelian field. That means any fermion field creates the
same structure of the A-field, if the lagrangean is to be SU(2) gauge invariant.
However there is a restriction for the charge (with charge we mean
the fermion-A-coupling constant) unlike to the abelian case: Suppose we
simply add to the lagrangean with the fermion field ψ with coupling g
another fermion field φ with a coupling ˜g = bg.
L = ψ [iγ µ D µ − m]ψ + φ[iγ µ D µ − m]φ − 1 4 F a µνF µν
a
D µ ψ(x) =
(
∂ µ + ig τa A a )
µ(x)
ψ(x)
2
58

(
D µ φ(x) = ∂ µ + igb τa A a )
µ(x)
φ ′ (x)
2
ψ(x) → ψ ′ (x) = exp(−i τa θ a
2 )ψ(x)
φ(x) → φ ′ (x) = exp(−ib τa θ a
2 )φ(x)
One likes to have one gauge field with the transformation properties (121)
However, this provides immediately problems: The gauge transformation
of the field φ can be written as having the angle bθ a (x). This is identical
to having generators of the group not τ a but bτ a . This, however, has
consequences since the bτ a have to fulfill commutation rules (in contrast
to the abelian case). They are given by (48) and should fulfill also
[ ] bτ
a
2 , bτb bτ c
= iɛ abc
2 2
This, however, requires b 2 = b or b = 1. Hence all particles coupling to
a given gauge field, must have the same charge. In an abelian theory the
r.h.s of these equations vanishes and hence there is no restriction on the
charge.
59

2. There is no mass term for the gauge field in the Lagrangean. If it was
there it should have the structure A a µA µ a and would be not gauge invariant
as one can see at eq.(121).
6 The QCD-Lagrangean
6.1 The gauge transformation
The Lagrangean of the QCD results from starting from the free Lagrangean of
fermions with flavour f:
N f
∑
L 0 = q f [iγ µ (D µ ) − m f ]q f
f=1
Quark fields: Spin- 1 2 fermions, 6 flavours (N f = 6). We demand now that
each quark of a particular flavour has three components with different colourquantum
number. The QCD-Lagrangean can be constructed by gauging the
colour degrees of freedom with a SU(3) − colour gauge transformation.
The gauge transformation, under which this Lagrangean is invariant, acts
on the colour degree of freedom and is ( for each flavour separately) following
eq.(51)
U(θ(x)) = exp(−i λa θ a (x)
) (126)
2
q f (x) → q ′ f(x) = U(θ(x))q f (x)
The λ a are Gell-Mann Matrices of the SU(3)-algebra. They defined by
eq.(48) and are given in eq.(52). The totally antisymmetric structure coefficients
f abc and the totally symmetric ondes d abc can be found in eq.(53) and
eq.(54), respectively. In the end the full QCD-Lagrangean is given by
N f
L QCD = − 1 2 Tr(G ∑
µνG µν ) + q f [iγ µ D µ − m f ]q f (127)
The colour-index is hidden, the flavour index is f . The covariant derivative
is defined as, with g being a dimensionless coupling constant:
(
D µ q f (x) = ∂ µ + ig λa A a )
µ(x)
q f (x)
2
Gluon field:
Massless, Spin-1 bosons, colour octet because the group contains eight generators,
flavour singulet (i.e. gluons are flavour-blind, not distinguishing between
f=1
61

different flavours of whatever object it acts.
A µ (x) =
8∑
a=1
A a µ(x) λa
2
Thus explicitely with flavour index f and colour index c and Dirac index d one
writes the QCD-lagrangean (127) as
L QCD = − 1 N
∑ f
2 Tr(Ga µνG µν
a )+ q fcd
[i(γ µ ) dd ′
f=1
) ]
(δ cc ′∂ µ + ig( λa
2 ) cc ′Aa µ(x) − δ dd ′δ cc ′ q fd′ c ′
We define a field strenght tensor in the usual nonabelian way
or explicitely
G µν (x) = ∂ µ A ν (x) − ∂ ν A µ (x) + ig [A µ (x),A ν (x)]
G a µν(x) = ∂ µ A a ν(x) − ∂ ν A a µ(x) − gf abc A b µ(x)A c µ(x)
with
G µν (x) = λa
2 Ga µν(x)
The factor in front of the kinetic energy of the gluons comes from the normalization
of the Gell-Mann matrices 49:
Tr(G µν G µν ) = 1 4 Tr(λ aλ b )G a µν(x)G µν
b (x) = 1 2 Ga µν(x)G µν
a (x)
The classical equations of motion are
(iγ µ D µ − m f )q f (x) = 0
D µ G a µν(x) = g ∑ f
q f (x) λa
2 γ νq f (x)
In the limit of vanishing coupling constant g → 0 these equations describe a very
simple world, i.e. free massive quarks and independently from that a gluonic
field. The gluon field is, however, quite complicated and not comparable with
the QED-field, because of the self interacing terms. The full theory is even
more complicated since the quarks get dressed with gluons etc , dynamic mass
generation happens and the experimental spectrum has nothing to do with free
quarks.
It is interesting to note, that the gluonic field tensor G a µν is not invariant
under the gauge tranformation. This is in contrast to the gauge property of
the QED. There the invariance of the field tensor was necessary since the corresponding
electric and magnetic field was observable via the Lorentz force. Here
in QCD it is different: The gluonic field is not observable directly. Observable
62

are only glueballs and complex quark-gluon systems. Glueballs are stable gluon
configurations which are observable since they are colour singuletts. They have
been ”observed” in lattice gauge calculations of pure gluon field.
One often finds a different form of the QCD-lagrangean. There one redefines
the gluon field in such a way thqt it absorbs the strong coupling constant:
gA µ → A µ . Then we have the lagrangean equivalent to (127) as
L QCD = − 1
N
2g 2 Tr(G ∑ f
µνG µν ) + q f [iγ µ (∂ µ + iA µ ) − m f ]q f
6.2 Strong coupling constant
f=1
We know that we have to renormalize the QCD as we have to renormalize QED
(read Mandl-Shaw for an elementary introduction). This means that e.g. the
following graphs have to be summed up in order to describe properly the quark
gluon vertex:
This summation (including higher order terms) results in an effective vertex
with an effective coupling constant ḡ = ḡ(Q 2 ). Thus the interaction between
63

the gluons themselves and the quarks and gluons is dependent on the process
considered, in particular dependent on the momentum transfer Q 2 of the process.
This is a result of renormalization techniques and the transition from g to
gdepending on the scale chosen Q 2 .
This is expressed in the “running coupling constant”. With the runing coupling
constant is defined as
α s (Q 2 ) = g2 (Q 2 )
4π
(128)
.Here the effective coupling constant ḡ(Q 2 ) is given by means of the betafunction,
which is calculated perturbatively:
with the beta-function
log( Q2 2
Q 2 ) =
1
β(α s ) = α s(Q 2 )
b 1 +
π
∫ ḡ(Q
2
2 )
ḡ(Q 2 1 )
dg
β(g)
( α
2
s (Q 2 )
π
)
b 2 + ... (129)
und
b 1 = −[ 11 3 N c − 2 3 N f] (130)
The fact that the beta-function is negative for small values α s is very important,
because due to this feature we have asymptotic freedom, i.e. for large
Q 2 we have a loragithically vanishing effective coupling constant. A picture of
α s (Q 2 ) is given as
64

The running coupling constant is given in QED equally to (128) as
α ( Q 2) = ē2 (Q 2 )
4π
(131)
65

with α ( Q 2 → 0 ) = 1
137
. However the variation of the electromagnetic running
coupling constant with Q 2 is extremely small compared to the variation of the
strong running coupling constant (see figures and eq.(132)). For QED the change
of α ( Q 2) is less than 0.1 % over the energies of the available accelerators.
As an easy example to obtain α s (Q 2 ) consider the process e + e − → q¯q divided
by e + e − → µ + µ − .
e+
q+
e−
q−
e+
µ+
e−
µ−
This ratio is given by(Leader, Predazzi Vol.2, p.134)
σ(e + e − → q¯q)
σ(e + e − → µ + µ − ) = 3∑ Q 2 f(1 + α s(s)
+ 1.411( α s(s)
π π )2 )
where s = (p e + + p e −) 2 , e.e. the energy of the colliding leptons. One performs
the measurement at various values of s and determines the αs(s)
π
by comparison
of the formula with experiment. Actually in practice there other, fully strong experiments,
which determine the α s . From the particle data booklet one extracts
for three-jet events in e + e − -collisions the value
α s (s = m 2 Z) = 0.1134 ± 0.0035
66

α s (s = 34GeV 2 ) = 0.1424 ± 0.018
Usually one does not plot α s in dependence of the Mandelstam variable s but
in dependence on the virtuality of the impinging particle (coming from deep
inelastic collisions: There one has the asymptotic form i.e. for large values
of Q 2 where large means much larger than 10GeV 2 .Perturbtion theory of first
order in the QCD-coupling constant gives
α s (Q 2 ) =
4π
( 11
3 N c − 2 3 N f)
ln
(
Q 2
) (132)
Figure on α s
Λ 2 QCD
times correction terms. The Λ QCD is a parameter, which must be determined
from the experimental values of α s . One obtains
Λ QCD = (150 − 300)MeV (133)
Apparently the value of Λ QCD is not well known, since it appears in the
logarithmus, which is a very slowly varying function. The values of α s (s) and
of α s (Q 2 ) are related by analytical continuation. Actually the determination of
such an important constant as Λ QCD is a bit tricky:
67

1) The processes to do experimentally are complicated and hence there are
non-negligeable experimental errors. Furthermore the Λ QCD appears in the
logarithm, where it is difficult to extract from the experimental data of α s .
2) Each extraction of Λ QCD is based upon a pertubative pQCD calculation
to some order, e.g. leading order or next to leading order and a renormalization
scheme. This means the α s used in the calculation must be small, and so must
be the α s resulting in the above formula. This is a consistency condition.
3) The value of Λ QCD dependes on the number of active flavours in the
process considered. Active flavours are those, whose mass is below the scale
of the reaction, i.e. below the momentum transfers or energies of the reaction.
This is analog to perturbation theory in non-relativistic quantum mechanics,
where there appears in the first order perturbation theory for the wave function
an energy denominator. This denominator is, in our context, twice the mass of
the quark. Hence the heavier the quarks, the less their Dirac sea is perturbed.
This means there are often threshold effects, whose treatment deservs other
techniques and where the above formula (132) cannot be used.
For Q 2 → ∞ the interaction vanishes, which is called “asymptotic freedom”.
At Q 2 ≤ 1GeV 2 the dependence of α s on Q 2 is very uncertain and in the limit
Q 2 → Λ QCD the above expression even diverges. This is not a problem: The reason
for divergence is that in the construction of α s (Q 2 ) one needs perturbative
QCD (i.e. Feynman perturbation theory). This perturbative theory assumes a
coupling constant α s ≪ 1, which is for Q 2 ≤ 1GeV 2 not the case. Hence to
use eq.(132) for small Q 2 is not consistent. There are modern theories, which
have arguments for another dependence of α s (Q 2 ) on Q 2 at small Q 2 . There
the α s (Q 2 ) grows monotonically for Q 2 → 0 but never exceeds a value of around
1.5. A suggestion has been made by Solovtsov and Shirkov: The curves a and b
show the 1-loop analytic results of their theory for Λ QCD = 200 and 400MeV .
The cuves c and d the corresponding perturbative results.
Important Remark: The very feature of QCD is the fact that at low energies
(few GeV) the physical degrees of freedom have nothing to do with the
elementary degrees fo freedom. This is quite different from QED, where we
always have electrons and photons. At low energies the physical degrees of freedom
in QCD are baryons and mesons. The elementary degrees of freedom are
quarks and gluons. The fact that they are so different is a consequence of the
non-abelian nature of the gauge symmetry and the fact that the coupling constant
is so large and that the quantum fluctuations are really important. This
feature will it make necessary to consider current algebras, which are purely
non-perturbative concelpts
6.3 Mandelstam variables
For general education it is interesting to consider the kinematics of the above
process a bit more in detail. Here we have two real incoming particles and
two real outgoing particles. That means the momenta at the particles are all
timelike: p 2 i = m2 i . The coordinates of the process are given by
68

c
d
b
a
Q (GeV)
69

Mandelstam Variables for the reaction a + b → A + B are generally defined
as properties:
s = (p a + p b ) 2 = (p A + p B ) 2 (134)
t = (p a − p A ) 2 = (p b − p B ) 2
with the subsidiary condition
u = (p a − p B ) 2 = (p b − p B ) 2
s + t + u = ∑ i
m 2 i
One has to distinguish this kinematics from the one, where one considers
a virtual particle hitting a nucleon as e.g in the determination of structure
functions or form factors: There we have
q raumartig
virtual particle
hitting a nucleon
p zeitartig
p’ zeitartig
7 Symmetries and anomalies
7.1 Mass terms
It will be very important and convenient, do distinguish in QCD betweeen light
and heavy quarks:
⎫
m u = mū = (4 ± 2)MeV ˜5MeV ⎬
m d = m ¯d = (8 ± 4)MeV ˜10MeV
⎭ m q ≪ Λ QCD :::: or ::::: M N ,M ρ
m s = m¯s = (164 ± 33)MeV ˜180MeV
(135)
70

m c = 1.4 : GeV
m b = 5.3GeV
m t = 175GeV
⎫
⎬
⎭ m q ≫ Λ QCD :::::: or :::::: M N ,M ρ (136)
One should realize, that the ratio of mu
mt
= 1/30000. The mass of the nucleon
is M N = 938MeV and of the Rho-Meson is M ρ = 770MeV . (With numbers
like this one must be a bit careful: The quark masses are scale dependent.)
By energy reasons we see immediately that the nucleon is governed by up and
down quarks and also a bit of strange quark antiquark admixtures. Thus for
the nucleon at reactions, with momentum transfers or energies smaller than
the masses of the heavy quarks, one can ignore the existence of the heavy
quarks, since their quark-antiquark excitation requires too much energy and
hence (by non-relativistic simple perturbation theory) those excitations are suppressed.However,
in changing the energy in an experiment one must be careful of
threshold effects. Altogether: For low energies one can consider the Lagrangean
with three flavours as the relevant QCD-Lagrangean:
L QCD =
3∑
f=1
q f [iγ µ D µ ]q f − 1 4 Ga µνG µν
a
−
3∑
¯q f m f q f (137)
The fact, that at low energies the heavy quarks do not play a role has the
consequence that certain symmeetries, which are valid for a massless theory,
like e.g. the chiral symmetry, is approximately valid for the nucleon. One has
now to do a decision, if one considers the strange quark as a “light” quark or as
a “heavy” quark. This means, one has to decide if one postulates approximately
a SU(2) or SU(3) global flavour invariance. Thus one rewrites (137) as
f=1
L QCD = ψ [iγ µ D µ ]ψ − 1 4 Ga µνG µν
a − ¯ψmψ (138)
In SU(2)-flavour the ψ is a doublet consisting of the up-spinor and down-spinor
( )
ψu (x)
ψ =
ψ d (x)
and the mass term is
or
with
( )
mu 0 ¯ψ
0 m d
ψ = m u + m d
2
(ūu + ¯dd) + m u − m d
(ūu −
2
¯dd)
( )
mu 0 ¯ψ ψ =
0 m ¯ψ [ m 0 I + m 3 τ 3] ψ (139)
d
m 0 = m u + m d
2
(140)
m 3 = m u − m d
2
(141)
71

In SU(3) we have
⎛
ψ = ⎝
ψ u (x)
ψ d (x)
ψ s (x)
and the mass term is
⎛
⎞
m u 0 0
¯ψ ⎝ 0 m d 0 ⎠ ψ = ¯ψ [m 1 I + m 3 λ 3 + m 8 λ 8 ]ψ (142)
0 0 m s
⎞
⎠
with (141) and
m 1 = m u + m d + m s
3
m 8 = m u + m d − 2m s
2 √ 3
(143)
(144)
7.2 Vector symmetries
7.2.1 Global fermion symmetry
In the Lagrangean of QCD (137) there exists a symmetry of the form
q f → q ′ f = exp(−iθ f )q f (145)
for each flavour separately and with constant θ f . This is obvious because the
gluons are not affected by this transformation. The corresponding current is
given by
j µ f = q fγ µ q f (146)
The corresponding charge of a given flavour corresponds to the number of quarks
minus the number of antiquarks. Since the invariance property holds for each
flavour separately one can form linear combinations like e.g:
with
j µ em = ψQ em γ µ ψ = 2 3 uγµ u − 1 3 dγµ d − 1 3 sγµ s (147)
⎛
Q em = ⎝
+ 2 3
0 0
0 − 1 3
0
0 0 − 1 3
The current (147) is the electromagnetic current. Apparently it originates as
conserved current from another symmetry of the Lagrangean, namely U(1) local,
but is conserved in QCD as well, however as consequence of the global flavour
symmetry.
⎞
⎠
72

7.2.2 Global (iso-)vectorial symmetry
SU(2): Consider up- and down-quarks in QCD (138) and assume them to have
the same mass. Then we have the global isospin invariance of (138):
(
u
ψ =
d
)
→ ψ ′ = exp(−iθ a τa
)ψ (148)
2
with the τ a being the tree Pauli-matrices (12). We know this invariance from
the studies around eqs. (81). The Noether current corresponding to the action
of (148) to the QCD lagrangean is a triple:
V µ = jµ a τ a
= ψγ µ ψ a = 1,2,3 (149)
2
This Noether-current is conserved if the masses of up- and down-quarks are
identical.
SU(3): The same symmetry reads in SU(3):
⎛
ψ = ⎝
u
d
s
⎞
⎠ → ψ ′ = exp(−iθ a λa
)ψ (150)
2
Here the λ a are the Gell-mann-matrices (), however not for color (as used to
define the colour gauge transformation) but for flavour. The Noether current
corresponding to the action of (150) to the QCD lagrangean is an octet
λ a
V µ = ψγ µ ψ a = 1,2,...,8 (151)
2
This Noether-current is conserved if the masses of up- and down- and s-
quarks are identical.
The SU(3) has three subgroups of SU(2)-nature.
This is the T-spin:
( ) ( )
u
ψ = → ψ ′ = exp(−iθ a τa u
d
2 ) d
This is the U-spin:
(
d
ψ =
s
This is the V-spin:
(
u
ψ =
s
)
→ ψ ′ = exp(−iθ a τa
2 ) (
d
s
)
→ ψ ′ = exp(−iθ a τa
2 ) (
u
s
)
)
Let us now consider the mass terms eq.(135) and their effect on the divergence
of the above currents (149,151). The masses are different from each
73

other, nevertheless the currents are the same. However their divergence is not
vanishing and one can derive from eq.(55) and (63) easily a general formula
∂ µ V µ
a = ¯ψi 1 2 [λ a,m]ψ a = 1,2,3,..,8 (152)
from which we obtain in the SU(2)-case with m u = m d and m 0 = (m u +m d )/2
given by eq.(140)
∂ µ V a µ = m 0 ¯ψi τa ψ a = 1,2,3 (153)
Compared to all masses of hadrons the up- and down-mass is always small,
hence the SU(2) T-spin-symmetry (isospin-symmetry) is a generally good approximation.
The U-spin and V-spin symmetries are eplicitely broken due to
the mass difference of about 170MeV between s-quarks and u,d-quarks. Hence
predictions with isospin symmetry work at the level of 1%, whereas the corresponding
ones with U-spin or V-spin symmetry in SU(3) work at 30% accuracy.
Explicitely we can write with
the following useful formulae
and in case of allowing m u ≠ m d
V µ + = V µ
1 + iV µ
2
V µ − = V µ
1 − iV µ
2
V µ + = ¯ψ u γ µ ψ d
V µ − = ¯ψ d γ µ ψ u
∂ µ V µ + = i(m u − m d ) ¯ψ u ψ d
∂ µ V µ + = i(m d − m u ) ¯ψ d ψ u
One often defines also the isospin current V µ
3 and the hypercharge current
J µ Y in the following way (V µ as such is called isovector current):
.
V µ
3 = 1 2 [ ¯ψ u γ µ ψ u − ¯ψ d γ µ ψ d ]
J µ Y = 2 √
3
V µ
8 = 1 8 [ ¯ψ u γ µ ψ u + ¯ψ d γ µ ψ d ] − 2 3 ¯ψ s γ µ ψ s
In this business the following formula is useful
[Aλ a ,Bλ b ] = 1 2 {A,B}[λ a,λ b ] + 1 2 [A,B]{λ a,λ b }
74

7.3 Axial (non-)symmetries:
7.3.1 Flavour symmetry and U A (1) anomaly
The iso-vector symmetry holds, if the quark masses are identical. If the quark
masses vanish, there are additional symmetries because in this limit left-handed
and right-handed components of the quark-fields are decoupled. They are both
coupled to the gluon field, however that does not carry flavour quantum numbers.
So we can write an expression for the quark fields in the QCD-lagrangean
analogous to eq.(72) Thus, for the massless QCD Lagrangean one can derive a
left and right hand flavour symmetry and also a global axial flavour symmetry
(76,77) under the axial transformation
with the current ()
q f → q ′ f = exp(−iθ f γ 5 )q f (154)
j µ 5f = q fγ µ γ 5 q f (155)
However, this symmetry exists only for the classical lagrangean, it does not
exist in a full quantum field theory (without proof). There the correspponding
current is not conserved. Instead we have so called axial anomaly, where the
divergence of the axial vector current is related to the gluon field:
with the definition
∂ µ j f 5µ = α s
8π Ga µν
µν ˜G a for eachflavour f separately (156)
˜G µν
a = ɛ µναβ G a αβ (157)
If one combines the fields with different flavour f into the usual dublets for
SU(2)-flavour and triplets for SU(3)-flavour it makes sense to consider the singlet
axial current and writes is in a different nomenclature for e.g. SU(3)-flavour
as
j (0)
5µ = ¯ψγ µ γ 5 ψ = ūγ µ γ 5 u + ¯dγ µ γ 5 d + sγ µ γ 5 s (158)
Here the anomaly equation reads as follows and is is called the U A (1)−anomaly:
∂ µ j (0)
5µ = N fα s
8π Ga µν
µν ˜G a (159)
with N f = 2 for SU(2) and N f = 3 for SU(3). The reason for this anomaly
is the following: If the renormalization procedure of QCD one has to regularize
the integrals in the Feynman diagrams and at the end of the procedure one lets
the cut-off parameter go to infinity or, if one does a dimensional regularization,
one lets the ε-parmaeter go to zero. Actually one can chose a regularization in
such a way that there is no U A (1)−anomaly, or exactly that we have an axial
singlet current (158) which vanishes in the massless limit. Unfortunately one
cannot avoid in this case that the singlet axial current j (0)
5µ is no longer gauge
invariant under the SU(3)-colour−transformation of QCD. Since, however, QCD
is constructed by demanding colour-gauge invariance such a situation is not
75

acceptable. Hence one must choose a regularization, in which the colour-gauge
invariance is preserved but then one cannot avoid to have the U A (1)-anomaly.
In fact this anomaly is responsible for the decay of the π 0 → γγ, which
is experimentally well known. The axial anomaly is also responsible for the
fact that the mass of the η ′ -Meson is considerably larger than the mass of the
η-meson (m η ′ = 958MeV,m η = 547MeV ). Without the U A (1)-Anomaly the
massless QCD would have a chiral U(3) R ◦U(3) L -symmetry, and its spontaneous
breakdown would lead to nine rather than eight Goldstone bosons and this
would include the η ′ -Meson. The axial anomaly removes the U A (1)-symmetry,
keeping SU(3) R ◦SU(3) L ◦ U(1) intact, which is spontaneously broken down to
SU(3) V ◦ U(1) V .
In case we consider quark masses the formula modifies to
∂ µ j (0)
5µ = 3α s
8π Ga µν
µν ˜G a + 2i ( )
m u ¯ψu γ 5 ψ u + m d ¯ψd γ 5 ψ d + m s ¯ψs γ 5 ψ s
Proof: Using eqs.(18,19) we can write
∂ µ j (0)
5µ = ∂µ [ ¯ψγ µ γ 5 ψ] = (∂ µ ¯ψ)γµ γ 5 ψ + ¯ψγ µ γ 5 (∂ µ ψ)
= (∂ µ ¯ψ)γµ γ 5 ψ − ¯ψγ 5 γ µ (∂ µ ψ)
(160)
= (im + iγ µ A µ ) ¯ψγ 5 ψ − ¯ψγ 5 (−im − iγ µ A µ )ψ = im ¯ψγ 5 ψ − ¯ψγ 5 (−imψ) qed
We note here that the finite quark masses do not modify the coefficient in font
of the anomaly term.
In reading al the arguments about regularization one realizes that they apply
equally well to QED if one replaces the strong coupling constant by the
electromagnetic one and the gluonic field tensor by the electromagnetic field
tensor.
7.3.2 Axial Vector symmetry and Axial Anomaly
Massless quarks: The QCD-Lagrangean is for massless quarks invariant under
the axial vector transformation, often called also chiral transformation:
ψ → ψ ′ = exp(−iθ a λa
2 γ 5)ψ (161)
It results in the axial current
A a µ = ¯ψγ µ γ 5
λ a
2 ψ (162)
For a = 0 we have the above case of a singlet axial current and it is U A (1)−anomalous
(see prev. subsubsection). For the cases a = 1,2,4,5,6,7 the divergence is zero
since the λ a are not diagonal and hence, written explicitely there are always
non-diagonal terms (e.g. of the sort ūγ µ γ 5 d) whose divergence does not show
an anomaly. For the cases a = 3,8 we also do not have an anomaly, but the
76

eason is different. For these a = 3,8 the λ-matrices are diagonal but traceless,
so we have e.g. for λ 8 a term like ūγ µ γ 5 u + ¯dγ µ γ 5 dūγ − 2¯s µ γ 5 s where the
U A (1)−anomaly cancels.
However, explicit calculations within a QCD and electromagnetic SU(3) color
⊗ U(1)
theory show, that this axial vector for a = 3,8 has another anomaly, not with
the gluon field G a µνbut with the photon field F µν :
∂ µ A 3 µ = N cα
12π F µν ˜F µν (163)
∂ µ A 8 µ = N cα
12π F µν ˜F µν (164)
Actually a direct consequence of this anomaly is the fact that the neutral pion
can decay intotwo photons: π 0 → γγ. This decay is much faster, since it is
connected with the electromagnetic coupling constant α, than the decay due
to the electroweak interaction (small weak coupling constant of Fermi), under
which the charged pions decay. So in practice the anomalous decay of the pion
π 0 → γγ is most important. The details of the pion decay will come later.
Explicitly we can write the divergence of the axial vector current in the
presence of finite quark masses. In addition to the anomaly term for a = 3 and
a = 8 one obtains for a = 1 → 8 :
∂ µ A µ a = ¯ψiγ 5 1 2 {λ a,m} ψ (165)
For isospin symmetry in the up-down sector this is very simple
∂ µ A µ a = m 0 ¯ψiγ 5 τ a ψ (166)
with m 0 = mu+m d
2
given in eq.(140). Explicitly we can write also the following
useful formulae
A µ + = A µ 1 + iAµ 2
A µ − = A µ 1 − iAµ 2
with
A µ + = ¯ψ u γ µ γ 5 ψ d
A µ − = ¯ψ d γ µ γ 5 ψ u
and
∂ µ A µ 3 = m ¯ψ u u iγ 5 ψ u − m d ¯ψd iγ 5 ψ d
∂ µ A µ + = (m u + m d ) ¯ψ u iγ 5 ψ d
∂ µ A µ − = (m u + m d ) ¯ψ d iγ 5 ψ u
77

7.4 Other symmetries, Theta-vacuum
7.4.1 Discrete symmetries
The standard model is a Lorentz invariant local quantum field theory.Its Lagrangean
is hermitean. It is invariant under the combined set of transformations
CPT. The QCD separately conserves C, P, and T separately (we assume that
there is no θ-term, see later).
7.4.2 Scale invariance and trace-anomaly
If the quarks were massless the QCD lagrangean would contain no dimensional
quantity. The action (not the Lagrangean) would therefore be invariant under
the scale transformations
ψ(x) ↦→ ψ ′ (x) = λ 3 2 ψ(λx) A
a
µ (x) → λA a µ(λx) (167)
Proof: Take a symbolic QCD-Lagrangean of the form L QCD = q [∂ + D µ ]q,
where we write down the action, and in the action we perform a change of
variable y = λx:
∫
S =
∫
=
∫
=
∫
=
d 4 y[q(y) d q(y) + q(y)A(y)q(y)]
dy
λ 4 d 4 x[q(λx) 1 d
q(λx) + q(λx)A(λx)q(λx)]
λ dx
λ 3 d 4 x[q(λx) d q(λx) + q(λx)λA(λx)q(λx)]
dx
d 4 x[λ 3 d
2 q(λx)
dx λ 3 3
3
2 q(λx) + λ 2 q(λx)λA(λx)λ 2 q(λx)]
One sees immediately that the above replacements concerve the action. q.e.d.
This is a continuous symmetry and hence it must have a conserved Noether
current. This current is given by (without proof)
J µ scale (x) = x νΘ µν (x) (168)
There the Θ µ ν is the energy momentum tensor (64) of QCD and given by:
Θ µν = −g µν (− 1 4 Gλσ a G a λσ + ψiγ λ D λ ψ) − G µλ
a G ν aλ + A ν a∂ λ G µλ
a + (169)
+ i 2 ψγµ ( ←− ∂ + −→ ∂ ) ν ψ + g µν ψmψ
The current is conserved on the classical level, because the above mentioned
invariance under a continuous scale transformation holds:
∂ µ J µ scale (x) = 0
78

One can calculate the divergence immediately:
∂ µ J µ scale = ∂ µ(x ν Θ µν ) = ∂ µ (g νρ x ρ Θ µν ) = g νρ δ ρµ Θ µν +g νρ x ρ ∂ µ Θ µν = g νµ Θ µν = Θ ν ν
where we have used the fact that the energy momentum tensor is conserved,
which is fulfilled for a time independent and homogeneous system, what QCD
is. .
∂ µ Θ µν (x) = 0
Thus a conserved scaling current has as consequence that the trace of the
energy momentum tensor vanishes:
Θ ν ν(x) = 0
If the scaling current would really be conserved such a situation with a vanishing
trace of the energy momentum tensor would have drastic consequences on the
theory, since all particles of this world (now built of massless quarks) would
then be massless, i.e. the proton would be massles, the pion as well, etc..To
proof this statemtne we forst show: For any hadron H the matrix element of
the energy-momentum tensor at zero momentum transfer is (assertion):
〈
H(p)
∣ ∣ Θ µν (0) ∣ ∣ H(p)
〉
= 2p µ p ν (170)
Proof:: We have on one hand for a given hadron
〈H(p ′ )| ̂P µ |H(p)〉 = p µ 2p 0 δ (3) (p − p ′ )(2π) 3
and on the other hand
∫
〈H(p ′ )| ̂P µ |H(p)〉 =
〈p ′ |Θ µ0 (x) |p〉 d 3 x
Using the shift operator (34) we obtain
〈H(p ′ )| Θ µ0 (x) |H(p)〉 = exp(+ix(p ′ − p)) 〈H(p ′ )| Θ µ0 (0) |H(p)〉
We can rewrite this by noting that the integral over exponential gives the deltadistribution
∫
∫
〈H(p ′ )| Θ µ0 (x) |H(p)〉 d 3 x = [ d 3 xexp(+ix(p ′ − p))] 〈H(p ′ )| Θ µ0 (0) |H(p)〉
= exp(+ix 0 (p ′ 0 − p 0 ) 〈H(p ′ )|Θ µ0 (0) |H(p)〉 (2π) 3 δ (3) (p − p ′ )
Since the 3-momenta are equal the 0-components are equal as well and the
exponential function equals One. Then one gets by comparison
〈H(p)| Θ µ0 (0) |H(p)〉 = p µ 2p 0
and by Lorentzs-invariance this holds not only for ν = 0 but for all ν. Thus one
gets the assertion eq.(170).q.e.d.
79

A vanishing trace would mean
〈
H(p)
∣ ∣ Θ
µ
µ (0) ∣ ∣ H(p)
〉
= 2p µ p µ = 0 = 2M 2 H
This is direct a problem of QCD since the proton is built of up and down
quarks, which are basically massless. On the contrary the experimental value
of the proton mass is 938 MeV. The very fact is, that this scale current J µ scale is
anomalous. The divergence of the quantum current is given by (without proof)
∂ µ J µ scale = Θν ν = β QCD
2g Ga µνG µν
a + m u ūu + m d ¯dd + ms ss (171)
Here β QCD is the beta-function of QCD, known from renormalization techniques
and perturbation theory of QCD, see (129) with (130).This term is the
anomalous term, which gives rise to the ”Scale anomaly”. The other terms are
proportional to the quark masses and are the result of explicit breaking of the
scale invariance. This means that e.g. for the nucleon we have
M N ū(p)u(p) = 〈 N(p) ∣ ∣ β QCD
2g Ga µνG µν
a + m u ūu + m d ¯dd + ms ss ∣ ∣ N(p)
〉
(172)
Since we have for the Dirac spinors the normalization
ū(p,r)u(p,s) = 2M N δ rs
we can write
M N = 1
2M N
〈
N(p)
∣ ∣
β QCD
2g Ga µνG µν
a + m u ūu + m d ¯dd + ms ss ∣ ∣ N(p)
〉
(173)
The term corresponding to the light masses m u ,m d is related to the famous
Σ πN −Term. It is measured and shows, that it contribute to about 45MeV or,
according to more modern measurements, to about 65MeV to the total mass
of the nucleon of m N = 938MeV . This leaves the bulk of the nucleons mass
to the gluons and to the s-quarks. This is very interesting because it says, that
the by far largest contribution to the nucleon mass comes from non-valence
particles! This is a model independent statement and is based purely on QCD
and experiment. There is one problem in this argument. If we consider heavy
quarks then their contribution is negligeable since < N(p)|m Q QQ ∣ ∣ N(p)
〉
/2MN
is smaller than 1MeV and hende negligeable even for the heavy top-quark. In
facht the condensate in the nucleon is that small Altogether, ist is fully justified
to ignore the contributions of heavy quarks to the nucleon mass, nucleon angular
momentum, nucleon momentum, etc.. The “intrinsic” charm, e.g., is small.
One can support these arguments quantitatively: The contribution of the s-
quarks to the mass of the nucleon is still under debate, but more and more people
agree, that it is not small because of the large mass m s = 180MeV . Estimates
in the literature are for nucleons with momentum p = 0, with m 0 = 1 2 (m u+m d )
from (140)
Σ πN = m 0
2M N
〈
N(p)
∣ ∣ mu ūu + m d ¯dd
∣ ∣N(p)
〉
= (45 ± 5)MeV (174)
80

1 〈 ∣
N(p) β QCD
∣
2M N 2g Ga µνG µν
a
∣N(p) 〉 = 634MeV :::: (764MeV ) (175)
1
2M N
〈
N(p)
∣ ∣ ms ss ∣ ∣ N(p)
〉
= 260MeV ::::: (130MeV ) (176)
In modern texts the value of the sigma-term is believed to be noticeaby larger,
i.e. Σ πN˜(65 ± 8)MeV , but the question is not completely settlet yet. Note
that the bulk contribution to the energy comes from the gluons, but the strange
quarks also contribute noticeably, in fact more clearly than the up- and downquarks.
This fact provides a problem to a naive interpretation of the nucleon as
composed of three constituent up- and down-quarks without internal structure.
Actually it shows that there is some gluon field and even ss-fields inside the
constituent quarks, which are the dominant carrier of the energy density. Thus
in the process of forming a constituent quark, the quark is “dressed” by gluonic
and strange fields.
The arguments leading to the above numbers can be understood after having
studied the quark model. Basically they are the following: One can introduce
a mass splitting operator führe dieses
L m−s = 1 3 (m 0 − m s )(ūu + ¯dd − 2ss)
später vor
Actually the mass-splittings between the hyperons are governed by this operator.
Thus one can show by very simple means involving SU(3)-symmetry and
perturbation to first order in m s :
δ s = 〈N|(m 0 − m s )(ūu + ¯dd − 2ss) |N〉
2m N
= 3 2 (m Ξ − m N ) = 574MeV
One should note that the structure of the nucleon is given by P = uud and
N = udd and of the Xi by Ξ 0 = uss ::::: Ξ − = dss. Multiplying this with the
small ratio m0
m s
and neglecting small terms yields the quantity
δ = m 0
〈N|ūu + ¯dd − 2ss |N〉
2m N
m 2 π
= 3 2 m 2 K − (m Ξ − m N ) ∼ = 25MeV (35MeV )
m2 π
where the number in brackets includes higher order chiral corrections (Gasser
87). Comparison of δ und σ yields, that the strange quark matrix element does
not vanish. Indeed, one requires
〈N|ss |N〉
〈N|ūu + ¯dd + ss |N〉 ∼ = 0.18 ::: (0.09)
if one uses these values one obtains the above numbers.
81

7.4.3 Theta-Vacuum
There are a great differences between an abelian gauge theory and a non-abelian
one. One of those is the Theta-Vacuum. The gauge transformations one usually
considers are those, which are connected to the identity operator in a continuous
manner. They are called small gauge transformations.. There are however
some, where this is not so, they are called large gauge transformations. For
the following we consider the temporal gauge, which is given by A a 0 = 0 for
a = 1,...,8. By definition the small gauge transformations do not change the
winding number, the large ones do. To study this we remember the gauge degree
of freedom (121), which is still there even if we have decided to work inside the
temporal gauge. We restrict ourselve to the SU(2) subgroup a = 1,2,3 within
the temporal gauge :
τ a A a j (x)
2
→ τa A a j (x)′
2
( τ a A a j
= U(θ(x))
(x)
2
)
U −1 (θ(x))+ i g [∂ jU(θ(x))] U −1 (θ(x))
where the θ(x) has no time-dependence, since otherwise ba the gauge transformation
an A 0 a ≠ 0 would be created.
Consider now the following function:
Λ 1 (x) = x2 − d 2
x 2 + d 2 + 2idτx
x 2 + d 2
One can formulate a gauge transformation in such a way that we have
A j a(x) λ a
2 → U(α (x) Aj a(x) τ a
2 U −1 (x) + ∆A j a(x)
τ a
2 ∆Aj a(x) = − i g [∇ jΛ 1 (x)]Λ −1
1 (x)
2d [
= − τj
g(x 2 + d 2 (d 2 − x 2 ) + 2x j (τ a x a ]
) − 2d(x × τ) j
)
The interesting fact is, that this gauge transformation transforms even a
trivial vacuum field A j a(x) = 0 into a finite gauge field. The d is an arbitrary
free parameter, thus even for d → 0 we do not meet the identity, since then
the Λ 1 (x) becomes singular at x = 0. These gauge potentials carry a so called
topological charge or the winding number (without proof):
n = ig3
24π 2 ∫
d 3 xTr( τa
2 Aa i (x) τb
2 Ab j(x) τc
2 Ac k(x))ɛ ijk (177)
If one constructs the gauge potentials out of the trivial vacuum then one can
show that for all values of d the gauge potentials carries a conseraved topological
charge for the above transformation
n = 1
82

If we apply the above gauge transformation several times one can successively
construct A-fields with winding numbers n = 1,2,3,.....and also with n =
−1, −2, −3,..... Since these are all obtained from the trivial vacuum by a gauge
transformation, their energy is always zero. Hence one can plot the energies of
the vacua against the winding number:
E(A)
0
n
Indeed, all possible vacua, having al vanishing energy, can be classified into
disjunct sectors |n〉 labelled by their winding number n. Inside the sector we
have small gauge transformations, if we transform from one sector to another
one we need large gauge transformations. And there exist operators,U 1 which
do the transformation of the sort
U 1 |n〉 = |n + 1〉
This implies that the real vacuum, being necessarily gauge invariant, requires
contributions from all classes, such as the coherent superposition
|θ〉 = ∑ n
exp(−iθn) |n〉
where θ is an arbitrary parameter. This vacuum is gauge invariant since we
have
U 1 |θ〉 = ∑ n
exp(−iθn) |n + 1〉 = exp(−iθ) ∑ m
exp(−iθm) |m〉 = exp(−iθ) |θ〉
This Theta-vacuum is interesting because of several reasons.
1) If θ ≠ 0 then the so formulated QCD is no longer invariant under parity
transformation and also under time-reversal transformation. Thus it violates
CP-invariance in addition to the usual mechanism caused by the Cabibbo-
Kobayashi-Maskawa matrix. CPT is always conserved. If these symmetries are
not conserved, then the neutron should have a electric dipole moment. There are
83

measurements of this but the upper limits are so small, that certainly θ ≤ 10 −6 .
Hence presently one does not know theta but only its upper limit.
2) If one performs for any observable a path integral over the gauge-field
A a µ(x) then the integral must extend over all gauge fields including those, which
lead from one “minimum” with winding number n to another “minimum” with
another winding number. Such fields do exist. They start off at t =-∞ as the
zero potential A a µ(x) = 0 , have some interpolating A a µ(x) ≠ 0 for intermediate
times, and end up at t =+∞ in the gauge equivalent configuration A j µ(x) =
∆A j a(x) and a vanishing 0-component. For those fields one can show that the
following integral is non-vanishing:
g 2 ∫
24π 2
d 3 xTr( τa
2 ∆Aa i (x) τb
2 ∆Ab j(x) τc
2 ∆Ac k(x))ɛ ijk
which means that the interpolating field configuration gives a change in the
winding number between asysmptotic gauge field configurations.
Ε
n
The contributions from those fields are truly non-perturbative since they are
not confined to one minimum, as Feynman diagrams are (this can be demonstrated,
but will not be shown here)
Actually the height of the barriers between the “minima” depends on the
spatial extension we allow the intermediate fields A(x,t) in the functional integral.
Such a relation is necessary in order to fulfill the scale invariance of the
(massless) QCD. Basically all sizes have to be considered, since the functional
integral is not restricted. Now the interestign feature is the following:
If we consider small extensions of the interpolating field contributing to the
functional integral (small compared to 0.1 fm) then the corresponding barrier is
high and for those configurations it is sufficient to approximate the functional
integral by a saddlepoint aproximation and zero-mode fluctuations and small
fluctuations around the saddle point. The configuration of the saddle point are
the so called instantons. They are the solutions of the classical equations of
motion in the Euclidean space. The zero mode fluctuations around them correspond
to symmetries of the instantons, are treated via collective coordinates
84

and can be treated exactly. The small fluctuations around the instantons correspond
to a perturbative treatment along the non-perturbative intermediate
field (instantons leading from one minimum to another). As long as they are
small such a perturbative treatment is justified and it can be done in fact.
If we consider large extensions of the interpolating field contributing to the
functional integral, then the corresponding barrier is low. Still the saddlepoint
approximation consists in taking the instantons and zero-mode fluctuations and
small fluctuations around them. However, for small barriers this is no longer a
good approximation: One has in principle to sove fully the functional integral
and the techniques for this are unknown. That is the reason, why QCD is not
fully solvable due to non-perturbative effects.
Actually, given this nontrivial vacuum structure, one requires three ingredients
to completely specify QCD: 1) the QCD lagrangean including all quark
masses, 2) the coupling constant, i.e. Λ QCD , 3) the vacuum label θ. One can
show that one can modify the QCD-Lagrangean, known to us so far, by adding
a simple term, which does not destroy the gauge invariance and leads for a given
θ to a corresponding Theta-vacuum:
˜G
µν
a
L QCD = L (θ=0)
QCD + θ g
64π 2 Ga µν
µν ˜G a
with given by eq.(157). So without proof we state, taht a correct procedure
for doing calculations involving θ-vacua is to follow the ordiönary path
integra method but with this modified Lagrangean. One sees clearly: A different
θ corresponds to a different Lagrangean and hence to a different theory
altogether. One further sees: The operator F ˜F is P-odd and T-odd, because
it corresponds to EB and this changes sign under Parity and Time-reversal.
Hence, as mentioned above, a finite θ introduces CP-violation.
8 Anomalies
9 Electroweak Interactions in the Standard Model
By now we have investigated symmetries of the lagrangeans and the corresponding
symmetry currents as e.g. V a
µ or A µ a. Some of these currents were
preserved,others had a non-vanishing divergence. The interactions of a hadronic
system, consisting of quark and gluons, with external fields of electroweak characterm,
is given by other currents, i.e. the electromagnetic current J µ EM , the
weak charged current J µ CC and the weak neutral current Jµ NC
. The interesting
point is that both sorts of currents are intimately related in such a way that the
interaction currents can be expressed in terms of the symmetry currents. To
see this we first repeat in the following the basic interactions of the Standard
Model. Actually, we do not consider colour currents interacting with an external
glluon filed since the latter one does not exist as an external field due to the
confinement mechnism.
Here one
should insert
Mario´s notes
on anomalies.
Check them ,
only very little
problem with
some technical
specifications
85

The Standard Model of electromagnetic and weak interactions is based on local
gauge symmetry under the group SU(2) L ×U(1). After spontaneous symmetry
breaking through a Higgs mechanism the interaction part of the Lagrangian
is:
L int = − eJ µ EM (x)A g
µ(x) − J µ NC
2cos θ (x)Z µ(x)
W
−
g
2 √ 2 Jµ CC (x)W† µ(x) + Hermitian conjugate.
(178)
It involves J µ EM , Jµ CC and Jµ NC , and their couplings to the photon field A µ, the
charged W-boson fields W ± µ and the neutral Z 0 -boson field Z µ . The electromagnetic
and weak couplings are related by the Weinberg angle θ W :
sin θ W = e g .
Its cosine gives the ratio of W- and Z-boson masses:
cos θ W = M W
M Z
.
The weak coupling constant g is usually given interms of the Fermi coupling
constant, G F , whic are related by
G F
√
2
=
g2
8M 2 W
At present the best values of these parameters are:
.
G F =1.16639 ± 0.00001 × 10 −5 GeV −2 ,
sin 2 θ W =0.2230 ± 0.0004,
M W =80.42 ± 0.06GeV; M Z = 91.188 ± 0.002GeV.
(179)
The electroweak currents in Eq.(178) include sums over all quarks and leptons.
We now specify the quark currents, starting from the electromagnetic ones.
9.1 Electromagnetic Quark Currents
The u-,d- and s-quarks are the building blocks of a fundamental representation
of the unitary flavor group SU(3) f . Together they form a triplet field ψ(x):
⎛ ⎞
ψ u (x)
ψ(x) = ⎝ψ d (x) ⎠
ψ s (x)
where x ≡ x µ = (t,⃗x) represents time and space coordinates. Each of the spin
1
2 quark fields ψ u, ψ d and ψ s is a four-component Dirac field which annihilates a
quark or creates an antiquark of a given flavor. (For the present discussion color
86

plays no role and we shall drop explicit color labels. We simply note that all
electroweak currents (e.g. Eq.(178) , below), being color blind, involve a trace
over color.)
The hypothesis that the quarks are elementary, pointlike Dirac particles with
charges Q u = + 2 3 and Q d = Q s = − 1 3
(in units of e,electrons having the charge
−e) immediately implies that their electromagnetic (EM) current is
J µ EM (x) = ¯ψ(x)Qγ µ ψ(x) (180)
with Dirac matrices γ µ (µ = 0,1,2,3), ¯ψ = ψ † γ 0 , and our conventions for the
metric, Dirac matrices and so on are summarized in one fo the first sections.
The charge matrix ⎛
2/3 0 0
⎞
Q = ⎝ 0 −1/3 0 ⎠ (181)
0 0 −1/3
takes into account the electric charges of the quarks. More explicitly,
J µ EM (x) = (+2 3 ) ¯ψ u (x)γ µ ψ u (x) + (− 1 3 ) ¯ψ d (x)γ µ ψ d (x) + (− 1 3 ) ¯ψ s (x)γ µ ψ s (x).
(182)
The J µ EM
(x) is conserved since each flavour component is conserved due to the
flavour symmetry. Note that the u- and d-quarks alone form an isospin I = 1/2
doublet, with
( )
⃗t 2 ψu
= 3 ( ) ( )
ψu ψu
, t 3 = 1 ( )
ψu
,
ψ d 4 ψ d ψ d 2 −ψ d
where ⃗t = 1 2 (τ 1,τ 2 ,τ 3 ) are the generators of the isospin SU(2) group . Their
EM current has an isoscalar and an isovector part. The charge operator for the
u- and d-quarks is
SU(2) : Q = 1 2 (B + τ 3) =
(
2/3 0
0 −1/3
)
, (183)
where the isoscalar part involves the quark baryon number, B = 1/3, and
the isovector part is proportional to τ 3 . When the s-quark is included, the
isoscalar part is generalized to incorporate its strangeness S = −1, and the
baryon number is replaced by the hypercharge, Y = B + S and Q = Y 2 + I 3
with
⎛
⎞
⎛ ⎞
Y = λ 8
√
3
=
1/3 0 0
⎝ 0 1/3 0 ⎠ , I 3 = λ 1/2 0 0
3
√ = ⎝ 0 −1/2 0⎠ , (184)
0 0 −2/3 2 0 0 0
in terms of the flavor SU(3) matrices λ 3 and λ 8 . The electromagnetic current
(182) can now be written as
J µ EM (x) = 1 2 Jµ Y + V µ
3
87

with the isospin (isovector) current
and the hypercharge current
V µ
3 (x) = ¯ψ(x)I 3 γ µ ψ(x) = 1 ( ¯ψu γ µ ψ u −
2
¯ψ d γ µ )
ψ d ,
J µ Y (x) = ¯ψ(x)Y γ µ ψ(x) = 1 3
( ¯ψu γ µ ψ u + ¯ψ d γ µ ψ d
)
−
2
3 ¯ψ s γ µ ψ s .
The generalization including the heavy (c, b and t) quarks is straightforward but
will not be of great relevance as long as we consider light mesons and baryons
build of u-, d- and s-quarks.
9.2 Weak Quark Currents
Charged current (CC) weak interactions occur in processes like neutron beta
decay, muon capture or neutrino scattering, i.e.:
¯ν µ + p → µ + + n,
Neutral current (NC) interactions can be studied, either through direct reactions
such as
ν µ + p → ν µ + p,
or through interference effects such as the measurement of parity violation in
electron scattering (performed in order to learn something about the strange
quark content of the nucleon). As a consequence we shall need all of the weak
currents associated with quarks.
Where it is necessary to distinguish between the hadronic and the leptonic
weak currents we shall add an appropriate subscript – e.g. J µ had,W
. This will be
dropped where it is obvious from the context which current is intended. The
full hadronic, weak current will be conveniently separated into neutral (NC) and
charged (CC) current pieces:
J µ had,W = Jµ NC + Jµ CC .
The NC is the easiest to write as it is diagonal in flavor:
(
J µ had,NC = ¯ψ 1
u γ µ 2 − 4 3 sin2 θ W − 1 )
2 γ 5 ψ u + (u → c) + (u → t)
+ ¯ψ d γ
(− µ 1 2 + 2 3 sin2 θ W + 1 )
2 γ 5 ψ d + (d → s) + (d → b).
(185)
Whereas the neutral current couplings are identical (within experimental
accuracy) for u,c and t as well as for d, s and b, the situation for the charged
couplings is considerably more complicated. The hadronic part of the charged
current CC is given by
( ) ( )
J µ had,CC = ¯ψ u → c u → t
u γ µ (1 − γ 5 )ψ d ′ +
d ′ → s ′ +
d ′ → b ′ + h.c. (186)
88

where h.c. stands for the Hermitian conjugate. The structure of this current
is characterized by ”V − A”, i.e. vector current minus axial current. In this
current there appear the the weak eigenstates d ′ , s ′ and b ′ obtained through
a unitary transformation involving the Cabibbo-Kobayashi-Maskawa (CKM)
matrix V applied to the quark mass eigenstates with (weak) isospin − 1 2 (i.e.
d, s and b), which are the states appearing in the propagator. The Cabibbo-
Kobayashi-Maskawa (CKM) matrix V :
⎛ ⎞ ⎛ ⎞
d ′ d
⎝s ′ ⎠ = V ⎝s⎠ , (187)
b ′ b
where V is expressed in terms of three ”Cabibbo angles”, θ 1,2,3 , and a phase
angle δ. The matrix V has the form
⎛
⎞ ⎛
⎞
c 1 −s 1 c 3 −s 1 s 3 V ud V us V ub
V = ⎝s 1 c 2 c 1 c 2 c 3 − s 2 s 3 e iδ c 1 c 2 s 3 + s 2 c 3 e iδ ⎠ = ⎝V cd V cs V cb
⎠ ,
s 1 s 2 c 1 s 2 c 3 + c 2 s 3 e iδ c 1 s 2 s 3 − c 2 c 3 e iδ V td V ts V tb
(188)
where for ease of writing we have abbreviated cosθ 1 as c 1 ,sin θ 2 as s 2 and so
on. (This is one out of several equivalent parametrizations of the CKM matrix.
The CKM-matrix is often also named as indicated in the RHS of the above
formula.) The reason for the existence of the Cabibbo-Kobayashi-Maskawa
matrix is, that in weak processes the W-bson couples only to the left handed
particles and right handed anti-particles. This means, it can only decay into
leptons or by W → u L ¯dR or W → d L ū R there is no decay of the sort W →
u R ¯dL . Thus in weak processes the d-, s- and b-quarks are generated only via
their left-handed component (weak eigenstates), whereas the eigenstates of the
hamiltonian, describing their free evolution in time (mass eigenstates), involves
the full wave function having both left- and right handed components. There
is nothing particular in all these down quarks, one can formulate the theory
equivalently in such a way that one mixes all quark states or one uses a CKMmatrix
only for u-, c- and t-quarks instead for d-, s- and b-quarks. This one
can see if one writes down a 6-plet Ψ = (ψ u ,ψ d ,ψ c ,..,ψ b ) and some sort of
CKM-matrix Ṽ yielding for the charged hadronic current
J had,CC = ¯Ψγ µ (1 − γ 5 )Ṽ Ψ (189)
One can apply the Ṽ to Ψ, or to ¯Ψ or write Ṽ = S† S and apply S to Ψ and S †
to ¯Ψ.
For practical purposes θ 2 and θ 3 are small. Thus for most applications in
this book we can ignore in the CKM-matrix (188)all but the usual Cabibbo
angle θ C = −θ 1 , writing
( d
′
s ′ )
( )( )
cos θC sinθ C d
∼= , (190)
−sin θ C cos θ C s
with sinθ C = 0.220 ± 0.002 and cos θ C = 0.975 ± 0.002. Sometimes we are
ignoring even this mixing and assume θ C = 0.
89

9.3 Leptonic currents
The leptonic elcetromagnetic current is clear and simple
J µ lep,EM = − ¯ψ e γ µ ψ e + (e → µ) + (e → τ) (191)
Finally, we shall also need the leptonic weak current, J µ lep,W
, which we summarize
here:
J µ lep,W = Jµ lep,NC + Jµ lep,CC .
Here the charged current is
J µ lep,CC = ¯ψ νe γ µ (1 − γ 5 )ψ e +
and the neutral current is
( )
νe → ν µ
+
e → µ
( )
νe → ν τ
+ h.c., (192)
e → τ
J µ lep,NC = ¯ψ e γ µ (g V − g A γ 5 )ψ e + (e → µ) + (e → τ)
+ 1 2 ¯ψ νe γ µ (1 − γ 5 )ψ νe + (ν e → ν µ ) + (ν e → ν τ ),
(193)
with
g A = − 1 2 ; g V = 2sin 2 θ W − 1 2 .
Equipped with these basic tools we are now ready to explore the structure of the
nucleon. We will in the following look in which way the interaction currents can
be replaced with symmetry currents, whose algebraic rules (current algebras)
we know.
10 Chiral symmetry breaking
10.1 Chiral Symmetry and Current algebras
In the limit of massless quarks the QCD lagrangean has the symmetry related
to the conserved right- or left-handedness (chirality) of zero mass spin-1/2 particles.
From this one can derive the invariance under the global SU(3) (iso-)vector
flavour transformation (89)
ψ(x) → ψ ′ (x) = exp(−iα a λa
2 )ψ(x)
and the SU(3) axial flavour transformation (90)
ψ(x) → ψ ′ (x) = exp(−iα a λa
2 γ 5)ψ(x)
One can now write down the algebras of the vector and axial currents, and
these expressions remind us of the expressions we had in the general chapter
90

on Noether currents: With the explicit expressions of the vector and axial currents()
and ()
V µ (a) = ¯ψγ τ a
µ
2 ψ
A (a)
µ = ¯ψγ τ a
µ γ 5
2 ψ
We also can define vector and axial vector charges
∫ ∫
Q V a (t) =
d 3 xV 0
a (x) =
∫ ∫
Q A a (t) = d 3 xA 0 a(x) =
d 3 xψ † (x) λa
2 ψ
d 3 xψ † (x)γ 5 λa
2 ψ
Actually one should note: The charge operator e.g. Q A a (t) is an operator in the
Hilbert space spanned by the creation and annihilation operators of the field
operators ψ of all flavours. They are not operators in the vlavour space where
the λ a act nor in the spinorspace, where the γ µ act.The charge operators are
time independent since the chiral currents are conserved. This holds only if the
fermions are considered massless. Actually we have a mass term in the QCD-
Lagrangean, thus the time independence holds only approximately. It can be
used in particular processes, where the momentum transfer is that large and
sets such a large scale that one can neglect the quark masses.
We have a closed algebra for the vector charges and an open algebra for the
axial charges:
[
Q
V
a ,Q V ]
b = ifabc Q V c
[
Q
V
a ,Q A ]
b = ifabc Q A c
[
Q
A
a ,Q A ]
b = ifabc Q V c
and similarly
[Q a (t),V µ
b (t,x)] = ifabc V µ
c (x)
[Q a (t),A µ b (t,x)] = ifabc A µ c (x)
[
Q
5
a (t),A µ b (t,x)] = if abc V µ
c (x)
and similarly
[
Q
5
a (t),V µ
b (t,x)] = −if abc A µ c (x)
δ(x 0 − y 0 ) [ V 0
a (x),V 0
b (y) ] = iδ(x − y)f abc V 0
c (x)
91

δ(x 0 − y 0 ) [ V 0
a (x),A 0 b(y) ] = iδ(x − y)f abc A 0 c(x)
δ(x 0 − y 0 ) [ A 0 a(x),A 0 b(y) ] = iδ(x − y)f abc V 0
c (x)
These relations are obtained directly by explicit calculations following the techniqus
explained below. They hold, even if there are finite quark masses, because
the definition of the currents involves only the kinetic energy of the quarks and
the transformations done on the system. We need the properties of SU(3)Lie
groups with the totally antisymmetric structure coefficients f abc
[ λ
a
2 , λb
2
]
abc λc
= if
2 ::::::::: Tr(λa λ b ) = 2δ ab
and the symmetric coefficients d abc with d abc = d bac = d acb etc:
{
λa
2 , λ }
b
= 1 2 3 δab + d abc λ c
2
The proofs of the above formulae are based on the following equations, which
one can prove by writing them down explicitely:
and
[AB,C] = A[B,C] + [A,C] B
[C,AB] = [C,A] B + A[C,B]
[AB,C] = A{B,C} − {A,C}B
[C,AB] = {C,A}B − A{C,B}
We will prove now the following formulae
[Q V a ,ψ(y)] = − λ a
2 ψ(y)
[Q V a , ¯ψ(y)] = λ a
2 ¯ψ(y)
[Q V a ,γ 5 ψ(y)] = − λ a
2 γ 5ψ(y)
[Q V a , ¯ψ(y)γ 5 ] = λ a
2 ¯ψ(y)γ 5
Before one starts to calculate one should note: The charge operator Q V a (t)
is an operator in the Hilbert space spanned by the creation and annihilation
operators of the field operators ψ of all flavours. It is are not an operator in
the flavour space where the λ a act nor in the spinorspace, where the γ µ act.
92

Proof: At equal times and with A = ψ † (x),B = λa
2
ψ(x),C = ψ(y) we have with
[AB,C] = A{B,C} − {A,C}B
∫
[Q V a ,ψ(y)] = d 3 x[ψ † (x) λa
2 ψ(x),ψ(y)]
∫
= d 3 x(ψ † (x){ λa
2 ψ(x),ψ(y)} − {ψ† (x),ψ(y)} λa
2 ψ(x))
∫
= − d 3 xδ 3 (x − y) λa
ψ(x) = −λa ψ(y) qed
2 2
Doing the hermitian conjugate we obtain the second formula:
[Q V a ,ψ(y)] † = −( λ a
2 ψ(y))†
−[Q V a ,ψ † (y)] = −(ψ † (y) λ a
2 )
−[Q V a ,ψ † (y)γ 0 ] = −ψ † (y)γ 0 λ a
2
−[Q V a , ¯ψ(y)] = − ¯ψ(y) λ a
2
[Q V a , ¯ψ(y)] = ¯ψ(y) λ a
2
We need the analogue formulae for axial quantities
[Q A a ,ψ(y)] = − λ a
2 γ5 ψ(y)
[Q A a , ¯ψ(y)] = − ¯ψ(y)γ 5 λ a
2
qed
and because γ 5 = γ † 5 also [Q A a ,γ 5 ψ(y)] = − λ a
2 ψ(y)
[Q A a , ¯ψ(y)γ 5 ] = − ¯ψ(y) λ a
2
The proof for this is similar to the above one and like that. We take
[AB,C] = A{B,C} − {A,C}B and use A = ψ † ,B = λa
2 γ5 ψ(x),C = ψ(y)
and equal times:
∫
]
d 3 x
[ψ † (x) λa
2 γ5 ψ(x),ψ(y)
∫
= d 3 x(ψ † (x){ λa
2 γ5 ψ(x),ψ(y)} − {ψ † (x),ψ(y)} λa
2 γ5 ψ(x))
∫
= − d 3 xδ 3 (x − y) λa
2 γ5 ψ(x) = − λa
2 γ5 ψ(y) qed
93

To get the commutator of Q A a with ¯ψ we need the hermitian conjugate (and
the fact that γ † 5 = γ 5and γ 5 anticommuting with γ 0 )
[Q A a ,ψ(y)] † = −( λ a
2 γ5 ψ(y)) †
−[Q A a ,ψ † (y)] = −(ψ † (y)γ 5 λ a
2 )
−[Q A a ,ψ † (y)γ 0 ] = −ψ † (y)γ 5 λ a
2 γ0
−[Q A a ,ψ † (y)γ 0 ] = ψ † (y)γ 0 γ 5 λ a
2
−[Q A a , ¯ψ(y)] = ¯ψ(y)γ 5 λ a
2
[Q A a , ¯ψ(y)] = − ¯ψ(y)γ 5 λ a
2
qed
And now we can derive two formulae, which we need later in the section on
PCAC and the vacuum condensates and sigma terms::
[
Q
V
a , ¯ψ(y)λ b ψ(y) ] = 2 ¯ψ(y)[ λ a
2 , λ b
2 ]ψ(y) = ifabc ¯ψλc ψ (194)
[
Q
A
a , ¯ψ(y)λ b γ 5 ψ(y) ] = −2 ¯ψ(y){ λ a
2 , λ b
2 }ψ(y) = −2 3 δ ab ¯ψ(y)ψ(y) − d abc ¯ψ(y)λc ψ(y)
(195)
[
Q
A
a , ¯ψ(y)λ b ψ(y) ] = −2 ¯ψ(y){ λ a
2 , λ b
2 }γ5 ψ(y) = − 2 3 δ ab ¯ψ(y)γ 5 ψ(y) − d abc ¯ψ(y)λc γ 5 ψ(y)
(196)
The proof is the following using the above simple commutators: At equal times
and with C = Q V a ,A = ¯ψ,B = λ b ψ one obtains from [C,AB] = [C,A] B +
A[C,B]
[
Q
V
a , ¯ψ(y)λ b ψ(y) ] = [ Q V a , ¯ψ(y) ] λ b ψ(y) + ¯ψ(y) [ Q V a ,λ b ψ(y) ]
= ¯ψ(y) λ a
2 λ bψ(y) − ¯ψ(y)λ b
λ a
2 ψ(y)
= 2 ¯ψ(y)[ λ a
2 , λ b
2 ]ψ(y) = λ c
ifabc ¯ψ(y) ψ(y) qed
2
The proof for the second formula is the following: At equal times and with
C = Q V a ,A = ¯ψ,B = λ b γ 5 ψ one obtains from [C,AB] = [C,A] B + A[C,B] the
94

following equations:
[
Q
A
a , ¯ψ(y)λ b γ 5 ψ(y) ] = [ Q A a , ¯ψ(y) ] λ b γ 5 ψ(y) + ¯ψ(y) [ Q A a ,λ b γ 5 ψ(y) ]
= − ¯ψ(y) λ a
2 γ5 λ b γ 5 ψ(y) − ¯ψ(y)λ b
λ a
2 γ5 γ 5 ψ(y)
= −2 ¯ψ(y){ λ a
2 , λ b
2 }ψ(y)
= − 2 3 δ ab ¯ψ(y)ψ(y) − d abc ¯ψ(y)λc ψ(y) qed
if one now inserts the explicit expression for e.g. λ 1 and with the only
non-zero relevant coefficient d 118 = 1 √
3
one obtains the above assertion:
[
Q
A
1 , ¯ψλ 1 γ 5 ψ ]
⎛
= − 2 3 ( ¯ψ u ψ u + ¯ψ d ψ d + ¯ψ s ψ s ) − d 118 ( ¯ψ u , ¯ψ s , ¯ψ s ) √ 1 ⎝
3
1 0 0
0 1 0
0 0 −2
= − 2 3 ( ¯ψ u ψ u + ¯ψ d ψ d + ¯ψ s ψ s ) − ( ¯ψ u , ¯ψ s , ¯ψ s ) 1 3 ( ¯ψ u ψ u + ¯ψ d ψ d − 2 ¯ψ s ψ s )
= −( ¯ψ u ψ u + ¯ψ d ψ d )
⎞
⎠
⎛
⎝
⎞
ψ u
ψ d
⎠
ψ s
.
We need in the following the special cases of this
[
Q
A
1 , ¯ψγ 5 λ 1 ψ ] = − ( ¯ψu ψ u + ¯ψ d ψ d
)
[
Q
A
8 , ¯ψγ 5 λ 8 ψ ] = − 1 3
( ¯ψu ψ u + ¯ψ d ψ d − 4 ¯ψ s ψ s
)
(197)
As example we give the derivation of the first of them. When we inserts the
explicit expression for e.g. λ 1 and with the only non-zero relevant coefficient
d 118 = √ 1
3
one obtains:
[
Q
A
1 , ¯ψλ 1 γ 5 ψ ]
⎛
= − 2 3 ( ¯ψ u ψ u + ¯ψ d ψ d + ¯ψ s ψ s ) − 2d 118 ( ¯ψ u , ¯ψ s , ¯ψ 1
s )
2 √ ⎝
3
1 0 0
0 1 0
0 0 −2
= − 2 3 ( ¯ψ u ψ u + ¯ψ d ψ d + ¯ψ s ψ s ) − ( ¯ψ u , ¯ψ s , ¯ψ s ) 1 3 ( ¯ψ u ψ u + ¯ψ d ψ d − 2 ¯ψ s ψ s )
= −( ¯ψ u ψ u + ¯ψ d ψ d )
One can also immediately derive the formula
[
Q
A
a , ¯ψ(y)ψ(y) ] = − ¯ψ(y)λ a γ 5 ψ(y)
⎞
⎠
⎛
⎝
⎞
ψ u
ψ d
⎠
ψ s
[
Q
A
a , ¯ψ(y)γ 5 ψ(y) ] = − ¯ψ(y)λ a ψ(y)
95

Proof: We use [C,AB] = [C,A] B + A[C,B] with C = Q A a ,A = ¯ψ(y),B = ψ(y)
and obtain immediately with the above formulae
[
Q
A
a , ¯ψ(y)ψ(y) ] = [ Q A a , ¯ψ(y) ] ψ(y) + ¯ψ(y) [ Q A a ,ψ(y) ]
[
Q
A
a , ¯ψ(y)ψ(y) ] = − ¯ψ(y)γ 5 λ a
2 ψ1944scri
y) − ¯ψ(y) λ a
2 γ5 ψ(y)
= − ¯ψ(y)λ a γ 5 ψ(y)
Actually, in his business the following formula is useful (without proof, which
however can be obtained with the above techniques):
[Aλ a ,Bλ b ] = 1 2 {A,B}[λ a,λ b ] + 1 2 [A,B]{λ a,λ b }
We will need also the following formulae (preliminary)
We have to proof the above expression of the double commutator (do not
use it,
Σ ab
πN = − [ Q a A, [ Q b A, L M
]]
Proof: We have [C,AB] = [C,A] B + A[C,B] and C = Q b A ,A = ¯ψ,B = ψand
[Q A a ,ψ] = − λa
2 γ5 ψand [Q A a , ¯ψ] 5 λa
= − ¯ψγ
2
and hence
[
Q
a
A , [ Q b A, ¯ψψ ]] = [Q a A,[C,AB]] = [Q a A,[C,A]B + A[C,B]]
= [Q a A,[Q b A, ¯ψ]ψ + ¯ψ[Q b A,ψ] ]
= −[Q a A, ¯ψλ b γ 5 ψ]
= 2 3 δ ab ¯ψ(y)ψ(y) + d abc ¯ψ(y)λc ψ(y) q.e.d.
For the following we need[C,AB] = [C,A] B + A[C,B] and C = Q a A ,A =
¯ψ,B = γ 5 ψ
[Q a A, ¯ψ γ 5 ψ ] = [ Q a A, ¯ψ ] γ 5 ψ + ¯ψ [ Q a A,γ 5 ψ ]
= − ¯ψγ 5 λ a
2 γ5 ψ + ¯ψγ 5 [Q a A, ψ]
= − ¯ψγ 5 λ a
2 γ5 ψ + ¯ψγ 5 (− λ a
2 γ5 ψ)
= − ¯ψλ a ψ
We have [C,AB] = [C,A] B + A[C,B] and C = Q b A ,A = ¯ψ,B = λ c ψ and
[
Q
A
a , ¯ψγ 5 ψ ] = − ¯ψλ a ψ and [ Q A a , ¯ψλ b γ 5 ψ ] = − 2 3 δ ab ¯ψψ − d abc ¯ψλ c ψ and hence
96

[
Q
a
A , [ Q b A, ¯ψλ c ψ ]] = [Q a A,[C,AB]] = [Q a A,([C,A]B + A[C,B])]
= [Q a A,([Q b A, ¯ψ]λ c ψ + ¯ψ[Q b A,λ c ψ]) ]
= [Q a A,(− ¯ψγ 5 λ b
2 λ cψ − ¯ψ λ c
λ b
2 γ5 ψ) ]
= −[Q a A,( ¯ψγ 5 λ b
2 λ cψ + ¯ψ λ c
λ b
2 γ5 ψ) ]
= − 1 2 [Qa A, ¯ψ{λ b , λ c }γ 5 ψ ]
So altogether we have (preliminary)
= − 1 2 [Qa A, ¯ψ(− 2 3 δbc )γ 5 ψ + ¯ψd bcd λ d γ 5 ψ ]
= − 1 3 δbc [Q a A, ¯ψ γ 5 ψ ] − 1 2 dbcd [Q a A, ¯ψ λ d γ 5 ψ ]
= − 1 3 δbc (− ¯ψλ a ψ) − 1 2 dbcd (− 2 3 δ ad ¯ψψ − d ade ¯ψλe ψ)
= − 1 3 δbc (− ¯ψλ a ψ) + 1 3 dbcd δ ad ¯ψψ +
1
2 dbcd d ade ¯ψλe ψ
= 1 6 ¯ψλ a ψδ bc + 1 3 dbca ¯ψψ +
1
2 dbcd d ade ¯ψλe ψ
[
Q
a
A , [ Q b A, ¯ψλ c ψ ]] = 1 6 ¯ψλ a ψδ bc + 1 3 dbca ¯ψψ +
1
2 dbcd d ade ¯ψλe ψ q.e.d.
10.2 Spontaneous symmetry breaking
We remember the situation of a spontaneously broken symmetry without explicit
breaking. There we have shown that we must have an excited state of
the vacuum with some particular properties, i.e. Goldstone bosonic state. We
consider SU(2) and will call this state ∣ ∣ π(p)
〉
and have then
m π = 0
〈0|π j (0) ∣ ∣ π k (p) 〉 = aδ jk
〈0|A j µ(0) ∣ ∣ π k (p) 〉 = bδ jk
〈0|∂ µ A j µ(0) ∣ ∣ π k (p) 〉 = 0
Attention: we denote field-operators by π k and the one boson field states by
∣
∣π k〉 . The first equation is merely an equation which denotes the normalization
of the boson field states, and we take for convenienct a = 1.We see what that
97

means if we take the expression of the quantized boson field (32) and use it at
x = 0
∫
d 3 k
[
φ(0) = √ a(k) + a(k) ∗]
(2π)3 2ω k
The a = 1 corresponds to
〈0| π j (0) ∣ ∣ π k (p) 〉 = δ jk
To get this the one-pion state must be normalized with a(p) |0 pion 〉 = 0 as
√
|π(p)〉 = 2ω p (2π) 3 a(p) † |0〉
or equivalently the overlap of two pion states is
〈
π k (p) ∣ ∣ π j (p ′ ) = 2ω p (2π) 3 δ (3) (p − p ′ )δ jk
Because of the Lorentz structure of the LHS the b of the second equation can
be rewritten as
〈0|A j µ(0) ∣ ∣ π k (p) 〉 = if π p µ δ jk (198)
The f π is the pion decay constant. It is not obvious at all that the f π is indeed
the pion decay constant. However, why this is so, this will be discussed in the
next subsection where we explicitely calculate the pion decay.
The fact that we have spontaneous chiral symmetry breaking has some very
simple but also very basic consequences: We know that the vacuum has the
properties
Q V a |0 >= 0 Q A a |0 >≠ 0
from which one can immeditely conclude using eq.(194) that
〈0| [ Q V a , ¯ψ(y)λ b ψ(y) ] |0〉 = if abc 〈0| ¯ψ(y) λ c
ψ(y) |0〉 = 0
2
If one uses all combinations of a and b and takes the few f abc which are nonzero
() one obtains immediately the following statements about the vacuum
condensates:
〈0| ¯ψ i ψ k |0〉 = 0 for i ≠ k (199)
and
〈0| ¯ψ u ψ u |0〉 = 〈0| ¯ψ d ψ d |0〉 = 〈0| ¯ψ s ψ s |0〉 (200)
One obtains from Q A a |0 >≠ 0 and
[
Q
A
1 , ¯ψγ 5 λ 1 ψ ] = − ( ¯ψu ψ u + ¯ψ d ψ d
)
[
Q
A
8 , ¯ψγ 5 λ 8 ψ ] = − 1 3
( ¯ψu ψ u + ¯ψ d ψ d − 4 ¯ψ s ψ s
)
the fact that the following condensates exist
〈0| ¯ψ u ψ u + ¯ψ d ψ d |0〉 ≠ 0
〈0| ¯ψ u ψ u + ¯ψ d ψ d − 4 ¯ψ s ψ s |0〉 ≠ 0
98

From these formulae we conclude that all three condensates are identical and
non-zero. As we shall see later in the section about PCAC and Gell-Mann-
Oakes-Renner we have
〈0| ¯ψ u ψ u |0〉 = 〈0| ¯ψ d ψ d |0〉 = 〈0| ¯ψ s ψ s |0〉 = −(225 ± 25) 3 MeV 3 (201)
The fact those condensates tells a lot about the vacuum |0〉 . We know the
expression for the fermion field (39). Due to translational invariance of the
vacuum we are allowed to consider
ψ(0) = ∑
r=1,2
∫
d 3 k
√
(2π)
3
and hence we have with normal ordering
〈0| ¯ψ u ψ u |0〉 = ∑ rs
[
]
c r (k)u r (k) + d † r(k)v r (k) : |0〉
√ m
[
]
c r (k)u r (k) + d †
E
r(k)v r (k)
k
∫ d 3 kd 3 k ′ √ m m
[
]
(2π) 3 〈0| : c †
E k E
r(k ′ )ū r (k ′ ) + d r (k ′ )¯v r (k ′ )
k ′
Suppose ∣ ∣˜0 〉 is be the vacuum of the operators c r and d r with c r
∣ ∣˜0 〉 = 0 and
d r
∣ ∣˜0 〉 = 0 then the above operator expression sandwiched between ∣ ∣˜0 〉 would
be obviously zero. Since it does not vanish we have |0〉 ≠ ∣ ∣˜0 〉 , which means that
|0〉 must be vacuum for another set of annihilation operators. The particles
which correspond to these new single particle operators are constituent quarks,
in contrast to the particles corresponding to the the operators c r and d r in the
quantized field expressions, which are the QCD-quarks or current quarks. The
latter name is obvious since their field operators ψ constitute the currents e.g.
A a µ(x) we are dealing within our current algebra.
10.3 Pion decay
Looking at the formulae relevant for the spontanesous breakdown of the chiral
symmetry we find that the pion decay constant f π plays a dominant role. Thus
we consider now the pion decay in nature. The purpose is to extract the f π
from experimental data.
The charged pions decay to 99% via the following channel
π + → µ + ν µ π − → µ − ν µ (202)
The neutral pion decays to 99% into two photons via the chiral anomaly
of the axial current (163). The decay of the charged pion is induced by the
electroweak interaction Lagrangean. At a scale of the W + -boson this can be
written as
99

u
µ+
W+
d
bar
ν
bar
Electroweak interaction
For Feynman-diagrams one should recall the Feynman amplitude of the process
e + e − → µ + µ − which is given by
M (2) (e + e − → µ + µ − ) = −iū (µ) (p ′ 2)γ α v (µ) (p ′ 1)D αβ
F (p1 + p 2 )¯v (e) (p 1 )γ β u (e) (p 2 )
For the present process the current ū (µ) (p ′ 2)γ α v (µ) is somewhat different since
we have a different interaction, but the structure current-propagator-current is
the same. In the present case of pion decay one one needs the propagator of the
W-boson, which is given by
(
iD αβ
−g αβ
F (k,m + k α k β /m 2 )
W
W) = i
k 2 − m 2 W + iε
Since the W + 1
is heavy (80 GeV) the propagator of it is merely since its
m 2 W
momentum dependence k = p 1 + p 2 can be ignored at low energies. If we
denote the fermion-W coupling constant by g ′ (analogous to e in the above
) 2
process) the diagram is proportional to (g′ , which is called in the definition
of the interaction Lagrangean
(
g
2 √ 2
m 2 W
) 2
.With the definition of GF as GF √
2
=
g 2
one obtains with the charged currents (186192) the corresponding effective
8MW
2
Hamiltonian:
H weak = G F
√
2
[J µ had,CC + Jµ lep,CC ][Jµ had,CC + Jµ lep,CC ]† (203)
This is the famous current-current interaction. Altogether, if we concentrate
on the decay of the π + into a pair of leptons we have to consider the matrix
element
M = 〈0 hadrons | 〈 l + ν l
∣ ∣ Hweak
∣ ∣π + 〉 |0 leptons 〉
with |0〉 = |0 hadros 〉 |0 leptons 〉. The only possible combination of fields is
here:
H weak = G F
√
2
V ud ¯ψd γ λ (1 − γ 5 )ψ u
[ ¯ψνe γ λ (1 − γ 5 )ψ e + (e ↔ µ) ]
100

with ( in another nomenclature) V ud = cos θ C = 0.975.This can be diagrammatically
represented as:
u
µ+
d
bar
ν bar
Fermi−interaction (weak)
The Lagrangean is now considered at a low scale and hence the value of G F is
taken from the decay of the free neutron and given by G F = 1.17⋆10 −5 GeV −2 .
Apparently the decay of the pion is governed by this very small coupling
constant, whereas the decay of the π 0 is govened by the electric coupling constant
α = 1
137 , which is much larger. Thus we expect the liftetime of the π+ to be
much larger than that of π 0 . This is indeed the case as the experimental numbers
indicate:
τ(π + ) = 2.6 ∗ 10 −8 s
τ(π 0 ) = 8.4 ⋆ 10 −17 s
The leptonic part of the above matirxelement is
〈
l + ν l
∣ ∣ Hweak |0 leptons 〉 = 〈 l + ν l
∣ ∣ ¯ψνe γ λ (1 − γ 5 )ψ e |0 leptons 〉
= ū ν γ λ (1 − γ 5 )v µ
This cannot be evaluated further since the neutrino and the lepton are physical
particles leaving the interaction region and being detected somewhere. The
hadronic part of this matrixelement is
〈0 hadrons | H weak
∣ ∣π + 〉 = G F
√
2
V ud
1 √2 〈0 hadrons | ¯ψ d γ λ (1 − γ 5 )ψ u [ ∣ ∣ π 1 (p) + iπ 2 (p) 〉 ]
To evaluate the non-trivial quark part one should use the following trivial identities:
¯ψ d γ λ ψ u = V1 λ − iV2
λ
¯ψ d γ λ γ 5 ψ u = A λ 1 − iA λ 2
The transition from the vacuum to the pion via the vector current vanishes,
because the pion is pseudoscalar and the current is a vector. For, if we denote the
101

parity transformation by P, we have 〈0 hadrons |P † = 〈0 hadrons |and P ∣ ∣ π k (p) 〉 =
− ∣ ∣ π k (p) 〉 and hence:
〈0 hadrons |V j µ(0) ∣ ∣ π k (p) 〉 = 〈0 hadrons |PV j µ(0)P † P ∣ ∣ π k (p) 〉 = − 〈0 hadrons |V j µ(0) ∣ ∣ π k (p) 〉
since. Thus for the strong part of the transition matrixelement the matrixelement
of the axial current is relevant, which we of course know (198). Thus we
have
Collecting terms we find
〈0 hadrons |V j µ(0) ∣ ∣ π k (p) 〉 = 0
〈0 hadrons |A j µ(0) ∣ ∣ π k (p) 〉 = if π p µ δ jk
〈0 hadrons | H weak
∣ ∣π + 〉 = G F
√
2
V ud
1 √2 〈0 hadrons |[A λ 1 − iA λ 2][ ∣ ∣ π 1 (p) + iπ 2 (p) 〉 ]
= G F 1
√ V ud √2 2if π p λ
2
This formula contains the f π simply as a number, which characterizes the
transition matrixelement from vacuum to pion via the axial current. For the
present process this is the only unknown quantity coming from the strong part.
Apparently this number can be fixed here by reproducing the decay of the real
pion. In the present nomenclature the f π is defined such that its experimental
value will be 93MeV . The decay has then the invariant Feynman amplitude
T π+ →µ + ν µ
= G F
√
2
V ud
√
2fπ p λ ū ν γ λ (1 − γ 5 )v µ
Using now the Dirac equation (γ µ p µ − m)u(p) = 0 one can simplify
getting the final expression
p λ ū ν γ λ (1 − γ 5 ) = mū ν (1 − γ 5 )
T π+ →µ + ν µ
= −G F V ud f π m µ ū ν (1 − γ 5 )v µ
An analogous expression holds for the decay π − → µ − ν µ .
One sees here the well known helicity suppression phenomenon. Besides the
decay of the pion into muons (202) one also has decay into electrons like
π + → e + ν e
π − → e − ν e
This decay is very much suppresed compared to the muon decay. The reason is,
that the weak leptonic current contains the left-handed chiral projection operator
(1−γ 5 ) which in the limit of massles leptons produces only left-handed particles
and right-handed antiparticles. In this limit “lefthandedness” means “negative
helicity” i.e. spin and momentum antiparallel, and “right-handedness”
102

means “positive helicity” i.e. spin and momentum parallel. If a pion in rest decays,
the resulting leptons must have opposite momenta. Suppose e.g. that in
the decay of π − the outgoing muon runs into the right direction. It is a particle,
has negative helicity, and so its spin points towards left. The antineutrino runs
into the left direction in order to conserve momentum, as antiparticle it must
have positive helicity and hence its spin points towards left. Thus the spins of
the muon and antineutrino add up to ONE pinting left. That contradicts the
fact that the pion has ZERO spin. Hence the pion decay is only possible because
the electron and myon are massive. Since the muon is 200 times more massive
than the electron the decay goes primarily through the muon channel.
Altogether one obtains for the decay rate
Γ π+ →µ + ν µ
= G2 F
4π |V ud| 2 fπm 2 2 µm π (1 − m2 µ
)
Before one can compare this with experiment one has to perform some radiative
corrections. Taking in the end V ud from the Kobayashi-Maskawa-Matrix and
using the experimental number
one obtains
Γ π+ →µ + ν µ
= 3.841x10 7 s −1
m 2 π
f π = (92.4 ± 0.2)MeV (204)
It is important to note, that the f π appears as number chacterizing the decay
of the pion. Simultaneously it characterzes the transition matrix element of the
axial current between vacuum and pion. The reason for this is: The pion is a
Goldstone boson, therefore This is a strong interaction matrix element.
10.4 PCAC: partial conservation of axial current
The concept of PCAC is very important, since it provides a direct link between
the QCD (quark fields) and effective theories (pion fields). The final PCAC
formula is given by
∂ µ A j µ(x) = m 0 ¯ψ(x)iγ 5 τ j ψ(x) = f π m 2 ππ j (x) (205)
This equation tells clearly that the pion field is identical to a certain combination
of quark and antiquark fields, it is only another name. This has the
consequence that it makes sense to forget the quark fields and to consider only
pionic fields. This is the philosophy of effecive theories. The main purpose for
the effective theories is, to derive equations for the π-field itself (meson field), by
this avoiding to consider equations for the ψ-field (fermion field), in such a way
that S-matrix elements or more generally observables in the effective theory and
in QCD are very similar.. Effective theories provide often advantages because
they are often simpler and nevertheless hit the correct physics. We know such
a theory, it is the chiral sigma model with Gell-Mann-Levy lagrangean. We will
103

learn about others, for instance the model independent lagrangean used e.g. in
chiral perturbation theory, or later on the Skyrme model, etc..
For the proof of PCAC (205) we calculate the divergence of the axial vector
current. In fact we had from spontaneous chiral symmetry breaking the basic
relation (198) In order to calculate now the matrixelement of the divergence of
the current explicitey we use the shift operator (34)
〈0|A j µ(0) ∣ ∣ π k (p) 〉 = 〈0| exp(−iPx)A j µ(x)exp(+iPx) ∣ ∣ π k (p) 〉 = 〈0|A j µ(x) ∣ ∣ π k (p) 〉 exp(+ipx)
or
〈0|A j µ(x) ∣ ∣π k (p) 〉 = 〈0|A j µ(0) ∣ ∣π k (p) 〉 exp(−ipx) = if π p µ δ jk exp(−ipx)
and hence
〈0|∂ µ A j µ(x) ∣ ∣π k (p) 〉 = (∂ µ exp(−ipx))if π p µ δ jk = f π p µ p µ δ jk exp(−ipx) = f π m 2 πδ jk exp(−ipx)
Apparently, for a conserved axial current (CAC) we have (this is an ideal case
of course)
conserved axial current ⇒ m 2 π = 0 and/or f π = 0
The fact is, that m 2 π = 0 agrees with having a massless Goldstone boson. Thus
the pion decay constant can still be non-zero, as it is in nature and as it is
required by sppontaneous symmetry breaking which requires a nonzere matrixelement
between the vacuum and the goldstone state.
One can generalize this considerations to the case of non-vanishing divergence
of the axial current because of the finite masses of the quarks. This
comes to the concept of PCAC (partially conserved axial current). We just
discussed the divergence of the axial current and can write it down again
〈0|∂ µ A j µ(0) ∣ ∣ π k (p) 〉 = f π m 2 π 〈0| π j (0) ∣ ∣ π k (p) 〉
Where we have replaced δ jk by the normalization expression of the pion field
〈0|π j (0) ∣ ∣ π k (p) 〉 = δ jk .On the other hand we know from the section about QCD
with m 0 = mu+m d
2
:
∂ µ A j µ(x) = m 0 ¯ψ(x)iγ 5 τ j ψ(x)
from which we can take the matrixelement
and obtain by comparison
〈0|∂ µ A j µ(0) ∣ ∣ π k (p) 〉 = m 0 〈0| ¯ψ(0)iγ 5 τ j ψ(0) ∣ ∣ π k (p) 〉
m 0 〈0| ¯ψ(0)iγ 5 τ j ψ(0) ∣ ∣ π k (p) 〉 = f π m 2 π 〈0|π j (0) ∣ ∣ π k (p) 〉
or generalized to an operator equation and using translation invariance of the
vacuum
∂ µ A j µ(x) = m 0 ¯ψ(x)iγ 5 τ j ψ(x) = f π m 2 ππ j (x)
This completes the proof of the PCAC relation.
104

10.5 The chiral condensate
We can now easily derive the existence of vacuum condensates and their values.
We will prove the famous Gell-Mann–Oakes–Renner formula,
f 2 πm 2 π = − 1 2 (m u + m d ) 〈0| ¯ψ u ψ u + ¯ψ d ψ d |0〉
which relates observable quantities like pion decay constant and pion mass to
QCD-quantities like the current masses of the quarks and the condensates.We
will derive from this the value of the condensate (201). The Gell-Mann–Oakes–
Renner formula makes a connection between the properties of the vacuum,
i.e. condensate, the properties of the quarks in the QCD-Lagrangean, i.e. the
masses, and properties of the physical and observable hadronic properties, i.e.
the mass and decay constant or the pion, being the goldstone boson of the
spontaneously broken chiral symmetry of the vacuum. In fact we have a huge
condensate, connecting a number of the order of 10 8 MeV 4 on the LHS to 10MeV
on the RHS. The condensate is connected with an energy density, which is about
1000 MeV per fm 3 .This energy has been released in the chiral phase transition
about 10 −6 sec after the Big Bang, where the Universe changed from something
like a quark-gluon plasma to our present hadronic phase. The energy was released
in terms of pions, which radiated into photons and W-bosons. The chiral
phase transition occured because the universe expanded and cooled such that a
hadronic vacuum state got energetically preferable.
For the proof we start from the divergence of the axial current (166)
∂ µ A 1 µ = m 0 ¯ψiγ 5 τ 1 ψ = 1 2 (m u + m d ) ¯ψiγ 5 λ 1 ψ
We know from charge algebras the formula (197)
[
Q
A
1 , ¯ψγ 5 λ 1 ψ ] = − ( ¯ψu ψ u + ¯ψ d ψ d
)
and can replace the RHS of the commutator yielding
〈0| [ Q A 1 ,∂ µ A 1 ] i
µ |0〉 = −
2 (m u + m d ) 〈0| ¯ψ u ψ u + ¯ψ d ψ d |0〉
We now insert a complete set of states into the commutator, which however
following the philosophy of spontaneous symmetry breaking, is exausted solely
by the pion state (attention: sum over a because the pionic goldstone state is
an iso-triplet and there are in SU(3) also the kaon and the eta as Goldstone
states).
∑
∫
d 3 p
2E p (2π) 3 |π a(p) >< π a (p)| = 1
a
105

In this way we obtain (only the term with a = 1 remains):
− ∑ a
∫
d 3 p
2E p (2π) 3 〈0| ∂µ A 1 µ(0)|π a (p) >< π a (p)|Q A 1 |0〉
< 0| [ Q A 1 ,∂ µ A 1 µ(0) ] |0 >
= ∑ ∫
d 3 p
2E
a p (2π) 3 〈0| QA 1 |π a (p) >< π a (p)|∂ µ A 1 µ(0) |0〉
One sees at this formula immediately that there is no contribution from the
vacuum |0〉 ,which is of course also part of the complete set, since both terms
are identical in this case. To calculate the matrixelements in this expresseion
we have used the above matrixelement of the axial current between vacuum and
pion state, being the definition of the pion decay constant,
〈0| A j µ(x) ∣ ∣π k (p) 〉 = if π p µ δ jk exp(−ipx)
and obtain from that
〈0| Q j ∣
A π k (p) 〉 ∫
=
d 3 x 〈0| A j 0 (x)∣ ∣ π k (p) 〉 = if π p 0 δ jk ∫
d 3 xexp(−ipx)
yielding
Furthermore we use
〈0|Q A a |π b (p) >= if π δ ab E p (2π) 3 δ 3 (p)
〈0| ∂ µ A j µ(0) ∣ ∣ π k (p) 〉 = f π m 2 πδ jk
One can continue now
〈0| [ Q A 1 ,∂ µ A 1 µ(0) ] |0〉 =
∑
∫
d 3 p
2E
a p (2π) 3 [if πE p (2π) 3 δ 3 (p) f π m 2 πδ a1
− ∑ ∫
d 3 p
2E
a p (2π) 3 f πm 2 πδ a1 (−i)f π E p (2π) 3 δ 3 (p)
= if 2 πm 2 π
The integral gives a ONE, lots of terms cancel, and we obtain the famous Gell-
Mann–Oakes–Renner formula,
f 2 πm 2 π = − 1 2 (m u + m d ) 〈0| ¯ψ u ψ u + ¯ψ d ψ d |0〉
which relates observable quantities like pion decay constant and pion mass to
QCD-quantities like the current masses of the quarks and the condensates. One
should note that the current masses and the condensate separately depend on
the renormalization scale, their product, however does not.:
106

Using f π = 93MeV and m π = 139MeV and m u +m d ≈ 14MeV (we learn
later how to get the masses of the current quarks) one obtains eq.(201)
〈0| ¯ψ u ψ u |0〉 = 〈0| ¯ψ d ψ d |0〉 = 〈0| ¯ψ s ψ s |0〉 = −(225MeV ) 3 = −1.5fm −3
This completes the proof of the Gell-Mann–Oakes–Renner formula and of
the magnitude of the chiral condensate.
10.6 LSZ-Reduction formulae
This section summarizes some useful formulae. In the following one normalizes
the pion field as mentioned above. Then the Fourier transform of the two-point
correlation function, considered as an analytic function of p 2 , has a simple pole
at the mass of the one-particle state. Thus in the limit p 2 → m 2 a one can write
(no sum over a):
∫
d 4 xexp(ipx) 〈0|T {π a (0)π a i
(x)} |0〉 =
p 2 − m 2 a + iɛ
For the following we need some formulae, which are known under the name
Lehman-Symanzik-Zimmerman reduction formulae, the derivation of which is
based on basic S-matrix properties and one can find e.g. in Peskin-Schroeder.
The formulae involve the invariant amplitude and physical particles, ie. those
which are on the mass shell, i.e. q 2 1 = q 2 2 = m 2 π, and p 2 1 = p 2 2 = m 2 N . Instead
of the nucleon states we can have also other states of physical particles, e.g.
the vacuum as well. In the formulae there appear in- and out-states. They
are relativistic analogues of the scattering states of non-relativistic quantum
mechanics, which correspond e.g. to outgoing waves from the interaction region.
Just to remind, each of these sets separately are complete. We remember:
Ψ out = exp(ikr) + f(φ) exp(+ikr)
r
Ψ in = exp(ikr) + f(φ) exp(−ikr)
r
We have the Lehmann-Symanzik-Zimmermann reduction formula (LSZ):
< π a (q 2 )N(p 2 );out ∣ N(p1 );in 〉 = i(2π) 4 δ (4) (p 1 − p 2 − q 2 )TN→πN
b
< π b (q 2 )N(p 2 );out ∣ N(p1 );in 〉 ∫
= i d 4 y exp(iq 2 y)(−q2+m 2 2 π) 〈 N(p 2 ) ∣ π b (y) ∣ N(p1 ) 〉
and the correponding process with q 2 1 = m 2 π, and p 2 1 = p 2 2 = m 2 N
< N(p 2 );out ∣ π a (q 1 )N(p 1 );in 〉 = i(2π) 4 δ (4) (p 1 + q 1 − p 2 )TπN→N
a
< N(p 2 );out ∣ π a (q 1 )N(p 1 );in 〉 ∫
= i d 4 y exp(−iq 1 y)(−q1+m 2 2 π) 〈 N(p 2 ) ∣ π a (y) ∣ N(p1 ) 〉
107

One can easily see the consistency in these formulae. Consider the following
case
∫
i d 4 y exp(iq 2 y)(−q2 2 + m 2 π) 〈 N(p 2 ) ∣ π b (y) ∣ N(p1 ) 〉
∫
= i d 4 y exp(iq 2 y)(−q2 2 + m 2 π) 〈 N(p 2 ) ∣ exp(iPy)π b (0)exp(−iPy) ∣ N(p1 ) 〉
∫
= i d 4 y exp(iq 2 y)exp(ip 2 y)exp(−ip 1 y)(−q2 2 + m 2 π) 〈 N(p 2 ) ∣ π b (0) ∣ N(p1 ) 〉
= i(2π) 4 δ (4) (p 1 − p 2 − q 2 )(−q2 2 + m 2 π) 〈 N(p 2 ) ∣ π b (0) ∣ N(p1 ) 〉
and hence
T b N→πN = (−q 2 2 + m 2 π) 〈 N(p 2 ) ∣ ∣ π b (0) ∣ ∣N(p 1 ) 〉
T b πN→N = (−q 2 1 + m 2 π) 〈 N(p 2 ) ∣ ∣ π b (0) ∣ ∣N(p 1 ) 〉
We get a reduction formula similarly for pion-nucleon scattering in the limit
q 2 1 → m 2 a = m 2 π and q 2 2 → m 2 b = m2 π :
< π b (q 2 )N(p 2 );out ∣ ∣ π a (q 1 )N(p 1 );in 〉 = i(2π) 4 δ (4) (p 1 + q 1 − p 2 − q 2 )T ab
πN→πN
< π b (q 2 )N(p 2 );out ∣ ∣ π a (q 1 )N(p 1 );in 〉
= (i) 2 ∫
d 4 xd 4 y exp(iq 2 y)exp(−iq 1 x)(−q 2 1 + m 2 a)(−q 2 2 + m 2 b) 〈 N(p 2 ) ∣ ∣ T {π b (y)π a (x)} ∣ ∣ N(p1 ) 〉
and :
We get a similar reduction formula :
〉 ∫
∣
< 0 ∣Ô(0)|πa (q 1 ) = i d 4 xexp(−iq 1 x)(−q1 2 + m 2 b) 〈0|T {Ô(0)πa (x)} |0〉
〉 ∫
< π b ∣
(q 2 ) ∣Ô(0)|0 = i
d 4 y exp(iq 2 y)(−q2 2 + m 2 b
b) 〈0|T {π (y)Ô(0)} |0〉
The formulae are somehow understandable, since the field operators consist of
creation and annihilation operators of the field quanta, and the physical particles
in the entrance and exit channels are created from the vacuum by application of
the creation operators. The particles appearing in the formulae are particular
since they are all on the mass shell . The detailled proof is given in field theory
books (e.g. Peskin-Schroeder, or Izykson-Zuber §5-1-3) .
10.7 Pion-Nucleon coupling constant
Another thing is important, that is the definition of the pion-nucleon-nucleon
coupling constant.
< π + (q 2 )N(p 2 ) ∣ ∣ N(p1 ) 〉 = i(2π) 4 δ (4) (p 1 − p 2 − q 2 )T + N→πN
108

with the pion emission amplitude
T + N→πN = i√ 2g πNN (q 2 )u(p 2 )γ 5 τ a u(p 1 )
This definition implies q 2 = m 2 π. Due to symmetry considerations the coupling
constants are independent on a. Actually one can rewrite this using the
reduction theorems:
(−q 2 + m 2 π) 〈p(k ′ )| π + (0) |n(k)〉 = √ 2g πNN (q 2 )u(k ′ )iγ 5 u(k)
Again this implies q 2 = m 2 π, however it is generalized to the definition of the
pion-nucleon-nucleon form factor. In this way the form factor g πNN (q 2 ) is defined
even for space-like q 2 by analytic continuation. This is possible since
the pion mass is small. Experimentally the pion-nucleon coupling constant is
measured in pion-nucleon scattering and nucleon-nucleon scattering in forward
direction. The varous kinematical regimes are connected by means of analytic
continuation. Altogether one has experimentally
with
g πNN = g πNN (q 2 = m 2 π)
g πNN = 14.6
The value is not extremely accurate since one must have pions as secondary
beam or must look at NN-scattering. Pions are generated by shooting protons
on nuclei, then pions are generated in forward direction. (They decay into
muons, which are then used for muon-nucleon scattering).
Pion−Nucleon Scattering
π (q)
π (q’)
N(p)
N(p’)
We have the follolwing on-shell conditions:
q 2 = (q ′ ) 2 = m 2 π
p 2 = (p ′ ) 2 = m 2 N
109

Hence the “nucleon” in the propagator is off-shell:
(p + q) 2 = 2m 2 N + 2pq ≠ m 2 N
10.8 Goldberger-Treiman relation and pion pole
The Goldberger-Treiman relation is
f π g πNN = m N g A (0)
It is an important relation relating strong and weak interactions and reflecting
the chiral spontaneous symmetry breaking. If the pion were massles (chiral
limit) then the Goldberger-Treiman relation would hold exactly. Since the pion
is massive it holds with 7% deviation. We had a similar relation already in the
effective linear chiral sigma model. However, now we do it fully.
In order to derive the Goldberger-Treiman relation consider the matrix element
of the axial current:
〈p(k ′ )|(A 1 µ + iA 2 µ)(0) |n(k)〉 = u p (k ′ ) [ γ µ γ 5 g A (q 2 ) + q µ γ 5 h A (q 2 ) ] u n (k)
One can show, that from Lorenze covariance the LHS must have the structure
as it is given on the RHS. Here is q = k − k ′ . Actually this matrix element is
measured in the beta decay of the nucleon. Experimentally one has g A (0) =
1.26. On can derive from this the matrix element of the divergence of the axial
current between proton and neutron. This goes in the following way analogous
to the matrixelement of the axial current between vacuum and one-pion state:
We know that (see section about quantization of boson field) for any operator,
depending on the elementary fields
F(x) = exp(iPx)F(0)exp(−iPx)
F(0) = exp(−iPx)F(x)exp(+iPx)
Inserting F(0) into the expression with the current and using e.g.
one obtains immediately
>From this we get
exp(iPx) |n(k)〉 = exp(ikx) |n(k)〉
〈p(k ′ )|(A 1 µ + iA 2 µ)(0) |n(k)〉
= 〈p(k ′ )| exp(−iPx)(A 1 µ + iA 2 µ)(x)exp(+iPx) |n(k)〉
= 〈p(k ′ )| (A 1 µ + iA 2 µ)(x) |n(k)〉 exp(+ikx − ik ′ x)
〈p(k ′ )|(A 1 µ+iA 2 µ)(x) |n(k)〉 = exp(i(k ′ −k)x)u p (k ′ ) [ γ µ γ 5 g A (q 2 ) + q µ γ 5 h A (q 2 ) ] u n (k)
110

Now we perform the divergence on both sides yielding:
〈p(k ′ )|(∂ µ A 1 µ + i∂ µ A 2 µ)(x) |n(k)〉 =
= i(k ′ − k) µ exp(i(k ′ − k)x)u p (k ′ ) [ γ µ γ 5 g A (q 2 ) + q µ γ 5 h A (q 2 ) ] u n (k)
or at x = 0:
〈p(k ′ )|(∂ µ A 1 µ + i∂ µ A 2 µ)(0) |n(k)〉 =
= i(k ′ − k) µ u p (k ′ ) [ γ µ γ 5 g A (q 2 ) + q µ γ 5 h A (q 2 ) ] u n (k)
We can rewrite this by using the Dirac equation (γ µ k µ − m)u(k) = 0 obtaining
〈p(k ′ )|(∂ µ A 1 µ + i∂ µ A 2 µ)(0) |n(k)〉 =
= iu p (k ′ )γ 5 u n (k) [ 2m N g A (q 2 ) + q 2 h A (q 2 ) ]
In order to relate this to the pion-nucleon coupling constant we have to consider
physical pion fields:
and using the PCAC relation
π + (x) = 1 √
2
(π 1 (x) + iπ 2 (x))
∂ µ A j µ(x) = f π m 2 ππ j (x)
one gets by sandwiching between proton and neutron states
〈p(k ′ )| (∂ µ A 1 µ + i∂ µ A 2 µ)(0) |n(k)〉 = √ 2f π m 2 π 〈p(k ′ )| π + (0) |n(k)〉
This can directly be related to the pion-nucleon coupling constant as it is known
to us from the previous subsection. There we had for g πNN (q 2 ) the following
√
2
〈p(k ′ )|π + (0) |n(k)〉 =
−q 2 + m 2 g πNN (q 2 )u(k ′ )iγ 5 u(k)
π
Combing this with the above formulae yields
〈p(k ′ )|(∂ µ A 1 µ + i∂ µ A 2 µ)(0) |n(k)〉 = √ √
2
2f π m 2 π
−q 2 + m 2 g πNN (q 2 )u(k ′ )iγ 5 u(k)
π
Comparison with the above expression for the divergence yields
f π m 2 2
π
−q 2 + m 2 g πNN (q 2 ) = [ 2m N g A (q 2 ) + q 2 h A (q 2 ) ]
π
This is a general relationship valid for all momentum transfers q.
111

If we set q 2 = 0 we have
f π g πNN (0) = m N g A (0)
Unfortunately the point g πNN (0) is not physical since neither pion nucleon
scattering nor nucleon-nucleon scattering with pion exchange has a vanishing
momentum transfer. One has therefore to assume a “smooth” and analytical
behaviour from q 2 = 0 to q 2 = m 2 π (which is rather small) such that we have
g πNN (0) ≃ g πNN (m 2 π) = g πNN
which gives then the famous Goldberger-Treiman relation:
f π g πNN = m N g A (0)
This relation is satisfied to about 7% in nature, what we can easily check using
the numbers f π = 93MeV,g πNN = 14.6,m N = 938MeV,g A (0) = 1.26 for yielding
1357˜1181. This example shows, that one can get really relations between
physical observables by utilizing the PCAC relation. On the other hand we
needed an extrapolation g πNN (0) ≃ g πNN (m 2 π) = g πNN which is believed to be
justified, because the pion mass is small on a hadronic scale. One sees clearly,
that one gets in trouble if one generalizes such an argumentation to SU(3). The
relationship of PCAC to physical quantities in that case requires extrapolation
from q 2 = 0 to q 2 = m 2 K , where the kaon mass is about 500 MeV. This large
kinematical region makes PCAC less usefull for SU(3).
One can see here immediately the famous pion pole term: Consider
f π m 2 2
π
−q 2 + m 2 g πNN (q 2 ) = [ 2m N g A (q 2 ) + q 2 h A (q 2 ) ]
π
and consider then the limit q 2 → m 2 π. Since the first term on the RHS ˜g A
remains finite and q 2 as well, we see that the h A (q 2 ) is completely determined
by the diverging term on the LHS. Thus in his limit we have the so called pion
pole.
Limes q 2 → m 2 π ::::⇒:::: h A (q 2 ) =
2f π
m 2 π − q 2 g πNN(q 2 ) :::
In fact, if we have a reaction, which is governed by some form factors in the
region of q 2˜m 2 π the reaction process is govened by h A (q 2 ) and this is given by
the pion-nucleon coupling constant, known from many pion-nucleon scattering
processes.
10.9 Vacuum Sigma Term and Gell-Mann–Okubo
The vacuum Sigma term allows to characterize the vacuum and and its condensate
by relating them to the Goldstone meson masses. That why it is an
important quantity. WE first consider it in SU(2). There it is defined by
SU(2):
σ ab
0 = − 〈0| [ Q a A, [ Q b A, L m
]]
|0〉
112

In fact, as we will see immediately, the diagonal numbers can be related directly
to the pseudo scalar meson mass m a and decay constants f a , which are of
course observable quantities. Thus one learns something about the vacuum by
considering σ0 ab and the meson masses. One basically learns something about
the product of current quark masses and the vacuum condensate.
The assertion is for a = 1,2,3
σ ab
0 = δ ab m 2 af a
In order to see this we start with the PCAC relation
∂ µ A a µ(x) = f a m 2 aπ a (x)
and the redction formula
< 0 ∣ π a (0)|π b (k) 〉 ∫
= i d 4 y exp(−iky)(−k 2 + m 2 b) 〈0|T {π a (0)π b (y)} |0〉
We insert the PCAC relation into this expression and obtain
δ ab = i (−k2 + m 2 b ) ∫
f a m 2 af b m 2 d 4 y exp(−iky) 〈0| T {∂xA ν a ν(0)∂ y µ A b µ(y)} |0〉
b
= i (−k2 + m 2 b ) ∫
f a m 2 af b m 2 d 4 y exp(−iky)∗
b
∗ 〈0| θ(−y 0 )∂xA ν a ν(0)∂ y µ A b µ(y) + θ(y 0 )∂ y µ A b µ(y)∂xA ν a ν(0) |0〉
Performing now partial integration yields −∂ µ y exp(−iky) = ik µ exp(−iky) and
apparently terms where the ∂ µ y is applied to the θ(y 0 ) and θ(−y 0 ).With the
equations
one obtains immediately
∂ y µθ(y 0 − x 0 ) = g µ0 δ(y 0 − x 0 )
∂ y µθ(x 0 − y 0 ) = −g µ0 δ(y 0 − x 0 )
δ ab f a m 2 a = i (−k2 + m 2 b ) ∫
f b m 2 {ik µ d 4 y exp(−iky) 〈0| T {∂xA ν a ν(0)∂ y µ A b µ(y)} |0〉
b
∫
− d 4 y exp(−iky) 〈0| δ(y 0 )[A b 0(y),∂ µ A a µ(0)] |0〉}
in the limit k → 0 we obtain a d 3 y-integral yielding an axial charge Q b A (x 0),
and the integral over the time component yields Q b A (0). Thus we obtain
δ ab f a m 2 a = − 〈0| [Q b A(0),∂ µ A a µ(0)] |0〉 for a = 1,2,3
In SU(2), what we have presently assumed, one can develope it further, as it
is often done. The divergence of the axial current is known for isospin symmetry
in the up-down sector
∂ µ A µ a = (m u + m d ) ¯ψiγ 5 1 2 τ aψ for a = 1,2,3
113

together with the equation from the section on chiral symmetry breaking
[
Q
A
a , ¯ψ(y)ψ(y) ] = − ¯ψ(y)τ a γ 5 ψ(y)
we obtain that
or
∂ µ A a µ(0) = − i 2 (m u + m d ) [ Q A a , ¯ψ(y)ψ(y) ] for a = 1,2,3
∂ µ A a µ(0) = −i [ Q A a , L u,d ]
m
such that we altogether obtain in the limit k → 0
δ ab f a m 2 a = − 〈0| [Q A b , [ Q A a (0), L u,d
m (0) ] ] |0〉 = σ0
ab
This is a well known formula written in many books. Often one writes H(0) =
−L u,d
m (0).
For SU(3) we have to go back to the formula with the divergence of the axial
current. In fact we have
δ ab f a m 2 a = − 〈0| [Q a A(0),∂ µ A a µ(0)] |0〉 for a = 1,...,8
since we never assumed in the derivation of this formula really SU(2). However
the step from ∂ µ A a µ(0) to L m (0) cannot be done since the strange mass is much
larger than the up- and down-mass. However one can directly insert the explicit
expression for the divergence of the axial current
∂ µ A µ a = ¯ψiγ 5 1 2 {λ a,m} ψ
with the mass matrix being m = [m 1 I + m 3 λ 3 + m 8 λ 8 ]. The commutator of the
axial charges with ¯ψiγ 5 1 2 {λ a,m} ψ can be explicitely calculated using known
formulae. This yields in the end important relations between the vacuum condensates
and the masses and decay constants of the correspponding Goldstone
bosons:
f π m 2 π = m u + m d
2
f K m 2 K = m u + m s
2
f η m 2 η = m u + m d
6
〈0| ¯ψ u ψ u + ¯ψ d ψ d |0〉
〈0| ¯ψ u ψ u + ¯ψ s ψ s |0〉
〈0| ¯ψ u ψ u + ¯ψ d ψ d |0〉 + 4m s
3 〈0| ¯ψ s ψ s |0〉
We can estimate now the current masses of the quarks in an approximate way:
We have derived in the massless limit the property for the vacuum condensates
as
〈0| ¯ψ u ψ u |0〉 = 〈0| ¯ψ d ψ d |0〉 = 〈0| ¯ψ s ψ s |0〉
Assuming massles limit also for the decay constants we have
f π = f K = f η
114

and one obtains immediately the famous Gell-Mann–Okubo mass relation
4m 2 K = 3m 2 η + m 2 π
which is well fulfilled, sinced we have m K = 494MeV,m η = 548MeV,m π =
139MeV.Using these numbers one obtains
4m 2 K − (3m2 η + m 2 π)
= 10%
4m 2 K
One can also obtain the quark mass ratio
m u + m d
2m s
=
m 2 π
2m 2 K − m2 π
≈ 1
25
Apparently the strange quark mass is very much larger than the up- and downquark
mass. If one assumes for m s = 180MeV then m u + m d = 14, which is
about right.
10.10 Pion-Nucleon Sigma Term Σ πN
The pion nucleon Sigma Term Σ πN is an important quantity. It relates the
mass-terms of the QCD Lagrangean L M to the pion-nucleon scattering. This is
interesting since the quarks are never free and hence we cannot measure their
mass directly. We can measure, however, the pion-nucleon scattering. So a
conclusion from such an observable quantity like the pion-nucleon scattering
amplitude to an not-observable quantity like L M , which however appears in the
QCD-Lagrangean, is extremely interesting. Therefore the Σ πN is discussed for
years already.
We consider the pion-nucleon sigma term first in SU(2). If we write down
the QCD Lagrangean then the mass terms of the quarks are given by
L m = − ¯ψMψ
with M = diag(m u ,m d ). The mass term is related to the operator of the Sigma
term Σ πN .This is defined as
[ [ ]]
Σ ab
πN = − Q a A, Q b A, L u,d
M
with the mass term
L M = −m 0 ¯ψ ψ
and can be shown to have the structure (proof below):
Σ ab
πN =
8∑
e=0
C ab
e (m u ,m d ,m s ) ¯ψλ e ψ
The term is measured in the pion-nucleon scattering
π b (q 1 ) + N(p 1 ) → π a (q 2 ) + N(p 2 )
115

One usually takes the kinematic variables
ν = 1 4 (p 1 + p 2 )(q 1 + q 2 )
ν B = − 1 2 q 1q 2
and we use from the section on Lehmann-Symanzik-Zimmerman reduction formulae
the equation for pion-nucleon scattering:
< π a (q 2 )N(p 2 );out ∣ ∣π b (q 1 )N(p 1 );in 〉 = i(2π) 4 δ (4) (p 1 +q 1 −p 2 −q 2 )T ab
πN→πN(ν,ν B )
in the limit q 2 1 → m 2 b = m2 π and q 2 2 → m 2 a = m 2 π
< π a (q 2 )N(p 2 );out ∣ ∣π b (q 1 )N(p 1 );in 〉
= (i) 2 ∫
d 4 xd 4 y exp(iq 2 x)exp(−iq 1 y)(−q 2 1 + m 2 b)(−q 2 2 + m 2 a) 〈 N(p 2 ) ∣ ∣ T {π a (x)π b (y)} ∣ ∣ N(p1 ) 〉
We know that
〈0|A a µ(x) ∣ ∣ π b (p) 〉 = ip µ f π δ ab exp(−ipx)
〈0|∂ µ A a µ(x) ∣ ∣ π b (p) 〉 = im 2 πf π δ ab exp(−ipx)
and PCAC
∂ µ A a µ(x) = m 2 πf π π a (x)
Take now the RHS of the above expression for < π a (q 2 )N(p 2 );out ∣ π b (q 1 )N(p 1 );in 〉
and replace the pion field by the divergence of the axial current. The expression
is defined at the points q1 2 = q2 2 = m 2 π . This is not sufficient for the following,
therefor take the analytical continuation to any q1 2 ≠ q2 2 ≠ m 2 π. This yields
i(2π) 4 δ (4) (p 1 + q 1 − p 2 − q 2 )TπN→πN(ν,ν ab
B ,q1,q 2 2)
2
1 1 (
= i m
2
m 2 af π m 2 b f b − q 2 ) (
1 m
2
a − q 2 )
2 ∗
π
∫
∗ d 4 xd 4 y exp(iq 2 x)exp(−iq 1 y) 〈 N(p 2 ) ∣ T {∂
x
ν A νa (x)∂µA y µb (y)} ∣ N(p1 ) 〉
Consider now the integral
∫
I = d 4 xd 4 y exp(iq 2 x)exp(−iq 1 y) 〈 N(p 2 ) ∣ T {∂
x
ν A νa (x)∂µA y µb (y)} ∣ N(p1 ) 〉
We can reformulate it (proof below) yielding by a complicated but straight
forward calculation the Ward Identity:
T {∂ x νA νa (x)∂ y µA µb (y)} = A + B + C
= ∂µ∂ y νT x {A µb (y)A νa (x)}
− δ(y 0 − x 0 ) [ A 0b (y),∂νA x νa (x) ]
+ ∂µ
y (
δ(y0 − x 0 ) [ A µb (y),A 0a (x) ])
116

the proof of this formula is given below: Actually, although this looks like a
complication, we rewrite this because with this new expression we can have
tremendous simplifications in the limit q µ 1 → 0 and qµ 2 → 0.The RHS is a sum of
three terms I = A+B+C, which have different properties if we consider the soft
pion limit, defined by q µ 1 → 0 and qµ 2 → 0. One sees this by shifting e.g. in A
the derivative ∂µ y in the integral over y by partial integration on the exponential
exp(−iq 1 y) yielding exp(−iq 1 y)∂µ y → −(−iq µ 1 )exp(−iq 1y) = iq µ 1 exp(−iq 1y) and
the same for exp(iq 1 y)∂ν x → −iq2 ν exp(iq 1 y). Thus we have A˜q µ 1 qν 2 and hence
A → 0 for q µ 1 → 0 or qµ 2 → 0. Similarly we have C˜ − iq 2ν, which also goes
to zero in the soft pion limit. The term B does not allow for such a simple
treatment, since the derivatives act also on the δ(x 0 − y 0 ) and hence do not
yield a simple q µ 1 or so. Thus in the soft pion limit the only remaining term is
∫
I B = − d 4 xd 4 y exp(iq 2 x)exp(−iq 1 y)δ(y 0 − x 0) [A 0b (y),∂νA x νa (x)]
This term will turn out to be basically the pion-nucleon sigma term. To see this
we have first to show that with equal up- and down-masses we have the useful
expression (proven already in the section about vacuum sigma term)
[
Q
a
A (x 0 ), ¯ψ(x)ψ(x) ] = − ¯ψ(x)γ 5 τ a ψ(x)
and
∂ µ A µ a = (m u + m d ) ¯ψiγ 5 1 2 τ aψ
which yields then after sandwiching with the nucleonic states
∫
B ′ = −i d 4 xd 4 y exp(iq 2 x)exp(−iq 1 y)δ(y 0 −x 0 ) 〈 N(p 2 ) ∣ [
] ∣∣N(p1 [A 0b (y), Q a A(x 0 ), L u,d
M (x) ) 〉
This expression is already quite similar to the sigma term, which is basically
the double commutator of the mass term with the axial charges. In fact the
integrals and the delta-function make out of the A 0b the axial charge as we will
see now. We know the relation F(x + y) = exp(iPx)F(y)exp(−iPx) which we
apply to the above expression:
〈
N(p2 ) ∣ ]
[A 0b (y),[Q a A(x 0 ), L u,d
M (x) ] ∣ N(p1 ) 〉
= 〈 N(p 2 ) ∣ [ ]
exp(iPx)[A 0b (y − x), Q a A(0), L u,d
M (0) ]exp(−iPx) ∣ ∣N(p 1 ) 〉
= exp(i(p 2 − p 1 )x) 〈 N(p 2 ) ∣ [ ]
[A 0b (y − x), Q a A(0), L u,d
M (0) ] ∣ N(p1 ) 〉
inserted in the above expression yields
∫
B ′ = −i d 4 xd 4 y exp(i(q 2 +p 2 −p 1 )x)exp(−iq 1 y)δ(y 0 −x 0 ) 〈 N(p 2 ) ∣ [ ]
[A 0b (y−x), Q a A(0), L u,d
M (0) ] ∣ N(p1 ) 〉
117

which is reformulated by changing the variables y −x = z and dy = dz .yielding
∫
B ′ = −i d 4 xd 4 z exp(i(q 2 + p 2 − p 1 )x)exp(−iq 1 x − iq 1 z)δ(z 0 ) 〈 N(p 2 ) ∣ [ ]
[A 0b (z), Q a A(0), L u,d
M (0) ] ∣ ∣N(p 1 ) 〉
∫
= −i(2π) 4 δ (4) (q 2 + p 2 − p 1 − q 1 ) d 4 z exp(−iq 1 z)δ(z 0 ) 〈 N(p 2 ) ∣ [ ]
[A 0b (z), Q a A(0), L u,d
M (0) ] ∣ N(p1 ) 〉
This can now be simplified to the axial charge, if we take again the limit q µ 1 → 0,
because
∫
∫
∫
d 4 z exp(−iq 1 z)δ(z 0 )A 0b (z) = d 3 z exp(iq 1
z ) dz 0 δ(z 0 )A 0b (z)
∫
= d 3 z A 0b (0,z) = Q b A(0)
Apparently the last step is only possible in the soft pion limit, otherwise in the
space integral we would have the phase and the result would be only approximately
the axial charge.
We can now collect the terms and have in the soft pion limit
with
B ′ = i(2π) 4 δ (4) (p 2 − p 1 )Σ ab
πN
Σ ab
πN = − 〈 N(p 2 ) ∣ [ ]
[Q
b
A (0), Q a A(0), L u,d
M (0) ] ∣ N(p1 ) 〉
Comparison with the expression for the pion-nucleon scattering amplitude yields
in the soft pion limit
i(2π) 4 δ (4) (p 2 − p 1 ) T ab
πN→πN(ν,ν B ,q 2 1 = 0,q 2 2 = 0) =
i(2π) 4 δ (4) (p 2 − p 1 ) Σ ab
or, since in the soft pion limit also the ν,ν B go to zero, we have finally
πN
TπN→πN(ν ab = 0,ν B = 0,q1 2 = 0,q2 2 = 0) = 1
fπ
2 Σ ab
πN
1 1
m 2 πf π m 2 (m 2 π − q1)(m 2 2 π − q1)
2
πf π
This is a direct relation between the pion-nucleon scattering amplitude and the
pion-nucleon sigma term. The unfortunate thing is, that one needs TπN→πN ab (0,0,0,0),which
cannot be measured directly since it is in a kinetatically forbidden region. The
experiment has always q1 2 = m 2 π and q2 2 = m 2 π and hence the pions in the
above TπN→πN ab (0,0,0,0) are off-shell. In addition we have in a real scattering
process at the threshold ν = m π m N ,ν B = − 1 2 m2 π . Therefore the point with
ν = 0,ν B = 0 has a special name, it is called the Cheng-Dashen-Point. Using
Mandelstam coordinates s = (p 1 + q 1 ) 2 and t = (q 2 − q 1 ) the Cheng Dashen
point is at t = 2m 2 π and s = m 2 N .Thus it cannot be reached since in a real scattering
process, one has at least s = m 2 N +m2 π. The extrapolation from the data
at kinematically accessible points to the Cheng-Dashen point and the off-shell
118

pions is approximately possible as we shall see below, since the pion mass is
only 139 MeV and hence small. For kaon- or eta-scattering the extrapolation
must be done over a larger mass scale and is therfore not that reliable. Thus the
sigma term involving components a > 3 and b > 3 is not that well destermined
by kano-nucleon or eta-nucleon scattering.
The extrapolation from TπN→πN ab (ν = 0,ν B = 0,q1 2 = m 2 π,q2 2 = m 2 π) to
T ab
πN→πN (0,0,q2 1 = 0,q2 2 = 0) can be done by means of the Adlers consistency
relation. This relation can be proved with the same techniques as above, but
this will not be done here (see Cheng-Li for that). In order to do that we
decompose the above amplitude in isospin space by (remember that the TπN→πN
ab
as operator act on the spinor or the nucleon, which is a doublet of proton and
neutron:
TπN→πN ab = δ ab T (+)
πN→πN + 1 [
τ a ,τ b] T (−)
πN→πN
2
The indices a,b act in the space of the pion and the 2x2-matrices τ a act in the
protono-neutron-space. Adler has shown the PCAC consistency conditions
T (+)
πN→πN (0,0,m2 π,0) = 0......in the limit q 2 2 → 0
T (+)
πN→πN (0,0,0,m2 π) = 0.....in the limit q 2 1 → 0
We can use these conditions in the following way;
T (+)
πN→πN (0,0,m2 π,m 2 π) = T (+)
πN→πN (0,0,0,0) + dT (+)
m2 π
dq1
2 + m 2 dT (+)
π
dq2
2 + o(m 4 π)
>From Adlers PCAC consistency condition follows
or
T (+)
πN→πN (0,0,m2 π,0) = T (+)
πN→πN (0,0,0,0) + dT (+)
m2 π
dq1
2 = 0
m 2 dT (+)
π
dq1
2 = −T (+)
πN→πN (0,0,0,0)
and the same for the derivative w.r. to q 2 2.If we replace the derivatives in the
expansion for T (+)
πN→πN (0,0,m2 π,m 2 π) we altogether get
T (+)
πN→πN (0,0,m2 π,m 2 π) = −T (+)
πN→πN (0,0,0,0) + o(m4 π)
= − 1
fπ
2 Σ πN + o(m 4 π)
with Σ ab
πN = Σ πNδ ab , if one ignores isospin breaking effects, i.e. if one assumes
m u = m d . If one calculates the double commutator, we get out in SU(2) in the
rest frame of the nucleon:
Σ πN = m 0
2M N
〈N| ¯ψ u ψ u + ¯ψ d ψ d |N〉
119

Thus the Σ πN measures the contribution of the non-vanishing quark masses
m 0 = mu+m d
2
to the nucleon mass m N .(Remember the section on scaling
anomaly, where this contribution was discussed.) What is left in the context
of pion-nucleon scattering is the extrapolation from the physical point to the
Cheng-Dashen point. This has been done first in a famous paper by Gasser,
Leutwyler and Sainio (chiral perturbation theory, 1-loop calculation) yielding
(this has to be done more in detail and show also that the mass-term of the
lagrangean comes out of the double commutator, or do this last step at the
beginning of the section)
Σ πN (2m 2 π) − Σ πN (0) ≃ 15MeV
this gives for the Sigma-term in the nomenclature of Gasser, Leutwyler and
Sainio
F 2 π ¯D (+) (2m 2 π) = Σ πN (2m 2 π) = (60 ± 8)MeV
yielding then for the Σ πN (0) ≃ (45 ± 8) MeV. These are the well known values
from 1991. There are new measurements in the last few years, which yield larger
values Fπ 2 ¯D (+) (2m 2 π) = (75±5) MeV or Σ πN (0) ≃ (60±5) MeV. The final value
is not settled yet.Nevertheless, since the Σ πN measures the contribution of the
non-vanishing quark masses m 0 = mu+m d
2
to the nucleon mass m N ,.all these
values show, that this contribution is small compared to the nucleon mass of
938 MeV.
From the sigma-term we can learn something about the strange quark content
of the nucleon. The term can exactly be written as
Σ πN = m 0 〈N| ¯ψ u ψ u + ¯ψ d ψ d − 2 ¯ψ s ψ s |N〉
2M N 1 − y
where the quantity y is a measure of the scalar strang quark content of the
nucleon:
2 〈N|
y =
¯ψ s ψ s |N〉
〈N| ¯ψ u ψ u + ¯ψ d ψ d |N〉
Assuming that the SU(3) symmetry breaking is small, one can derive from this
expression and the octet mass relations (see later) the expression
(1 − y)Σ πN =
m 0
m s − m 0
(M Ξ + M Σ − 2M N ) (206)
An expression of this sort is not surprising at all. First it resembles the situation
in the vacuum, where we had pseudoscalar meson masses related to the
vacuum sigma term. For the baryons the situation is similar since the mass
term 〈N| ¯ψ u ψ u + ¯ψ d ψ d − 2 ¯ψ s ψ s |N〉 is the only term in the QCD-lagrangean by
which the masses of the baryons can differ. Without this all multiplets and all
members in the multiplets (to be discussed in the section about quark model)
would be identical. In one uses m s = 25m 0 one obtains with the known masses
(1−y)Σ πN = 26MeV. Together with the empirical value of Σ πN = (45±8)MeV
one gets y = 0.2 ± 0.2. With the new and larger values of Σ πN = 79MeV one
120

obtains a value of y no longer compatible with Zero. However by chiral perturbation
theory the connection between Σ πN and the masses of the baryons is
somewhat different and one gets there (1 − y)Σ πN = (36 ± 7)MeV so that one
gets altogether somehow
y = 0.2 → 0.6
and from there a contribution to the mass of the strangeness contribution to
nucleon mass: m s 〈N| ¯ψ s ψ s |N〉 = (150 → 480)MeV
The reasoning for the last point goes as follows. We start from a SU(3) theory
without symmetry breaking, i.e. where all quark masses are equal. To calculate
the masses of the hyperons one does at least first order perturbation theory
and hence one has to calculate something like < Σ nonbroken |m s¯ss|Σ nonbroken >
where we can write
⎛ ⎞
¯ss = 1 2 ¯ψ
1 0 0 √
⎝ 0 1 0 ⎠ 3
ψ −
2 ¯ψλ 8 ψ
0 0 1
In order to calculate the mass splitting of the unbroken hyperons due to strange
quark mass we need the Wigner Eckart theorem of the SU(3) group like
〈a|O b |c〉 = B 1 f abc + B 2 d abc
where each of the states is a members of a multiplet and B 1 and B 2 are dynamical
numbers, which are valid for the corresponding whole multiplet. This is in
the cartesian representation and we need it in the spherical representation since
there the particles are defined with diagonal charge. If one does all this one is
able to express the B 1 and B 2 through mass differences. This yields the above
formula eq.(206).
We have still to prove the Ward-Identity: For this we use
And now we want to proof that:
∂ y µθ(y 0 − x 0 ) = g µ0 δ(y 0 − x 0 )
∂ y µθ(x 0 − y 0 ) = −g µ0 δ(y 0 − x 0 )
T {∂µA y µb (y)∂νA x νa (x)} = ∂µ∂ y νT x {A µb (y)A νa (x)}
− δ(y 0 − x 0 ) [ A 0b (y),∂νA x νa (x) ]
+ ∂µ
y (
δ(y0 − x 0 ) [ A µb (y),A 0a (x) ])
121

we do it by evaluating the first term on the RHS:
∂µ∂ y νT x {A µb (y)A νa (x)} =∂µ∂ y ν x {θ(y 0 − x 0 )A µb (y)A νa (x) + θ(x 0 − y 0 )A νa (x)A µb (y)}
= ∂µ[∂ y νθ(y x 0 − x 0 )A µb (y)A νa (x) + θ(y 0 − x 0 )A µb (y)∂νA x νa (x)
+ ∂νθ(x x 0 − y 0 )A νa (x)A µb (y) + θ(x 0 − y 0 )∂νA x νa (x)A µb (y)]
= ∂µ∂ y νθ(y x 0 − x 0 )A µb (y)A νa (x) + ∂νθ(y x 0 − x 0 )∂µA y µb (y)A νa (x)
+ ∂µθ(y y 0 − x 0 )A µb (y)∂νA x νa (x) + θ(y 0 − x 0 )∂µA y µb (y)∂νA x νa (x)
+ ∂µ∂ y νθ(x x 0 − y 0 )A νa (x)A µb (y) + ∂νθ(x x 0 − y 0 )A νa (x)∂µA y µb (y)
+ ∂µθ(x y 0 − y 0 )∂νA x νa (x)A µb (y) + θ(x 0 − y 0 )∂νA x νa (x)∂µA y µb (y)
∂µ∂ y νT x {A µb (y)A νa (x)} = ∂µ∂ y νθ(y x 0 − x 0 )A µb (y)A νa (x) + (−)g ν0 δ(y 0 − x 0 )∂µA y µb (y)A νa (x)
+ g µ0 δ(y 0 − x 0 )A µb (y)∂νA x νa (x) + θ(y 0 − x 0 )∂µA y µb (y)∂νA x νa (x)
+ ∂µg y ν0 δ(y 0 − x 0 )A νa (x)A µb (y) + g ν0 δ(y 0 − x 0 )A νa (x)∂µA y µb (y)
+ (−)g µ0 δ(y 0 − x 0 )∂νA x νa (x)A µb (y) + θ(x 0 − y 0 )∂νA x νa (x)∂µA y µb (y)
∂µ∂ y νT x {A µb (y)A νa (x)} = ∂µ∂ y νθ(y x 0 − x 0 )A µb (y)A νa (x) + (−)δ(y 0 − x 0 )∂µA y µb (y)A 0a (x)
+ δ(y 0 − x 0 )A 0b (y)∂νA x νa (x) + θ(y 0 − x 0 )∂µA y µb (y)∂νA x νa (x)
+ ∂µδ(y y 0 − x 0 )A 0a (x)A µb (y) + δ(y 0 − x 0 )A 0a (x)∂µA y µb (y)
+ (−)δ(y 0 − x 0 )∂νA x νa (x)A 0b (y) + θ(x 0 − y 0 )∂νA x νa (x)∂µA y µb (y)
∂µ∂ y νT x {A µb (y)A νa (x)} = −∂µδ(y y 0 − x 0 )A µb (y)A 0a (x) + (−)δ(y 0 − x 0 )∂µA y µb (y)A 0a (x)
+ δ(y 0 − x 0 )A 0b (y)∂νA x νa (x) + θ(y 0 − x 0 )∂µA y µb (y)∂νA x νa (x)
+ ∂µδ(y y 0 − x 0 )A 0a (x)A µb (y) + δ(y 0 − x 0 )A 0a (x)∂µA y µb (y)
+ (−)δ(y 0 − x 0 )∂νA x νa (x)A 0b (y) + θ(x 0 − y 0 )∂νA x νa (x)∂µA y µb (y)
∂µ∂ y νT x {A µb (y)A νa (x)} = θ(y 0 − x 0 )∂µA y µb (y)∂νA x νa (x) + θ(x 0 − y 0 )∂νA x νa (x)∂µA y µb (y)
+ (−)∂µδ(y y 0 − x 0 )A 0b (y) (y)A νa (x) + (−)δ(y 0 − x 0 )∂µA y µb (y)A 0a (x)
+ δ(y 0 − x 0 )A 0b (y)∂νA x νa (x) + (−)δ(y 0 − x 0 )∂νA x νa (x)A 0b (y)
+ ∂µδ(y y 0 − x 0 )A 0a (x)A µb (y) + δ(y 0 − x 0 )A 0a (x)∂µA y µb (y)
∂µ∂ y νT x {A µb (y)A νa (x)} = T {∂µA y µb (y)∂νA x νa (x)}
− ∂µδ(y y 0 − x 0 )A µb (y)A 0a (x) + (−)δ(y 0 − x 0 )∂µA y µb (y)A 0a (x)
+ δ(y 0 − x 0 )[A 0b (y),∂νA x νa (x)]
+ ∂µδ(y y 0 − x 0 )A 0a (x)A µb (y) + δ(y 0 − x 0 )A 0a (x)∂µA y µb (y)
122

We have and reformulate
∂µ∂ y νT x {A µb (y)A νa (x)} = T {∂µA y µb (y)∂νA x νa (x)}
− ∂µδ(y y 0 − x 0 )[A µb (y),A 0a (x)]
+ δ(y 0 − x 0 )[A 0b (y),∂νA x νa (x)]
+ δ(y 0 − x 0 )[A 0a (x),∂µA y µb (y)]
∂µ∂ y νT x {A µb (y)A νa (x)} = T {∂µA y µb (y)∂νA x νa (x)}
− ∂µδ(y y 0 − x 0 )[A µb (y),A 0a (x)]
+ δ(y 0 − x 0 )[A 0b (y),∂νA x νa (x)]
− δ(y 0 − x 0 )[∂µA y µb (y),A 0a (x)]
∂µ∂ y νT x {A µb (y)A νa (x)} = T {∂µA y µb (y)∂νA x νa (x)}
− ∂µ(δ(y y 0 − x 0 )[A µb (y),A 0a (x)])
+ δ(y 0 − x 0 )[A 0b (y),∂νA x νa (x)]
or finally
T {∂ y µA µb (y)∂ x νA νa (x)} = ∂ y µ∂ x νT {A µb (y)A νa (x)} =
+ ∂ y µ(δ(y 0 − x 0 )[A µb (y),A 0a (x)])
− δ(y 0 − x 0 )[A 0b (y),∂ x νA νa (x)]
qe.d
We get the above formula by realizing that T {∂ x νA νa (x)∂ y µA µb (y)} = T {∂ y µA µb (y)∂ x νA νa (x)}
The proof is by writing down
T {∂ y µA µb (y)∂ x νA νa (x)} = θ(y 0 − x 0 )∂ y µA µb (y)∂ x νA νa (x) + θ(x 0 − y 0 )∂ x νA νa (x)∂ y µA µb (y)
T {∂ x νA νa (x)∂ y µA µb (y)} = θ(y 0 − x 0 )∂ y µA µb (y)∂ x νA νa (x) + θ(x 0 − y 0 )∂ x νA νa (x)∂ y µA µb (y)
This finalizes the proof for the above Ward identity.
10.11 Ward-Identities and Low energy pion-nucleon theorems
We remember the definition for isospin even and isospin-odd part of the pionnucleon
scattering T-matrix:
TπN→πN ab = δ ab T (+)
πN→πN + 1 [
τ a ,τ b] T (−)
πN→πN
(207)
2
123

We repeat Adler’s PCAC consistency conditions
T (+)
πN→πN (0,0,m2 π,q 2 2) = 0......in the limit q 2 2 → 0
T (+)
πN→πN (0,0,q2 1,m 2 π) = 0.....in the limit q 2 1 → 0
With similar techniques as used in the section about pion-nucleon sigma
term one can show several other Ward identities and low energy theorems (for
indication of proofs see Cheng and Li):
ig 2 Aν 1 2
[
τ a ,τ b] = −ifπT 2 πN→πN ab + iν 1 [
τ a ,τ b] − iδ ab Σ πN
2
lim 1
ν → 0 ν T (−)
πN→πN (ν,0,0,0) = 1 − g2 A
with the axial coupling constant g A = 1.26. We furthermore use the unsubtracted
dispersion relation
from which we get
or
1
ν T (−)
πN→πN (ν,0,0,0) = 2 π
1
g 2 A
1 − g 2 A
f 2 π
= 2 π
∫ ∞
ν 0
∫ ∞
= 1 + 2M2 N
πg πNN
∫ ∞
ν 0
f 2 π
(−)
′ Im T
πN→πN
dν (ν,0,0,0)
ν ′2 − ν 2
(−)
′ Im T
πN→πN
dν (ν,0,0,0)
ν ′2
ν 0
(−)
′ ImT
πN→πN
dν (ν,0,0,0)
ν ′2
with ν 0 = m π m N .Using a smoothness assumption and from the optical theorem
Im T (−)
πN→πN
(ν,0,0,0) = νσ(−)
tot (ν) = ν[σ π− N
tot
(ν) − σ π+ N
tot (ν)]
we get the famous Adler-Weissberger relation, which is well fulfilled.
with
1
g 2 A
= 1 + 2M2 N
πg 2 πNN
∫ ∞
ν 0
N
dν [σπ− tot
(ν) − σ π+ N
tot (ν)]
ν
Finally we have the Kroll-Rudermann-Theorem:
σ(γN → πN) = 4π |−→ q |
ω
(|E 0+ | 2 + 2 |M 1+ | 2 + |M 1− | 2 )
E 0+ (γN → πN) = e√ 2g A
8πf π
We get also for neutral pion photoproduction
E 0+ (γp → π 0 p) = eg [
A
− m π
+ m2 π
8πf π m N 2m 2 (3 + κ p ) + m2 π
N
16fπ
2 + o( m3 π
m 3 )
N
124
]

These low energy theorems are all based on Ward-identities which are generalizations
of Noethers theorem. For the low energy theorems Ward identities are
used which involve relations between the matrix elements of currents and the
matrix elements of divergences. One Ward Identity we know already and have
used it in the section about the pion-nucleon sigma-term.
There is also for s-wave pion -nucleon scattering another low-energy theorem:
Weinberg-Tomozawa Theorem. If one starts from the decomposition of the
t-matrix eq.(207) then the corresponding s-wave scattering lengths are
a ± = 1 (
1 + m ) −1
π
T ± | threshold
4π M N
Since at threshold th isospin-odd amplitudebecomes
T − = ω
2f 2 π
with q 2 = ω 2 + −→ q 2 . This implies at threshold that the isovectors cattering length
is given by (
1 + m )
π
a − = m π
M N 8πfπ
2 ∼ 0.13fm
and hence it vanishes in the chiral soft-pion limit m π −→ 0. On the other hand
the isoscalar scattering length
(
1 + m π
M N
)
a + = m2 π
4πf 2 π
(
ΣπN
m 2 π
)
+ d − g2 A
4M N
is suppressed in comparison with a − by an additional power of m π and in volves
a subtel cancellation bewtween the sigma term and the parameter d, which must
be frixed by comparison with experiment. The empirical pion-nuclen scattering
length can be taken from the analysis of Ericson, Loisier and Thomas and yields
in agreement with the above estimates
10.12 Isospin-Violation
α + = −(0.003 ± 0.002) fm
α − = +(0.128 ± 0.002) fm
We remember the symmmetry breaking mass-term in the QCD hamiltonian
H sbr−m
QCD
= −[m uūu + m d ¯dd + ms¯ss]
In the context of the sigma-term we considered only the dase of m u = m d and
ignored completely the fact that there are also electromagnetic self energies,
which contribute to the mass of a particle. Such a electromagnetic contribution
is by no means small. On the contrary, its contribution is of the same order
of magnitude as the contribution due to the different quark masses. Thus one
125

must carefully distinguish between the hadronic and electromagnetic isospin
splitting of the masses. For example: The pion is built like π + = u ¯d and
π 0 = 1 √
2
(uū + d ¯d). Thus the quark masses contribute basically equal to both
sorts of pions, whereas the electromagnetic selfenergy of the π 0 is zero and of
π + is nonzero. The mass difference of both pions can be read off from the table
particle mass (MeV) structure
π + 139.6 u ¯d
π 0 135.0 uū + d ¯d
K + 493.7 ¯su
K 0 497.7 ¯sd
η 547.5 ūu + ¯dd − 2¯ss
p 938.3 uud
n 939.6 udd
Actually one has to coplement the above symmetry breaking term by an
electromagnetic symmetry breaking term. It can be written as
∫
= e2 d 4 xT (J µ elm (x)Jν elm(0)) Dµν(x)
γ
H sbr−γ
QCD
where Dµν(x) γ is the photon propagator and J µ elm
(x) is the electromagnetic current.
We will not need in the following the particular expression. We can now
directly generalize the expressions of the previous subsection. We obtain: insert the
with
f a m 2 aδ ab = σ0m ab + σ0γ
ab
σ ab
0m = 〈0| [ Q 5a , [ Q 5b ,H m QCD]]
|0〉
[ [ ]]
σ0γ ab = 〈0| Q 5a , Q 5b ,H γ QCD
|0〉
with a = 1,2,3 corresponding to pion, a = 4,5,6,7 corresponding to kaon and
a = 8 corresponding to eta. For the electrically neutral axial charge operators
we have [ ]
Q 5a ,H γ QCD
= 0
and hence we obtain (Dashen 1979)
σ 0γ (π 0 ) = σ 0γ (K 0 ) = σ 0γ (K 0 ) = σ 0γ (η 0 ) = 0
Since in the terms σ 0γ there are no masses involved we have for them exact
U-spin-invariance one also, i.e. invariance under
( )
( )
d
ψ = → ψ ′ = exp(−iθ a τ a d
)
s
s
and hence one gets exactly (and this defines the constant B)
σ 0γ (π + ) = σ 0γ (K + ) = B
previous expressions
without
isosopin
breakint
126

For the estimation of the total isospin splitting, originating from quark masses
and electromagnetic effects, we proceed now similarly as we know. We assume
〈0| ūu |0〉 = 〈0| ¯dd |0〉 = 〈0| ¯ss |0〉 = A
Then a simple generalization of our previous calculation yields:
f π m 2 π + = (m u + m d )A + B
f π m 2 π 0 = (m u + m d )A
f K m 2 K + = (m u + m s )A + B
f K m 2 K 0 = (m u + m s )A
f η m 2 η 0 = 1 3 (m u + m d + 4m s )A
Assuming all the decay constants to be identical we obtain by linear combinations
the generalized Gell-Mann-Okubo relation:
4m 2 K + = 3m2 η + m 2 π +
and
and
We obtain again
and
4m 2 K 0 = 3m2 η + m 2 π 0
m 2 π 0
m 2 K 0
= m u + m d
m s + m d
m 2 K + − m 2 K 0 − m 2 π +
m 2 K 0 − m 2 K + + m 2 π + − 2m 2 π 0
m u + m d
2m s
≃ 1
25
m d
m u
∼ = 1.8
= m d
m u
The considerations we just have done had a great impact about 30 years
ago. At that time it was not clear, that the quarks up and down had different
masses and one attributed the mass difference of the physical particles solely
to electromagnetic effects. The consequences would be: The charged pion has
a larger mass than the neutral one, due to the repulsive electric forces in the
charged pions. One expected the same for the kaon. This had lead to the so
called Dashen-sum-rule m 2 K + − m 2 K 0 = m 2 π + − m 2 π 0 However, if one looks
at the experimental masses one notices that the neutral kaon is heavier than
127

the charged one. Today one explains this discrepancy with the mass-difference
between up- and down-quark.
Altogether we obtain for the current quark masses (QCD-quarks):
m u ≃ 6MeV m d ≃ 10MeV m s ≃ 180MeV
Comment: One should clearly distinguish these current masses from the
constituten masses, which we find in simple non-relativistic hadron models.
Those masses are about 350-450 MeV
11 Group theory, Flavour structure and Quark
model
We remember the Lagrangean of the QCD (127). In the following we consider
the QCD-Lagrangean restricted to up-, down- and strange fields. This is of
course an approximation, but if one restricts oneself to the calculation of light
mesons and baryons this is in view of their masses (i.e. around and smaller
than 1-2 GeV) probably not too bad. This mutilated QCD–Lagrangean is
SU(3)-flavour invariant (vector symmetry), if all fields are assumed to have the
same mass parameter in the Lagrangean. In fact this is true as far as up- and
down-quark fields are concerned, where it is a good approximation to assume
the masses to be the same. If onc includes the strange quark mass in the Lagrangean
one one has to treat it in a different way. One way is, to incorporate
it by perturbation theory. It will be shown later that this is indeed a good
way and that first order is already enough. Thus in the following we will work
out eigenstates of the SU(3)-flavour symmetry for up- down- and strange quark
fields (they form the so called SU(3)-multiplets) and after that treat the term
m s¯ss by first order perturbation. In fact the particles in nature correspond to a
good approximation to the states in the multiplets. We particularly consider the
isovector SU(2) and isovector SU(3)-symmetry and derive mesonic and baryonic
states, which belong to multiplets of this symmetry. So far we are talking only of
the QCD-Lagrangean and its approximate symmetries. However the quantum
numbers of the multiplets can be equally well be constructed by a coupling of
three quarks, if one considers them as the particles of the fundamental representation
of SU(3)-Flavour and if one associates with them three colour degrees of
freedom in a totally antisymmetric state, and spin- and flavour-degrees of freedom
in symmetric states. Then practically all currently known hadrons can be
classified as either q¯q states (mesons) or qqq states (baryons). This is the reason
for the success of the quark model. However, one should mention, that all the
SU(3)-algebra holds without assuming the existence of quarks. In fact, when
Gell-Mann and Zweig invented the SU(3)-algebra they thought of an abstract
classification of the symmetries and not of particles. For the following, however,
it often makes life easier, if one thinks of quarks as real particles forming a
(small) many-body state with the structure as either q¯q states (mesons) or qqq
states (baryons). In the first sections of this chapter this is not necessary since
128

we deal solely with symmetries and fields. The last step builds a bridge to the
quark model and there one makes assumptions going beyond simple symmetry
algebra.
The detailed structure of the baryons and mesons requires a study of the
colour group, the rotational group and the flavour group. Colour turns out
mostly as a trivial quantum number and requires mostly only a trace. The
rotational group O(3) is known and it is known that it is isomorphic to SU(2).
It will be shortly reviewed. The flavour group, since it is an SU(3) group, will
be discussed in detail.
11.1 Elements of Lie-group theory
We consider the transformation of the Lie group
|ψ〉 → |ψ ′ 〉 = U(θ a ) |ψ〉
U(θ a ) = exp(−iθ a T a )
Here |ψ〉 is an abstract Hilbert vector. The θ a are real numbers, a =
1,2,...,N 2 − 1. The operators U(θ a ) are members of a the Lie-group SU(N).
The operators T a are hermitian and called “generators of the Lie-group SU(N)”
and by definition they fulfill the following commutation rules
[
T a ,T b] = if abc t c Tr(T a T b ) = 1 2 δab
The numbers f abc are called structure functions of the Lie-group. They determine
the group properties uniquely. The Lie-Group SU(N) can be shown to be
of rank r = N − 1. This by definition means that among the N 2 − 1 generators
T a there exist r = N − 1 generators, which commute with each other.
Each Lie-group has N −1 Casimir-operators C i with 1 = 1,2,...N −1, which
are functions of the generators and which are defined by the property that each
of them commutes with each of the generators and hence with each member of
the group:
[C i ,T a ] = 0 i = 1,2,...N − 1 a = 1,....,N 2 − 1
The N − 1 Casimir operators and the r commuting generators have common
eigenvectors. They are grouped in multiplets |C 1 ,...C N−1 ,α 1 ,....,α r 〉, where
all eigenvectors within a multiplet are characterized by r = N − 1 quantum
numbers. The multiplets will be explicitely constructed below. If the Hamilton
operator of a system shows the symmetry corresponding to the group considert,
then it commutes with the group operators U(θ a ), then it commutes also with
the generators
[H,T a ] = 0
and in particular with the r = N − 1 generators, which commute with each
other. Thus the states of the multiplet are also eigenstates of the Hamiltonian.
129

In this case they all have the same energy. Since they are eigenstates of hermitian
operators they form a basis. In this basis the commuting generators are diagonal,
the non-commuting ones are non-diagonal.
It is important to define a representation of a Lie-group: A representation
of a group is a mapping of the group element U(θ a ) on a unitariy nxn matrix
D(U(θ a )):
U(θ a ) ↦→ D(U(θ a ))
where the mapping is such that the group properties are fulfilled:
and
D(U 1 U 2 ) = D(U 1 )D(U 2 )
D(U)D(U) † = 1
The simplest representation, with dimension larger than one, is called “fundamental
representation”.
11.2 SU(2)-algebra:
We know as example the rotational operators:
U(θ a ) = exp(−iθ a J a ) a = 1,2,3
with
[
J a ,J b] = iɛ abc J c
They form a SU(2) group. Its fundamental representation has the dimension 2
and it is given by the Pauli-matrices. The rank of the SU(2) group is given by
r = N − 1 = 1 and hence has one casimir operator, which we know, of course
C = J 2 = J 2 1 + J 2 2 + J 2 3
By definition J 2 commutes with all generators J a , however for the construction
of multiplets one choses always J 3 as the one, whose eigenstates group into
multiplets. The multiplets are charakterized by being eigenstates of J 2 ,J 3 and
indicated as |j,m〉, where the various multiplets are distinguished by different
values of j and the states within a multiplet are distinguished by differen values
of m with −j ≤ m ≤ +j. The various multiplets can be constructed easily. We
repeat this simple exercise because in the case of SU(3) it is pricipally the same,
but technically more complicated:
Define for this raising and lowering operators:
with the following properties
J ± = J 1 ± iJ 2
(J + ) † = J −
(J − ) † = J +
130

Then we have immediately the following formulae:
J 2 = 1 2 (J +J − + J − J + ) + J 2 3
[J + ,J − ] = 2J 3
[J ± ,J 3 ] = ∓J ±
One can immediately construct the multiplets. Consider normalized eigenstates
|λ,m〉 of J 2 and J 3 :
J 2 |λ,m〉 = λ |λ,m〉
J 3 |λ,m〉 = m |λ,m〉
One can easily show that the states |λ,m〉 have the following properties under
the action of J ± :
J ± |λ,m〉 = a ± (λ,m) |λ,m ± 1〉
because e.g:
J 3 (J ± |λ,m〉) = ±J ± |λ,m〉 + J ± J 3 |λ,m〉
= ±J ± |λ,m〉 + mJ ± |λ,m〉 = (m ± 1)J ± |λ,m〉
That gives the name for the raising and lowering operators since their application
to |λ,m〉 raises and lowers the quantum number m by one. For a given λ the
values of m are bounded:
λ − m 2 > 0
because 0 ≤ 〈λ,m| J 2 1 |λ,m〉 + 〈λ,m| J 2 2 |λ,m〉 = 〈λ,m| J 2 − J 2 3 |λ,m〉. For a
given λ we consider the largest m and call this j. Then we have, since m is
largest
J + |λ,j〉 = 0
From this we get by simple calculation
Or
0 = J − J + |λ,j〉 = (J 2 − J 2 3 − J 3 ) |λ,j〉 = (λ − j 2 − j) |λ,j〉
λ = j(j + 1)
If onc starts from the smallest value of m, we call it j ′ , then we can show by
similar means that
λ = j ′ (j ′ − 1)
From this follows that
j(j + 1) = j ′ (j ′ − 1)
131

or either j ′ = −j or j ′ = j + 1. Since we have assumed that j is the largest
value, we must have
j ′ = −j
By application of J − to |λ,m〉 one changes m → m − 1. By a finite number of
applications of J − to |λ,j〉 we must end up at |λ,j ′ 〉. Hence j − j ′ must be an
integer. The j − j ′ is known and it is j − j ′ = 2j Hence
j = integer or half-integer
Now we can easily evaluate the coefficients a ± (λ,m) for an arbitrary m in the
following way: We know that
We can calculate explicitely
〈λ,m|J − J + |λ,m〉 = |a + (λ,m)| 2
〈λ,m|J − J + |λ,m〉 = 〈λ,m|J 2 −J 2 3 −J 3 |λ,m〉 = 〈λ,m| j(j +1) −m 2 −m |λ,m〉
with the result that
a + (λ,m) = ((j − m) (j + m + 1)) 1 2
a − (λ,m) = ((j + m) (j − m + 1)) 1 2
We can summarize and have then the multiplet:
J 2 |j,m〉 = j(j + 1) |j,m〉
J 3 |j,m〉 = m |j,m〉
J ± |j,m〉 = ((j ∓ m) (j ± m + 1)) 1 2
|j,m ± 1〉
The different multiplets are now those which we know well:
j dim= 2j + 1
0 1
1
2
2
1 3
3
2
4
2 5
132

11.3 SU(2)-multiplets in nature
We consider the QCD-Lagrangean and consider the SU(2)-flavour transformation
between up- and down-quark fields or quarks. If we consider these quark
fields to be of equal current mass, the Lagrangean is invariant, as it is well
known. We then have the global isospin invariance of the system:
( )
u
ψ = → ψ ′ = exp(−iθ a τ a )ψ
d
This SU(2) is a symmetry of the QCD-Hamiltonian and there should be eigenstates
of the QCD-Hamiltonian which form SU(2) multiplets. The eigenstates
of the QCD-Hamiltonian are particles, which we find in nature. Thus we expect
the some of the particles to group into multiplets, whose members all have the
same mass. Actually those multiplets are called isospin-multiplets and they are
well known. The difference in mass of the various members of the multiplet is
much less than 1%, only for the pion we have about 5%, because it is rather
light:
particles dimension isospin-multiplet T J mass (MeV)
1 1
nucleon 2 p,n
2 2
938
Sigma 3 Σ + ,Σ 0 ,Σ − 1
1
2
1189
Xi, Cascade 2 Ξ − ,Ξ 0 1 1
2 2
1315
Delta 4 ∆ ++ ,∆ + ,∆ 0 ,∆ − 3 3
2 2
1230
Lambda 1 Λ 0 1
0
2
1115
Pion 3 π + ,π 0 ,π − 1 0 139
Rho 3 ρ + ,ρ 0 ,ρ − 1 1 768
Kaon 2 K 0 ,K + 1 2
0 494
Kaon 2 K ∗0 ,K ∗+ 1 2
1 892
Eta 1 η 0 0 547
Omega 1 ω 0 1 783
Thus we see that there are systems in nature, which fulfill the SU(2)-
symmetry.
11.4 SU(3)-algebra
The SU(3)-Lie-Group is given by the commutation relations
[
F a ,F b] = if abc F c Tr(F a F b ) = 1 2 δab
with given and well documented structure coefficients f abc . The fundamental
representation is 3-dimensional. It is characterized by the Gell-Mann matrices.
The Gell-Mann matrices are a basis for all traceless hermitean matrices of dimensin
3x3. Any hermitian 3x3 matrix can be written as a linear combination
of the unity matrix and the eight Gell-Mann matrices. One can constrauct the
133

multipletts in a way completely analogous to the SU(2) case. For this one needs
the following step-up and step-down operators:
T ± = F 1 ± iF 2 T 3 (
= F 3
V ± = F 4 ± iF 5 V 3 = 1 3
2 ( 2 Y + T 3)
U ± = F 6 ± iF 7 U 3 = 1 3
2 2 Y − T 3)
Y = √ 2 3
F 8
with the properties
[T 3 ,T ± ] = ±T ± [T + ,T − ] = 2T 3 T-Spin SU(2)
[U 3 ,U ± ] = ±U ± [U + ,U − ] = 2U 3 U-Spin SU(2)
[V 3 ,V ± ] = ±V ± [V + ,V − ] = 2V 3 V-Spin SU(2)
and
[Y,T ± ] = 0 [Y,U ± ] = ±U ± [Y,V ± ] = ±V ±
[T + ,V + ] = [T + ,U − ] = 0 [U + ,V + ] = 0 [T 3 ,Y ] = 0
[T + ,V − ] = −U − [T + ,U + ] = V + [U + ,V − ] = T −
[T 3 ,V ± ] = ± 1 2 V ±
Ther hermiticity properties are given by
T + = (T − ) † U + = (U − ) † V + = (V − ) †
The Casimir are defined as operators which commute iwth each generator
and hence with each group element. They are given in each group theory book
(see F. Stancu). They are denoted as
C 1 =
8∑
i=1
F 2 i = 2i √
3
∑
fijk F i F j F k
C 2 = ∑ d ijk F i F j F k
Here the d ijk are the totally symmetric structure coefficients.
The SU(3) group has rank 2, thus there are 2 commuting generators. These
are Y and T 3 with
[Y,T 3 ] = 0
This corresponds to the fact that λ 3 and λ 8 commute. Since there are also
the two Casimir operators, which commute with each generator, the multiplet
is characterized by a complete system of commuting operators in the following
way.
C 1 |αβT 3 Y 〉 = α |αβT 3 Y 〉
C 2 |αβT 3 Y 〉 = β |αβT 3 Y 〉
T 3 |αβT 3 Y 〉 = T 3 |αβT 3 Y 〉
Y |αβT 3 Y 〉 = Y |αβT 3 Y 〉
134

By direct calculation, using the commutator rules, one can show now that
the following equations hold
V ± |αβT 3 Y 〉 = a(T 3 ,Y ) ∣ α,β,T3 ± 1 2
,Y ± 1〉
U ± |αβT 3 Y 〉 = b(T 3 ,Y ) ∣ ∣α,β,T 3 ∓ 1 2
,Y ± 1〉
T ± |αβT 3 Y 〉 = c(T 3 ,Y ) |α,β,T 3 ± 1,Y 〉
The multiplets are now certain figures in the T 3 −Y −plane. Due to symmetries
they show a 6-plet structure. Because the operators T ± ,T 3 obbey a SU(2)
algebra, the T 3 is integer or half-integer. The units and limits of Y are not
fixed yet. Their difference must be Unity. One yields in the end a structure like
this, which is characterized by several symmetries which all originat from the
sdymmetries of the SU(2) subgoups:
Reflection symmetry w.r. to straight line given by T 3 = 0
Reflection symmetry w.r. to straight line given by U 3 = 0 corresponding
(see above formulae) to Y = 2 3 T 3
Reflection symmetry w.r. to straight line given by V 3 = 0 corresponding (see
above formulae) to Y = − 2 3 T 3
In general a multiplet looks like this:
U
+
Y
V
+
T
3
The explicit construction of a multilplet works now in the following way:
Chose a point, call this |ψ max 〉 and this should be the state of the largest T 3 -
value and som Y -value. Apparently we have
T + |ψ max 〉 = 0
V + |ψ max 〉 = 0
U − |ψ max 〉 = 0
This point defines the right upper end of the multiplet. In order to draw the
outer most borderlines one applies V − to |ψ max 〉. This is, say, possible p-times.
135

Then we have
(V − ) p+1 |ψ max 〉 = 0
Then apply to this point T − as often as possible. This will go q-times, i.e:
(T − ) q+1 (V − ) p |ψ max 〉 = 0
The rest of the multiplet is given by symmetries. The inner points one can
construct accordingly. The scheme is given in the following way:
Y
ψ
max
V
−
T
3
T
−
One can also calculate the multiplicity of each point, i.e. the number how
often in this construction a certain point is generated. In this way one can
construct various multiplets, from which we note the smallest ones:
notion dimension (p,q)
[1] 1 (0,0)
[
[3]
]
3 (1,0)
3 3 (0,1)
[8] 8 (1,1)
[6] 6 (2,0)
[10] 10 (3,0)
[
10
]
10 (0,3)
11.5 SU(3)-multiplets in nature
One noteces immediately that the particles in nature group themselves according
to the lowest dimensional multiplets of SU(3) if one links the hypercharge with
the charge of the particle by means of the famous Gell-Mann-Nishijima relation,
which is a phenomenological relation at the time, when it was discovered:
Q = 1 2 Y + T 3
136

η
0
T
3
Actually one proceeds in the following way: One takes a particle, looks in
the data for its isospin partners, which are nearly mass-degenerate, from the
number of isospin partners one extracts the T 3 . Then one looks for the charge
of the particle, extracts from it and from the now known T 3 the hypercharge
Y according to Gell-Mann-Nishijima relation and then one has its quantum
numbers T 3 and Y and its place in the multiplet. In this way one constructs the
following correspondences of particles with multiplets, where we give already
the angular momentum J without telling yet, wlhere it comes from:
Meson-singlet,J=0,T=0, bessere Bezeichnung η 1 , weil Singulett:
Meson-octet: pseudoscalar mesons, J=0, follows as:
K
0
Y
+
K
−
π
0
π
8
η
+
π
T
3
−
K
0
K
Meson-octet, J=1, follows as:
137

−
ρ
* 0 Y
* +
K
K
0
ρ
ρ
+
ω
* 0
*
K
− K
T
3
Baryon-octet, J=1/2, follows as:
Y
0
n
p
− 0
+
Σ Σ Σ
−
Ξ
0
Ξ
T
3
Baryon-decuplet, J=3/2, follows as:
138

Y
−
∆
0
∆
∆
+
++
∆
∗ −
Σ
∗
Σ
Σ
∗+
T
3
*
−
Ξ
0
*
Ξ
Ω
One should note that in constructing the multiplets we never made anny
assumption about the existence of quarks. In fact they are not needed at all in
order to have multiplets. That is the reason why there are models (e.g. Skyrmemodel
and all its variants), which work reasonably well and do not have quarks
at all but are quantum theories of interacting meson fields with some particular
topological structure.
11.6 Quarks and SU(3)
11.6.1 Quarks and fractional charges
We have seen that the mesons and baryons group themselves in a natural way
in SU(3)-multiplets, if one interpretes Y as hypercharge, which is related to the
charge by the Gell-Mann-Nishijima relations:
Q = 1 2 Y + T 3
The smallest representation of SU(3) is [1] = D(0,0). This trivial representation
is in fact realized in nature (e.g. Λ and η). If quarks are perhaps the building
blocks of the light baryons and mesons from etsthetical points one would like
them to be in the smallest representation. This is not true, howevber one can
mathch the phenomenology if one assumes three quarks of different flavour to
be represented by the smallest NON-trivial representation, i.e. the fundamental
representation of SU(3)-flavour. Achtually SU(3) has rank 2 and hence 2
139

fundamental repesentations:
[3] = D(1,0)
[¯3] = D(0,1)
We will see, that the quarks form [3] and the anti-quarks form [¯3]. This can
graphically be represented as:
ψ
2
ψ
1
ψ
*
3
ψ
3
ψ * ψ *
1 2
Quarks: [3] = D(1,0)
Antiquarks: [¯3] = D(0,1)
We can proceed mathematically as: Consider ψ 1 = ψ max and another ψ 2
with (ψ 1 ,ψ 2 ) forming an isospin dublett, in which then by definition we have
ψ 2 = T − ψ 1 . and
T 3 ψ 1 = 1 2 ψ 1 T 3 ψ 2 = − 1 2 ψ 2
The ψ 3 forms an isospin singlet, i.e. T 3 ψ 3 = 0ψ 3 .
The ψ 2 ,ψ 3 form an U-Spin dublett, i.e. ψ 2 = U − ψ 3
The ψ 1 forms an U-Spin singlet, i.e. U 3 ψ 1 = 0
The ψ 1 ,ψ 3 form an V-Spin dublett, i.e. ψ 3 = V − ψ 1
The ψ 2 forms an U-Spin singlet, i.e. V 3 ψ 2 = 0
These simple features have direct consequences for the charge of the states
ψ i and ( hence for the particles corresponding the these states: We have with
U 3 = 1 3
2 2 Y − T )
3 from above the equation
and hence
U 3 = 1 4 (3Y − 2T 3)
or
Y ψ 1 = 1 3 (4U 3 + 2T 3 )ψ 1 = 2 3 T 3ψ 1 = 1 3 ψ 1
Qψ 1 = ( 1 2 Y + T 3)ψ 1 = ( 1 6 + 3 6 )ψ 1 = 2 3 ψ 1
and in a similar way charges of ψ 2 and ψ 3 . If we recall now these results we have
alltogether (where we have renamed the quark flavours in the usual way):
Qψ 1 = 2 3 ψ 1
ψ 1 = u
140

Qψ 2 = − 1 3 ψ 2
ψ 2 = d
Qψ 3 = − 1 3 ψ 3
ψ 3 = s
Thus we get automatically fractional charges for the quarks. We remember:
The assumption was that they occupy the fundamental representation fo the
SU(3)-flavour. In the exactly similar way we construct the antiquark triplett:
Qψ 1 = − 2 3 ψ 1
ψ 1 = ū
Qψ 2 = + 1 3 ψ 2
ψ 2 = ¯d
Qψ 3 = + 1 3 ψ 3
ψ 3 = ¯s
The fundamental representations [3] and [¯3] are two totally different representations.
They cannot be transformed into each other by transformation of one
or several generators. Thus we obtain fractional charges of the quarks if we
demand SU(3) symmetry and the Gellman-Nishijima relation and assume that
the quarks occupy the smallest non-trivial representation.
11.6.2 Construction of higher multiplets by triplets, Confinement
(phenomenological)
Coupling of Fundamental representation We will see, that one can construct
the other multiplets in nature outgoing from these two fundamental representations,
if one couples these fundamental representationsby means of SU(3)-
Clebsh-Gordan coefficients and constructs successively the higher dimensional
multiplets. We can give examples:
In SU(2), which we take as an example, we couple
−→ 1
2 ⊕ −→ 1
2 = −→ −→ −→
1 1 ⊕ 1
2 = −→ 3
2
−→ 1
2
−→ −→ −→
0 0 ⊕
1
2 = −→ 1
2
In the usual nomenclature this is written as
[2] ⊕ [2] = [3] + [1]
[2] ⊕ [2] ⊕ [2] = [4] + [2] + [2]
In SU(3) all multiplets can be constructed by coupling in a particular way
the quark triplet and quark anti-triiplet. The general construction is
141

[3] ⊕ [3] ⊕ [3] ⊕ [3] ⊕ [3] ⊕ [ 3 ] ⊕ [ 3 ] ⊕ [ 3 ] = D(p,q)
p-times [3] and q-times 3
The simplest coupling consists in coupling an quark triplet with an quarkantitriplet:
[3] ⊕ [ 3 ] = [1] + [8]
Although the fundamental triplets do not occur freely in nature, the coupled
ones exist and their members are the well known pseudoscalar mesons, i.e. a
meson octet and meson singlet.
The interesting point is, that not all possible multiplets are realized in nature.
It is mathematically possible to construct
[3] ⊕ [3] = [6] + [ 3 ]
However the corresponding particles do not exist. In terms of quarks the resulting
particles should then consist of 2 quarks. This is not possible because of
confinement: The 3 colour degrees of freedom for each of the 2 quarks cannot
be coupled to a colour singlet. And the concept of confinement demands that
physical states are exclusively colour singlets.
The concept of confinement is not yet fully understood. One observes confinement
everywhere and no experiment is known which contradicts the confinement.
Confinement means first, that physical states, i.e. states which can be
detected in a detector, must be in a colour singlet. This is realized in the quark
model (see below) by explicitely constructing a colour singlet from three quarks
or from quark-antiquarks. It means second, that all three quarks must be close
together in a small volume of about 1 fm diameter. Actually, the demand of
colourless physical states is not a consequence of the colour gauge invariance
of the QCD-Lagrangean. One can see this immediately by an analogy to the
QED, which is invariant under abelian gauge transformation. There we have
charged states of different charges and apparently this is compatible with gauge
invariance.
Thus, if we couple the above non-existing multiplets of [3] ⊕[3] with another
quark triplet, we get physical baryonic (B=1) states, because three quarks with
different colours can couple to a colour singlet as it is given e.g. in eq.(211) for
the ∆ ++ :
[3] ⊕ [3] ⊕ [3] = [6] ⊕ [ 3 ] ⊕ [3] = [8] + [8] + [10] + [10]
Thus the phenomenological baryons and mesons are properly grouped in
SU(3) flavour -multiplets if those are constructed by an appropriate coupling of
of quark triplets and antiquark-triplets, where the quarks and anti-quarks have
the following quantum numbers:
142

quark T T 3 Q Y S B
1 1 2 1 1
q 1 = u
2 2 3 3
0
3
1
q 2 = d
2
− 1 2
− 1 1 1
3 3
0
3
q 3 = s 0 0 − 1 3
− 2 1
3
−1
3
1
q 1 = u
2
− 1 2
− 2 3
− 1 3
0 − 1 3
1 1 1
q 2 = d
2 2 3
− 1 3
0 − 1 3
1 2
q 3 = s 0 0
3 3
1 − 1 3
Thus we obtain finally the flavour structure of the mesons and baryons. In
fact one obtains Clebsh-Gordan coefficients which yield the combinations of the
group elements in the coupling process to generate higher multiplets.
Mesons:
J=0 singlet η 1 = ūu + ¯dd + ¯ss
J=0 octet K 0 ∼ ¯sd K + ∼ ¯su 494MeV
π − ∼ ūd π 0 ∼ ūu − ¯dd π + ∼ ¯du 139MeV
η 8 ∼ ūu + ¯dd − 2¯ss
K − ∼ ūs K 0 ∼ ¯ds J=0 494MeV
ii
J=1 singlet ω 1 = ūu + ¯dd + ¯ss(= φ)
J=1 octet K ∗0 ∼ ¯sd K ∗+ ∼ ¯su 892MeV
ρ − ∼ ūd ρ 0 ∼ ūu − ¯dd ρ + ∼ ¯du 768MeV
ω 8 ∼ ūu + ¯dd − 2¯ss
K ∗− ∼ ūs K ∗0 ∼ ¯ds J=1 892MeV
Baryons:
J=1/2 octet n ∼ udd p ∼ uud 938MeV
Σ − ∼ dds Σ 0 ∼ (ud + du)s Σ + ∼ uus 1189MeV
Λ 0 ∼ (ud − du)s
1115MeV
Ξ − ∼ dss Ξ 0 ∼ uss J=1/2 1315MeV
∆ − ∼ ddd ∆ 0 ∼ udd ∆ + ∼ uud ∆ ++ ∼ uuu 1230MeV
Σ ∗− ∼ dds Σ ∗0 ∼ uds Σ ∗+ ∼ uus 1385MeV
Ξ ∗− ∼ dss Ξ ∗0 ∼ uss 1530MeV
Ω − ∼ sss octet J=3/2 1672MeV
In the above tables there are the masses of the particles given as well. This
is an antiacipation: At the present level of the discussion the members of a
multiplet are degenerate. We will calculate the splitting due to the finite current
mass of the strange quark in one of the next sections.
Mixing of η and ω There is one comment in order about the pseudoscalar
η-mesons. Actually we have one η as a member of the meson octet: η 8 =
143

(uū+d ¯d−2s¯s) and another η as the only member of a singlet: η 1 = uū+d ¯d+s¯s.
Since both particles have the same quantum numbers and since the pure SU(3)-
symmetry is only an approximation (since we neglect the fact that the strange
quark has a finite mass), in nature these two eta-states mix, such that there are
two new eta-states of the following form:
η = η 8 cos θ − η 1 sin θ
η ′ = η 8 sinθ + η 1 cos θ
However one should note: There is the axial anomaly, and due to this we get
an additional mixing mechanism between η and η ′ creating two new particles,
which as a surprise are again closer to the original ones, i.e. η 8 and η 0 yielding
η = 0.58(uū + d ¯d) − 0.57s¯s
η ′ = 0.40(uū + d ¯d) + 0.82s¯s
547MeV
958MeV
Thus after these two mixings the s¯s-content of η is similar to its (uū + d ¯d)-
component, which bears a strong similarity to the original structure of η 1 . Similarly
s¯s-content of η ′ is about twice as large as its (uū + d ¯d)-componen, which
reminds of the structure of η 8 .
A similar mixing happens for the pseudoscalar vectormesons, however without
the effect of the anomaly. There is the singlet ω 1 and the octett ω 8 . These
mesons show the so called ideal mixing, which is defined in such a way, that the
s¯s component decouples completely from the (uū+d ¯d)-component. Thus in the
end we have for the physical mesons in a good approximation the structures
ω = uū + d ¯d
φ = s¯s
782MeV
1020MeV
These are the famous multiplets of mesons and baryons, which have been
popular for more than 40 years.
OZI-rule In the 1960’s an empirical property, called the Okubo-Zweig-Iizuka
rule (OZI-rule) was developed for mesonic decays and coupling constants. Its
usual statement is that flavour disconnected pocesses are suppressed compared
to those in which quark lines are nonnected. Unfortuntely the phenomenological
and theoretical status of this so-called rule is abibuous and several violations are
known. Nevertheless in the present language we can express it by the following
picture
Each straight line represents a quark. Thus OZI-allowed transitions are
those, where the quarks of the decaying particle end up as quarks of the daughter
particles. If the flavour structure of the daughter particles is so different from
the mother particle, then intermediate gluons are required. However, each of
the gluons couples with the strong coupling constant α s (Q 2 ), which is mostly
144

145

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s
¤ ¤ ¤ ¤ ¤
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¥ ¥ ¥ ¥
¥ ¤ ¤ ¤ ¤ ¤
s
u−bar
¤ ¤ ¤ ¤ ¤
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¥
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K−minus
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smaller than one. Thus the right graph is suppressed compared to the left graph.
This argument is not very convincing since the α s (Q 2 ) is not small for small
Q 2 . Thus kinematic properties enter as well. In fact there are several exceptions
known, where this OZI rule does not work. There is some justification also in
the limit of large N c .
An example is the decay of the φ-meson, which is basically s¯s:
and
It decays to 49% into K + K − and only to 10 −3 % into π + π − , which follows
the OZI rule as the two pictures show.
Matrix representation of octets This group representation is not always
favourable. There exist effective theories which are not QCD in terms of gluons
and quarks but treat solely the mesonic and baryonic degrees of freedom. The
Lagrangean of these theories expressed in terms of U = exp[iλ a φ a ] and this
gives rise to the following set of matrices depending separately on the parity
and angular momentum. This yields then for pseudoscalar mesons the field
matrix
φ PS (x) = 1 √
2
λ a φ a PS(x) =
⎛
⎜
⎝
π 0
√
2
+ η8 √
6
π + K +
π −
− π0 √
2
+ η8 √
6
K 0
K − ¯K0 − 2η8
√
6
⎞
⎟
⎠ (208)
146

Here the η meson field is the octet component of the η field, i.e.η 8 , and not the
observed η meson. This is a bit confusing as far as the notation is concerned.
All upper indices indicate the charge of the particle but in case of the eta, where
it indicates the group component. For the vector mesons we have similarly
φ V (x) = 1 √
2
λ a φ a V (x) =
⎛
⎜
⎝
ρ 0
√
2
+ ω8 √
6
ρ + K ∗+
− √ ρ0
2
+ √ ω8
6
K ∗− ¯K∗0 − 2ω0
ρ −
and for the octet baryons the field matrix
⎛
Ψ B (x) = 1 √
2
λ a ψ a B(x) =
⎜
⎝
Σ 0
√
2
+ Λ0 √
6
Σ + p
Σ −
⎞
K ∗0
√
6
− √ Σ0
2
+ √ Λ0
6
n
Ξ − Ξ 0 − 2Λ0
⎟
⎠ (209)
√
6
⎞
⎟
⎠ (210)
If one is interested in properties of a particular particle field, e.g. the field Σ + ,
one easily finds out the combination of λ a which yields this field. Actually one
should note that the normalizations of these matrices differ in the literature
and often the √ 2 are hidden somewhere else. Here we take the normalization
of Donoghue, Golowich, Holstein. There is a clear transformation between the
above multiplets and these new fields.
11.7 Fock states and non-relativistic quark model
We have considered fields by now. We also can consider single particle quantized
states of the fields. This is not exactly done, since the field can e.g. be
interacting, but it is done in the sense of the Lehman-Symanzik-Zmmermann
reduction formalismus. In this sense we can have |π + 〉 = b † π + |0〉 , which can be
considered as the result of canonical quantization of the field π + and the application
of the creation operator b † π + to the vacuum. Since, however, the field
π + is obtained by the product 3x¯3 and one can associate with the fundamental
triplet a quark, the final hadron Fock state is at least composed of a quark and
a anti-quark creation operator. If one assignes phenonemological masses to the
quarks and anti-quarks one can couple quark-antiquark paires to J = 0,1 and
three quarks can be coupled to J = 1/2,3/2. In all practical quark models an
assumption is made which greatly simplifies subsequent steps in the analysis,
namely that the spatial, spin, flavour and colour degrees of freedom factorize, or
in other words, the total wave function is a product of a spatial wave function,
times a spin-flavour Hilbert vector, and times a colour Hilbert vector. This
assumption allows us to write the creation and annihilation operators or the
quarks in terms of the spatial (n) spin (s = 1/2,m s ),flavour(q = u,d,s) and
color (k = 1,2,3) degrees of freedom. Furthermore it is assumed that the colour
degrees of freedom are fully antisymmetrized, such that the product of the other
degrees must be symmetric, in order to reflect the fermionic character of the
quarks. In the following we consider the quarks as sitting in the same orbital 1sstate
e.g. of some potential. Therefor the spatial degree of freedom is given by
147

an orbital wavefunction with n = 0,which is by construction symmetric. Thus
the spin-flavour wave function must be symmetric.
Altogother we have creation operators for quarks and antiquarks of the form
∆ ++
3/2
q † k,m s
= b † (n = 0,q,m s ,k)
¯q † k,m s
= d † (n = 0,q,m s ,k)
All this means that e.g. the following non-relativistic state of the delta-isobar
is written as (sum over double colour indices i,j,k)
∣
∣∆ ++
3/2
〉
= 1 6 ε ijku † i↑ u† j↑ u† k↑
|0〉 (211)
Due to the factorization only in this way a reasonable non-relativistic wave
function for the ∆ ++
3/2
can be constructed, which is known from experiment to
be in all respects consistent with a state of of three up-quarks only. From this
one concludes even that we have exactly 3 colours. In a similar way we obtain
the state of the rho-+ meson (J=1,T 3 = 1) as
∣ ρ
+
1
〉
=
1 √3 u † i↑ ¯d † i↑ |0〉
In a coordinate representation the creation operators create a state with a certain
wave function in coordinate space, given by e.g. a harmonic oscillator or
a bag or some other apropriate potential. Since this coupling does not change
the flavour structure one can explicitely write down the non-relativistic wave
function of the meson or baryon constructing it in a way that the colour is totally
antisymmetrized and hence the spin-flavour part fully symmetrized. The
flavour part is already constructed properly, the spin part has still to be treated.
This can be done that one performs Clebsh-Gordan sums over the 3-component
of the spin of the quark fields, coupling first two spins to J=0 and J=1 and
coupling these states then with the third quark to J=1/2. In fact it does not
matter in which way the coupling is performed as long as one considers symmetric
spin-flavour wave functions. Inthis way one obtains e.g. for some of the
mesons with J = 0 and some of the baryons with J = 1/2 the following 3-quark
wave functions (sum over colour indices i,j,k included):
state vectors of the pseudoscalar octet and singlet mesons (J=0)
∣ π
+ 〉 = √ 1 [u † ¯d †
6
i↑ i↓ − u† ¯d † i↓ i↑ ] |0〉
∣ π
− 〉 = 1 √
6
[d † i↑ū† i↓ − d† i↓ū† i↑ ] |0〉
∣ K
+ 〉 = 1 √
6
[u † i↑¯s† i↓ − u† i↓¯s† i↑ ] |0〉
etc.
148

und
state vectors of baryon spin-1/2 octet
|p ↑ 〉 = √ 1 ε ijk [u †
18
i↓ d† j↑ − u† i↑ d† j↓ ]u† k↑ |0〉
|n ↑ 〉 = 1 √
18
ε ijk [u † i↓ d† j↑ − u† i↑ d† j↓ ]u† k↑ |0〉
|Λ ↑ 〉 = 1 √
12
ε ijk [u † i↓ d† j↑ − u† i↑ d† j↓ ]s† k↑ |0〉
〉
∣
∣Ξ − ↑
= √ 1 ε ijk [s †
18
i↓ d† j↑ − s† i↑ d† j↓ ]s† k↑ |0〉
etc.
As one sees the sum over colours (indices=i,j,k)corresponds for the mesons
to a trace, where colour in the quark is paired with anti-colour in the antiquark,
and corresponds for the baryons to an expression involving the totally
antisymmetrized tensor ε ijk . The orbital wave functions are obtained in the
simplest way by the s-state of a harmonic oscillator, whose width is adjusted
to some experiments and where the masses of the quarks are about 300-400
MeV. The outcome is then the famous non-relativistic quark-model. Such a
149

model had many succsses and we just describe briefly its magnetic properties in
comparison with experiment. Let us assume that the quark magnetic moments
are given by their standard Dirac values
µ q = e q
2M q
whith the quark electric charges e u = 2 3e, etc. (electron has charge −e). The
baryonic magnetic moments are easily evaluated as matrix elements of the nonrelativistic
magnetic moment operator
µ B = 〈B ↑|m z |B ↑〉
−→ m = µu
∑
kss ′ 〈s| −→ σ |s ′ 〉 u † sk u s ′ k + (d) + (s)
where k indicates the colour over which is summarized. Using the above Fockstates
the results for the proton and neutron are simply
µ p = 4 3 µ u − 1 3 µ d
µ n = 4 3 µ d − 1 3 µ u
and in the limit of exact isospin symmetry with m u = m d one finds the famous
relation
µ p
µ n
= − 3 2
150

which is close to the observed value µp
µ n
= −1.46, and was and is considered
as a great success of the quark model. Fixing the constituent mass M u of the
up-quark and the (identical) down quark by the proton magnetic moment
µ p = 4 e u
− 1 e d
=
e [ 4 2
3 2M u 3 2M d 2M u 3 3 + 1 ]
1
3 3
gives a value for the quark masses
= e
2M u
=
[
2.79 e
2M p
]exp
M u = M d = 938MeV = 340MeV
2.79
Obviously these values are quite different from the numbers of 7MeV or 10MeV,
which we have quotet previously. In fact these big masses are called ”constituent
quark masses” and they canbe understood as the result of the spontaneous chiral
symmetry breaking, which is connected with dynamical mass generation.
Similar considerations for the Λ give us an estimate for the strange constituent
quark masse. The magnetic moment µ Λ of the Λ equals the magnetic
moment of the strange quark, µ s , since the ud-pair in the Λ is coupled to spin
zero (see its wavefunction). The µ s depends on the constituent mass of the
strange quark, and we obtain:
µ Λ = µ s = e q
= − 1 e
=
2M q 3 2M s
which implies with M s = 938MeV
3∗0.61
the values
[
−0.61 e
2M p
]exp
M s = 510MeV and M s − M u = 150MeV ∼ m s (212)
One sees here already that one expects a mass splitting between quark-states
consisting of up- and down-quarks only and those which contain one or two s-
quarks. This mass splitting will be presently ignored, however it will be treated
in the next subsection.
The quark model also establishes a simple relationship between the proton
magnetic moment and the matrix element for the magnetic diple (M 1 ) transition
between the nucleon and the Delta-isobar. The matrix element
µ p∆ = 〈p ↑|m z
∣ ∣∣∆ +
J z=+ 1 2
enters the photoproduction amplitude for the γp → ∆ + transition. THis is
easily evaluated with the above non-relativistic wave functions yielding
µ p∆ = 2 3√
2µp
This results holds only if one assumes that the momentum transfer is small
compared to the nucleon or delta mass. However in practice the photo excitation
of the delta requires an energy transfer of about 300 MeV since the mass of the
delta is 1232 MeV and the mass of the nucleon is 938 MeV. So in fact recoil
corrections are not negligeable. A full comparison of the magnetic moments of
the octet baryons with experiment is given in the table:
〉
151

11.7.1 Mass splittings
By now we have considered pure SU(3)-flavour times SU(2)-rotation and in
principle all states within a given multiplet are degenerate. This is because
we have ignored any mass difference between up-, down- and strange-quarks.
For up- and down-quarks this is well fulfilled in nature, since e.g. the mass
splitting between Neutron and Proton is only 2 MeV compared to the proton
mass of 938 MeV. For the strange quarks this is no longer a good approximation
since its mass is about 150-180 MeV, as we have seen at eq.(212). Therefore
we have to take it into account. Today we know that the term m s¯ss in the
QCD-Lagrangean is the only term, which can be responsible for the splitting
inside the octet and inside the decuplet. The simplest way to treeat it, and it
will be shown that this is indeed sufficient, consists in first order perturbation
theory, what we will do now. This will lead to the famous Gell-Mann-Okubo
mass formula.
One can easily show that we have
⎛
m s¯ss = m s¯q ⎝
0 0 0
0 0 0
0 0 1
⎞
⎠ q = m s¯q
[ 1
3 I − √ 1 ] [ ] 1
λ 8 q = m s¯q 3 3 I − Y q
thus the perturbation is proportional to m s and consists of a constant term and
a term linear in the hypercharge. Thus we have equal spacings between the
various isospin multiplets, since they differ linearly by Y . Thus we have for the
lowest baryon masses
M Σ = M N + M s
M Ξ = M N + 2M s
M Λ = M N + M s
152

or the Gell-Mann-Okubo relation:
M Σ + 3M Λ = 2(M N + M Ξ )
[2.23GeV ] exp
= [2.25GeV ] exp
This relation is experimentally well satisfied as one sees. Of course there is also
equal spacing between the members of the decuplet. This actually lead to the
prediction of the Ω − particle, which was then identified experimentally and for
which Gell-Mann obtained the Nobel prize. One should note, that this this
relation was discovered at a time, where no QCD was known yet. That means
the perturbation operator was not known. Thus its algebraic structure had to
be inferred from the experiment.
Following SU(3) and its breaking by the strange mass there is no relationship
between the splitting in the octet and the splitting in the decuplet. This just
does not follow from the quark model. However, Guadagnini has extracted
from the Skyme model and phenomenological considerations an equation, which
relates the masses of the octed and decuplett to each other. The Guadagnini
formula is
8(M Ξ ∗ + M N ) + 3M Σ = 11m Λ + 8M Σ ∗
and both sides differ only by less than 1%. The chiral quark soliton model gets
this result again by first order perturbation theory in m s in a natural way. This
is by no means trivial, since e.g. the Skyrme model does not provide it.
All these considerations show, that the strange mass m s can be considered
as ”small” such that first order perturbation theory is justified.
11.7.2 Quark model: Calculations
The quark model has a long history and it has been improved several times
by using wave functions of more and more sophisticated potentials. This was
altogether rather successfull in describing the energies of the various baryons.
Not only those with the quarks in the orbital s-state of some 3-dimensional
harmonic oscillator but also states which correspond to quarks in one or two
single particle excited states. We quote below the first pioneering calculations
by Isgur and Karl:
Their hamiltonian is given by
with
H =
3∑
i=1
(
M i + p2 i
2M i
)
+ K 2
∑
i

The first figure shows the spectrum of 1ħω-negative parity nucleon and delta
resonances. the shaded regions indicate the empirical positions and widths, the
solid bars are the predictions of the Isgur-Karl model.
The second figure shows the spectrum of the 2ħω-positive parity nuccleon
and delta resonances.
One should note, however, that there is no fundamantal reason for the potentials
used in those models and also for the use of the 3-quark wave functions.
There are several inner inconsistencies in the model:
First, the average momentum of a constituent quark inside the harmonic
oscillator potential can easily be calculated and is of the same order of magnitude
as its constituent mass, which means that a relativistic description is
needed. There are those extensions, however their conceptual background is
not there, since they are not relativistic field theories. Second: The assumption
of a three-quark state with constituent masses is a pure assumption which
cannot be derived from QCD. In QCD one has dynamical mass generation due
to spontaneous breakdown of chiral symmetrie, however this mass is momentum
dependent and serves also as a quark-pion coupling constant. Furthermore,
effects like the strange content of the nucleon - a rather topical problem - or
strange contributions to parton distribution functions are not described in the
quark models.
However, the virtue of those crude models lies in their phenomenological
simplicity. They work to some extent, in particular for excitation energies of
the mesons and baryons, although one does not really know, why. Altogether no
rasoning in terms of a quantum field theory is known which justifies the whole
154

non-relativistic quark model or its ”relativised” variants.
12 Effective bosonic Lagrangean
The objective of this section is to construct an effective theory, which describes
the interaction between Goldstone-bosons. The term “effective” means here,
that the fields entering the Lagrangean are not the quark or gluon fields but
the fields of the Goldstone bosons, which are in fact composite particles. The
construction of the effective Lagrangean will be based purely on the concept
of spontaneously broken chiral symmetry. Whenever useful we will make connections
to QCD. However, one should note, that never the deatilled kowledge
of QCD in terms of quarks and gluons is required, only the notion of chiral
symmetry and its spontaneous breakdown. At the end we want a Lagrangean
of the form o
L = L eff (GoldstoneFields,ExternalFields)
and a clear prescription how to use this Lagrangean in order to calculate a certain
process between Goldstone Bosons. The Goldstone Bosons, although being
composite particles, are treated as fields. On the level of QCD the operations
we will do correspond to a SU(3)-flavour QCD, i.e. a QCD where the heavy
quark degrees of freedom have already been integrated out.
155

12.1 Review: Spontaneous breakdown of chiral symmetry
12.1.1 Current algebra and transformation properties
We recollect the spontaneous chiral symmetry breaking of the QCD and as we
see it in nature: The QCD Lagrangean is in the massless limit invariant under
SU(3) L ⊗SU(3) R ⊗U(1) V . This yields conserved left-handed and right-handed
currents and the fermion current (which will be ignored in the following). We
have then left-handed and right-handed charges Q a L and Qa R and vector- and
axial-vector combinations:
Q a V = Q a R + Q a L
with the vector algebra
Q a L = Q a R − Q a L
[
Q a (t),Q b (t) ] = if abc Q c (t)
The ground state of the system is invariant under the vector transformation
exp(−iα a Q a V ) |0〉 = |0〉 Q a V |0〉 = 0 a = 1....8
>From Coleman´s theorem it follows then that the Hamilton operator H of the
system commutes with the vector charge:
[Q a V ,H] = 0
and hence the energy eigenstates can be grouped into SU(3)-multiplets where
all members of a certain multiplet have the same mass.
However, the ground state is not invariant under the corresponding axial
transformations
exp(−iα a Q a A) |0〉 ≠ |0〉 Q a V |0〉 ≠ 0 a = 1....8
In fact for each Q a A there exists a Goldstone-Boson, which is massless, has the
quantum numbers of the axial current (pseudoscalar, spinn zero). Actually
there are 8 generators of SU(3) L and also 8 generators of SU(3) L i.e. altogether
16 generators. Since the subgroup SU(3) V has 8 generators there must be
(8 + 8) − 8 = 8 Goldstone bosons. The goldstone bosonic fields π a (x) a =
1....8 transform under the subgroup SU(3) V of SU(3) L ⊗SU(3) R like an octett
representation: [
Q
a
V ,π b (x) ] = if abc π c (x)
12.1.2 Chiral Quark Condensate
Actuall there is a direct connection between the appearance of the Goldstoe
bosons and the scalar quark condensate 〈0| ¯qq |0〉 = 〈0|ūu+ ¯dd+ ¯ss |0〉. In order
to recall this we define the scalar quark density
s a (x) = ¯q(x)λ a q(x)
156

√
with λ a being the flavour Gell Mann matrices and λ 0 =
now show by direct calculation that we have
[
Q
a
V ,s 0 (x) ] = 0 a = 1....8
2
3
diag(1,1,1) One can
[
Q
a
V ,s b (x) ] = if abc s c (x) a,b,c = 1.....8
The proof for this needs the simple facts that Q a V = ∫ d 3 x¯q(x)λ a q(x) and
the simple fact that for an operator A(x) = q †(x)Âq(x) and B(x) = q† (x) ̂Bq(x)
with  and ̂Bsome Dirac-Flavour-Colour-matrices, ]
we have [A(x,t),B(y,t)] =
δ (3) (x − y)q † (x)[Â, ̂B q(x). With help of the relation f abc f abd = 3δ cd we immediately
obtain
s a (x) = − i 3 f [
abc Q
b
V ,s c (x) ]
For an SU(3) V invariant vacuum we have therefore
〈0| s a (x) |0〉 = 〈0| − i 3 f [
abc Q
b
V ,s c (x) ] |0〉 = 0 a = 1.....8
For 〈0|s 0 (x) |0〉 there does not exist an equation like this and hence in general
we have 〈0|s 0 (x) |0〉 ≠ 0. Because of〈0|s 3 (x) |0〉 = 〈0|s 8 (x) |0〉 = 0 follows then
a finite condensate.
0 ≠ 〈0| ¯q(x)q(x) |0〉 = 3 〈0|ū(x)u(x) |0〉 = 3 〈0| ¯d(x)d(x)s |0〉 = 3 〈0| ¯s(x)s(x) |0〉
This is the direct connection between spontaneous symmetry breaking and the
existence of a chiral condensate.
12.2 Representations of the chiral boson fields
12.2.1 Linear Sigma Model
We know the linear Sigma-Model without explicit fermion mass term:
L = ψ [iγ µ ∂ µ ]ψ + 1 2
(
∂λ π a ∂ λ π a + ∂ λ σ∂ λ σ ) ::::::::::::::::::::
:::::::::::::::::::: −gψ(σ + iτ a π a γ 5 )ψ − µ2
2 (σ2 + π a π a ) − λ 4 (σ2 + π a π a ) 2
We know that this model exhibits the spontaneous symmetry breaking. In fact
it has been constructed to do so In treating this Lagrangean it is useful to
rewrite the mesons in terms of a matrix field
Σ = σ + iτ a π a
157

such that
σ 2 + π a π a = 1 2 Tr(Σ† Σ)
Then we obtain the well known form
L = ψ R [iγ µ ∂ µ ]ψ R + ψ L [iγ µ ∂ µ ]ψ L + 1 4 Tr(∂ µΣ∂ µ Σ † )
+ 1 4 µ2 Tr(ΣΣ † ) − λ [
Tr(ΣΣ † ) ] 2 (
− g ψL Σψ R + ψ R Σ † )
ψ L
16
The above lagrangean
⊗
has separate Left- and right-invariances, i.e. it is invariant
under SU(2) L SU(2)R :
ψ R (x) → ψ ′ R(x) = Rψ R (x)
ψ L (x) → ψ ′ L(x) = Lψ L (x)
Σ ′ = LΣR †
with R and L being two arbitrary SU(2) matrices.
R(α R ) = exp(−i τa αR
a )
2
L(α L ) = exp(−i τa αL
a )
2
In the case of spontaneous symmetry breaking we separate the vacuum value v
of the sigma field and write
L = ψ [iγ µ ∂ µ − gv]ψ + 1 2
(
∂λ˜σ∂ λ˜σ − 4C 2 v˜σ 2) + 1 2 ∂ λπ a ∂ λ π a + :::::::::
::::::::::::::::::::::: −gψ(˜σ + iτ a π a γ 5 )ψ − vC 2˜σ(˜σ 2 + π a π a ) + C 2 (˜σ 2 + π a π a ) 2 . The
interactions in the Lagrangean are polynomic couplings without derivatives.
We will se immediately that this structure of the Lagrangean corresponds to a
particular choice of parametrizing the fields. One can write down a Lagrangean
with a completely different form with completely different interactions, still
describing the same physics.
12.2.2 Non-linear Sigma Model
We can rewrite the linear Sigma model to the non-linear one by using an exponential
representation, which in fact corresponds to a polar representation:
Σ = σ + iτ a π a = v + ˜σ + iτ a π a = (v + S)U
with
U = exp( i v ⃗τ⃗π′ )
158

If you do a Taylor expansion of the U-field then one obtains an expansion in
povers of ( 1 v )n . In detail:
Σ = v + ˜σ + iτ a π a = (v + S)(1 + i v ⃗τ⃗π′ + 1 2 ( i v )2 (⃗τ⃗π ′ ) 2 + ....
= v + S + i −→ τ −→ π + higher − Terms − in − 1 v
Or
π a = (π ′ ) a + ∑ ( 1 v )n F n ((π ′ ) a ,S)
with F n ((π ′ ) a ,S = 0) = 0. Hence, asymptotically in a large distance from the
interaction region, where S → 0, we have π a = (π ′ ) a . This means that the free
particles in their asysmptotic scattering states are identical. The lagrangean
of the linear and non-linear Sigma model are only rewritten in a way, that the
free particles are identical. A particular S-matrix element must not depend on
the way we define the fields, as long as the physical particles are not redefined.
Thus both representations describe the identical physics.
Actually the non-linear representation will play in the following a dominant
role, since it simplifies certain parts of the formalism tremendously. In fact, it
will turn out to be the corner stone of the effective low energy theory of the
QCD. It is interesting to note the following point. If we perform a left- and
right-handed transformation the U-field changes according to
Is this correct
U → U ′ = RUL †
and we obtain explicitely
L = 1 2 (∂ µS∂ µ S − 2µ 2 S 2 ) +
(v + S)2
Tr(∂ µ U∂ µ U † ) − λvS 3 − λ 4
4 S4 +
¯ψiγ µ ∂ µ ψ − g(v + S)( ¯ ψ L U † ψ R + ¯ ψ R Uψ L )
Apparently the bosonic part of the Lagrangean is written down in a rather
compact form, i.e. Tr(∂ µ U∂ µ U † ) and that is in the end one of the reasons
why one prefers this representation. On the other hand, if one expands the
exponentials, one obtains immediately derivatives acting on the physical pion
fields. Nevertheless, as mentioned above already, both Lagrangeans describe the
same physics, S-matrixelements for a certain process are bound to be identical.
This has been explicitely shown in the book by Donoghue et al p.100.
12.2.3 Generalization to finite Goldstone Boson mass
We will later spply the chiral perturbation theory in a realistic way and hence
we have to consider finite pion masses. We want to study this in the nonlinearrepresentation,
which we write down without the scalar field S:
L = 1 4 Tr(∂ µΣ∂ µ Σ † )
159

+ 1 4 µ2 Tr(ΣΣ † ) − λ [
Tr(ΣΣ † ) ] 2
16
This Lagrngean is invariant under the transformation
Σ → LΣR †
We add now a symmetry breaking term, wich will do what we want, i.e.
L sb = ɛTr(Σ + Σ † )
Apparently the term is invariant only under a vector transformation, i.e. if
we have then e.g.
L = R = V
Tr(Σ + Σ † ) → Tr(V ΣV † ) = Tr(ΣV † V ) = Tr(Σ)
However the term is variant under the axial transformation, where L † = R = A:
Tr(Σ) → Tr(AΣA) = Tr(ΣA 2 ) ≠ Tr(Σ)
We again change to the non-linear or polar representation)
Σ = σ + iτ a π a = v + ˜σ + iτ a π a = (v + S)U
for which we can write down the mass term as
L sb = ɛ 4 (v + S)Tr(U + U † )
If one expands now the symmetry breaking Lagrangean in powers of π a one
obtains
L sb = ɛ 4 (v + S)Tr(2 − τa π a
) 2 + ....
F 0
or
L sb = ɛ (v + S) −
ɛ
4
which correspondes to a pion mass of
m 2 π = ɛ
F 0
2F 0
π a π a ....
Hence the total Lagrangean for the Goldstone fields (non-linear Sigma model)
can be written in the simple form
L = f2 π
4 Tr(∂ µU∂ µ U † ) + m2 π
4 f2 πTr(U + U † )
We will see later that this is indeed the lowest order term of the effective Lagrangean
160

12.3 QCD and chiral fields
Let us consider QCD and let us perfom the usual transformations
R(α R ) = exp(−i τa αR
a )
2
L(α L ) = exp(−i τa αL
a )
2
acting on the left- and right-handed quark fields separately. In doing the transformations
the Goldstone boson fields transform themeselves like an octett, as
we know. One can now easily define the QCD-analogue of the matrix U of the
non-linear sigma model:
U(x) = exp i φ(x)
F 0
with
φ(x) = λ a π a (x)
and a = 1,2,3 corresonding to pion, a = 4,5 corresonding to K + ,K 0 , and
a = 6,7 corresponding to K − , ¯K 0 , and a = 8 corresponding to η. Then we have
⎛
√ √
π 0 + 1
⎞
√
3
η 8 2π
+ 2K
+
⎜ √ √
φ(x) = ⎝ 2π
−
−π 0 + √ 1 3
η 8 2K
0 ⎟
√ √
⎠
2K
− 2K
0
− √ 2
3
η 8
Actually: If the quarks are transformed as described above, the transformation
law of U(x) turns out to be extremely simple. In fact we have
U(x) → U ′ (x) = RU(x)L †
The proof goes as follows: We will first show, how the bosonic fields π a in
the above expression for U transform under the vector transformation if this is
applied to U. Then we compare this with the old expressions for the transformation
of boson field (we know: like an octet) and show that they are identical.
Thus:
For a simple vector transformation we have R = L. Put for simplicity
R = L = V and then we have
U → U ′ (x) = V U(x)V †
The expansion of the exponential function yields
U(x) = 1 + i φ(x) − 1 F 0 2
→ U ′ = V
(
1 + i φ(x) − 1 F 0 2
φ 2
F0
2
φ 2
F0
2
+ ...
)
+ ... V †
161

→ U ′ = V V † + iV φ F 0
V † − 1 2 V φ F 0
V † V φ F 0
V † + ...
→ U ′ = 1 + iV φ V † − 1 F 0 2 V φ2
F0
2 V † + ...
>From the lase expression we can read off that the φ(x) transformes as
φ(x) → V φ(x)V †
>From this one can show easily, that the π a -fields transform like an octet. Take
for this the usual parametrization
V (α V ) = exp(−i λa α a V
2
Then we obtain
[ ] λ
φ(x) = λ b π b (x) → V φ(x)V † = φ(x) − iαV
a a
2 ,λb π b (x) + ...
= φ(x) + F abc α a V π b (x)λ c + ....
This expression can be compared with the one, we had already: Because we
know
[
Q
a
V ,π b (x) ] = if abc π c (x)
That actually ment, that the Goldstone fields transform like members of the
octett. This simple and dimportant expression can be immediately generalized
to
]
exp(i αa V Qa V
2
)(λ b π b (x))exp(−i αa V Qa V
2
)
) = λ b π b (x) + iα a V
= φ(x) + f abc α a V π b (x)λ c + ....
[ λ
a
2 ,λb π b (x)
+ ...
Apparently both expressions , one derived from U and the other one derived
from the known octett transformation, are identical henc the transformation
law of U is proven by this.
We can also consider an axial transformation. There we have R = L † and
we set R = L † = A Now we can repeat the above steps and transform
U(x) → Aφ(x)A = AA + iA φ(x) A − 1 F 0 2 Aφ(x) φ(x)
A + ...
F 0 F 0
Here we have AA ≠ 1 in contrast to above V V † = 1 and hence the transfomation
properties of φ(x) and of the pionic filds π a (x) are not that simple. This
corresponds to the fact that the vacuum is not invariant under the axial transformation.
We also cannot insert a term AA between the two φ-fields, because
AA ≠ 1.
162

Altogether we can summarize this subsection: We have an appropriate field
U(x) whose transformation properties are simple and known. The U(x) is
parametrized in the physical fields (Goldstone fields). The ground state of
the system is described by U(x) = 1. This state is invariant under vector
transformation, since V V † = 1, however it is not invariant under the axial
transformation since AA ≠ 1.
12.4 The minimal effective bosonic Lagrange-density
12.4.1 Massless case
We consider a system of Goldstone bosons and their interactions. The simplest
Lagrange density describing this is given by
L eff = F 2 0
4 Tr(∂ µU∂ µ U † )
with
U = exp( i τ a π a )
F 0
In fact it is the simplest Lagrangean, if we restrict ourselves to 2 derivatives.
For more derivatives see later. The F 0 = f π is the pion decay constant in
the massles limit. It is taken equal for pions and kaons, which correspondes
approximately to the empirical values of f π = 93MeV adn f K = 113MeV . The
term F0 2 takes care that the kinetic energy of the Goldstone bosons have the
standardform (remember: Tr(λ a λ b ) = 2δ ab )
T = 1 2 ∂ µπ a ∂ µ π a
One sees this explicitely if one expands the exponentials in U and writes down
the various terms in the Goldstone fields π a . One obtains in SU(2):
L eff = 1 2 ∂ µπ a ∂ µ π a 1
1 + 1 π a π a
2 F0
2
= 1 2 ∂ µπ a ∂ µ π a (1 − 1 π a π a
2 F0
2 + −....)
Apparently there are many more terms behind the kinetic energy. They will be
used indeed in order to dewscribe properly e.g. pion-pion-scattering or pionkaon-scattering
etc.
We will prove all these statements, but before we do so we will generalize
the above Lagrangean by including a term, which gives the Goldstone bosons a
finite mass. We know from the non-linear Sigma model, how to do this. Thus
the final simplest effective Lagrangean with finite Goldstone boson mass is (this
will be proven immediately):
L eff = F 2 0
4 Tr(∂ µU∂ µ U † ) + F 2 0
2 B 0Tr(MU † + UM † )
Actually the B 0 has to do with the chiral vacuum condensates and the current
quark masses. If we consider SU(2) and the case m u = m d because then the
163

mass term can be written interms of the pion mass and is then identical to the
one of the non-linear Sigma model
F 2 0
2 B 0Tr(MU † + UM † ) → m2 π
Tr(U + U † )
We will proceed with the prof by considering first the massless limit.
Proof: There are other expressions of second order in the derivatives, they
are however equivalent to the above one and can berewritten as e.g:
Tr((∂ µ ∂ µ U)U † ) = ∂ µ
[
Tr(∂ µ UU † ) ] − Tr(∂ µ U∂ µ U † )
The first term is a total derivative and hence irrelevant for the Lagangean.
A term of the sort Tr(U + U †) contributes to the masses of the Goldstone
bosons and will be considered separately and later. A term of the sort Tr(U−U †)
does not contribute to the mass, however it is not allowed since it has the wrong
behaviour under parity transformation. Thus:
We have to show finally, that terms with ONE derivative do not contribute.
The reason is simple because
Tr ( ∂ µ UU † = 0 )
However we will show this explicitely. Consider for this the case F 0 = 1. Then
we have
U = exp(iφ) = 1 + iφ + 1 2 (iφ)2 + ...
f 2 π
Tr [∂ µ U] = exp(iφ)i∂ µ φ
Tr [ ∂ µ UU †] = Tr [ i∂ µ φUU †] = tr [i∂ µ φ]
= Tr [i∂ µ π a (x)λ a ] = 0
since we have Tr(λ a ) = 0.
Currents:
We construct explicitely the Noether-currents for the above effective Lagrangean
We will show that
L eff = F 2 0
4 Tr(∂ µU∂ µ U † )
J µ,a
L
(x) = if2 π
4 Tr [ λ a ∂ µ U † U ]
and hence
J µ,a
V
J µ,a
R
(x) = if2 π
4 Tr [ λ a U∂ µ U †]
= J µ,a
R
+ Jµ,a L
= −if2 π
4 Tr ( λ a [ U,∂ µ U †])
164

J µ,a
A
= Jµ,a R
− Jµ,a L
= −if2 π
4 Tr ( λ a { U,∂ µ U †})
The proof for these expressions for the currents goes as follows: We parametrize
in the konwn way
L = exp(−i θa L (x) λ a )
2
and we take θR a = 0 and we take these parameters x-dependent in order to derive
currents. For the infinitesimal transformation we have
)
U → U ′ = RUL † = U
(1 + i λa
2 θa L
and therefrom
U † → (U ′ ) † = LU † R † =
) (1 − i λa
2 θa L U †
∂ µ U → ∂ µ U ′ = (∂ µ U)(1 + i λa
2 θa L) + 1 2 U(i∂ µθ L )λ a
∂ µ U † → ∂ µ (U ′ ) † = (1 − i λa
2 θa L)(∂ µ U † ) + 1 2 (−i∂ µθ L )λ a U †
Herewith we obtain for δL (only terms linear in θL a are kept):
(
δL = f2 π
4 Tr Ui∂ µ θL
a λ a
)
2 ∂µ U † + ∂ µ U(−i∂ µ θL) a λa
2 U †
If one uses trace-properties one obtains
(
δL = f2 π λ
4 i(∂ µθL)Tr
a a [
∂ µ UU † − U † ∂ µ U ])
2
= f2 π
4 i(∂ µθ a L)Tr ( λ a ∂ µ U † U )
For an internal symmetry the Noether current is then obtained by
J µ,a
L (x) = ∂
∂(∂ µ θ a L (x))L( ̂φ i ,∂ µ ̂φi )
which yields immediately the above formula.
165

12.4.2 Massive case
We will now consider the case with massive Goldstone bosons due to some
explicit symmetry breaking by quark masses. In the reality we have to add
terms to the above effective Lagrangean, which correspond to the mass terms
of the QCD. We know those;
L QCD
sb
= −q¯
R Mq L − q¯
L M † q R
where the mass matrix is given by M = diag(m u ,m d ,m s ). Apparently in QCD
this term is invariant under SU(2)xSU(2). We know from the non-linear Sigma
model which term to add to the effective Lagrangean in order to obtain a proper
mass term:
L sb
eff = f2 π
4 B 0Tr ( MU † + UM †)
Actually the constant B 0 is related to the chiral quark condensate (without
proof):
3f 2 πB 0 = − 〈0 |¯qq|0〉 = − 〈 0 ∣ ∣ūu − ¯dd − ¯ss
∣ ∣ 0
〉
One can easily see that this mass term is correct and does what it should. We
expand the exponential function in U to second order in φ with φ = λ a π a and
obtain:
L sb
eff = fπB 2 0 (m u + m d ) − B 0
2 Tr ( φ 2 M )
We use the fact that M is diagonal and we take the explicit form of φ:
⎛
√
2K
+
φ(x) =
⎜
⎝
√
π 0 + √ 1
⎞
3
η 8 2π
+
√ √
2π
−
−π 0 + √ 1 3
η 8 2K
0 ⎟
√ √
⎠
2K
− 2K
0
− √ 2
3
η 8
giving
Tr(φ 2 M) = m u
[
(π 0 + 1 3 η)2 + 2π + π − + 2K + K − ]
[
+m d 2π − π + + (−π 0 + √ 1 ]
η) 2 + 2K 0 ¯K0 3
+m s
[
2K − K + + 2 ¯K 0 K 0 + 4 3 η2 ]
= 2(m u + m d )π + π − + 2(m u + m s )K + K − + 2(m d + m s )K 0 ¯K0
+2(m u + m d )π 0 π 0 + 2 √
3
(m u − m d )π 0 η + 1 3 (m u + m d + m s )η 2
166

Apparently we can identify the mass terms and compare them with the experimental
data.
Assume first isospin symmetry, i.e. m u = m d . In the mesonic language
wesay “We ignore the pion-eta mixing”. In this approximation we note down
the quadratic terms of the above expression and we obtain up to corrections
quadratic in ¯m.:
m 2 π = 2B 0 ¯m
m 2 K = B 0 ( ¯m + m s )
m 2 η = 2 3 B 0 ( ¯m + 2m s )
Apparently these are exactly the Gell-Mann-Oakes-Renner-Relations. The masses
also fulfill the Gell-Mann-Okubo-Relation
4m 2 K == 3m 2 η + m 2 π
Apparently the above expression for the symmetry breaking term in the bosonic
language works.
Without further information on B 0 we cannot extract absolute values for
the quark masses. The reason is simple, because we meet always the product
B 0 ¯m and B 0 m s and never the B 0 or the mass as such. However we know the
empirical values for the ratios
m 2 K
m 2 π
= ¯m + m s
2 ¯m → m s
m π
≃ 25.9
m 2 η
m 2 π
= ¯m + 2m s
3 ¯m → m s
m π
≃ 24.3
Several comments are in order: 1) In the above expansion
L sb
eff = f 2 πB 0 (m u + m d ) − B 0
2 Tr ( φ 2 M )
we have the first term without boson field φ It describes the vacuum energy
which is caused by the spontaneous symmetry breaking
E vac = f 2 πB 0 (m u + m d )
2) In the expansion of the effective symmetry breaking Lagrangean
L sb
eff = f2 π
4 B 0Tr ( MU † + UM †)
in terms of the goldstone boson fields there appear terms of higher than second
order in the boson field. They are of the sort (⃗π⃗π) 2 and hence contribute to the
boson-boson interaction. Thus the mass terms change the interaction of the
bosons amongst each other.
167

12.4.3 Higher order terms
We have considered by now the most general effective Lagrangean exhibiting
spontaneous broken chiral symmetry but restricted to the lowest number of
derivatives. For a complete theory of the interaction between goldstone bosons
one should envisage the most general effective Lagrangean without such a restriction.
So one should consider higher numbers of derivatives and hence one
should add to the simple 2.derivative effective Lagrangean terms of the form
Tr ( ∂ µ U∂ ν U †) Tr ( ∂ µ U∂ ν U †)
Tr ( ∂ µ U∂ µ U †) Tr ( ∂ ν U∂ κ ∂ κ ∂ ν U †)
and other terms. There are several similar terms with 4 derivatives and several
with 6 derivatives etc etc. In the end there is a growing number of unknnown
coefficients, which must be determined by experiment. For a theoretical consideration,
however, it is more appropriate and even sufficient to write down the
Lagrangean purely organized by the dimensionality of the operators:
There appear unknown coefficients:
L eff = L 2 + L 4 + L 6 + ....
L 2 → g 2 = f π
L 4 → g (1)
4 ,g(2) 4 ,g(3) 4 ,.....
L 6 → g (1)
6 ,g(2) 6 ,g(3) 6 ,.....
which have, like e.g. the masses of the mesons, to be determined by experiment.
It would be useless, if the above series would never end, because then one would
need innumerable many coefficients g (k)
i which must result from innumerable
numbers of experiments. That means the above formalism makes only applicable
if the above series concverges more or less rapidly such that we have only few
coefficients to determine and few terms to consider in our calculation (which
we still do not yet know how to do!). Tlhis is in fact true as the following
consideration will lshow us:
Consider e.g. pion-pion scattering and consider the mandelstam variables:.
168

u
B
s
a
t
A
Mandelstam−Variables
In the follwing all Mandelstam variables of this process are called q and
are all assumed to be small compared to a typical hadronic scale Λ hadr , i.e.
for reactions involving mesons the mass or the Rho-Meson (700 MeV) or for
reactions involving baryons the mass of the proton (900 MeV). The mass of the
pion or kaon is NOT taken, because these bosons are Goldstone bosons and have
mass zero. Since U is dimensionless and f π has the dimension Λ 2 the effective
Lagrangean L 2 has the dimension Λ 4 :
[L 2 ] = Λ 4
This is consistent with
∫
S =
d 4 xL 2 = 1
If we consider the dimension of the higher terms then each term of the Lagrangean
has obviously the dimension Λ 4 but if we write
L 2n = C 2n Tr ( ∂ µ U......∂ ν U †)
with 2n derivative terms, then we can write
[C 2n ] =∼ Λ 4−2n
Suppose we consider now an arbitrary vertex with 2n derivatives, then the
contribution of this vertex to an S-matrix element is proportional to
q 2n
Λ 2n−4 ∼ ( q Λ )2n 1 Λ 4
Thus, if the reaction is of low energy, i.e. if each Mandelstam variable is smaller
than Λ then the vertex contributes less, if it has more derivatives. This is
the basic feature: We can chose the momentum of the reaction partners in
the experiment small enough such that higher order vertices do not contribute
much. In this way we can force the system to converge.
169

The argument above is not quite correct: If the Mandelstam variable of the
reaction are small, it does not mean that the momenta of loops are small too
and hence it can be that we have few external lines with small q and several
internal lines with large q and hence our argument is no longer true.
Fortunately we need not worry, because there is the famous power counting
theorem by Weinberg:
Consider an arbitrary diagram based on the effective Lagrangean. if one
rescales the momenta of the externl mesons, i.e. the mesons, which are physical
and take part in the reaction, then we have for the invariant Feynman amplitude
the following feature:
M(tp,t 2 M 2 ) = t D M(p,M 2 )
D = 2 + ∑ 2(n − 1)N 2n + 2N L
Here M is the mass of the Goldstone boson, the N 2n is the number of vertices
in the diagram, which come from L 2n herrühren. and N L is the number of loops
in the diagram. Hence, if in a certain reaction we reduce the momenta, then the
diagrams with vertices of a higher number of derivatives contribute less and less
to the Feynman amplitude and the diagrams with more loops contribute also
less and less to the amplitude. In the ende the lower the incoming momenta
the more dominating are the TREE DIAGRAMS coming from L 2 ! That is the
secret and that is why perturbation theory works even in the non-perturbative
region. One does perturbation with effective and physical fields, pions, and not
with quarks and gluons. uvertices withof higher of the effeem, becausere is only
one pointcto be able to apply the morezs nzHence, If we have hence a higher
ordnn, whiched. The incoming pions have the eWe are saved
To illustrate the above arguments we consider a simple exemple: Take
L 2 = gφ 2 ∂ µ φ∂ µ φ
Taking the usual rules for Feynman diagrams and considering a tree- and a loopdiagram
we can write down the following formulae for the Feynman amplitude:
The corresponding Feynman amplitude is:
M(p 1 ,p 2 ,p 3 ,p 4 ) = 4ig [(p 1 + p 2 )(p 3 + p 4 ) − p 1 p 2 − p 3 p 4 ]
170

which obviously scales as
M(tp 1 ,tp 2 ,tp 3 ,tp 4 ) = t 2 M(p 1 ,p 2 ,p 3 ,p 4 )
Comparison with Weinbergs formula yields:
M(tp,t 2 M 2 ) = t D M(p,M 2 )
D = 2 + ∑ 2(n − 1)N 2n + 2N L
Apparently we have here: D = 2 corresponding to N L = 0 and N 2 = 1 since we
have considered ONE vertex, which comes from L 2 . The scaling feature can be
contrasted with the one of a digaram, which contains one loop:
L
2
D=4
The Feynman rules give here:
M(p 1 ,p 2 ,p 3 ,p 4 ) =
= 16g 2 ∫
d 4 k
(2π) 4 [(p 1
1 + p 2 )(p 3 + p 4 ) − (p + p 2 − k)k − p 3 p 4 ]
k 2 − M 2 + iɛ
1 [
(p1
(p 1 + p 2 − k) 2 − M 2 + p 2 ) 2 − p 1 p 2 − (p 1 + p 2 − k)k ]
+ iɛ
If we do now the scaling like p i → tp i and M 2 → t 2 M 2 and the variablesubstitution
k = tl we can write down
M(tp 1 ,tp 2 ,tp 3 ,tp 4 ) =
= 16g 2 ∫ t 4 d 4 l
(2π) 4 [
t 2 (p 1 + p 2 )(p 3 + p 4 ) − t 2 (p + p 2 − l)l − t 2 p 3 p 4
] 1
t 2 (l 2 − M 2 + iɛ)
1 [
t 2
t 2 ((p 1 + p 2 − l) 2 − M 2 (p 1 + p 2 ) 2 − t 2 p 1 p 2 − t 2 (p 1 + p 2 − l)l ]
+ iɛ)
We obtain a scaling with D = 4 which corresponds to Weinbergs formula with
N L = 1 and N 2 = 2. Weinbergs formula yields the same value:
D = 2 + ∑ 2(n − 1)N 2n + 2N L
171

If we replace in this diagram in the right vertex the L 2 by and L 4 then we
have the following picture:
L
2
L
4
D=6
And if both vertices haven an L 4 we get
L
4
D=8
Thus with Weinbergs rule we can immediately estimate the relevance of the
graph for the calculation of the process.
Here we have done tacitley an important step. We have considered the mass
of the Goldstone Boson in the scaling procedure lieke an external momentum.
This makes indeed sense, because the momenta of the external particles obbey
p 2 = M 2 . This step is essenttial, because without this we would not have a
clear scaling and hence no clear decision, which diagrams to consider and which
to ignore. In this step it is also clear, that we cannot have mesons in the game,
which are not goldstone bosons, like e.g. σ-mesons or ρ-mesons. They would
occur only in inner loops and would destroy the power counting.
172

Instead of saying “the diagram scales with D = 4 or t 4 one say also “the
diagram is of order E 4 . And one sees immediately: The diagrams can be ordered
according to energy, we have, in fact an energy expansion. And vertices, which
originat from L 2 , which contribute e.g. as a tree diagram to O(E 2 ) contribute
to O(E 2 ) via a loop and to O(E 6 ) via two loops etc.
Hence: The lower the energy of the process considered, the more dominate
tree diagrams with low dimensional L 2n i.e. low n.
We can summarize the points:
1) L-Loop diagrams are suppressed by powers E 2L .
2) From the series
L eff = L 2 + L 4 + L 6 + ....
we have the following contributions
O(E 2 ): tree diagrams with L 2 -insertions
O(E 4 ): tree diagrams with one L 4 -insertion plus 1-loop graphs with only L 2
insertions.
O(E 6 ): tree diagrams one with L 6 -insertion plus 1-loop graphs with one
L 4 -insertion plus 2-loop graphs with only L 2 -insertions.
Altogether we have: ENERGY EXPANSION = EXPANSION IN LOOPS
Take an example: ππ-scattering: Again we verify Weinbergs counting rule.
For exampe the lower left diagram has 3L 2 -vertices (none of them contributes
to the D of the Weinberg rule) and 2 Loops, such that D = 2 + 2 ∗ 2:
L
2
L
4
L
6
D=2
D=4
D=6
The more accurate one wants to calculate the higher one goes in the energy
expansion, the more graphs one has to consider. However, one is clear: it is a
systematic expansion which can be improved in a systematic way. The higher
one goes in energy expansion the more unknown coefficients are there, see above:
:
L 2 → g 2 = f π
L 4 → g (1)
4 ,g(2) 4 ,g(3) 4 ,.....
L 6 → g (1)
6 ,g(2) 6 ,g(3) 6 ,.....
173

which have to be determined by experiment. I.e. one calculates the cross section
of a certin experiment in a certain degree of energy expansion and adiusts the
coefficients g (i)
k
to the experimental data. One should obtain the result, that this
expansion is well converging. However, it is not always the case, for example in
the vicinity of resonances.
12.5 Application: pion-pion scattering
12.5.1 Calculation:
The scattering process
π + π → π + π
is the purest process to test the ideas of ppure chiral dynamics. It is also the
easiest process on the theoretial side, however, it is not easy to measure. We
consider L 2 with mass term
L 2 = F 2 0
4 Tr(∂ µU∂ µ U † ) + F 2 0
2 B 0Tr(MU † + UM † )
For the calculation we have to select that part of L 2 , which will appear in the
graphs for pion -pion scattering. These are the ones which allow 2 incoming and
2 outgoing pions. Apparently L 2 has only even powers of φ so we can write
L 2 = L 2φ
2 + L4φ 2 + L6φ 2 + ...
Vertices with 3 bosons do not exist so for the pi-pi-scattering with D = 2 we
must only consider a contact interaction with 4 goldstone bosons: To indentify
this we expand
U = 1 + i φ F 0
− 1 2
φ 2
F0
2
− i 6
φ 3
F0
3
+ 1
24
φ 4
F0
4
+ ....
U † = 1 − i φ F 0
− 1 2
φ 2
F0
2
+ i 6
φ 3
F0
3
∂ µ U = i
F 0
∂ µ φ + ....
+ 1 24
∂ µ U † = − i
F 0
∂ µ φ + .....
This yields in the expansion to fourth order
φ 4
F0
4
+ ....
L 2φ
2 = 1 4 Tr(∂ µφ∂ µ φ − 2BMφ 2 )
and then after some calculation we can identify
L 4φ
2 = 1
24F0
2 Tr ([φ,∂ µ φ]φ∂ µ φ) + 1
24F0
2 B 0 Tr(Mφ 4 )
174

For the pi-pi-scattering we have to reduce this on expressions involving solely
pion fields π a and for this we have to use φ(x) = τ a π a (x). To treat this we need
some simple formulae like
φ = ⃗τ⃗π
τ a τ b = δ ab + iɛ abc τ c
This yields then
φ 2 = π a π b τ a π b = ⃗π 2 + iɛ abc τ c π a π b
∂ µ φ 2 = 2π a ∂ µ π a + iɛ abc τ c (∂ µ π a π b + π a ∂ µ π b )
φ 3 = τ a π a π b π b + iɛ abc π a π b π c
∂ µ φ 3 = iɛ abc (∂ µ π a π b π c + π a ∂ µ π b π c + π a π b ∂ µ π c ) + τ a (∂ µ π a π b π b + 2π a π b ∂ µ π b )
We insert these expressions in the above formula and use the identities
Tr ([(⃗τ⃗π),(⃗τ∂ µ ⃗π)] (⃗τ⃗π)(⃗τ∂ µ ⃗π)) = 4(⃗π∂ µ ⃗π)(⃗π∂ µ ⃗π) − (⃗π⃗π)(∂ µ ⃗π∂ µ ⃗π)
Tr ( Mφ 4) = mTr ( (⃗τ⃗π) 4) = 2m(⃗π⃗π) 2
Hence with m 2 π = 2B 0 m with m u = m d = m we get that part of L 2 which
enters the pion-pion scattering:
L 2π
2 = 1 [
∂µ π a ∂ µ π a − m 2
2
ππ a π a]
L 4π
2 = 1
6F 2 0
((⃗π∂ µ ⃗π)(⃗π∂ µ ⃗π) − (⃗π⃗π)(∂ µ ⃗π∂ µ ⃗π)) + m2 π
24F0
2 (⃗π⃗π) 2
One can rewrite this to cartesian isospin-indices and then we obtain for the
scattering
π a (p a ) + π b (p b ) → π c (p c ) + π d (p d )
the Feynman-amplitude M
with
−i6F 2 0 M(p a ,p b ,p c ,p d ) = 2A + B
A = δ ab δ cd (−ip a − ip b ) (ip c + ip d ) + δ ac δ bd (−ip a + ip c )(−ip b + ip d )
+δ ad δ bc (−ip a + ip d ) (−ip b + ip c ) − 4(δ ab δ cd ((−ip a )(−ip b ) + (ip c )(ip d ))
175

+δ ac δ bd ((−ip a )(ip c ) + (−ip b )(ip d )) + δ ad δ bc ((−ip a )(ip d ) + (ip b )(ip c )))
B = m2 π
4 8( δ ab δ cd + δ ac δ bd + δ ad δ bc)
If one takes into account that the incoming and outgoing particles are all on the
mass shell one can change to
−i3F 2 0 M(p a ,p b ,p c ,p d ) =
δ ab δ cd ((p a + p b ) 2 + 2p a p c − 2p c p d + m 2 π)
+δ ac δ bd ((p a − p c ) 2 − 2p a p c − 2p b p d + m 2 π)
+δ ad δ bc ((p a − p d ) 2 − 2p a p d − 2p b p c + m 2 π)
and then one obtains finally after some direct and explicit calculation:
M(p a ,p b ,p c ,p d ) = i
F 2 0
[
δ ab δ cd (s − m 2 π) + δ ac δ bd (t − m 2 π) + δ ad δ bc (u − m 2 π) ]
where we have the Mandelstam variables (where the arrow indicates, how they
look in the center of mass system:
s = (p a + p b ) 2 = (p c + p d ) 2 → 4(q 2 + m 2 π) = W
t = (p a − p c ) 2 = (p d − p b ) 2 → −2q 2 (1 − cos(Θ scatt ))
u = (p a − p d ) 2 → −2q 2 (1 + cos(Θ scatt ))
s + t + u = ∑ p 2 i = 4m 2 π
Here Θ scatt is the scattering angle in the cm-system and q is the momentum of
the incoing pion in the cm-system. We have used relations like 2p a p b = 2p c p d =
s − 2m 2 π. We have of course assumed that the incoming and outgoing pions are
on the mass shell. This result has also been obtained purely by means of current
algebra by Weinberg (1966), where the formalism was much more complicated.
176

12.5.2 Comparison with experiment
One writes the scattering amplitude (or Feynman amplitude) in the following
way:
T abcd = δ ab δ cd A(s,t,u) + δ ac δ bd A(t,s,u) + δ ad δ bc A(u,t,s)
It is remarkable that the pion-pion scattering amplitude depends on one
function A(s,t,u) only:
A(s,t,u) = s2 − m 2 π
f 2 π
Actually pions have isospin 1 and hence it is customary to characterize the
input- and output channels by the total isospin. Since the isospin is a conserved
quantum number (assuming equal masses of up and down quarks) the isosopin
of the entrance channel and the one of the exit echannel are identical. We can
have isospins of T = 0,1,2. Starting from the cartesian representation we can
change it into a representation with good isospin yielding then for the scattering
amplitudes
T (0) = 3A(s,t,u) + A(t,s,u) + A(u,t,s)
T (1) = A(t,s,u) − A(u,t,s)
T (2) = A(t,s,u) + A(u,t,s)
In practice one decomposes these scattering amplitudes into partial waves with
definite isospin and dfinite angular momentum l:
T (I)
l
(s) = 1
64π
∫ +1
−1
d(cosΘ)P l (cosΘ)T (I) (s,t,u)
(Attention: t,u depend on the scattering angle Θ). We have also
T (I) (s,t,u) = 32π
∞∑
P l (cosΘ)T (I) (s)
l=0
The scattering amplitudes T (I) and the S-matrix are both charakterized by the
phase shifts:
S = exp{2iδ (I)
T (I)
l
(s) =
√
s
s − 4m 2 π
√ s 1
=
s − 4m 2 π 2i
l
)
exp[iδ (I)
l
]sin δ (I)
l
[
exp(2iδ (I)
l
) − 1
For a comparison with experiment one investigates the behaviour of the scattering
amplitude T (I)
l
in the vicinity of the threshold by doing an expansion in
terms of the pion momentum q 2 . This yields
[ ]
ReT (I)
l
= q 2l a (I)
l
+ q 2 b (I)
l
+ ...
]
177

where a (I)
l
is called “scattering length” and b (I)
l
is called range parameter, and
both are called threshold parameters. The predictions, as calculated above, are
for the s- and p-waves (i.e. l = 0,1):
T (0)
0 (s) = 1
32πfπ
2 (2s − m 2 π).................s − wave
T (1)
1 (s) = 1
96πfπ
2 (s − 4m 2 π)...............p − wave
T (2)
0 (s) = 1
32πfπ
2 (2m 2 π − s)................s − wave
There are no d, f,.... waves due to the simple form of A(s,t,u) ∼ αP 0 (cosΘ) +
βP 1 (cosΘ) Since the s = 4(q 2 + m 2 π) we can evaluate the threshold parameters
yielding:
a (0)
0 = 7m2 π
32πfπ
2
b (0)
0 = 1
4πfπ
2
a (1)
1 = 1
24πfπ
2
a (2)
0 = − m2 π
16πfπ
2
b (2)
0 = − 1
8πfπ
2
Now we can perform the comparison with the data. In the table we include for
infromation the results of the calculation to lowest order and to first two orders
(book: Donoghue, Golowich, Holstein): For the scattering length and the range
parameter we get
ππ-scatt. a (0)
0 b (0)
0 a (1)
1 b (1)
1 a (2)
0 b (2)
0 a (0)
2 a (2)
2
L 2 0.16 0.18 0.030 0 −0.045 −0.089 0
L 4 0.20 0.26 0.036 0.043 −0.041 −0.070 20x10 −4 3.5x10 −4
Experiment 0.26 0.25 0.038 − −0.028 −0.082 (17 ± 3)10 −4 1.3 ± 3)10 −4
exp. error ±0.05 ±0.03 ±0.002 − ±0.012 ±0.008 ±3x10 −4 ±3x10 −4
Here the values originate e.g. from a (0)
0 = 7
32π (139 93 )2 i.e. they are given in
appropriate units of m π . The agreement between theory and experiment is
amazing! The calculation assumes just chiral symmetry and is fully analytic. It
does not include any higher order terms in the number of derivatives.
Actually the derivation of the a (0)
0 and b =(0) is easy: We have
Tl (I) = 1
64π
∫ +1
−1
dxP l (x)T (I) (s,t,u)
178

and
and we have
>From this we get
which gives
T (0)
0 = 1
64π
ReT (I)
l
= q 2l [ a (I)
l
]
+ q 2 b (I)
l
T (0) = 3A(s,t,u) + A(t,s,u) + A(u,t,s)
∫ +1
−1
dx 1
fπ
2 P 0 (x) [ 3s − 3m 2 π + t − m 2 π + u − m 2 ]
π
T (0)
0 = 1 [ ]
3s + t + u − 5m
2
64πfπ
2 π 2
,where the 2 comes from the integral over P 0 (x) = 1 . Using s + t + u = 4m 2 π
yields
T (0)
0 = 1 [ ]
2s − m
2
32πfπ
2 π
. Further we have s = 4(q 2 + m 2 π) and hence
T (0)
0 = 1 [
7m
2
32πfπ
2 π + 8q 2]
which yields a (0)
0 = 7m2 π
32πf
and b (0)
π
2 0 = 1
4πf
. One can easily show, that in lowest
π
2
order we have b (1)
1 = 0 .
Above the threshold, which is of course at 2m 2 π we only have the phase shift
of the s-wave and this starts at zero exacxtly at the threshold. If we there
approximate exp(iδ (I)
l
sin(δ (I)
l
) ≃ δ (I)
l
( √ s) then we obtain
√
δ (I)
l
≃
1 − 4m2 π
s T (I)
l
(s)
For the s-wave we see the comparison with the experimental data (Rosselet et
al.)in the following graph:
179

12.6 Chiral perturbation theory and QCD: Functional integrals
12.6.1 Internal symmetries and Ward-Identities
We are interested generally in theories described by lagrangeans L = L 0 + L 1 ,
where L 0 is invariat under global internal symmetry G and L 1 explicitely breaks
the symmetry. Furtherore we assume the bare fields φ i in L form a basis for
some definite representation R of the symmetry group of L 0 . That means, if we
perform an infinitesimal transformation of the fields we have
δφ i = −iθ a T a ijφ j = θ a δ a φ i
with δ a being an operator. The bare currents
jµ(x) a ∂L
= i
∂(∂ µ φ i ) T ijφ a j
which are conserved currents of the G symmetric classical theory described by
L 0 remain unchanged when L 0 → L if L 1 does not contain derivatives of the
fields present in the j a µ(x), and we assume this to be the case.
180

Under an infinitesimal global symmetry transformation
φ ′ i(x) = φ i (x) + θ a δ a φ i (x)
the lagrangean L transformes as follows
L ′ (x) = L(x) + θ a δ a L(x) = L(x) + θ a δ a L 1 (x)
since the lagrangean L is considered invariant. Using the classical equations of
motion
∂ µ j a µ(x) = δ a L 1 (x)
In the following we shall also need the variation of the same lagrangean L under
the local transformation
φ ′ i(x) = φ i (x) + θ a (x)δ a φ i (x)
Remembering that L is invariant under global transformations and that L 1 does
not contain derivatives of the fields we get
an hence
δL(x) = θ a (x)δ a L(x) + j a µ(x)∂ µ θ a (x)
j a µ(x) = ∂(δL)
∂(∂ µ θ a ) (x)
We want to derive Ward identities for bare regularized or for renormalized
Greens functions followint from the symmetries of the lagrangean. We assume in
this section thath the UV regfularization of the theory does preserve its classical
symmetries. Thus we ignore the problem of anomalies. They can be treated in
fact.
We can now proceed to derive Ward identities using the functional integral
formulation of quantum field theory. Consider the generating functional W [J]
with external sources J i (x) for the fields φ i (x):
∫ [ ∫
]
Z [J] = N Dφ i exp i d 4 x(L(φ i ) + J i φ i )
This integral is invariant under the change of integration variables defined above
(local transformation). If the integration measure is invariant under the change
of integration variables (no anomaly)
φ i (x) → φ ′ i(x) = φ i (x) + θ a (x)δ a φ i (x)
we get with the above expression for δL(x) = θ a (x)δ a L(x) + j a µ(x)∂ µ θ a (x)
∫
0 =
∫
d 4 x
( ∫
Dφ i exp iS + i
d 4 xJ i φ i
)(θ a (x)δ a L
+j a µ(x)∂ µ θ a (x) + J i (x)θ a (x)δ a φ i (x)
181

One defines the Greens functions by
{
〈0|T A(x) ∏ } ∫
φ i (y i ) |0〉 ∼
Dφ i A(x) ∏ φ i (y i )exp(iS)
where A(x) denotes one of the jµ(x),δL,δ a a φ i (x). This are here more general
Greens functions, namely not only of the fields but also of composite operators.
One should remember that the lhs of this equatin is just a short hand notation
of the rhs. One obtains now the general expression for the Ward identity:
∂ µ 〈0|T
−i ∑ i
{
j a µ(x) ∏ φ i (y i )
}
|0〉 = 〈0|T
{
δ a L ∏ φ i (y i )
⎧
⎫
⎨
δ(x − y) 〈0| T
⎩ δa φ i (y i ) ∏ ⎬
φ j (x j )
⎭ |0〉
j≠i
}
|0〉
One obtains the Ward identity by the following steps: by 1) integrating the
above eq. 0 = ∫ d 4 x... by parts to get rid of the ∂ µ θ a (x) and 2) differentiating
it functionally with respect to the sources J i and 3) setting the J i (x) to zero.
One sees that this Ward identity is general concept, which can be formulated
for any field theory, in particular for QCD. This ward identity is valid
irrespective of whether the symmetry is spontaneously broken or not. If the
symmetry G is an exact global symmetry of the lagrangean thei first term of
the rhs vanishes. Up tonow the Ward identity is a concept which is based on
LOCAL symmetry. They represent the symmetry properties of the theory on
the level of the Green functions.
The Ward Identities are intimately connected with the global symmetries of
the Lagrangean. One can intimately connect them also to the local symmetries
of the system. In fact one can express the existence of the Ward Identities in
this way in a very compact forem. For this we conswider even Ward identities
for Greens functions invelving several symmetry currents jµ(x) a and/or the symmetry
breaking operator L 1 . We take L 1 to have well defined transformation
properties under the symmetry group. The most compact way to derive such
Ward identities is based on the general method to introduce background fields to
get, by construction, an action which is exactly invariant under the considered
symmetry transformations. In our case the background fields are the sources
J i ,A a µ and K for the operators φ i ,jµ a and L 1 , respectively. The new step consists
now in introducing them in such a way that the action S [ φ i ,J,A a µ,K ] is
invariant under the LOCAL transformation θ a (x). This is achieved if
∫
S [φ,J,A,K] = S 0 [φ,A] + d 4 x ( J i )
φ i + KL 1
where S 0 (φ,A) = ∫ d 4 xL 0 (φ i (x),D µ φ i (x)) with the covariant derivative D µ =
∂ µ − iA a µT a . This action can be shown to be invariant under the local transformation
if simultaneous transformations on A µ (x), J(x),K(x) are defined as
δA a µ(x) = −∂ µ θ a (x) + f abc θ b (x)A c µ(x)
182

δJ i (x)φ i (x) = −J i (x)δφ i (x)
δKL 1 = −KδL 1
The generating functional is then invariant under the above transformations of
the background fields, i.e.
Z [A,J,K] = Z [A ′ ,J ′ ,K ′ ]
Actually: One can show (Leutwyler) that this equation contains all the information
which is contained in the Ward identities (which originally come from the
global symmetries). The interesting point is, that these Ward identities (which
follow from the global symmetry of the original theory) can be equivalently expressed
by demanding LOCL symmetry Z [A,J,K] = Z [A ′ ,J ′ ,K ′ ] . One can
see this by performing an infinitesimal transformation of Z [A,J,K] yielding
∫
0 = d 4 x
[ δZ
δJ i (x) δJi (x) +
δZ
δA a µ(x) δAa µ(x) +
δZ ]
δK(x) δK(x)
The proof will not be given here.
Important is: If one demands for the background fields invariance under local
transformations, then the Ward identities are automatially satisfied. Or in
other words: The invariance of the generating functional under “gauge transformations”
of the external fields expresses the symmetry properties of the theory
on the level of the Green functions, i.e. the Ward identities. This feature represents
the basic ingredient of the follwoing analysis, while the specific properties,
which the theory may otherwise have, do not play any role.
12.6.2 External fields and Greens functions
In order to translate the above ideas to our concrete problem of embedding the
effective Lagrangean into QCD we first introduce certain external hermitean
fields into the QCD-Lagrangean. This is a method which goes back to Gasser
and Leutwyler Ann.Phys. 1984.
L QCD = L 0 QCD − ¯qγ µ 1 2 (1 + γ 5)l µ q − ¯qγ µ 1 2 (1 − γ 5)r µ q
or
−[ q¯
L (s + ip)q R − q¯
R (s − ip)q L ]
L QCD = L 0 QCD − q¯
L γ µ l µ q L − q¯
R γ µ r µ q R
or
−[ q¯
L (s + ip)q R − q¯
R (s − ip)q L ]
L QCD = L 0 QCD − ¯qγ µ v µ q − ¯qγ µ γ 5 a µ q − [ q¯
L (s + ip)q R − q¯
R (s − ip)q L ]
183

with
v µ = r µ + l µ
a µ = r µ − l µ
In the present SU(3) formalism all these fields are decomposed according to
v µ (x) = λa
2 va µ(x)
a µ (x) = λa
2 aa µ(x)
s(x) = λa
2 sa (x)
p(x) = λa
2 pa (x)
where the sum goes over a = 0,1,...,8
If one considers the generating functional with the external fields one calculates
basically how the system developes in the presence of the external field.
That means one observs the response of the strong system to the external field
and how the responding system developes in time. This is the way, how one can
learn something about the system and that is why the derivatives of the generating
function with respect to the external fields contain all the information on
the system, which is then formulated in terms of Greens functions.
The embedding of the previous formalism into QCD consists now in writing
down the master integral for the generating function of QCD:
∫
∫
Z QCD [l µ ,r µ ,s,p] = DqD¯qDA a µ exp d 4 xL QCD (q, ¯q A a µ,l µ ,r µ ,s,p)
and identifying this with
∫
Z eff [l µ ,r µ ,s,p] =
∫
DU exp
d 4 xL eff (U,l µ ,r µ ,s,p)
The effective chiral action L eff can be expanded in energy terms as we know
from above:
L eff = L 2 + L 4 + L 6 + ....
One has to specify what “identifying” in the above formula. Suppose one
calculates a correlation function of axial currents A a µ(x) and vector currents
V a µ (x) like e.g:
δ δ
〈0| T {V µ (x 1 )A ν (x 2 )} |0〉 QCD
=
δv µ (x 1 ) δa ν (x 2 ) Z QCD [l µ ,r µ ,s,p]
184

δ δ
〈0| T {V µ (x 1 )A ν (x 2 )} |0〉 eff
=
δv µ (x 1 ) δa ν (x 2 ) Z eff [l µ ,r µ ,s,p]
Then we perform fourier transformations of the correlation functions:G(
∫
G QCD (p 1 ,p 2 ) = d 4 x 1 d 4 x 2 exp(−ip 1 x 1 − ip 2 x 2 ) 〈0|T {V µ (x 1 )A ν (x 2 )} |0〉 QCD
∫
G eff (p 1 ,p 2 ) = d 4 x 1 d 4 x 2 exp(−ip 1 x 1 − ip 2 x 2 ) 〈0|T {V µ (x 1 )A ν (x 2 )} |0〉 eff
Then we expand in Z eff [l µ ,...] the L eff = L 2 +L 4 +.... and denote the functional
integral by means of Feynman diagrams with L 2 -insertions and L 4 -insertions
etc. and with 1-Loops and 2-Loops etc etc. Thus altogether we have expanded
G eff (p 1 ,p 2 ) in powers of the momenta and the mass of the goldstone bosons,
all this in the way discussed above, a technically well defined expansion, which
can and is really perfored (remember the pion-pion scattering, think that the
functional integral is only a short hand writing of the Feynman perturbation series).
We then perform (only in our head!) an expansion of the G QCD (p 1 ,p 2 ) in
poweres of the momenta, which allows us to compare momentum by momenta.
All this we do only for axial currents A a µ(x) and vector currents V a µ (x), not
for any arbitrary current. Then the following is ment with “identifying”: The
smaller the external momenta, the higher the expansion in L 2n , the higher the
scaling properties t D the more the effective calculation agrees with the QCDcalculation.
And in the limit of decreasing external momenta we obtain at the
threshold really an identity. Actually this sort or argument cannot only be applied
to the above two-point green functiosn but to any Ward identity involving
axial and vector currents and symmetry breaking mass terms. This very feature
is a consequence of demanding for the external background fields LOCAL symmetries,
as we know from above. Thus the identity of Ward identities, achieved
by demanding local symmetry is that, what is called embedding of the effective
theory into QCD.
Besides embedding the effective theory fully into QCD the procedure of
coupling external fields to the QCD-Lagrangean or the effective Lagrangean has
distinct advantageous properties:
1) Electromagnetic interactions are automatically included with
and Q = diag( 2 3 , −1
3 , −1
3
r µ = l µ = eqA µ (x)
) because then we have
−eA µ ( q¯
L Qγ µ q L + q¯
R Qγ µ q R ) = −eA µ¯qQγ µ q =
−eA µ ( 2 u − 1 3ūγµ 3 ¯dγ µ q − 1 3¯sγµ s)
2) Semileptonic weak interactions are automatically included with
e
l µ = −√ (W µ + T + + h.c.)
2 sin ϑW
185

with
⎛
T + = ⎝
0 V ud V us
0 0 0
0 0 0
here the V ud ,V us are Cabibo-Kobayashi-Maskawa-Elements, the Weinberg-Angle
is related to the Fermi-weak coupling constant by
√
2e
2
G F =
8MW 2 sinϑ W 2
⎞
⎠
and we have
W + µ = 1 √
2
(W 1 µ + iW 2 µ)
If you insert this in the Lagrange density of the QCD we have the interaction
term, well known:
2 [
−√ W
+
µ (V ud ūγ µ (1 − γ 5 )d + V us ūγ µ (1 − γ 5 )s + h.c.) ]
2 sin ϑW
The numbers are known from various experiments:
V ud = 0.9744 ± 0.0010
V us = 0.220 ± 0.004
3) S-matrix elements and general Greens functions of quark currents can be
obtained directly from the partition function by functional differentiation.
Thus the well known gauge symmetries U(1) and SU(2) L of the Standard
Model are automatically transferred to the effective theory and we know there,
how to couple external fields to the effective theory. Actually we have demanded
local chiral symmetry of the full lagrangean including the external fields and this
has given us the transformation properties of the external fields. One may ask
if we have demanded too much, since altogether only the global symmetry is
known to exist. In some sense we have done more than we needed since it
would be sufficient to couple only the electromagnetic field (U(1)) and the weak
field SU(2)-left as real gauge fields. This would provide technical difficulties
and hence it is more conveninet to demand local chiral symmetry. Since in the
end we consider only physical processes which are by definition associated with
physical fields, i.e. photon or W ± , the final formulae only invovle those fields
and notheing of those, which exist only due to the larger chiral local symmetry.
Thus: One does more than needed, this is technically simpler, and one does
nothing wrong, if on calculates physical processes.
12.6.3 Local chiral symmetry
We know that the QCD-Lagrangean is invariant under global chiral transformations.
Consider now the coupling of an external field, lets say r µ (x) to the
186

QCD-Lagrangean. Since the Lagrangean has the symmetries, it can happen that
we meet the situation that two different [ external ] fields [ r µ (1) (x) and ] r µ (2) (x) have
the same generating functional Z l µ ,r µ (1) ,s,p = Z l µ ,r µ (2) ,s,p . Apparently
the generating functional Z [l µ ,r µ ,s,p] must have a very particular structure, in
order to reflect this property. The transition r µ (1) (x) → r µ (2) (x) can in this case
be compensated by a transition U (1) (x) → U (2) (x), which is a chiral transformation
such that there exists a R(x) and L(x) with U (2) (x) = R(x)U (1) (x)L † (x).
One can revert this argument: For a transformation U (1) (x) → U (2) (x) =
R(x)U (1) (x)L † (x) there exists a transformation r (1)
µ (x) → r (2)
µ (x) which compensates
the effect of the chiral transformation on U and leaves the generating
functional Z [l µ ,r µ ,s,p] invariant. In a famous paper by Gasser and Leutwyler
Ann.Phys. Vol.158 (1984)p.142 und Ann.Phys. Vol.235 (1994) p.165 it has been
shown that the compensating transformation looks formally like some sort of a
chiral gauge transformation for the external fields: Under the transformations
applied to U
R(α R (x)) = exp(−i τa αR a (x) )
2
L(α L (x)) = exp(−i τa αL a(x)
)
2
the external fields behave like
r µ → Rr µ R † + iR∂ µ R †
l µ → Ll µ l † + iL∂ µ L †
s + ip → R(s + ip)L †
s − ip → L(s − ip)R †
and then the Z [l µ ,r µ ,s,p] remains unchanged. Now: We know from the previous
section that this particular structure of Z [l µ ,r µ ,s,p] guarantees, that the
Ward-identities, which provide links between the divergences of currents and
the currents, are fulfilled, if one concentrates on vector and axial currents. In
detail:
In QCD you have Ward-Identities. These are identities, which connect the
divergences of currents with the currents either at certain momenta or at certain
coordinates. One can derive those ward identities directly from the QCD (see
above). There are certain Ward identities connected with vector currents and
with axial currents, i.e. directly with the symmetries which we consider here and
which are connected with spontaneous broken chiral symmetry. In Gasser and
Leutwylers famous paper (1984) it has been shown, that generating functionals,
where the external fields follow the above transformation laws, reproduce these
187

Ward identities. Thus in order to handle this formally we introduce like in a
gauge theory covariant derivatives:
which obbeys the law
(D µ U)(x) = ∂ µ U(x) + iU(x)l µ (x) − ir µ (x)U(x)
D µ U → RD µ UL †
We also define field strength tensors (not to be used for the definition of kinetic
terms fo the background fields) as:
F R µν(x) = ∂ µ r ν (x) + ∂ ν r µ (x) − i[r µ (x),r ν (x)] → R(x)F R µν(x)R † (x)
F L µν(x) = ∂ µ l ν (x) + ∂ ν l µ (x) − i[l µ (x),l ν (x)] → L(x)F L µν(x)L † (x)
With Tr(l µ ) = Tr(r µ ) = 0 (trace over the lambda matrices) follows immediately
that
Tr(F L µν) = Tr(F R µν) = 0
Following Gasser and Leutwyler we introduce the combination
χ(x) = 2B 0 (s(x) + ip(x))
We write down the chiral counting rules: We define the U as of order
U = Ord(E 0 )
and the other terms according their number of derivatives and each derivative
we count as Ord(E 1 ). This means that in the process considered we must have
all Mandelstam variables small compared to a hadronic scale, i.e. Rho-Mass.
U = Ord(E 0 )
D µ U r µ l µ = Ord(E 1 )
L µν R µν s r = Ord(E 2 )
Altogether we can summarize: The simplest effective Lagrangean with two
derivatives reads
L 2 = F 2 0
4 Tr(D µUD µ U † ) + F 2 0
2 Tr(χU † + Uχ † )
188

12.7 Application: Pion decay
We consider in this section the decay of the ion in the chiral perturbation theory
using the simplest approach, i.e. L 2 with covariant derivatives. The decay of
the pion (see previous section) happens via an intermediar W − -boson, which
is coupled one one side to the leptons, and on the other side to the pion. The
coupling to the leptons reads
L coupl =
e
2 √ 2sin(θ W ) (W + µ ν (µ) γ µ (1 − γ 5 )µ − + W − µ ¯µ − γ µ (1 − γ 5 )ν (µ)
The coupling of the W-bosons to the pion is obtained by the fact that the W-
bosonic field appears in the field l µ (x) , which is in the covariant derivative. We
have r µ (x) = 0 and
and
e
l µ = −√ (W µ + T + + h.c.)
2 sin ϑW
L 2 = F 2 0
4 Tr(D µUD µ U † ) + F 2 0
2 B 0Tr(MU † + UM † )
D µ U = ∂ µ + iUl µ
Similar as in the pion-pion-scattering we have to extract from this lagrangean
the term which corresponds to the pion decay. The simplest term is just linar
in l µ (x) and we disentengle the L 2 in order to find this:
F 2 0
4 Tr(D µU(D µ U) † ) = F 2 0
4 Tr ( (∂ µ + iUl µ )(∂ µ − il µ U † ) )
= .... + i F 0
2 Tr ( l µ ∂ µ U † U )
We set
l µ (x) = λa
2 la µ(x)
If we compare with the well known expression for the left handed current of the
effective Lagrangean, i.e.
then we can write
J µ,a
L
(x) = if2 π
4 Tr [ λ a ∂ µ U † U ]
L coupl = lµ(x)J a µ,a
L
(x)
We have now the term linear in l µ (x) and want to expand it in order to find the
term linear in the pion field. For this we expand the current J µ,a
L
(x) up to first
order in φ:
J µ,a
L = iF 2 0
4 T ( λ a ∂ µ U † U ) = F 0
4 Tr (λa ∂ µ φ) + ord(φ 2 )
189

Since we have Tr(λ a λ b ) = 2δ ab we can simplify
With this we have in particular
J µ,a
L = F 0
2 ∂µ φ + ord(φ 2 )
< 0|J µ,a
L (0)|φb (p) >= F 0
2 < 0|∂µ φ a (0)|φ b (p) >= ip µ F 0
2 δab
Now we insert l µ
e
l µ = −√ (W µ + T + + h.c.)
2 sin ϑW
and we obtain
L coupl = F 0 e F
2 ∂µ φ = −√ 0
2 sin ϑW 2 Tr((W µ + T + + Wµ − T − )∂ µ φ)
= − e
sinϑ W
F 0
2 (W + µ (V ud ∂ µ π − + V us ∂ µ K − ) + W − µ (V ud ∂ µ π + + V us ∂ µ K + ))
We know the propagator for the W-bosons, which can be approximated due to
the large mass of the W-boson:
−g µν + kµkν
M 2 W
k 2 − M 2 W
= g µν
M 2 W
+ ord( k2
MW
4 )
we get then for the invariant amplitude of the pion decay:
( )
e
M = −i√ ū (µ) γ µ ig µν e F0
(1 − γ 5 )v¯ν(µ) −i
2 sinϑW sin(θ W ) 2 V ud(−ip ν )
Here the p is the four-momentum of the pion and G F = 1.166x10 −5 GeV −2 .
Following Bjorken-Drell sec. 10.14 one obtains for the decay rate (replace √ a 2
=
V ud F 0 ) the expression
( )
1
τ = G2 F V ud
2
4π F 0 2 m π m 2 µ 1 − m2 µ
m 2 π
Actually, we had this expression already in the previous section on pion decay,
however, there it was simply taken from the literature, here we have derived it
in the chiral perturbation theory. Aplparently the constant F 0 is the pion decay
constant, shich in the present papproximatian is identical to the Kaon decay
constant. In nature we have 93MeV for the pion one and 113MeV for the Kaon
decay constant.
M 2 W
190

12.8 The chiral Lagrange-Density in ord(p 4 ) or ord(E 4 ) and
renormalization.
12.8.1 The Lagrangean
We quote the most general Lagrange-Density of ord(p 4 ) as it is presented by
Gasser and Leutwyler in Nucl.Phys. B250 (1985) 465:
with
L = L 2 + L 4
L 4 = L 1 Tr(D µ U(D µ U) † ) 2 + L 2 Tr(D µ U(D ν U) † )Tr(D µ U(D ν U) † )
+L 3 Tr(D µ U(D µ U) † D ν U(D ν U) † ) + L 4 Tr(D µ U(D µ U) † Tr(χU † + Uχ † )
+L 5 Tr(D µ U(D µ U) † (χU † + Uχ † )) + L 6 (Tr(χU † + Uχ † )) 2
+L 7 (Tr(χU † − Uχ † )) 2 + L 8 Tr(Uχ † Uχ † + χU † χU † )
−iL 9 Tr(F R µνD µ U(D ν U) † + F L µν(D µ U) † D ν U) + L 10 Tr(UF L µνU † F µν
R )
+H 1 Tr(F R µνF µν
R + F L µνF µν
L ) + H 2Tr(χχ † )
The terms H 1 and H 2 are chirally invariant without involving the matrix
U. They do not generate any couplings to goldstone bosons and hence are not
of great phenomenological interest. However, if one were to use the effective
Lagrangean to describe correlation funcitions of the external sources, these two
operators can generate contact terms (see below the renormalization program).
The values of the so called low energy constants (Leutwyler coefficients) L i
cannot be determined purely from chiral symmetry breaking, as it was possible
in L 2 with F 0 = f π and B 0 related to m π . They are dynamical coefficients,
whcih should be determined from QCD directly, it this was possible. They
should perhaps be calculated in some QCD-inspired model as it is possible with
the Chiral Quark Soliton MOdel. In practice they are determined by adjusting
them to experimental data as described below.
Remark: The effective lagrangean may be used in the context of either
chiral SU(2) of SU(3). Because SU(2) is a subgroup of SU(3) the general SU(3)
Lagrangean is also valid for chiral SU(2). However, the SU(2) version has fewer
low energy constants, so that only certain combinations of the L r i will apear in
purely pionic processes. If one is dealing with pions at low energy the kanos
and the eta are considered as heavy particles and may be integrated out. THis
procedure produces a shift in the values of the low energy renormalized constants
such tha the coefficients of a purely SU(2) lagrangean and an SU(3) one will
differ by a finite amount, which can be exactly calculated. Actually the SU(2)
coefficients can be found by first performing calculations in the SU(3) limit and
then treating the masses of kaons and eta as going to infinity.
191

12.8.2 Chiral perturbation theory, renormalization program
The way physics is extracted from the effective Lagrangean is called “Chiral
perturbation theory”. It works in order(E 4 ) in the following way, which contains
three ingredients:
• The general Lagrangean L 2 which is to be used both at tree level and in
loop diagrams.
• The general lagrangean L 4 which is to be used only at tree-level.
• The renormalization program which describes how to make physical predictions
at one-loop level.
• Generalization of all that to order(E 6 )
D=4
Above: Renormalization of loops with trees. Attention: There are various
different 1-loop-graphs with L 2 insertions since L 2 has to be expanded in the
physical fields. There are also various different tree graphs of L 4 due to the
same reason.
Actually for the last point needs some explanation: Consider the calculation
of an actual process. In lowest order one should use L 2 for that. When expanded
in terms of the meson fields π a it specifies a set of interaction vertices. These
can be used to calculate treee-level and one-loop diagrams for any process of
interest. This result is added to the contribution which comes from the vertices
contained in the lagrangean L 4 , treated at tree-level only (because then it has
the same dimension in Weinbergs counting). If we consider the one-loop graphs,
we realize that they are infinite.
One sees this immediately if one considers a typical 1-loop-diagram, which
always contains the following term
∫
d 4 k
(2π) 4
∫
i
k 2 − M 2 + iɛ ∼
k 3 1
dk
k 2 − M 2 + iɛ → ∞
Divergent parts in loops are not a problem, because there are well known
techniques to handle those infinities. One is the dimensional regularization,
where one considers the Feynman integrals not in a space with dimension d = 4
192

ut in a general space with dimension d > 4, does the integrals, and then performs
the limit d → 4. This procedure allows to identify from each Feynman
diagram the part which is finite and which contains the physics, and the part
which is infinite and is just there, because the integrals extend to infinite momenta,
where that part of the physics, which is described by the low energy
Lagrangean, is no longer valid. Actually we will use this dimensional regulariization.
There are various conventions for doing that, we take the one of
Donoghue, Golowifh and Holstein and write
∫
d 4 ∫
k
(2π) 4 → µ4−d
d d k
(2π) d
where the scale µ has to be introduced in order to keep the dimension of the
expression for any d. The µ is an auxiliary parameter, the so called renormalization
point or subtraction point. The final observalbe does not depend on it.
Thus we will meet always in the loops the following integral,
∫
I(M 2 ,µ 2 ) = µ 4−d d d [ ]
k i
(2π) d k 2 − M 2 + iɛ = M2
16π 2 R + log( M2
µ 2 )
with
R = 2
d − 4 − [log(4π) + Γ′ (1) + 1]
where the Γ ′ (x) is the derivative of the Γ-function. Actually the result of the
integration is proportional to a Gamma function Γ(− d 2
), which diverges for
d
2
= 1,2,3,... but converges for values in between. Using an expansion of Γ(x)
at the point x = − d 2
and at d = 4 one obtains the derivative of the Gamma
function due to the expansion and then the final result from above. One shoud
note that Γ ′ (1) = −γ = −0.5772 = d dz Γ(z + 1)| z=0.
Taking this into account at this stage (after calculating tree and 1-loop for
L 2 and tree for L 4 ) the result contains both bare parameters L i and divergent
loop integrals. One needs to determine the parameters from experiment. If the
Lagrangea is indeed the most general one possible, relations between observables
will be FINITE when expressed in terms of physical quantities. Thus all the
divergences can be absorbed in redefining the low energy coefficients L i . (We
will show this explicitely in the example below, where we calculate the meson
masses in one-loop form). This is called “renormalization” of the low energy
coefficients, i.e. one says: The coefficient in L 4 consists of two terms, one is
the coefficient L r i , containing the physics and determining the observables and
being finite, plus a term which compensates the divergent part from the one-loop
diagrams of L 2 . Thus one has
L i = L r i + Γ i
32π 2 R i = 1,.....8
H i = H r i + ∆ i
32π 2 R i = 1,2
193

The constants Γ i and ∆ i are in the table below. The renormalized low energy
coefficients are dependent on the renormalization scale µ. The coefficients for
two different scales are related to each other by
L r i(µ 2 ) = L r i(µ 1 ) + Γ i
16π 2 log(µ 1
µ 2
)
The low energy coefficients are given vor SU(3) in the following table. They
are real numbers and given below in units of 10 −3 at the renormalization scale
of the Rho-mass (see Bijnens, Ecker and Gasser, The socond DAΦNE physics
Handbook Nucl. Phys. The conventions are taken from the book by Donoghue,
Golowich and Holstein “Dynamics of the Standard model”.
coeff emp. value experiment Γ i
L r 3
1 0.4 ± 0.3 ππ → ππ
32
L r 3
2 1.35 ± 0.3 ππ → ππ
16
L r 3 −3.5 ± 1.1 ππ → ππ 0
L r 1
4 −0.3 ± 0.5 ∼ 0 Zweig-Regel
f K
8
3
fπ 8
L r 5 1.4 ± 0.5
L r 11
6 −0.2 ± 0.3 Zweig-Regel
144
L r 7 −0.4 ± 0.2 Gell-Man-Okube,L 5 ,L 8 0
L r 8 0.9 ± 0.3 M K 0 − M K +,L 5 ,(2m s − m u − m d) : (m d − m u )
5
48
L r 9 6.9 ± 0.7 isovector radius of the pion
1
4
L r 10 −5.5 ± 0.7 π → eνγ − 1 4
Remark: There exists always an ambiguity of what finite constants should
be absorbed into the renormalized low energy coefficients. This lambiguity
does not affect the relationship between observables, but only influences the
numerical values quaoted for the low energy constants. One should keep this
in mind, if one compares sets of constants of different authors. Also there are
various regularization procedures for handling the divergent integrals. They also
influence not the observalbes but the numerical values of L i .
Erklaere was
die Zweig
Regel ist.
12.9 Application in order(p 4 ): Masses of Goldstone bosons
12.9.1 Aim of this section
We have used once the L 2 in tree level to calculate the masses of the Goldstone
bosons. There it meant that we looked at the Lagrangean of the Goldstone boson
and we identified the mass of the particle with the coefficient of the quadratic
term. We learned that Gell-Mann-Okubo and Gell-Mann Oaks Renner etc. were
all fulfilled, which shows that such a procedure is not stupid. In this section
we calculate the masses of the Goldstone bosons again, however, we will do it
systematically better. We will use tree- and 1-loop-diagrams of L 2 and treediagrams
of L 4 . This corresponds systematically to all terms with d = 4. We
will demsonstrate in this calculation the following features, which are the basic
points in the chiral perturbation theory:
194

• The connection between the bare coefficients L i and the renormalized ones
L r i is done in such a way (see above formula) that the infinities of the 1-
loop diagrams of L 2 are compensated by the tree-diagrams of L 4 .
• The scale dependence (dpendence on µ 2 ) is such, that the expressions for
observables are independent on the choice of µ.
We consider L =L 2 +L 2 but do not take into account external fields. We
consider also isospin symmetry, i.e. assume up and down quarks to be degenerate.
First we have to learn, how to calculate masses beyond tree-level. Actually
there are several definitions of a mass of a particle. One is the so called “Pole
mass”. This means: Consider the particular example of the Propagator of a
meson field . The propagator is the Fourier-transform of the Greens 2-point
function:
∫
i∆(p) = d 4 xexp(−ipx) < 0|T [Φ(x)Φ(0)] |0 >
In lowest order of perturbation theory this propagator reads
i∆ 0 (p) =
i
p 2 − M 2 0 + iɛ
If we calculate from the Lagrangean it we find out that the M 0 of the propagator
is exactly the quadratic term in the Lagrangean, because this part defines the
“free” lagrangean and higher powers of the field φ are considered as perturbation,
which we treat in this example in lowest order. Thus in this approximation
(lowest order) the mass in the Lagrangean and the mass M 0 in the propagator
are the same. The M 0 is called pole mass, since the propagator has a pole at
p 2 = M 2 0.
12.9.2 Calculation of the pole mass
We want now to calculate the propagator in a better approximation. If one
follows standard books on field theory one can write down the full propagator
i∆(p) of the interacting field as a sum of all connected diagrams. In practice
one does not calculate this explicitely bit uses a simple relationship between
this and the self-energy Σ(p), which is ths sum of all one-particle-irreducible
diagrams.
∆(p) −1 = ∆ −1
0 (p) − Σ(p)
or equivalently
i∆(p) =
i
p 2 − M 2 0 + iɛ +
i
i
p 2 − M0 2 + ))
iɛ(−iΣ(p2 p 2 − M0 2 + iɛ
i
i
i
+
p 2 − M0 2 + ))
iɛ(−iΣ(p2 p 2 − M0 2 + ))
iɛ(−iΣ(p2 p 2 − M0 2 + ......
+ iɛ
i
=
p 2 − M0 2 − Σ(p2 ) + iɛ
195

Since Σ(p 2 ) consists of one-particle-irreducible diagrams we must apply to it the
chiral counting. That means it consits of a sum of tree-diagram with L 4 plus a
1-loop-diagram with L 2 since the tree diagram with L 2 is already contained in
M 0 .
So in the chiral counting the Σ(p 2 ) is of order ord(E 4 ) or ord(p 4 ) because
D = 2 + ∑ 2(n − 1)N 2n + 2N L = 4. If we look to the above expansion, we have
the series
ord(E −2 )+ord(E −2 )ord(E 4 )ord(E −2 )+ord(E −2 )ord(E 4 )ord(E −2 )ord(E 4 )ord(E −2 )+......
or ord(E −2 ) + ord(E 0 ) + ord(E 2 ) + .... Thus we have to stop the series after
the first non-trivial term. However, this is correct, if we are interested in the
propagator at momenta noticeably different from the physical mass, i.e. p 2 ≠
M 2 . When we are interested in the pole, that is in the behaviour at p 2 ≈ M 2
then we must be more careful. Then we have to consider that between the poleposition
to zeroth order (i.e. p 2 = M 2 0) and the pole position to first order (i.e.
p 2 = M 2 0 + Ωord(E 4 ) with Ω being a constant, which should be determined.
This corresponds exactly to the chiral counting, because the D = 4 for the self
energy insertion Σ(p 2 ). Thus we have
i∆ 0 (p) =
i
p 2 − M0 2 + iɛ = i
M0 2 − M2 0 − ord(E4 ) = ord(E−4 )
This has a drastic consequence: Now we have the order counting of the series
for the full propagator like:
ord(E −4 )+ord(E −4 )ord(E 4 )ord(E −4 )+ord(E −4 )ord(E 4 )ord(E −4 )ord(E 4 )ord(E −4 )+......
Thus one sees that the combination (−iΣ(p 2 )) is of order unity. The
p 2 −M0 2 +iɛ
consequence is, that it is consistent to sum up all contributions like this and
cannot chop of the series after the first non-trivial term. To make it clear, this
is consistent order counting in chiral perturbation theory. Actually, as we will
see below, one can handle with this fact, because one indeed can sum up the
series exactly and even easily.
We will explicitely do the calculation below, but lets first consider, what we
do with it once we know the Σ(p 2 )., i.e. how do we extract the pole-mass of the
field. This is actually very simple, because we have to look for the zero of the
i
196

inverse pole. We look for that p 2 for which ∆ −1 (p) = p 2 − M 2 0 − Σ(p 2 ) = 0. If
we call this solution p 2 = M 2 , the equation to solve reads
M 2 − M 2 0 − Σ(M 2 ) = 0
This is an implicit equation. In order to solve it and to get en explicit expression
for the physical mass we bring the expression of ∆(p) as close to a form of ∆ 0 (p 2 )
as possible, because from this one we know how to read off the mass.
In order to achieve this we first expand as a mathematical exercise the Σ(p 2 )
around a point µ 2 , which is considered arbitrary for the moment.
Σ(p 2 ) = Σ(µ 2 ) + (p 2 − µ 2 )Σ ′ (µ 2 ) + ˜Σ(p 2 )
where we collect all the rest in the ˜Σ(p 2 ), which can be easily shown to have
the property
˜Σ(µ 2 ) = ˜Σ ′ (µ 2 ) = 0
The propagator can now be written as
i
i∆(p) =
p 2 − M0 2 − Σ(µ2 ) − (p 2 − µ 2 )Σ ′ (µ 2 ) − ˜Σ(p 2 ) + iɛ
We replace now the M 2 0 by the pole-equation and choose as renormalization
point µ 2 = M 2 . Then the propagator reads
i
i∆(p) =
p 2 − M 2 − (p 2 − M 2 )Σ ′ (M 2 ) − ˜Σ(p 2 ) + iɛ
The popagator has still not yet the convenient form. However we get this if we
introduce the wave function renormalization constant
1
Z φ =
1 − Σ ′ (p 2 )
and then we can rewrite
i
i∆(p) =
(p 2 − M 2 ) (1 − Σ ′ (p 2 )) − ˜Σ(p 2 ) + iɛ = iZ φ
p 2 − M 2 − Z φ˜Σ(p2 ) + iɛ
If we define now renormalized fields like
φ R = √ φ
Zφ
then we have for the renormalized propagator
∫
i∆(p) = d 4 xexp(−ipx) < 0|T [Φ R (x)Φ R (0)] |0 >
with
i
=
p 2 − M 2 − Z φ˜Σ(p2 ) + iɛ
˜Σ(M 2 ) = 0
Hence in the vicinity of the physical mass the full propagator has the same
structure as the the free propagartor with the bare mass.
197

12.9.3 Calculation of the self energy
We have to calculate the self energy applying strictly the chiral counting rules.
That is we have to use
L 4φ
2 for the vertices in the loop diagram
L 2φ
4 for tree diagram
It is clear what L 4φ
2 is, because we have derived it in the section about elastic
pion scattering
L 4φ
2 = 1
24F0
2 Tr ([φ,∂ µ φ]φ∂ µ φ) + 1
24F0
2 B 0 Tr(Mφ 4 )
For L 2φ
4 we have to consider in detail the structure of L 4 . Since we want to
have expressions with two fields, we see immediately that only the terms with
L 4 ,L 5 ,L 6 and L 7 have to be considered. Take e.g. the L 4 -term. We can rewrite
it as
L 4 Tr(∂ µ U∂ µ U † Tr(χU † + Uχ † )
= L 4
2
F 2 0
(
∂µ η∂ µ η − ∂ µ π 0 ∂ µ π 0 + 2∂ µ π + ∂ µ π + + 2∂ µ K + ∂ µ K + + 2∂ µ K 0 ∂ µ ¯K0 ) (4B 0 (2m+m s )
We proceed analogously for the other terms and obtain in the end
L 2φ
4 = 1 2 (a η∂ µ η∂ µ η − b η η 2 ) + 1 2 (a π∂ µ π 0 ∂ µ π 0 − b π π 0 π 0 ) + a π ∂ µ π + ∂ µ π −
−b π π + π − + a k ∂ µ K + ∂ µ K − − b k K + K − + a K ∂ µ K 0 ∂ µ ¯K0 − b K K 0 ¯K0
with m = 1 2 (m u + m d ) and the the constants
b η = 64B 0
3F 2 0
a η = 16B 0
F 2 0
((2m + m s )L 4 + 1 3 (m + 2m s)L 5
)
(
(2m + ms )(m + 2m s )L 6 + 2(m − m s ) 2 L 7 + (m 2 + m 2 s)L 8
)
a π = 16B 0
F0
2 ((2m + m s )L 4 + mL 5 )
b π = 64B2 (
0 (2m + ms )mL 6 + m 2 )
L 8
F 2 0
a K = 16B 0
((2m + m s )L 4 + 1 )
2 (m + m s)L 5
b K = 32B2 0
F 2 0
F 2 0
((2m + m s )(m + m s )L 6 + 1 2 (m + m s) 2 L 8
)
With these expressions the self energies are of the form
Σ φ (p 2 ) = A φ + B φ p 2
198

Here indicates the index φ the various contributions separately from pions, kaons
and eta. Apparently we have only diagonal terms in the self energy, where the
incoming and outgoing boson are the same. One can show this structure easily:
Each of the coefficients A and B consists of two contributions: One coming
from the Tree-graph with the vertex based upon L 2φ
4 and another coming from
the 1-loop graph with a vertex based on L 4φ
2 . If we look at the tree graph of
L 2φ
4 and at the explicit expression for it (see above) we see that the terms in
L 2φ
4 there are either two derivatives or none, corresponding to p 2 or p 0 . So for
this the structure of Σ(p 2 ) is obvious. Indeed, the following example shows this
explicitely for the η-terms of L 2φ
4 , it yields
1
−iΣ tree
η (p 2 ) = i2(
2 a η(ip µ )(−ip µ ) − 1 )
2 b η = i(a η p 2 − b η )
For the terms of the 1-loop contribution of L 4φ
2 the argument for the structure
Σ(p 2 ) = A + Bp 2 goes as follows: L 4φ
2 has either two derivatives (symbolically
φφ∂φ∂φ) or no derivative (symbolically φ 4 ). The first term is proportional to
M 2 if the φ are contracted to external lines, and it is proportional to p 2 , if the
∂φ are connected with external lines. The second term does not yield external
momenta. Thus we obtain the structure Σ(p 2 ) = A + Bp 2 .
We show now, how the loop diagrams are actually calculated. For this
we consider first the vertex of L 2 which we actually know from the pion-pion
scattering: there we had for the Feynman-amplitude M the following expression
(We do not use the second expression of the pion-pion-vertex, as we have done
in the elastic pion scattering, because that expression assumes physical particles
in all four legs). Thus we have
with
−i6F 2 0 M(p a ,p b ,p c ,p d ) = 2A + B
A = δ ab δ cd (−ip a − ip b ) (ip c + ip d ) + δ ac δ bd (−ip a + ip c )(−ip b + ip d )
+δ ad δ bc (−ip a + ip d ) (−ip b + ip c ) − 4(δ ab δ cd ((−ip a )(−ip b ) + (ip c )(ip d ))
+δ ac δ bd ((−ip a )(ip c ) + (−ip b )(ip d )) + δ ad δ bc ((−ip a )(ip d ) + (ip b )(ip c )))
B = m2 π
4 8( δ ab δ cd + δ ac δ bd + δ ad δ bc)
Let us consider this for the pion-loop to the self energy of π 0 , i.e. with
a = 3 p a = p b = j p b = k c = 3 p c = p p d = k d = j
199

We obtain for the 1.loop contribution
Loop = 1 2
∫
d 4 k
(2π) 4
i
3F 2 0
3∑
i
(X j + Y j + Z j )
k 2 − m 2 π + iɛ
j=1
X j = δ 3j δ 3j [ (p + k) 2 + 2pk + m 2 ]
π
Y j = δ 33 δ jj [ (p − p) 2 − 2p 2 − 2k 2 + m 2 ]
π
Z j = δ 3j δ 3j [ (p − k) 2 − 2pk − 2kp + 5m 2 ]
π
For the evaluation we also need the divergent loop integral and the expression
∫
d n ∫
k k µkν
i
(2π) n k 2 − M 2 + iɛ = M2
n g d n k i
µν
(2π) n k 2 − M 2 + iɛ = M2
n µn−4 I(M 2 ,µ 2 )
performing the integrals yields:
loop = 1 ∫
d 4 k i
2 (2π) 4
3F 2 0
[
−4p 2 − 4k 2 + 5m 2 π] i
k 2 − m 2 π + iɛ
= i
6F0
2 (−4p 2 + m 2 π)I(m 2 π,µ 2 )
Apparently we have the structure, which we predicted i.e. Σ φ (p 2 ) = A φ +B φ p 2 .
If one performs the calculations with these 1-loop integrals, one obtains:
(
A π = m2 π
F0
2 − 1 6 I(m2 π) − 1 6 I(m2 η) − 1 )
3 I(m2 K) + 32[(2m + m s )B 0 L 6 + mB 0 L 8 ]
(
A K = m2 K 1
F0
2 12 I(m2 η) − 1 4 I(m2 π) − 1 2 I(m2 K) + 32
[(2m + m s )B 0 L 6 + 1 ])
2 (m + m s)B 0 L 8
A η = m2 η
(− 2 )
3 I(m2 η) + 16m 2 ηL 8 + 32(2m + m s )B 0 L 6
+ m π
F 2 0
B K = 1 I(m 2 η)
4
F 2 0
( 1
6 I(m2 η) − 1 2 I(m2 π) + 1 3 I(m2 K))
+ 128
9
B0(m 2 − m s ) 2
(3L 7 + L 8 )
B π = 2 I(m 2 π)
3 F0
2 + 1 I(m 2 K )
3 F0
2 − 16B 0
F0
2 [(2m + m s )L 4 + mL 5 ]
F 2 0
+ 1 I(m 2 π)
4 F0
2 + 1 I(m 2 K )
2 F0
2 − 16B 0
F0
2
B η = I(m2 K )
F 2 0
F 2 0
[(2m + m s )L 4 + 1 2 (m + m s)L 5
]
− 8m2 η
F0
2 L 5 − 16B 0
F0
2 [2m + m s ]B 0 L 4
Here the appearing masses (e.g. m 2 π are “abbreviations” for the terms
m 2 π = m 2 π,2 = 2B 0 m
200

m 2 K = m 2 K,2 = B 0 (m + m s )
m 2 η = m 2 η,2 = 2 3 B 0(m + 2m s )
We determine the masses by solving the equation
M 2 − M 2 0 − Σ(M 2 ) = 0
We do this in the following careful way for each boson separately (omit the
index φ):
M 2 = M 2 0 + A + Bp 2
or
Then
or approximately
This yields
M 2 = M 2 0 + A + BM 2
M 2 (1 − B) = M 2 0 + A
M 2
1 + B ≈ M2 0 + A
M 2 ≈ M 2 0(1 + B) + A(1 + B)
We can (and must) simplify this expression further by considering carefully
the ordering in the chiral perturbation approach. We have M 2 = ord(E 2 ),
M 2 0 = ord(E 2 ), and since Σ(p 2 ) = A + Bp 2 = ord(E 4 ) we have A = ord(E 4 )
and B = ord(E 2 ). Thus, if we collect in the above equation all terms up to
ord(E 4 ) we obtain the result:
M 2 = M 2 0(1 + B) + A
Now we are close to the end: With this prescription we write down the masses
and replace simultaneously the bare low energy coefficients by the renormalized
ones (plus the divergent integral of course), i.e. inserting
L i = L r i + Γ i
32π 2 R i = 1,.....8
H i = H r i + ∆ i
32π 2 R i = 1,2
This yields then the final expression up to terms of ord(E 4 ):
[
m 2 π,4 = m 2 π,2
1 + m2 π,2
32π 2 F 2 0
log( m2 π,2
µ 2 ) − m2 η,2
96π 2 F 2 0
]
log( m2 η,2
µ 2 )
+ 16m2 π,2
F 2 0
[(2m + m s )B 0 (2L r 6 − L r 4) + mB 0 (2L r 8 − L r 5)]
201

+ 16m2 K,2
F 2 0
+m 2 π,2
m 2 η,4 = m 2 η,2
+ 16m2 η,2
F 2 0
[
m
2
η,2
96π 2 F 2 0
m 2 K,4 = m 2 K,2
[
1 + m2 η,2
48π 2 F 2 0
]
log( m2 η,2
µ 2 )
[
(2m + m s )B 0 (2L r 6 − L r 4) + 1 ]
2 (m + m s)B 0 (2L r 8 − L r 5)
[
[
1 + m2 K,2
16π 2 F 2 0
log( m2 K,2
µ 2 ) − m2 η,2
24π 2 F 2 0
]
log( m2 η,2
µ 2 )
(2m + m s )B 0 (2L r 6 − L r 4) + 8 m2 η
F0
2 (2L r 8 − L r 5)
log( m2 η,2
µ 2 ) − m2 π,2
32π 2 F 2 0
+ 128
9F 2 0
log( m2 π,2
µ 2 ) + m2 K,2
48π 2 F 2 0
[
(m − ms )B 2 0(3L r 7 + L r 8) ]
]
]
log( m2 K,2
µ 2 )
These are the final formulae for the masses of the Goldstone bosons up to order
ord(E 4 ) in terms of the masses of the Goldstone bosons up to order ord(E 2 ),
where the latter ones are given in terms of unknown QCD-quark-masses (see
above). One notices the following important features:
The masses of the Goldstone bosons vanish if the quark masses go to zero,
since then also the masses to ord(E 2 ) go to zero. This is gratifying, since in case
of no symmetry breaking this is actually what should happen. We woud have a
bad theory if in the limit of exact chiral symmetry the goldstone bosons would
have mass. If one looks into the literature it often happens that this point is
not done right.
The expressions for the masses contain analytic terms ∝ m q and non-analytic
terms ∝ m q log(m q ) . The latter ones are multiplied with the low energy coefficients
and involve no new paramters. This illustrates the general theorem of Li
and Pagels that a symmetry, which is realized in the Nambu-Goldstone mode,
leads in perturbation theory to analytic as well as non-analytic trms.
Looking to the above formulae it seems, that the physical masses are dependent
on the scale µ 2 (scale dependence). However the low energy coefficients L r i
are also scale dependent, and their scale dependence is such that it is exactly
compensated by the scale dependence of the chiral logarithms. This is most
easily shown by performing the derivative of the above formulae with respect
to µ , which turns out to vanish. Thus the physical observables are not scale
dependent, as it should be.
13 Lattice Gauge Theory
13.1 Introduction
202

We assume that the QCD is the theory for strong interactions and that mesons
and baryons are mainly determined by this forces. In the previous chapters
we have never considered this fact, except in studying symmetries, but looked
always for phenomenological descriptions in terms of hadronic currents, quark
degrees of freedom, etc. In this chapter we will try to solve QCD directly as
a non-abelian gauge theory. The idea for this goes back to K. Wilson und
Wegener. The idea is to formulate QCD on a discrete space-time lattice (1+3
dimensions) written down in terms of altogether 4 Euklidean dimensions with
lattice constant a.
Hereby the information on the quark fields will reside on littice sites (points)
and the information on the gluon fields will reside on links between sites. The
final formulation maintains exact local gauge invariance. There is a natural
cut-off at p ∼ 1 a
. That means things smaller than a cannot be described and
momenta larger than the cut-off cannot be described either. The formulation is
suitable for (rather extensive!!) computer simulations in large supercomputers
and require usually large collaborations. The idea is to calculate observables,
in order to compare with experiment, or to provide benchmark results in accurate
calculations which simpler models with properly chosen effective degrees
of freedom have to reproduce. On the first view the lattice QCD techniques
sound very fundamental and exact. However in practice several shortcomings
and technical problems are encountered (finite a, finite lattice, too large current
quark mass) such that often the approach is by no means that fundamental as
it seems and often a comparison with experiment is difficult.
13.2 Quantum Mechanics: Transition amplitudes and path
integrals
We consider 1-dimensional quantum mechanics as an explicit and didactic example.
There the system is described in the Schrödinger picture by the wave
function ψ(x,t) which corresponds to a time dependent state vector in a Hilbert
203

space |ψ(t)〉. We know the state vector |x〉 which is eigenstate of the coordinate
operator ˆx |x〉 = x |x〉 and similarly the state vector |p〉 which is eigenstate of
the momentum operator ˆp |p〉 = p |p〉. The operators fulfill the commutation rule
[ˆx, ˆp] = iħ and since ˆx and ˆp are hermition we have the completeness relations
∫
1 = dx|x >< x|
∫
1 = dp|p >< p|
. The wave functions are defined as ψ(x,t) =< x |ψ(t)〉 and ψ(p,t) =< p |ψ(t)〉
.The Fourier transform relates both:
∫
< x |ψ(t)〉 = dp < x |p〉 < p |ψ(t)〉
with
< x |p〉 = 1 √
2πħ
e ipx/ħ (213)
The wavefunctions are described by the Schrödinger-eq. with Hamiltonian H
and the time evolution is given by (if the H is time-independent)
|ψ(t)〉 = e −iHt/ħ |ψ(0)〉
The fact that we assume H to be time independent is no limitation. We
will chose lateron small time steps such that at each time intervall the H can
be considered time independen. For the formulation of quantum mechanics
and field theory on the lattice the concept of a propagator or the correlator is
important. On gets in a natural way like this: We have
< x f |ψ(t f )〉 =< x f |exp [−iHt f /ħ] |ψ(0)〉
=< x f |exp [−iHt f /ħ] exp[+iHt i /ħ] |ψ(t i )〉
and inserting the completeness relation we get
∫
< x f |ψ(t f )〉 = dx i < x f |exp [−iH(t f − t i )/ħ] |x i 〉 〈x i |ψ(t i )
or
∫
ψ(x f ,t f ) =
dx i K( x f , t f ,x i ,t i )ψ(x i ,t i )
with the propagator K given by
K( x f , t f ,x i ,t i ) =< x f |exp [−iH(t f − t i )/ħ] |x i 〉 =< x f (t f ) |x i (t i )〉
204

13.2.1 Free Motion (Propagator)
In this subsection we calculate te propagator of a free particle with H = p2
2m
and we will introduce a formulation with discrete space and time points. We
have:
∫
]
< x f |exp [−iH(t f − t i )/ħ] |x i 〉 = dp < x f |p〉 exp
[−i p2
2mħ (t f − t i ) < p |x i 〉
We can insert eq.(213) and perform the integration. For this we use the well
known formulae
∫ +∞
√ ∫ π +∞
√ π
dx exp(−λx 2 ) = or dx exp(−λx 2 +2λx¯x) =
λ
λ exp( λ¯x 2)
−∞
−∞
however, we use analytical continuation to get rid of the i , integrate and continue
analytically back. Then we obtain (H for free motion)
( ) [ ]
1/2
m
im (x f − x i ) 2
< x f |exp [−iH(t f − t i )/ħ] |x i 〉 =
exp
2πiħ(t f − t i ) 2ħ (t f − t i )
(214)
The exponenet on RHS can be written as
exp
[
im
2ħ
(x f − x i ) 2
(t f − t i )
]
[ ]
i 1
= exp
ħ 2 mv2 (t f − t i ) = i ∫ tf
Ldt = i ħ t i
ħ S cl(x f t f ;x i t i )
Here is S cl the action (classical, i.e. no operators) between the space-time points.
Hence in this example of free motion of a massive point the propagator is known.
To get closer to the formulation on a lattice we rewrite the above expression
in a trivial way. For this we introduce an arbitrary intermediate space-time
point x 1 ,t 1 with t f > t 1 > t i :
∫
ψ(x f ,t f ) = dx 1 K( x f , t f ;x 1 ,t 1 )ψ(x 1 ,t 1 )
∫
ψ(x 1 ,t 1 ) = dx i K( x 1 , t 1 ;x i ,t i )ψ(x i ,t i )
and with that we have
∫
K( x f , t f ;x i ,t i ) =
dx 1 K( x f , t f ;x 1 ,t 1 )K( x 1 , t 1 ;x i ,t i )
where the integrand is proportional to
]
i
∝ exp[
ħ (S cl(x f t f ;x 1 t 1 ) + S cl (x 1 t 1 ;x i t i ))
We introduced one intermediate point in the time, which resulted in an integral
over the x-coordinate at that time. We will introduce now N intermediate
205

points, with N large, and will have at each time-point the integral over the
x-coordinate:
ε = t f − t i
N
t j = t i + jε
j = 0,1,...,N
Then we have for large N a small ε and can write down explicitely
[ ]
< x f |exp −iĤ(t f − t i )/ħ |x i 〉 =
[
< x f |exp f − t N−1 )/ħ
] [
exp N−1 − t N−2 )/ħ
∫
]
= dx 1 dx 2 ....dx N−1 < x f |exp
... < x 2 |exp
[−iĤε/ħ ] [−iĤε/ħ ]
|x 1 〉 < x 1 |exp |x i 〉
]
|x N−1 〉 < x N−1 |exp
[ ]
...exp −iĤ(t 1 − t i )/ħ |x i 〉
[−iĤε/ħ
]
|x N−2 〉 ...
These are N factors and N − 1 integrations. We cannot put them together
without a little thinking. We have (not for our free motion but) in general the
Hamiltonian written in terms of the coordinate and momentum operators
and we know that
Ĥ = ˆp2
2m + V (ˆx)
eÂe
ˆB =
eÂ+
ˆB+ 1
2[Â, ˆB]
The  and ˆB are ∝ ε, whereas the commutator is in our example ∝ ε 2 . Thus
for sufficiently small ε we can neglect the commutator. In this case (what must
206

always be guaranteed by proper lattice techniques) we have
[−iĤε/ħ
]
exp ≈ exp [ −iεˆp 2 /2mħ ] exp[−iεV (ˆx)/ħ]
and hence
< x k+1 |exp
[−iĤε/ħ ]
|x k 〉 ≈< x k+1 |exp [ −iεˆp 2 /2mħ ] [
|x k 〉 exp −iεV ( 1 ]
2 (x k+1 + x k )/ħ
The first term on RHS is known from our example of free motion eq.(214) and the
second term is a number. So we get altogether with ε = t f −t i
N
= t k+1 − t k = ∆t
the expressiion
[−iĤε/ħ [
] ( m
) ( ) 1/2 2
imε xk+1 − x k
< x k+1 |exp |x k 〉 = exp − iεV (x k ) /ħ]
2πiħε 2ħ ε
(215)
and with x 0 = x i and x N = x f we can write:
[ ]
< x f |exp −iĤ(t f − t i )/ħ |x i 〉 = lim
(216)
N→∞
⎡
∫
{
D (N) x exp ⎣ i N−1
ħ ε ∑
( ) } ⎤ 2
m xj+1 − x j
− V (x j ) ⎦
2 ε
j=0
where we have abbreviated as (we recognize N factors and N − 1 integrations)
(
) N/2
D (N) m
x =
dx 1 dx 2 ....dx N−1 (217)
2πiħ(t f − t i )/N
For vanishing potential we discover in the exponent the Lagrangean of the free
motion. If the potential is time dependent one has to take the time corresponding
to the coordinate, i.e. V (x j ) → V (x j ,t j ). The expression looks formally
difficult, however, it is easier as a working prescription: One cuts the time intervall
(t f − t i ) into many equidistant time subintervalls, at each intermediate
time t k one integrates along the x k −axis and multiplies with a weight according
to the exponent in eq.(216).
One can view this much simpler, and this is the view in practice: Suppose
you perform all these integrations simultaneously and let the x j run from −∞
to +∞, then do at a certain moment a snapshot. At this moment the snapshot
shows you a set of coordintes ¯x 1 ,...¯x N−1 between x i (t i ) and x f (t f ) which appears
to be in the looking glas a certain zick-zack-path ¯x(t). Since ε = ∆t the
sum in the exponent appears to be equal to i ħ ∆t ∑ { ( ) 2
N−1 m d¯xj
j=0 2 dt − V (¯xj )}
=
∫
i
ħ dtL(¯x(t)) =
i
ħS [¯x(t)] . Thus the action along this particular zick-zack-path
has to be calculated. See the figure for that:
The multitude of snapshots yields in the end all possible zick-zack-paths
between x i (t i ) and x f (t f ). Thus, because of the integrals between -∞ and +∞
207

one has to sum over all these paths giving each path of them in this huge sum a
weight exp ( i
ħ S[x]) . This sum over all possible zick-zack-paths is called ∫ t f
t i
Dx
. The result is the propagator, which in fact contains the full information about
the system.
Hence we can write down finally
∫ tf
[ ] i
< x f |exp [−iH(t f − t i )/ħ] |x i 〉 = lim D (N) xexp
N→∞ t i
ħ S [x] (218)
where x(t) is the path for t moving from t i −→ t f with
x(t i ) = x i
x(t f ) = x f
and the classical action (no operators)
S [x] =
∫ tf
t i
dtL(x,ẋ) =
∫ tf
t i
[ 1
dt
2 mẋ(t)2 − V (x(t))]
Presently it does not look very efficient for studying a quantum mechanical
system. However, when we consider quantized field theories, we will see, that
there are tremendous advantages of this formalism over a standard canonical
one. In fact, for non-abelian gauge theories as e.g. the QCD the path integral
formalism is the only one, which yields a quantized field theory and is
managable.
As we have seen the path integrals give us simple transition amplitudes
208

[which are often written as < x f (t f ) |x i (t i )〉]
< x f |exp [−iH(t f − t i )/ħ] |x i 〉 =
∫ xf
[ ∫ i
tf
]
= lim D (N) xexp dtL[x,ẋ]
N→∞ x i
ħ
This important result generalizes to more complicated amplitudes (without
proof), from which one can altogether extract information about our system.
We have e.g. for t i < t 1 < t 2 < t f the following identity
∫ xf
[ ] i
< x f (t f )|x(t 2 )x(t 1 ) |x i (t i )〉 = lim D (N) xx(t 2 )x(t 1 )exp
N→∞
ħ S [x] (219)
13.2.2 Propagators ( Harmonic oscillator)
x i
We will calculate another example, namely the harmonic oscillator, which is
less trivial, however one can learn more. We will learn that one can discretize
the time and will nevertheless get the proper continum limit, and that one can
extract directly observables and not only abstract quantities like propagators.
We will not do it any more in such a technical detail. There the action for the
one-dimensional harmonic oscillator is
∫ tf
[ 1
S[x] = dt
2 mẋ(t)2 − 1 ]
2 mω2 x(t) 2
t i
In order to solve the path integral (218) it is advisable to select from all possible
paths in the pathintegral only the classical path, i.e. the path which is the result
of solving the classical equations of motion, which in turn arise from demanding
δS = 0. For this particular path we have ẍ(t) + ω 2 x(t) = 0 . If we rewrite the
above S[x] by partial integration, i.e. using ∫ dtẋ 2 = xẋ − ∫ dtẍx, and if we
replace ẍ by ẍ(t) = −ω 2 x(t) we have simply
t i
S[x] = 1 2 mx(t)ẋ(t)|t f
ti
Since the classical path is known, namely
x cl (t) = x f sin[ω(t − t i )] − x i sin[ω(t − t f )]
sin(ωT)
we obtain easily
S[x cl ] =
mω [(
x
2
2sin(ωt) i + x 2 ]
f)
cos(ωT) − 2xi x f
with T = t f − t i
To calculate the full path integral it is convenient to write the paths by deviation
from the classical path: x(t) = x cl (t) + y(t) with y(t i ) = y(t f ) = 0. Thus we
obtain
∫ tf
[ 1
S[x] = S[x cl ] + dt
t i
2 mẏ(t)2 − 1 ]
2 mω2 y(t) 2
209

and therefore (please have a look at the limits of the integrals)
∫ xf
x i
[ ] [ ] ∫ i i 0
[ ] i
Dxexp
ħ S[x] = exp
ħ S[x cl] Dy exp
0 ħ S[y]
We can discretise now in the well known way following eq.(216) way modifying
slightly the potential part in a justified way for small ε = (t f − t i )/N = T/N:
∫ 0
0
[ ] i
( m
Dy exp
ħ S[y] = lim
N→∞ 2πiħε
⎡
∫ N−1 ∏
dy k exp
k=1
) N/2
⎣ 1 N−1
ħ ε ∑
j=0
{
m
2
( ) } ⎤ 2 yj+1 − y j
− 1 1 2 mω2 2 (y2 j + yj+1) 2 j
⎦
where y 0 = y N = 0. (Remember: N factors and N − 1 integrations). This
can be written in a vector and matrix form with y = (y 1 ,y 2 ,...,y N−1 ) and
M = (N − 1) · (N − 1)-matrix as
lim
N→∞
∫ 0
0
⎡
⎤
[ ] i
( m
) N/2
∫ N−1 ∏
N−1
D (N) y exp
ħ S[y] ∑
= lim
dy k exp ⎣
−m
N→∞ 2πiħε
2iħε yT My⎦
k=1 j=0
Explicitely we have
⎛
⎞
2 − ε 2 ω 2 −1 0 0 ... 0
−1 2 − ε 2 ω 2 −1 0 ... 0
0 −1 2 − ε 2 ω 2 −1 ... 0
M =
0 −1 2 − ε 2 ω 2 −1 ... 0
⎜ ... ... ... ... ... ...
⎟
⎝ 0 0 0 −1 2 − ε 2 ω 2 −1 ⎠
0 0 0 0 −1 2 − ε 2 ω 2
with M being real and symmetric, i.e. we can find a matrix A which diagonalizes
M:
from which follows
M = ADA −1 where A T A = 1 and det(A) = 1
exp [ −αy T My ] = exp [ −αy T ADA −1 y ] = exp [ −αu T Du ]
This is simple since D is diagonal with eigenvalues d i . So we√ have Gaussian
integrals over u yielding (if α were real...) a product of terms π
αd i
. Thus we
get
lim
N→∞
∫ 0
0
[ ] √
i
D (N) y exp
ħ S[y] = lim
N→∞
ε
m
2πiħεdet(M)
210

Actually the calculation of det(M) is not simple, one gets (without proof) with
the abbreviation δ = ωε the recursion
and hcnce
det(M) nxn = (2 − δ 2 )det(M) (n−1)x(n−1) − det(M) (n−2)x(n−2)
det(M) 1x1 = 2 − δ 2
det(M) 2x2 = 3 − 4δ 2 + δ 4
...
det(M) nxn = (n + 1) − δ2
6
n(n + 1)(n + 2) +
δ4
120 (n − 1)n(n + 1)(n + 2)(n + 3) + O(δ6 )
In the end we have to go with N to very large values. We use this to simplify the
above expression collecting only those terms which are leading in N,compared
to which all subleading terms are negligeable:
det(M) NxN = N − δ2
6 N3 + δ4
120 N5
We recall now that δ = ωε and rewrite this leading term using δN = ωεN = ωT
)
det(M) (N−1)x(N−1) = N
(1 − (ωT)2 + (ωT)4
6 120 + O((ωT)6 )
= N sin(ωT)
ωT
Thus we obtain finally (with N = T/ε ) for the path integral over the fluctuations
around the classical path of the harmonic oscillator
∫ 0
[ ] √ i
Dy exp
ħ S[y] mω
=
2πiħsin(ωT)
0
and thus finally
∫ f
[ ] i
K(x f ,T;x i 0) = lim D (N) xexp
N→∞ i
ħ S[x] = (220)
√
[
mω
=
2πiħsin(ωT) exp imω [(
x
2
2ħsin(ωt) i + x 2 ) ] ]
f cos(ωT) − 2xi x f
This result is exact and extremely didactical. We have obtained it by discretizing
space-time and considered its continuum limit (by using the terms leading in
N). In the present approach it was simple to do so, however when we consider
QCD the continuum limit will be a problem.
We remember that our path integral calculates the propagator K(x f t f ;x i t i ).
We can check our result by applying it e.g. to a wave function of Gaussian type
at t=0:
1
ψ(x i ,0) = √√ exp
[− (x i − ¯x) 2 ]
2πσ 4σ 2
211

yielding
ψ(x f ,T) =
∫ +∞
−∞
dx i K(x f T;x i 0)ψ(x i ,0)
Inserting our path integral eq.( 220) for the propagator we obtain again a gaussian,
which writes (apart of a phase).
[
1
ψ(x f ,T) = √√ exp − (x i − ¯x cos(ωT) 2 ]
2πσ
′ 4σ ′2
with
σ ′ = σ
√
cos 2 (ωT) + ħ2
2σ 2 sin2 (ωT)
One can easily check, that this ψ(x f ,T) satisfies indeed the Schroedinger equation
and is hence a true time evolved state from our initial gaussian packet.
13.2.3 Extraction of information, Euklidean time
By now we had integrations with imaginary exponents, which are highly oscillating.
They are not appropiate for a numerical treatment. To achieve this we
go to Euclidean time. This not only makes life easier we also will show that
one can extract directly physical information from the Euklidean correlation
functions. In fact there is some information, we can extract from the Euklidean
formulation (wave functions, transition elements, certain matrixelements), there
is another type of information (processes, cross sections) which require formulation
in Minkowski space and hence are not accessible by present time techniques.
Euklidean time is defined as
Then we have
t = −iτ (τ > 0)
〈x f |exp[−iH(t f − t i )/ħ] |x i 〉 −→ 〈x f |exp [−H(τ f − τ i )/ħ] |x i 〉 (221)
The transition from Minkowski to Euklidean space is easy, take as example the
propagtor of free motion eq.(214), which reads in Euklidean coordinatew
( ) [ ]
1/2
m
im (x f − x i ) 2
exp
2πiħ(t f − t i ) 2ħ (t f − t i )
( ) [
]
1/2
m
→
exp − m (x f − x i ) 2
2πħ(τ f − τ i ) 2ħ (τ f − τ i )
and the exponent is equal to − 1 ħ S E(x f ,τ f ;x i ,τ i ). If we have in general a Hamiltonian
with kinetic and potential terms ge get
∫ f
[
〈x f |exp [−H(τ f − τ i )]/ħ |x i 〉 = Dxexp − 1 ]
ħ S E[x]
212
i

with
S E [x] =
∫ τf
τ i
[ 1
dτ
2 mẋ(τ)2 + V (x)]
In order to see the equivalence of a continuum formulation and a lattice formulation
we will compare now the RHS with the LHS of the following formula,
where the LHS is calculated analytically (continuum) and the RHS numerically
(lattice):
∫ f
[
〈x f |exp [−H(τ f − τ i )/ħ] |x i 〉 | analyt = lim D (N) xexp − 1 ]
N→∞ i
ħ S E[x] | numerical
(222)
We take now a special case, at which we can see, how to extract physics out
of the above equations. We have with the choice x = x f = x i and T = τ f − τ i
the following formula after inserting a the complete set of eigenstates of the
Hamiltonian H:
LHS = 〈x|exp [−HT/ħ] |x〉 = ∑ [
< x|n > exp − E ]
nT
< n|x >
ħ
n,m
If we integrate over all x we obtain the famous partition function Z, which we
know from statistical mechanics:
∫
Z = dx 〈x|exp[−HT]/ħ |x〉 = ∑ [
exp − E ]
nT
ħ
n
In order to obtain explicit information about the system take now the limit
T → ∞.This makes all excited states contribute less and less with increasing T
to the partition function than the ground state. So in the end at T → ∞ only
the ground state remains in the sum:
〈x|exp[−HT]/ħ |x〉 −→ exp[−E 0 T/ħ] | < x|0 > | 2 (223)
Thus we can extract immediately the energy of the ground state E 0 and the
wave function of the ground state < x|0 > from the Euklidean form of the
correlation function.
So far the LHS, i.e. the analytic treatment of eq.(222). The numerical
treatment (RHS)goes in a well known way, τ j = τ i + ja with j = 0,1,...,N and
a = T N
. Then we have with eq.(217)
∫
D (N) x −→
( m
) N/2
∫ +∞
2πħa
−∞
...
∫ +∞
−∞
dx 1 dx 2 ...dx N−1
These are N factors and an N − 1 dimensional integration, where x j = x(τ j )
and with the choice x = x i = x f . In each intervall we have to perform the
213

integration over the Euklidean Lagrangean L E = −L in the exponential, in
order to yield the action, i.e.
∫ τj+1
∫ τj+1
[ 1
dτL E = dτ
τ j τ j
2 mẋ(τ)2 + V (x)]
≈
[ ( ) 2
1
a
2 m xj+1 − x j
+ 1 a 2 (V (x j+1) + V (x j ))]
With this we obtain the Euklidean action of the lattice
S latt [x] = − 1 ħ
N−1
∑
j=0
and hence we obtain for the RHS of eq.(222)
RHS =
∫ f
i
⎡
exp ⎣− 1 ħ
[ m
2a (x j+1 − x j ) + aV (x j )]
[
Dxexp − 1 ] ( m
) N/2
∫ +∞
ħ S E[x] =
2πħa
N−1
∑
j=0
[ m
] ⎤
2a (x j+1 − x j ) + aV (x j ) ⎦
−∞
...
∫ +∞
−∞
dx 1 dx 2 ...dx N−1
Actually we have integrals to solve with rather simple integrands, but it is a multiple
integral. There are sophisticated techniques to do multiple integrals, which
we will not discuss now. In fact these techniques correspond as well to select
all possible paths beween the fixed end points x i and x f and to integrate over
these paths weighted with the exponent of the corresponding classical action.
One might worry about approximating ẋ with (x j+1 − x j )/a in our formula
for the lattice action S latt [x]. It is not obvious that this is a good approximation
given that x j+1 − x j can be arbitrarily large in our path integral; that is,
paths can be arbitrarily rough. While not so important for our one-dimensional
problem, this becomes a crucial issue for four-dimensional field theories. It is
dealt with using renormalization theory, which we discuss in later sections.
If we consider x i = x f = x, and integrate over x then we have a formula
for the partition function, when we slightly change the definition of (217) from
a (N − 1)- dimensional integral to a N-dimensional one, i.e. N factors and
an N-dimensional integrations. Thus we obtain with this new definition of Dx
without the index N: ∫ [
Z = Dxexp − 1 ]
ħ S E[x]
13.2.4 Correlators
With the example of the harmonic oscillator we show how one can calculate
correlators, which are a bit more complicated than propagators, however, each
of them allows to extract other information from the system. By now we have
extracted from the propagator the ground state energy and the ground state
214

wave function. Now we consider (we put ħ = 1) the following correlator as
example (1-particle Euklidean Greens function)
N >= 〈x f |exp [−Hτ f ]ˆx(τ ′ )exp[+Hτ i ] |x i 〉
with a normalization factor N to be defined later. In the Heisenberg-picture we
have
ˆx(τ ′ ) = exp[Hτ ′ ]ˆx exp[−Hτ ′ ]
and hence
N >= 〈x f |exp[−H(τ f − τ ′ )]ˆx exp[−H(τ ′ − τ i )] |x i 〉
∫
= dy 〈x f |exp [−H(τ f − τ ′ )]ˆx|y >< y|exp[−H(τ ′ − τ i )] |x i 〉
∫
= dy 〈x f |exp[−H(τ f − τ ′ )]|y > y < y|exp[−H(τ ′ − τ i )] |x i 〉
Under the integral we have the product of two correlators, each of the being
represented by a path integral. Thus we obtain
∫ {∫
} {∫
}
N >= dy Dxexp[−S y−→x f
E
] y Dxexp[−S xi−→y
E
]
We can rewrite this in a familiar way, since in fact we have just added to all the
discretized paths another discrete point at the time τ ′ . Thus we obtain
∫
N >= Dxx(τ ′ )exp[−S xi−→x f
E
]
In full analogy we obtain an expression for 2-particle Greens function. We have
for τ 2 > τ 1
∫
〈x f |exp [−Hτ f ]ˆx(τ 2 )ˆx(τ 1 )exp[+Hτ i ] |x i 〉 = Dxx(τ 2 )x(τ 1 )exp[−S xi−→x f
E
] for τ 2 > τ 1
In order to extract information about the system we proceed as in the case of
the propagator. There we considered the limit of large time separations. To do
this we put x i = x f = x and define accordingly
∫
Dxx(τ2 )x(τ 1 )exp[−S E [x]]
>= ∫
Dxexp[−SE [x]]
where we have integrated also over x. This defines also the above normalization
factor
∫
N = Dxexp[−S E [x]]
More in general we define
>=
∫
DxΓ[x]exp[−SE [x]]
∫
Dxexp[−SE [x]]
215

These correlators help to extract information on the system from the path
integrals. To see this we consider the complete set of eigenstates of the Hamiltonian
(see e.g. harmonic oscillator) H|n >= E n |n > . Then we can insert
∫ completeness relations into the expression for the correlator yielding with
dx < n|x >< x|m >= δnm e.g.:
∫
dx 〈x|exp [−H(τ f − τ ′ )]ˆx exp[−H(τ ′ − τ i )] |x〉
∫
= dx 〈x|n > exp [−E n (τ f − τ ′ )] < n|ˆx|m > exp[−E m (τ ′ − τ i )] < m|x >
= ∑ n
exp[−E n (τ f − τ i )] < n|ˆx|n >
For symmetric potentials, like e.g. the harmonic oscillator, this expression is
not very interesting since due to parity reasons it vanishes.
However we obtain a very relevent expression if we take the 2-particle Greens
function:
∫
G E (τ 2 ,τ 1 ) = dx 〈x|exp [−H(τ f − τ 2 )]ˆx exp[−H(τ 2 − τ 1 )] ˆxexp[−H(τ 1 − τ i )] |x〉
= ∑ n
exp[−E n T < n|ˆx exp[−(H − E n )t]ˆx|n > with T = τ f − τ i and t = x 2 − x 1
oder
∑
n
>=
exp[−E nT] < n|ˆx exp[−(H − E n )t]ˆx|n >
∑
n exp[−E nT]
Now we can consider the limit for large T. The leading term is the one which
survives. It is
lim >= exp[−E 0T] < 0|ˆx exp[−(H − E 0 )t]ˆx|0 >
T −→∞ exp[−E 0 T]
There is some cancellation and we can write after inserting the completeness
relation
lim
T −→∞ >= ∑ m
exp[−(E m − E 0 )t < 0|ˆx|m >< m|ˆx|0 >
When we take also t −→ ∞, but always with T ≫ t. In this particular limit we
obtain
>= exp[−(E 1 − E 0 )t] |< 0|ˆx|1 >| 2 (224)
Actually from this expression one can extract immediately the energy of the
first excited state. Denote for a moment the LHS as G(t). Then one obtains
immediately
[ ] G(t)
log
≈ a(E 1 − E 0 ) (225)
G(t + a)
216

One can also obtain information on the transition matrix element < 0|ˆx|1 >, if
one inserts the energy difference into the correlator at large t.
One can redo the analysis for many different correlators als there are >, or >, or >.
All these correlators contain some specific information on the system. We should
remind, that this physical information originates from the Euklidean form of the
correlators.
13.2.5 Partition function
We know from the 1-dimensional example the partition function
Z = ∑ ∫ [
exp (−E n T) = Dx(τ) exp − 1 ]
ħ ST E[x]
n
(226)
with
S T E [x] =
∫ T
0
dτ
[
m
2
( ) 2 dx
+ V (x(τ))]
dτ
This expression can be used to illustrate several important things.
First, there is some similarity to statistical mechanics. There one implies
1
that T =
kT T emp
= β Temp , where T Temp is the temperature. In particle physics
T is the observation (Euklidean) time. In one is interested only in the ground
(or vacuum) state E 0 , one gets it by taking the temperature T Temp −→ 0 or
observation time T −→ ∞.
Second. The various paths x(τ) enter the Z with a weight given by their
action S [x(τ)] . In the classical limit, i.e. for ħ −→ 0, only the path with the
smallest action survives, i.e. the one with δS = 0. This is Hamiltons principle
and the corresponding variational Euler-equation are the Lagrange equations.
Third: For small but finite ħ one expects naively in eq.(226) that the path
with the smallest (minus sign in the exponent !) action in fact dominates the
path integral . However this is only one single path, which corresponds in the
figure just to the point (remember x 1 = x 2 ) at the bottom of the potential. If one
allows more paths which in some random way ”oscillate” around the minimum
then one has immediately lots of paths contributing to the path integral. Those
paths have actions larger than the minimal one, however since there are many
of them one deviates immediately from the classical limit. Thus one loses in
action but one gains in entropy, thats why quantum fluctuations are important.
Fourth: The paths, which contribute, e.g. for an harmonic oscillator and a
two-well harmonic oscillator can be depicted as shown in the figure (the figure
does not really show that x 1 = x 2 ). One realizes that paths extend from one
valley to the other. These paths are called ”Instantons”. They describe tunneling
processes between the valleys. They play a tremendous role in QCD, where
they are known to be responsible for the spontaneous chiral symmetry breaking.
217

13.2.6 Metropolis Formalism
We could evaluate the path integrals in 〈〈Γ[x]〉〉 using a standard multidimensional
integration code, at least for one-dimensional systems. However this is
unfeasable for realistic lattices. For eample a lattice gauge calculation in QCD
with 40 lattice points in every direction we have 4 ⋆ 10 4 so called link-variables
and due to the group SU(3) we have 81920000 real variables. That should be
intractable for conventional quadratures even in the future. Here, instead, we
employ a more generally useful Monte Carlo procedure. Noting that
∫
DxΓ[x]exp[−S[x]]
〈〈Γ[x]〉〉 = ∫
Dxexp[−S[x]]
is a weighted average over paths with weight exp(−S[x]), we generate a large
number, N cf , of random paths or configurations,
x (α) ≡ {x (α)
0 x (α)
1 ...x (α)
N−1 }
α = 1,2...N cf,
on our grid in such a particular way that the probability P[x (α) ] for obtaining
any particular path x (α) is
P[x (α) ] ∝ exp[−S[x (α) ]] (227)
218

e -S(x)
x
Then an unweighted average of Γ[x] over this particular set of paths approximates
the weighted average over uniformly distributed paths:
〈〈Γ[x]〉〉 ≈ Γ ≡ 1
N
∑ cf
Γ[x (α) ].
N cf
Γ is our “Monte Carlo estimator” for 〈〈Γ[x]〉〉 on our lattice. In fact this provides
us with a large number of points in the important regions of the integral,
improving the accuray drastically
Of course the estimate will never be exact since the number of paths N f will
never be infinite. The Monte Carlo uncertainty σ¯Γ in our estimate is a potential
source of error; it is estimated in the usual fashion:
{
σ 2¯Γ ≈ 1
N
}
1 ∑
Γ 2 [x (α) ] − Γ 2 . (228)
N cf
This becomes
N cf
α=1
α=1
σ 2¯Γ = 〈〈Γ2 〉〉 − 〈〈Γ〉〉 2
N f
for large N f . Since the numerator in this expression is independent of N f (in
principle, it can be determined directly from quantum mechanics), the statistical
uncertainties vanish as 1/ √ N f when N f increases.
We need some sort of specialized random-vector generator to create our set
of random paths x (α) with probability (227). Possibly the simplest procedure,
though not always the best, is the Metropolis Algorithm . In this procedure, we
start with an arbitrary path x (0) and modify it by visiting each of the sites on
the lattice, and randomizing the x j ’s at those sites, one at a time, in a particular
fashion that is described below. In this way we generate a new random path
from the old one: x (0) → x (1) . This is called “updating” the path. Applying the
algorithm to x (1) we generate path x (2) , and so on until we have N cf random
paths. This set of random paths has the correct distribution if N cf is sufficiently
large.
The algorithm for randomizing x j at the j h site is:
219

• generate a random number ζ, with probability uniformly distributed between
−δ and δ for some constant δ;
• replace x j → x j + ζ and compute the change ∆S in the action caused by
this replacement (generally only a few terms in the lattice action involve
x j , since lagrangians are local; only these need be examined);
• if ∆S < 0 (the action is reduced) retain this new value for x j , and proceed
to the next site;
• if ∆S > 0 accept the new value x j + ζ with probability exp(−∆S). This
means: generate a random number η unformly distributed between 0
and 1; retain the new value for x j if exp(−∆S) > η, otherwise restore
the old value; proceed to the next site. Acutally, to generate a random
number on the computer is a non-trivial task, since the computer cannot
generate a real random number. There are sophisticatead techniques to
do that, which we will not discuss here.
There are two important details concerning the tuning and use of this algorithm.
First, in general some or many of the x j ’s will be the same in two
successive random paths. The amount of such overlap is determined by the parameter
δ: when δ is very large, changes in the x j ’s are usually large and most
will be rejected; when δ is very small, changes are small and most are accepted,
but the new x j ’s will be almost equal to the old ones. Neither extreme is desirable
since each leads to very small changes in x, thereby slowing down the
220

numerical exploration of the space of all important paths. Typically δ should
be tuned so that 40%–60% of the x j ’s are changed on each pass (or “sweep”)
through the lattice. Then δ is of order the typical quantum fluctuations expected
in the theory. Whatever the δ, however, successive paths are going to be quite
similar (that is “highly correlated”) and so contain rather similar information
about the theory. Thus when we accumulate random paths x (α) for our Monte
Carlo estimates we should keep only every N cor -th path; the intervening sweeps
erase correlations, giving us configurations that are statistically independent.
The optimal value for N cor depends upon the theory, and can be found by trial.
It also depends on the lattice spacing a, going roughly as
N cor ∝ 1 a 2 .
Other algorithms exist for which N or grows only as 1/a when a is reduced, but
since our interest is in large a’s we will not discuss these further.
The second detail concerns the procedure for starting the algorithm. The
very first configuration used to seed the whole process is usually fairly atypical.
Consequently we should discard some number of configurations at the beginning,
before starting to collect x (α) ’s. Discarding 5N cor to 10N cor configurations is
usually adequate. This is called “thermalizing the lattice.”
To summarize, a computer code for a complete Monte Carlo calculation of
〈〈Γ[x]〉〉 for some function Γ[x] of a path x consists of the following steps:
• initialize the path, for example, by setting all x j ’s to zero;
• update the path 5N cor –10N cor times to thermalize it;
• update the path N cor times, then compute Γ[x] and save it; repeat N cf times.
• average the N cf values of Γ[x] saved in the previous step to obtain a Monte
Carlo estimate Γ for 〈〈Γ[x]〉〉.
A Monte Carlo estimate Γ of some expectation value 〈〈Γ〉〉 is never exact;
there are always statistical errors that vanish only in the limit where infinitely
many configurations are employed (N cf → ∞). An important part of any
Monte Carlo analysis is the estimation of these statistical errors. There is a
simple but very powerful method, called the “statistical bootstrap,” for making
such estimates.
In the previous exercises, for example, we assemble an “ensemble” of measurements
of the propagator G (α) , one for each configuration x (α) . These are
averaged to obtain G, and, from it, an estimate for ∆E n (generalization of
Eq. (224)). An obvious way to check the statistical errors on this estimate for
∆E n is to redo the whole calculation, say, 100 times, each time with different
random numbers to generate different random paths. With 100 copies of the
entire calculation, we could analyze the distribution of the 100 random ∆E n ’s
obtained, and deduce the statistical uncertainty in our original estimate. This,
however, is exceedingly expensive in computer time. The bootstrap procedure
221

provides new, almost zero-cost random ensembles of measurements by synthesizing
them from the original ensemble of N f measurements.
Given an ensemble {G (α) ,α = 1...N cf } of Monte Carlo measurements, we
assemble a “bootstrap copy” of that ensemble by selecting G (α) ’s at random
from the original ensemble, taking N cf in all while allowing duplications and
omissions. The resulting ensemble of G’s might have two or three copies of
some G (α) ’s, and no copies of others. This new ensemble can be averaged and a
new estimate obtained for ∆E n . This procedure can be repeated to generated
as many bootstrap copies of the original ensemble as we wish, and from each
we can generate a new estimate for ∆E n . The distribution of these ∆E n ’s
approximates the distribution of ∆E n ’s that would have been obtained from
the original Monte Carlo, and so can be used to estimate the statistical error in
our original estimate.
Another useful procedure related to statistical errors is “binning.” At the
end of a large simulation we might have 100’s or even 100,000’s of configurations
x (α) , and for each a set of measurements like G (α) , our propagator. The
measurements will inevitably be averaged, but we want to save the separate
G (α) ’s for making bootstrap error estimates and the like. We can save a lot of
disk space, RAM, and CPU time by partially averaging or binning the measurements:
For example, instead of storing each of
we might instead store
G (1) G (2) G (3) G (4) G (5) ...
G (1) ≡ G(1) + G (2) + G (3) + G (4)
4
G (2) ≡ G(5) + G (6) + G (7) + G (8)
4
...
The G (β) ’s are far less numerous but have the same average, standard deviation,
and other statistical properties as the original set. Typically the bin size is
adjusted so that there are only 50–100 G (β) ’s.
13.3 Boson Quantum Field on the lattice
13.3.1 Quantum Field Theory with functional integrals
Now we are going to translate the above representation of quantum mechanics
in terms of path integrals to field theory. We consider a scalar field φ(x), where
x = (⃗x,t) labels space-time coordinates, and the time evolution of φ(⃗x,t) is
given by
φ(⃗x,t) = e iHt φ(⃗x,t = 0)e −iHt .
The objects of interest in field theory are vacuum expectation values of (time
ordered) products of field operators, i.e. the Greens functions:
〈0|φ(x 1 )φ(x 2 )...φ(x n )|0〉, t 1 > t 2 > · · · > t n .
222

Prominent examples are propagators
〈0|φ(x)φ(y)|0〉.
The Greens functions essentially contain all physical information. In particular,
S-matrix elements are related to Greens functions, e.g. the 2-particle scattering
elements can be obtained from
〈0|φ(x 1 )...φ(x 4 )|0〉.
Instead of discussing the functional integral representation for quantum field
theory from the beginning, we shall restrict ourselves to translating the quantum
mechanical concepts to field theory by means of analogy. To this end we would
like to translate the basic variables x i (t) into fields φ(⃗x,t). The rules for the
translation are then
x i (t) ←→ φ(⃗x,t)
i ←→ ⃗x
∏
dx i (t) ←→ ∏ dφ(⃗x,t) ≡ Dφ
t,i
t,⃗x
∫
∫
S = dt L ←→ S = dtd 3 x L,
where S is the classical action.
For scalar field theory we might consider the following Lagrangian density
(φ 4 -theory):
L = 1 2
(
( ˙φ(x)) 2 − (∇φ(x)) 2) − m2 0
2 φ(x)2 − g 0
4! φ(x)4
= 1 2 (∂ µφ)(∂ µ φ) − m2 0
2 φ(x)2 − g 0
4! φ(x)4 .
The mass m 0 and coupling constant g 0 bear a subscript 0, since they are bare,
unrenormalized parameters. This theory plays a role in the context of Higgs-
Yukawa models, where φ(x) is the Higgs field.
In analogy to the quantum mechanical path integral we now write down
a representation of the Greens functions in terms of what one calls functional
integrals:
〈0|φ(x 1 )φ(x 2 )...φ(x n )|0〉 = 1 ∫
Dφ φ(x 1 )φ(x 2 )...φ(x n )e iS
Z
with
∫
Z =
Dφ e iS .
These expressions involve integrals over all classical field configurations.
As mentioned before, we do not attempt any derivation of functional integrals
but just want to motivate their form by analogy. Furthermore, in the case
223

of quantum mechanics we considered the transition amplitude, whereas now we
have written the formula for Greens functions, which is a bit different.
The formulae for functional integrals give rise to some questions. First of all,
how does the projection onto the groundstate |0〉 arise Secondly, these integrals
contain oscillating integrands, due to the imaginary exponents; what about their
convergence Moreover, is there a way to evaluate them numerically
In the following we shall discuss, how the introduction of imaginary times
helps in answering these questions.
Procedure: Summary In fact the task to compute observables on a lattice
inolves the following steps
i) Any renormalizable quantum field theory, such as QCD requires an ultraviolet
(short distance) regularization in order to eliminate infinities from calculated
physical quantities. Here this cutoff procedure is introduced by defining
the fields on a mesh of discrete lattice points which replaces the space-time
continuum. Ultimatly one must ensure that the dependence of any calculated
observable on the lattice spacing satisfies the appropriate renormalization group
equation. That is, the calculated values of physical observables should scale correctly
with a −→ 0. This meas, if one identifies the inverse lattice spacing a −1
with the UV-cut-off Λ cutoff , then the calculated values should have the proper
Λ cutoff -dependence given by renormalization group equations.
ii) An infrared (long-wavelength) cutoff is imposed by working on a finite
lattice volume. This is a techical necessity in order to be able to perform the
numerical calculation of the integrals, which unavoidably requires a finite number
of variables. Again, one must extrapolate the value of any observable to the
infinite volume limit.
iii) In general the functional integrals in Minkowski space are complex and oscillate
rapidly, so that they cannot be treated numerically in this form. In order
to provide a sound basis, both for formal mathematical reasons and for practical
numerical purposes, it is normal to calculate the path integrals for imaginary
times: t = x 0 = −ix 4 = iτ. This change of variables, together with the appropriate
analytic continuation of the Green functions, defines a Euclidean quantum
field theory: that is, the fields now live on a four dimensional Euclidean space.
Lattice QFT is then represented in terms of well defined functional integrals
taken over the Euclidean lattice hypercube Λ = { x; xµ a
∈ Z;µ = 1,2,3,4}
iv) In practice all these limits have to be studied and justified.
v) In fact there are even more problems if one wants to calclate e.g. nucleon
properties in terms of QCD: There are so called fermion doublings which are
caused by the fact that on a discrete lattice automatically several fermions
are calculated simultaneously and special techniques have to be developed to
minimize the errors caused by that. Furthermore present day lattices have
tremendous problems using quarks on the lattice which have physical values
of their masses. In practice all the used quark masses are too large and one
has to think about extrapolation methods towards physical values of current
quark masses. Very often not full lattice calculations involving fermions are
224

t
t t 1
2
performed but so-called quenched calculations. For large fermion masses they
seem to be alright, however for values closer to their pysical value the quenched
approximation does not seem to work well.
13.3.2 Euklidean Field Theory
Let us return to quantum mechanics for a moment. Here we have introduced
Greens functions, e.g.
G(t 1 ,t 2 ) = 〈0|ˆx(t 1 )ˆx(t 2 )|0〉, t 1 > t 2 .
and we have demonstfated that these Greens functions are related to quantum
mechanical amplitudes at imaginary times by analytic continuation. We had
expressed the Greens function at imaginary times,
G E (τ 1 ,τ 2 ) = 〈0|ˆxe −H(τ1−τ2)ˆx|0〉,
can be expressed as a path integral
G E (τ 1 ,τ 2 ) = 1 ∫
Z
Dx x(τ 1 )x(τ 2 )e −SE ,
where
∫
Z =
Dx e −SE
The Greens function at real times, which we were interested in originally,
can be obtained from G E by means of analytical continuation, G(t 1 ,t 2 ) =
G E (it 1 ,it 2 ). The analytical continuation has to be done in such a way that
all time arguments are rotated simultaneously counter-clockwise in the complex
t-plane. This is the so-called Wick rotation, illustrated in thei Fig..
Now we turn to field theory again. The Greens functions
G(x 1 ,...,x n ) = 〈0|Tφ(x 1 )...φ(x n )|0〉,
225

continued to imaginary times, t = −iτ, are the so-called Schwinger functions
G E ((⃗x 1 ,τ 1 ),...,(⃗x n ,τ n )) = G((⃗x 1 , −iτ 1 ),...,(⃗x n , −iτ n )).
In analogy to the quantum mechanical case their functional integral representation
reads
G E (x 1 ,...,x n ) = 1 ∫
Dφ φ(x 1 )...φ(x n )e −SE
Z
with
∫
Z =
Dφ e −SE
and e.g. in the above example of φ 4 -theory:
∫ {
S E = d 3 1
xdτ
2
∫
=
( ) }
2 dφ
+ 1 dτ 2 (∇φ)2 + m2 0
2 φ2 + g 0
4! φ4
{ 1
d 4 x
2 (∂ µφ) 2 + m2 0
2 φ2 + g }
0
4! φ4 . (229)
As can also be seen from the kinetic part contained in S E , the metric of
Minkowski space
−ds 2 = −dt 2 + dx 2 1 + dx 2 2 + dx 2 3
has changed into
dτ 2 + dx 2 1 + dx 2 2 + dx 2 3,
which is the metric of a Euclidean space. Therefore one speaks of Euclidean
Greens functions G E and of Euclidean functional integrals. They are taken
as starting point for non-perturbative investigations of field theories and for
constructive studies.
As S E is real, the integrals of interest are now real and no unpleasant oscillations
occur. Moreover, since S E is bounded from below, the factor exp(−S E )
in the integrand is bounded. Strongly fluctuating fields have a large Euclidean
action S E and are thus suppressed by the factor exp(−S E ). (Strictly speaking,
this statement does not make sense in field theory unless renormalization
is taken into account.) This makes Euclidean functional integrals so attractive
compared to their Minkowskian counterparts.
To illustrate the coordinate transformation to imaginary time, there is a
little exercise. Consider the Feynman propagator and it is easy to show that
∆ E F (x) =
∫ d 4 p
(2π) 4
e ipx
p 2 + m 2 ,
0
(where px is to be understood as a Euclidean scalar product), is obtained by
correct Wick rotation. To be more precise,
∆ F (⃗x,t) = lim
φ→π/2 ∆E F (⃗x,te iφ ),
226

p 2 + m 2
p 0
p
4
with ∆ F the Feynman propagator in Minkowski-space
∫ d 4 p e −ip∗x
∆ F (⃗x,t) = i
(2π) 4 p 2 − m 2 0 + iɛ,
where all scalar products in the last expression are defined with Minkowski
metric. An important feature of the Wick-rotated propagator is the absence of
singularities on the p 4 -axis in Euclidean space, see Fig.
One might think that in the Euclidean domain everything is unphysical and
there is no possibility to get physical results directly from the Euclidean Greens
functions. But this is not the case, as we might guess already from the simple
quantum mechanical exerzises. For example, the spectrum of the theory can be
obtained in the following way. Let us consider a vacuum expectation value of
the form
〈0|A 1 e −Hτ A 2 |0〉,
where the A i ’s are formed out of the field φ, e.g. A = φ(⃗x,0) or A = ∫ d 3 x φ(⃗x,0).
Now, with the familiar insertion of a complete set of energy eigenstates, we have
〈0|A 1 e −Hτ A 2 |0〉 = ∑ n
〈0|A 1 |n〉e −Enτ 〈n|A 2 |0〉.
In case of a continuous spectrum the sum is to be read as an integral. On the
other hand, representing the expectation value as a functional integral leads to
∫
1
Dφ e −SE A 1 (τ)A 2 (0) = ∑ 〈0|A 1 |n〉〈n|A 2 |0〉e −Enτ .
Z
n
This is similar to the ground state projection at the beginning of this chapter.
For large τ the lowest energy eigenstates will dominate the sum and we can thus
227

obtain the low-lying spectrum from the asymptotic behaviour of this expectation
value. One should note, that the excited states of a field theory are the
1-particle-2-particle states, which are theFock-states compose by the creation
operators of the quantized field operator. Inorder to obtain the mass of the
lowest lying excited state, i.e. the lightest particle in rest, one has to choose
A 1 ,A 2 suitably, e.g. for
∫
A ≡ A 1 = A 2 = d 3 x φ(⃗x,0),
such that 〈0|A|1〉 ≠ 0 for a one-particle state |1〉 with zero momentum ⃗p = 0
and mass m 1 , we will get
∫
1
Dφ e −SE A(τ)A(0) = |〈0|A|1〉| 2 e −m1τ + ...,
Z
which means that we can extract the mass of the lightest particle.
From now on we shall remain in Euclidean space and suppress the subscript
E, so that S ≡ S E means the Euclidean action.
Lattice discretization
One central question still remains: does the infinite dimensional integration
over all classical field configurations, i.e.
Dφ = ∏ x
dφ(x),
make sense at all How is it defined
13.3.3 Scalar boson field: Discretization of space-time
Remember the way we derived the path integral representation of quantum
mechanics. It was obtained as a limit of a discretization in time τ. As in field
theory the fields depend on the four Euclidean coordinates instead of a single
time coordinate, we may now introduce a discretized space-time in form of a
lattice, for example a hypercubic lattice, specified by
see Fig.
x µ = an µ , n µ ∈ Z,
The quantity a is called the lattice spacing for obvious reasons. The scalar
field
φ(x), x ∈ lattice, −→ φ j
is now defined on the lattice points only. Partial derivatives are replaced by
finite differences,
∂ µ φ −→ ∆ µ φ(x) ≡ 1 a (φ(x + aˆµ) − φ(x)) = 1 a
(
φj+1µ − φ j
)
,
228

and
∂ µ ∂ µ φ −→ ∑ µ
1 [ ]
a 2 φj+1µ − 2φ j φ j − φ j−1µ
and space-time integrals by sums:
∫
d 4 x −→ ∑ x
a 4 = a 4 ∑ j
.
The action of our discretized φ 4 -theory, Eq. (229), can be written as
S = ∑ {
}
a 4 1
4∑
(∆ µ φ(x)) 2 + m2 0
2
2 φ(x)2 + g 0
4! φ(x)4 .
x
µ=1
In the functional integrals the measure
Dφ = ∏ x
dφ(x) −→ ∏ j
dφ j
involves the lattice points x only. So we have a discrete set of variables to
integrate. If the lattice is taken to be finite, we just have finite dimensional
integrals.
Discretization of space-time using lattices has one very important consequence.
Due to a non-zero lattice spacing a cutoff in momentum space arises.
The cutoff can be observed by having a look at the Fourier transformed field
˜φ(p) = ∑ x
a 4 e −ipx φ(x).
The Fourier transformed functions are periodic in momentum-space, so that we
can identify
p µ
∼ = pµ + 2π a
229

and restrict the momenta to the so-called Brillouin zone
− π a < p µ ≤ π a .
The inverse Fourier transformation, for example, is given by
φ(x) =
We recognize an ultraviolet cutoff
∫ π/a
−π/a
d 4 p
(2π) 4 eipx ˜φ(p).
|p µ | ≤ π a .
Therefore field theories on a lattice are regularized in a natural way.
In order to begin in a well-defined way one would start with a finite lattice.
Let us assume a hypercubic lattice with length L 1 = L 2 = L 3 = L in every
spatial direction and length L 4 = T in Euclidean time,
x µ = an µ , n µ = 0,1,2,...,L µ − 1,
with finite volume V = L 3 T. In a finite volume one has to specify boundary
conditions. A popular choice are periodic boundary conditions
φ(x) = φ(x + aL µ ˆµ),
where ˆµ is the unit vector in the µ-direction. They imply that the momenta are
also discretized,
p µ = 2π a
l µ
L µ
with l µ = 0,1,2,...,L µ − 1,
and therefore momentum-space integration is replaced by finite sums
∫ d 4 p
(2π) 4 −→ 1 ∑
a 4 L 3 .
T
l µ
Now, all functional integrals have turned into regularized and finite expressions.
Of course, one would like to recover physics in a continuous and infinite
space-time eventually. The task is therefore to take the infinite volume limit,
L,T −→ ∞,
which is the easier part in general, and to take the the continuum limit,
a −→ 0.
Constructing the continuum limit of a lattice field theory is usually highly nontrivial
and most effort is often spent here.
230

The Eucclidean continuum action of the Klein-Gordon-Field is given by
S E [φ] = 1 ∫
d 4 xφ(x) ( −∂ µ ∂ µ + M 2) φ(x)
2
If we define
˜φ j = aφ j
˜Mj = aM
then we get
with
S E = 1 ∑
˜φ j K jk ˜φk
2
jk
[ ]
δj+1µ,k − 2δ jk + δ j−1µ,k + ˜M2 δ jk (230)
K jk = − ∑ µ
An importante technique in the quantum field theory is to use the generating
functional, since by functional derivatives one can obtain the greens functions :
∫ [ ∫ ]
Z [J] = Dφexp −S E [φ] + d 4 xJ(x)φ(x) (231)
which reads in the discretized case
⎡
∫
Z [J] = Π l d˜φ l exp ⎣− 1 ∑
˜φ j K jk ˜φk + ∑ 2
j
jk
J j ˜φj
⎤
⎦
and now the Greens function becomes just
∫ [
Πl d˜φ l ˜φr ˜φs exp
∑jk ˜φ
]
j K jk ˜φk
< 0|˜φ r ˜φs | >=
∫
Πl d˜φ l exp[
− 1 2
− 1 2
∑ ˜φ
] = ∂2 Z [J]
jk j K ∂J jk ˜φk r ∂J s Z [0] | J=0
In order to solve functional integral we remember the harmonic oscillator formulae,
which looked identical. There is indeed no difference, since the integral
over the fields is here reduced to an integral over the magnitude of the field at
a certain space-time point. Thus we obtain analogously to eq.() the integral
⎡
⎤
∫
Π l d˜φ l exp⎣− 1 ∑
˜φ j K jk ˜φk + ∑ (√ )
⎡
⎤
N
2π
J j ˜φj ⎦ = √ exp ⎣− 1 ∑
J j (K −1 ) jk J k
⎦
2
jk
j det K 2
jk
(232)
The proof of eq.(232) is simple: We can write the LHS as
∫ [
LHS = Π l d˜φ l exp −
2−→˜φ 1 T K −→˜φ ] ∫ [
−→ + J
−→˜φ = Π l d˜φ l exp −
2−→˜φ 1 T ADA −→˜φ ]
−→ + J
−→˜φ
231

since K is real and symmetric and we have then det(A) = 1. We define −→ u =
A −1−→˜φ and we can rewrite this expression as
∫ [
LHS = Π l du l exp −
2−→ 1 u T D −→ u + −→ J T A −→ ]
u
We use now the well known expression
∫ ∞
−∞
du exp ( −αu 2 + βu ) =
√ ( ) π
a exp − β2
4α
and obtain immediately eq.(232).
From second order derivatives with respect to J k of eq.(232) one obtains
immediately
< 0|˜φ r ˜φs | >= ( K −1) rs
The matrix K −1 has to be calculated on the lattice. The best is to do this in
momentum space. There we have the representation of the delta-function
δ nm =
∫ +π
−π
d 4ˆk [ ]
(2π) 4 exp iˆk(n − m)
where ˆk = (ˆk 1 , ˆk 2 , ˆk 3 , ˆk 4 ) = k/a is dimensionless. Thus we have to calculate the
K(ˆk) of
∫ +π
d 4ˆk [ ]
K nm =
(2π) 4 K(ˆk)exp iˆk(n − m)
−π
Starting from eq.(230) a straightforward and explicit but slightly complicated
calculation yields
[ ]
δj+1µ,k − 2δ jk + δ j−1µ,k + ˜M2 δ jk
K jk = − ∑ µ
=
and we find from
then also
∫ +π
−π
(K −1 ) rs =
d 4ˆk
[
(2π) 4 4
) ]
4∑
(ˆkµ
sin 2 +
2
˜M
[ ]
2 exp iˆk(j − k)
µ=1
∑
K rq (K −1 ) qs = δ rs
q
∫ +π
−π
d 4ˆk
(2π) 4
[ ]
exp iˆk(r − s)
4 ∑ 4
µ=1 sin2 ( ˆkµ
2
)
+ ˜M 2 (233)
It is useful and easy to check that this expression on the lattice has the proper
continuum limit. To show this we go back to dimensional quantities, defining
r = x/a and s = y/a and ˜M = aM and k = ˆk/a then we have
(K −1 ) rs = a 4 ∫ +π/a
−π/a
d 4 k
(2π) 4
exp [ik(x − y)]
4 ∑ 4
µ=1 sin2 ( ak µ
2
)
+ a 2 M 2
232

In the continuum limit a → 0 we can expand the sinus and retain only the first
term. This yields then
< 0|˜φ r ˜φs | >= ( K −1) rs → a2 ∫ +∞
−∞
d 4 k exp[ik(x − y)]
(2π) 4 k 2 + M 2
Except of the normalization factor a 2 this is the well known expression for the
Greens function in Euklidean space.
13.4 Fermion Quantum fields on the lattice
First we consider free massive fermions without any gauge field. The free Dirac
equation reads
[iγ µ ∂ µ − M]ψ(x) = 0
or [
iγ 0 ∂ t + i −→ γ −→ ]
∇ − M ψ(x) = 0
Their Minkowsky-action is given by
∫
S F = d 4 x ¯ψ(x)(iγ µ ∂ µ − M) ψ(x)
Again it is convenient to go to Euklidean space, t → −iτ, i.e. changing from
iγ 0 ∂ t = −γ 0 ∂ τ . Thus the Dirac equation reads
[
−γ 0 ∂ τ + i −→ γ −→ ]
∇ ψ(x) = 0
For a consistent writing it is useful to introduce Euklidean gamma matrices:
γ E 0 = γ 0 γ E 1 = −iγ 1 γ E 2 = −iγ 2 γ E 3 = −iγ 3
which satisfy
{
γ
E
µ ,γν
E }
= 2δµν
Then the Euklidean Dirac equation reads
[
−γ E 0 ∂ τ − −→ ∇ −→ γ E] ψ(x) = 0
and the Euklidean action is
∫ ( 4∑
)
SE F = d 4 x ¯ψ(x) γµ E ∂ µ + M ψ(x)
µ=1
From now on we will drop the index E and will always work in the Euklidean
space. The Greens functions are given by path integrals as (α,β,... are Dirac
indices)
∫
DψD ¯ψψα (x).....ψ β (x)......exp [ [
−SE
F ψ, ¯ψ]]
< 0|ψ α (x).....ψ β (x)......|0 >= + ∫ [ [
DψD ¯ψ exp −S
F
E
ψ, ¯ψ]]
233

However, one should notice, due to the anticommutation rules of the fermion
fields the path integral must be formulated in a special way. We descretize again
∫
x µ → j µ a d 4 x → a ∑ 4 j
and
and
ψ α (x) → ψ αj
¯ψα (x) → ¯ψ αj j=Euklidean 4-vector
∂ µ ψ α (x) → ψ α,j+1 µ−ψ α,j−1µ
2a
D ¯ψ(x)Dψ(x) → ∏ ∏
¯ψ αj ψ βk
α,j
Actually the ψ αj , ¯ψ αj are Grassmann variables, which takes into account that
the fermionic quantum fields are anticommuting. Its properties will be discussed
now.
13.4.1 Grassmann-algebra
The reasoning is as follows: First remember scalar fields in the continuum.
Classical fields are just ordinary functions and satisfy
[φ(x),φ(y)] = 0,
which can be considered as the limit → 0 of the quantum commutation relations.
Fermi statistics implies that fermionic quantum fields have the well-known
equal-time anticommutation relations
β,k
{ψ(⃗x,t),ψ(⃗y,t)} = 0.
Motivated by this, we might introduce a classical limit in which classical fermionic
fields satisfy
{ψ(x),ψ(y)} = 0
for all x,y. Classical fermionic fields are therefore anticommuting variables,
which are also called Grassmann variables.
We would like to point out that the argument above is just a heuristic
motivation. More rigorous approaches can be found in the literature.
In general, a complex Grassmann algebra is generated by elements η i and
¯η i , which obey
{η i ,η j } = 0
{η i , ¯η j } = 0
{¯η i , ¯η j } = 0.
234

An integration of Grassmann variables can be defined by
∫
dη i (a + bη i ) = b
for arbitrary complex numbers a,b.
In fermionic field theories we have Grassmann fields, which associate Grassmann
variables whith every space-time point. With the above rules of integration
and differentiation one can easily derive the important result
⎡ ⎤
∫
∏ N N∑
fermion fields: d¯η k dη k exp ⎣− ¯η i F ij η j
⎦ = det(F) (234)
k=1
which is to be compared to
⎡
∫
∏ N
boson fields: du k exp ⎣−
k=1
N∑
i,j=1
i,j=1
⎤
u i F ij u j
⎦
1
∼ √ (235)
det(F)
Thus the difference between Fermion- and Boson fields is just the power of
det(F).
13.4.2 Fermionic path integral
We write down now the Fermionic action in descretized form and proceed similarly
to the case of bosonic fields.
∫ ( 4∑
)
SE F = d 4 x ¯ψ(x) γµ E ∂ µ + M ψ(x)
µ=1
We rescale
∂ µ ψ α (x) → ψ α,j+1 µ−ψ α,j−1µ
2a
˜ψ α,j = a 3/2 ψ αj
˜¯ψα,j = a 3/2 ¯ψαj
˜M = aM
and we obtain the discretized action
S F E = ∑
and
α,β,j,k
˜¯ψ αj K αβ
jk ˜ψ βk
4∑
K αβ
jk = 1
2 (γ [ ]
µ) αβ δj+1µ,k − δ j−1µ,k + ˜Mδjk δ αβ
µ=1
With the experience we have, we can do now the path integral. We have to go
to the same steps as we did in the scalar field theory case and determine the
235

Greens functions. If we do this we find expressions very similar to those we had
before:
< 0| ˜ψ α,r˜¯ψβ,s |0 >= ( K −1) αβ
rs
the inverse matrix is defined by
∑
qλ
K αλ
rq
(
K
−1 ) λβ
qs = δ rsδ αβ
r,s=Euklidean 4-index
We can calculate it by going to the Fourier space and we find for our fermion
field
[
∫ +π
(K −1 ) αβ d 4ˆk −i ∑ 4
µ=1 γ µ sin(ˆk µ ) + ˜M
]
[ ]
αβ
rs =
−π (2π) 4 ∑ (ˆkµ )
4
µ=1 sin2 + ˜M
exp iˆk(r − s) (236)
2
compared to the expression of the scalar field eq.(233)
[ ]
∫ +π
(K −1 d 4ˆk exp iˆk(r − s)
) rs =
(2π) 4
−π
4 ∑ 4
µ=1 sin2 ( ˆkµ
2
)
+ ˜M 2
Although both expressions for fermions and bosons look rather similar there is a
fundamental difference. Actually in the continuum limit the fermion expression
(236) goes not into the usual fermion Greens function. which is to be seen if
one compares the denominators. If we go back for (236) to non-scaled variables
we have the following relevant expressions
sin (ak µ )
a
(fermion)
vs.
( )
2sin a kµ 2
a
(boson)
The argument of the sine-function of the boson case is only half of that of the
fermion case. As we shall see immediately, this makes a big difference and is the
origin of the so called ”fermion doubling” problem, which has hampered lattice
calculations with fermions for decades and still provides problems. The relevant
intervall is [−π,+π] and we can plot it
Consider a case with M = 0, which is plotted in the figure.Then the propagator
of the field should have a zero at k = 0, sinc the pole of the propagator
indicates the mass of the particle (pol mass). The bosonic case has indeed only a
zero at k = 0 representing the physical particle. This pole exists in the fermionic
case as well. However there are poles also at the edges of the Briolloin zone at
k µ = ± π a
due to the periodicity of the denominator. These two poles correspond
to a particle with mass ( π
a) 2
and do not correspond to a physical particle. So
the discretized version of the fermionic action corresponds (in four dimensions)
to 2 4 = 16 particles rather than 1 particle. This problem is called euphemistically
”fermion doubling”. These 16 fermions are pure discretization artefacts,
because of which the fermion theory does not have a proper continuum limit, if
one does not modify it properly.
236

13.4.3 Wilson fermions, quenched lattice. etc.
There have several remedies been considered to get rid of the problem of fermion
doubling. The first relevant one was the introduction of Wilson femions. Wilson
added an ”irrelevant” term to the fermion action, which cured the problem i
the continuum limit.
S F,Wilson = S F − ar
2
∫
d 4 x ¯ψ(x)∂ µ ∂ µ ψ(x)
with r a dimensionless quantity. The discretized version of the action becomes
now
S F,Wilson = ∑ ˜¯ψ αj W αβ ˜ψ jk βk (237)
with
W αβ
jk = 1 2
4∑
µ=1
α,β,j,k
[ ]
(γµ − r) αβ δ j+1µ,k − (γ µ + r) αβ δ j−1µ,k +( ˜M +4r)δjk δ αβ (238)
The corresponding inverse matrix is now
[
∫ +π
(W −1 ) αβ d 4ˆk −i ∑ 4
µ=1 γ µ sin(ˆk µ ) + ˜M(ˆk)
]
[ ]
αβ
rs =
−π (2π) 4 ∑ (ˆkµ )
4
µ=1 sin2 + ˜M(ˆk)
exp iˆk(r − s)
2
with
˜M(ˆk) = ˜M + 2r ∑ µ
sin 2 (ˆk µ
2 )
237

going to unscaled quantities with ˜M = aM and k = ˆk/a we get
M(k) = M + 2r
a
∑
µ
sin 2 ( ak µ
2 )
In the continuum limit a → 0 the second term vanishes for any k µ ≠ π a as
one sees at the expansion in powers of akµ
2
,where all powers are high enough
to compensate the 1/a in front. In these cases M(k) approaches the given M
for vanishing a. However the mass M(k) diverges for k µ = π a
where for all a
the sinus gives a finite value, i.e. one. Thus the doublers at the border of the
Brillouin zone become infinite heavy in the continuum limit, so that they do not
disturb the rest of the lattice. At a first glance this looks nice, there are however
problems with these Wilson-Femions. For a genuin vanishing fermion mass the
original fermion action was chirally symmetric. This is no longer the case due
to the additional Wilson term, since this one is proportional to ¯ψ(x)∂ µ ∂ µ ψ(x)
which behaves like a mass term under chiral transformations, and we know, that
mass terms destroy chiral symmetry. Thus the use of Wilson fermions is not
very suitable for problems where chiral symmetry is important. Unfortunately
all low energy hadronic physics is of this sort. Furthermore Wilson fermions
show large discretization errors, which however in practice can be compensated
by adding further terms.
There are also other and more advanced suggestions to cure the doubling
problem: Kogut-Susskind fermions (or staggered fermions), and Ginsparg-Wilson
fermions (or domain wall fermions). We will not discuss these points.
13.5 Gauge Fields on the Lattice
We derive the formalism first for abelian gauge fields and then generalize in a
straight forward way to non-abelian ones.
13.5.1 Abelian gauge fields: QED
We have been considering simple scalar boson fields, then fermion fields, and
now we consider a field theory which has a gauge degree of freedom. First we
take an abelian gauge theory and repeat quickly the basics. The action for a
free Dirac field is ∫
S F = d 4 xψ(x)[iγ µ ∂ µ − M]ψ(x)
it becomes invariant under local U(1) transformations by introducing a fourvector
potential A µ (x) changing to the covariant derivative
D µ ψ(x) = (∂ µ + ieA µ (x))ψ(x)
THen we obtain for the action
∫
S F = d 4 xψ(x)[iγ µ D µ − M]ψ(x)
238

which is now invariant under simultaneous transformations
with
Under this transformation we have
ψ(x) → ψ ′ (x) = G(x)ψ(x)
¯ψ(x) → ¯ψ ′ (x) = ψ(x)G −1 (x)
A µ (x) → A ′ µ(x) = A µ (x) + i e ∂ µα(x)
G(x) = exp[−iα(x)]
D µ → G(x)D µ G −1 (x)
We need also the action of the kinetic term of the gauge field
S A = − 1 ∫
d 4 xF µν F µν
4
with
F µν = ∂ µ A ν − ∂ ν A µ
which behaves under the gauge transformation as
F µν → F ′ µν = G(x)F µν G −1 (x)
Going to Euklidean time we get
∫ [ 4∑
]
SE F = d 4 xψ(x) γ µ D µ + M ψ(x)
µ=1
S A E = + 1 4
∫
d 4 x
4∑
µ,ν=1
F µν F µν
So far we have repeated simple abelian gauge theory (in the continuum limit)
rewritten in the Euklidean space.
To formulate a gauge theory on the lattice one does not proceed by taking
the above Lagrangean and place all the fields on the lattice points. Instead one
proceeds in the way that one takes the discretized Fermion theory and gauges
it directly in the discretized form, i.e. on the lattice. Thus, in order to obtain
gauge ivariance we are led to modify the expression
S F,Wilson = ∑ j
( ˜M + 4r)˜¯ψj ˜ψj + 1 2
∑
j
4∑
˜¯ψ j (γ µ − r) ˜ψ j+1µ − ˜¯ψj+1µ (γ µ + r) ˜ψ j
µ=1
(239)
There we realize immediately something special: Consider the term ˜¯ψj (γ µ −
r) ˜ψ j+1µ , which is corresponding to something like ¯ψ(x)ψ(y) . This is in fact a
non-local product which, however, is not gauge invariant, but transforms as
¯ψ(x)ψ(y) → ¯ψ ′ (x)ψ ′ (y) = ¯ψ(x)G −1 (x)G(y)ψ(y)
239

C
x
y
Actually such terms will occur, if we write down a kinetic term in lattice field
theory. Therefore, we need matrices U(x,y) ∈ SU(N) which transform as
U(x,y) −→ G(x)U(x,y)G −1 (y), such that ¯ψ(x) · U(x,y)ψ(y) would be invariant.
There is a solution to this problem. Take a arbitrary path C from x to y
and define
U(x,y; C) ≡ exp
[
ie
∫ y
where the integral is taken along the path C.
x
] [∫ sy
A µ (z)dz µ = exp dsA µ (c(s)) dc ]
µ(s)
, (240)
s x
ds
Then U(x,y; C) transforms as desired, i.e. it is inself an element of the gauge
group U(1), i.e. it transforms under the gauge transformation of the A-field as
since
U ′ (x,y; C) = exp{ie
or (q.e.d.):
U(x,y) → U ′ (x,y) = G(x)U(x,y)G −1 (y)
∫ y
x
[A µ (z)+ 1 e ∂ µα(z)]dz µ } = exp{ie
U ′ (x,y; C) = exp [−iα(x)] exp{ie
∫ y
Due to this it fulfills the goal that we have the invariance
x
∫ y
x
A µ (z)dz µ } exp [+iα(y)]
A µ (z)dz µ + iα(y) − iα(x)}
¯ψ(x)U(x,y)ψ(y) → ¯ψ(x)G −1 (x)G(x)U(x,y)G −1 (y)G(y)ψ(y) = ¯ψ(x)U(x,y)ψ(y)
The expression in eq.(240) is called gauge link or parallel transporter. The
latter expression is in analogy to similar objects in differential geometry, which
map vectors from one point to another along curves. The parallel transporters
depend not only on the points x and y but also on the chosen curve C. They
obey the composition rule
U(x,y; C) = U(x,u; C 1 ) · U(u,y; C 2 ),
240

where the path C is split into two parts C 1 and C 2 .
We will apply the above gauge procedure to the Wilson action for Fermions
eq.(239) based on (237,238). Thus we change
˜¯ψ j (γ µ − r) ˜ψ j+1µ
˜¯ψ j+1µ (γ µ + r) ˜ψ j
→
→
˜¯ψ j (γ µ − r)U(j,j + 1 µ ) ˜ψ j+1µ
˜¯ψ j+1µ (γ µ + r)U(j + 1 µ ,j) ˜ψ j
with U(j +1 µ ,j) = U(j,j +1 µ ) † . Hence the discretized Fermionic action is now
invariant under simultaneous transformations
˜ψ j → G j ˜ψj (241)
˜¯ψ j → ˜¯ψj G −1
j
U(j,j + 1 µ ) → G j U(j,j + 1 µ )G −1
j+1 µ
U(j + 1 µ ,j) → G j+1µ U(j + 1 µ ,j)G −1
j
where G j is an arbitrary local gauge transformation specified at each lattice
site j (being a 4-vector). Thus the Fermionic action is written in terms of
˜¯ψ j , ˜ψ j ,U(j,j +1 µ ),U(j +1 µ ,j). Obviously the fermion fields are defined at the
lattice sites, wheras the gauge fields are so called ”link variables”, which are
definde on links connecting the lattice sites. We can picture them
It is useful to have a slightly condensed way of writing these link variables:
link from site j in µ-direction by 1 step: U µ (j) = U(j,j + 1 µ )
link to site j in − µ-direction by 1 step: U † µ(j) = U(j + 1 µ ,j) = U −µ (j + 1 µ )
Using the trapezoidal rule for integration [ ∫ b
a f(x)dx ≈ 1 2
(f(b) + f(a)) (b − a)]
and with the definition of the gauge link U(x,y) = exp [ ie ∫ y
x A µdz µ] in eq.(240)
we can write for small a
U µ (j) = exp
[
ie a ]
2 (A µ,j + A µ,j+1µ )
(242)
When we will analyse the discretized and gauge invariant Wilson action
we will find out that the so called ”plaquettes” are the relevant quantities to
describe the gauge field. The simplest one is given by
241

with in the end (in fact we have 4 directions ond not only two as in the plot)
P µν = U µ (j)U ν (j + 1 µ )U † µ(j + 1 ν )U † ν(j) (243)
The reason for the importance of the plquettes is simple and very basic for
the lattice gauge theory: The plaquettes play the role of the guage field tensors
F µν , as we will show immediately.
The plaquette is gauge invariant because of eq.(241) and hence
U µ (j) → G j U µ (j)G −1
j+1 µ
and U µ(j) † → G j+1µ U µ(j)G † −1
j
using eq.(242) one obtains for P µν
P µν = exp{ie a 2
[
Aµ,,j + A µ,,j+1µ + A ν,j+1µ+A ν,,j+1µ+1 ν
− A µ,,j+1µ+1 ν−A µ,,j+1ν−A ν,j+1ν−A ν,j
]
}
To interpret the plaquette use now the fact that we always consider small lattice
constant a, such that
A µ,j − A µ,j+1ν → −a∂ ν A µ,j
such that
F µν (j) = 1 a
[
(Aµ,j − A µ,j+1ν ) − (A ν,j − A µ,j+1µ ) ]
and by Taylor expansion to first order in a
P µν (j) → exp [ iea 2 (∂ µ A ν,j − ∂ ν A µ,j ) + O(a 4 ) ]
= 1 + iea 2 F µν,j − a4 e 2
2 F µν,jF µν,j + O(a 6 )
where the last line ist obtained by expanding the exponential function. Apparently
the plaquette starting from the same point j but with all arrows inverted
242

yields instead of P µν (j) now a P νµ (j), which is identical to P µν (j) except in the
sign of the imaginary part. Hence adding both these plaquettes yields only a
real part if one neglects of order O(a 6 ). And this real part is exactly the kinetic
energy of the gauge field 1 4 F µν,jF µν,j . Thus one obtains an important result:
One has to sum up the contributions over all plaquettes to get the action of the
gauge field.
So finally the lattice action for gauge fields becomes
S A = 1 ∑ ∑
[
e 2 1 − 1 (
Pµν (j) + P †
2
µν(j) )]
j
P
j
µ

This is in fact not a complication since we had written the plaquettes (243)
already in coorrect ordering: P µν = U µ (j)U ν (j + 1 µ )U µ(j † + 1 ν )U ν(j), † see also
figure. Furthermore one has to take into account that for the quark fields an
additional sum appears since they have an additional colour index. If one writes
things carefully down the lattice action of QED (244) modifies into
S QCD
[
U,ψ, ¯ψ] =
6
g 2 ∑
with
+ ∑ j,ρ
+ 1 2
j
∑
P
[
1 − 1 6 Tr ( P(j) + P † (j) )] (245)
( ˜M (ρ)
+ 4r) ˜ψ j
˜ψ (ρ)
j
∑
4∑
j,ρ,σ µ=1
TrP µν = ∑ U (κ,ρ)
[ ˜ψ(ρ) j (γ µ − r)U µ (ρ,σ) (σ) (ρ)
(j) ˜ψ j+1 µ
− ˜ψ j+1 µ
(γ µ + r)U µ (ρ,σ)† (j)
µ (j)U ν
(ρ,σ)
(j + 1 µ )U µ (σ,τ)† (j + 1 ν )U ν
(τ,κ)† (j)
Altogether the non-abelian theory on the lattice is just technically more complicated
than the abelian one without bringing new conceptual problems. The
trace in the pure Yang-Mills part S latt denotes taking the trace of U µ i.e. the
sum of the 3 diagonal elements. S latt sums over all plaquettes of all orientations
on the lattice.
Actually the coupling constnt g is the so called ”bare coupling constant”. It
is dependent on the lattice spacing since this is connected with a momentum
cut-off. Usually one does not write g(a) but β. This is
β = 6 g 2 g = g(a) = bare coupling constant (246)
Actually the β the single input parameter for a QCD calculation involving only
gluon fields (whether on the lattice or not). Notice that the lattice spacing is
not explicit anywhere, and we do not know its value until after the calculation.
(This is a difference from the quantum mechanical example of the previous
section where we had to choose and input a value for a.) This is explained
in the next subsection. The value of the lattice spacing depends on the bare
coupling constant, or vice versa, the value of the bare coupling constant depends
on the lattice spacing. Typical values of β for current lattice calculations using
the Wilson plaquette action are β ≈ 6. One should note that the continuum
limit is reached for β → ∞.
13.5.3 Continuum limit
Although in the laattice formulation of the quantum field theory all quantities
have precise mathematical meaning, one must not foget the goal is to define
a quantum system incontinuous space-time, since there the experiments are
done. For this one must proceed to the continuum limit, letting the lattice
spacing a go to zero. Here one should note that the lattice spacing a plays
]
(σ) ˜ψ j
244

a double role. First it is there in order to have a discrete theory such that
one can handle it on the computer. Furthermore the finite a means also that
one can handle only finite momenta smaller than Λ cutoff ∼ 1/a in the lattice,
which means, that all quantities get automatically regularized with a finite
momentum cut-off Λ cutoff . In fact the lattice is a non-perturbative way of
regularizing and renormalization. In this context it is important to understand
the connection of β = 6
g
of eq.(246) to physics in nature and to the lattice
2
spacing a in the following way (we do this in the jargon of QCD including quarks
although we formally consider a pure gluon theory in the first place): The g is
the dimensionless bare coupling constant of QCD, it is a number and it is the
only quantity in the Lagrangean which tells anything about the magnitude of
any physical quantity to be calculated. Suppose we want to calculate the mass
M N of the nucleon. We do it at finite ´a and hence we have M N = M N (g,a),
however in the limit a −→ 0 we should have M N (g,a −→ 0) = 938MeV , at
least this value should be finite [i.e. not zero and not infinite, such that by a
proper normalization one could adjust it to 938MeV]. To achieve this one must
have the g a function of a, i.e. g = g(a), and in the limit a −→ 0 the function
g(a) should be such that ”M N (g(a),a −→ 0) = finite” is achieved. This
feature should hold for any physical observable, namely that it has a finite limit
for a → 0.Actually several functions g(a) can be thought of, which do that.
However for a → 0 we also have Λ cutoff → ∞, and in this limit perturbative
QCD holds. Thus one can read off the g(a) from a few Feynman diagrams of
first order in g. This is possible since QCD is an asymptotically free theory
whose effective coupling constant ḡ(µ) or α s (µ) (132) goes logarithmically to
zero for µ ∼ Λ cutoff ∼ 1/a → ∞. For small g first order perturbation theory is
sufficient and the relation between g and a can be shown to be governed by the
equation
a dg(a)
da = β 0g 3 (247)
where β 0 is the famous beta-function (129)
β 0 = 11 3 N 1
c
16π 2
with N c equal the number of colours in the theory. The differential equation
(247) is solved by
a = 1 (
exp − 1 )
Λ L 2β 0 g 2 (248)
We can rewrite
g(a) 2 1
= −
2β 0 log(Λ L a)
The Λ L is unknown in principle. It has to be adjusted for one observable to
its physical value, e.g. for the nucleon mass 938MeV: Then it is fixed and
one can calculate with this Λ L and the above g(a) other physical quantities.
Typical values of β = 6/g 2 for current lattice calculations (e.g. using the Wilson
plaquette action to get the area law and the quark-antiquark potential) are
245

β ≈ 6. This corresponds to a ≈ 0.1fm. Smaller values of β give coarser lattices,
larger ones, finer lattices. 3
In fact one should go even further. One knows from renormalization theory
the asymptotic (renormalization scale µ −→ ∞) expression of any observable.
E.g. for the nucleon mass one has
8π 2 ]
or expressed in terms of β :
M N = C a exp [−
M N a = C exp
11
3 N cg 2 (a)
] [− 4π2 β
11N c
This functional dependence on β one can in principle explicitely check by performing
calculations for various β, and in fact one should do so to see if one is
really close enough to the continuum limit to allow comparison of lattice results
with experiment. However the values of β and its range are mostly too small
by technical reasons and hence this check is not always done in a convincing
way. Usually people prefer to plot ratios of observables in dependence on β, and
they proceed in the following way: One chooses a value of β and the number of
lattice points, e.g. N 4 , and one performs a calculation for a correlation function
corresponding e.g. to the nucleon mass. The only input is this β and the
number of lattice sites N. In fact one has not to specify a lattice spacing a, it is
always finite and in fact equal ONE. If the lattice is large enough the correlation
function at large T is given by exp (−M N T) = exp (−[M N a]T/a) , where T/a
is the number of lattice points N and M N a is a dimensionless number. [This is
basically the only thing you can calculate, namely dimensionless numbers.] Now
one does this also for another observable mass, e.g. for the ρ-meson mass, and
one can extract the ratio MNa
M ρa = MN
M ρ
. One does this for several increasing values
of β and if beyond a certain β one obtains for the ratio MN
M ρ
no β-dependence
any more then one assumes that one is in the asymptotic region and close to
the continuum limit. In principle for sufficiently large β one is always in the
asymptotic region, however in practice it may happen that one cannot perform
such a calculation since the number of lattice points N is limited and one may
not be in the region where the exponential form of the correlation function holds
already.
If one goes beyond pure glue theory (Yang Mills) and includes quarks, i.e.
performs a full QCD calculation on the lattice, one can rewrite the above formulae
assuming now in eq.(248) a β 0 = ( 11
3 N c − 2 3 N f)
/16π 2 . If one identifies the
inverse of the lattice spacing with the renormalization scale µ [i.e. if one identifies
g(1/a) with ḡ(µ)] and if one identifies the Λ L with Λ QCD one obtains after
3 Actually one can go beyond first order perturbation theory for the relation between a and
g. If one does that one obtains more complicated expressions (see Rohde sect. 9.2).
246

a direct rewriting from eq.(248) the expression for the strong QCD-coupling
constant (132). In fact these identifications are correct and correspond to the
role of a as a regularization parameter.
Altogether one has to remember that the following limiting cases have to be
under control for a lattice simulation:
Continuum limit a → 0
Lattice size V → ∞
Euklidean time T → ∞
Light quark masses or pion mass agains the physical value m π → 139MeV
In addition one has to cope with
Fermion doubling
13.6 Practical applications of lattice QCD
13.6.1 Metropolis algorithm in pure gauge theory
The application of the Metropolis algorithm to a pure Yang-Mills action is
completely analogous to simple quantum mechanics. The only difference is that
we now have link variables, which are SU(3) elements. The update process
here means to propose a change in a link variable by multiplying it by a random
SU(3)-matrix and then to calculate the the difference in the action and to accept
it with some probability.
The algorithm for randomizing the link variable U µ (j) at the j h site in
direction µ is:
• generate a random SU(N c )−matrix M. Analog to the 1-dimensional case
in quantum mechanics one can chose in some way the size ζ of this matrix.
The ζ must be a random number, with probability uniformly distributed
between some limits −δ and δ for some given constant δ;
• replace U µ (j) → MU µ (j) and compute the change ∆S in the action caused
by this replacement. Generally only a few terms in the lattice action
involve U µ (j). Since the Yang-Mills lagrangians are local, only these need
to be examined. One should note, that each of the 4 ∗ N 4 links is shared
by 6 plaquettes. S, if one link changes, there are 6 terms in the action
that change.
• if ∆S < 0 (the action is reduced) retain this new value for U µ (j), and
proceed to the next link;
• if ∆S > 0 accept the new value MU µ (j) with probability exp(−∆S).
This means: generate a random number η unformly distributed between 0
and 1; retain the new value for U µ (j) if exp(−∆S) > η, otherwise restore
the old value; proceed to the next link. Acutally, to generate a random
SU(3)-matrix on the computer is a non-trivial task, since the computer
cannot generate a real random number. There are sophisticatead techniques
to do that, which we will not discuss here.
247

13.6.2 Pure gauge calculations
Wilson loop, area law. string constant Pure gauge calculations are obtained
by the previous formalism by putting all fermion fields to zero i.e. using
as action
S Y M [U] = 6 ∑ ∑
[
g 2 1 − 1 6 Tr ( P + P †)] (249)
j
P
which corresponds to a pure Yang-Mills theory. The most interesting quantity
to calculate is the Wilson loop given in the figure
To ‘measure’ this in pure gluon QCD, we generate configurations of gluon
fields with probability e −SY M where S Y M is the discretisation of the pure gluon
QCD action given by Equation (249). On each of these configurations we calculate
the r ×t Wilson loop, averaging over all positions of it on the lattice. We
then average over the results on each configuration in the ensemble to obtain a
final answer with statistical error, for lots of values of r and t. We have
W(r,t) = 1 ∫
DU
1
Z < 0|1 3 Tr(UU..U) 3
rt|0 >=
Tr(UU..U) ∫
rte −SY M
DUe
−S Y M
Typically we need many hundreds of configurations in an ensemble for a small
statistical error at large r and t.
This quantity is interesting since one can show (only plausible arguments
given below) that it is related to the static potential between a heavy quark and
a heavy anti-quark at a distance r in the limit of infinite mass of the quarks. In
fact, it is interesting to note that one can calculate this quark-quark-potential
without having explicit quarks in the lattice. We have
W(r,t) ∝ exp [−V (r)t]
The practical calculations yield curves like
In the figure a parameter R 0 appears. This is the so called Sommer-scale,
which is explained below, it is about 0.5 fṁ. It is The β = 6
g 2 , where g is the bare
248

coupling constant as it appears in the QCD lagrangean. In the limited region
of coupling constant considered all curves lie on top of each other. In the region
where V (r) ∝ r we see the so called Wilsons area law, which is interpreted as a
sign of confinement.
W(r,t) ∝ exp [−σA]
Here A = rt is the area of the Wilson loop and σ is the so called string constant.
In general we have [ ] W(r,t)
log = aV (r)
W(r,t + a)
At small r we have a Coulomb like behaviour, i.e. altogether we have
V (r) = − b r + σr
The ensemble average of O is related to the heavy quark potential by similar
arguments to those used for the operator x(t 1 )x(t 1 ) in section about harmonic
oscillator. One end of the Wilson loop creates a set of eigenstates of the Hamiltonian
that are based on a massive quark-antiquark pair. These eigenstates
are produced with different amplitudes by the Wilson loop operator and have
different energies. In this case there is no kinetic energy, so the energies are
those of the heavy quark potential. The different eigenstates propagate for time
249

t and, if t is large, the ground state eventually dominates.
W(r,t) = Ce −aV (r)t + C ′ e −aV ′ (r)t + ....
By fitting the results as a function of the euklidean time length, t, in lattice
units, the heavy quark potential in lattice units, aV (r), is obtained. The V ′ is
some kind of excitation of the potential which we will not be interested in here.
The heavy quark potential at short distances should behave perturbatively and
take a Coulomb form. At large distances we expect a ‘string’ to develop which
confines the quark and antiquark and gives a potential which rises linearly with
separation. We can therefore fit the lattice potential to the form
aV (ρ = ra) = − 4 α s (ρ)
+ σa 2 r +
3 r
˜C
where ˜C is a ‘self-energy’ constant that appears in the lattice calculation. If
the results for aV (R) are plotted against R, the slope at large r is the ‘string
tension’, σ, in lattice units, i.e. σa 2 . Phenomenological models of the heavy
quark potential give values for √ σ of around 440 MeV. Using this value for σ
and the result from the lattice of σa 2 , gives a value for a. This is often quoted
as a value for a −1 in GeV. Note that a −1 in GeV = 0.197/(ainfm).
Figure 1: The heavy quark potential in units of the parameter, r 0 , as a function
of distance, r, also in units of r 0 . The calculations were done in quenched (pure
glue) QCD at a variety of different values of the lattice spacing, corresponding
to the different values of β quoted. (Bali 2000)
250

The results for aV can then be multiplied by a −1 to convert them to physical
units of GeV and, having removed the constant ˜C, the results can be plotted as
a function of the physical distance, r in fm. If discretisation errors are small,
then results at different values of the lattice spacing should be the same. 1
shows results for the heavy quark potential on relatively fine lattices at different
values of β using the Wilson plaquette action for S g (Bali 2000). The V (r) is
not in fact given in GeV here, nor is r in fm, but both are given in terms of a
parameter called r 0 (Sommer 1994). This is the value of r at which r 2 ∂V/∂r
= 1.65 and is a commonly used quantity to determine the lattice spacing (or at
least relative lattice spacings), rather than the string tension. The r 0 is not a
physical parameter, and as such is not available in the Particle Data Tables. We
shall see later that there are good hadron masses to use for the determination
of a and these can also be used to determine the value of r 0 . Meanwhile,
r 0 ≈ 0.5fm. The results at different values of β lie on top of each other and this
gives us confidence that the discretisation errors are small.
The fact that The above curve is indeed some sort of potential energy between
quark and anti-quark with the separation r. This feature can be understood
in a Born-Oppenheimer concept: Consider as example a molecule with
two nuclei at distance R with charges Z 1 and Z 2 . An electron moving in the
field generated by the two nuclei obbeys the Schroedinger equation
{
− 1
}
2m ∇2 r + −e2 Z 1
+ −e2 Z 1
ψ R (r) = E R ψ R (r) (250)
|r| |r − R|
Due to the existence of the electrons and due to the relative repulsion of the
two nuclei we have a potential energy between the nuclei and for various R
altogether a potential energy surface,
V (R) = E R + Z 1Z 2 e 2
in which the two nuclei move and can vibrate against each other. Apparently
the two nuclei are hold together due to the existence of E R .The stationary states
of the motion of the nuclei is described by
{
− 1 }
2M ∇2 R + V (R) Ψ(R) = EΨ(R)
yielding the ground state and the excited states of the molecule.
In the present case we have the analogon ”Two nuclei ⇐⇒Heavy quark
and antiquark” and ”Electrons ⇐⇒Gluons, Glueballes, any gluonic structure
binding the heavy quarks to each other”. The analogue equation to eq.(250)
is an equation which describes the gluonic field between two heavy quarks of
relative distance R. If those are infinitely heavy they can be replaced by two
coloured charges creating the gluon field. In this case one really does not need
quarks but only a gluon field facing the two colour charges, which makes life
very much simpler. In the present field theoretical formalism this is not a
Schroedinger equation but a path integral which calculates the ground state of
R
251

t
p
1
p
2
eq.(250). To get the ground state of the gluonic field one has to evaluate path
integrals with large Euklidean time t, and see that there is some exponential
fall of with increasing t. That means one has to consider a Wilson loop with
space width r and time width t, as it is shown in the above picture. This yields
then the analogue of E r . In the present case V (r) = E r because there is no
gluon-less interaction between the heavy quarks. Thus the analogue of Z1Z2e2
R
is missing in the QCD case.
Actually the above picture is only correct if one considers the quarks as
heavy. For light quarks we have a different situation. There instanton effects
appear which exist only in the limit of vanishing quark mass and we have spontaneous
chiral symmetr breaking which does not exist with heavy qurks. As
a consequence with increasing distance r the quarks create a pion field around
them which shields them and the potential does not rise any more linearly but
flattens out.
Glueball-masses In a pure Yang-Mills theory we can calculate so called plaquettes
correlations. These are expectation values of the product of two spatial
plaquettes p 1 and p 2 , separated by a time t as illustrated in the Fig.
2.
file=plaqucor.eps,height=2cm
Figure 2: Plaquette–plaquette correlations
In non-abelian lattice gauge theories one finds that these correlations fall off
exponentially according to
〈Tr(U(p 1 ))Tr(U(p 2 ))〉 c ∼ exp(−mt)
From our general discussion about correlators e.g. in eq.(224) it follows
that m is the lowest particle mass in the theory since the ground state has
vanishing energy. Since there are only gluonic degrees of freedom present, the
corresponding massive particle is called glueball. In fact those calculations have
been performed for decades an now the numbers stabilize. Some results are
given in the figure
252

253

Although this is an extremely interesting object, it has not unambiguously
be identified experimentally. The reason is, that there are quark-antiquarkexcitations
of the vacuum, which have the same quantum numbers, and hence
mix with the pre glueball states. Thus the identification of the lattice results
with the experimental data (Crystal Barrel) is a bit dubious.
13.6.3 Problems with Fermions
Quenched-unquenched: If one looks at the actions of QED (244) or QCD
(245) one realizes that the Fermion fields in the exponent are quadratic or
linear. Hence one can perform an exact integral over the Fermion fields following
eq.(234): This integration can be done analytically and one obtains the so called
fermion determinant det(F), that depends only on the gauge fields U. This
means we have
∫
DUD ¯ψDψ exp [ −S [ U, ¯ψ,ψ ]] ∫
= DU det (F [U]) exp[−S eff [U]]
So far these expressions are equivalent. However path integrals over fermion
fields are costly and so are costly also the evaluation of the fermion determinant,
since F(U) involves the whole lattice simultaneously. There fore people were
very happy to formulate a so called quenched approximation:
quenched approximation: det (F [U]) = 1
This appreximation corresponds to ignore all diagrams with virtual quarks.
Missing out sea quarks entirely is known as the quenched approximation. It
is clearly wrong, but for a long time the presence of other systematic errors
and poor statistics obscured this fact. More recently it has become clear that
the systematic error in the quenched approximation is around 10-20%. When
light quark vacuum polarisation (det(F)) is included the calculation is said to
be ‘unquenched’ or ‘dynamical’. The sea quarks are then also called dynamical
quarks. We will see in the results section that it is now possible to include
realistic quark vacuum polarisation effects and the quenched approximation can
be laid aside at last.
Light quarks Manipulations of the matrix F are computationally costly.
Even though it is a sparse matrix (with only a few non-zero entries) it is very
large. There are various computational techniques for calculating the F −1 factors
and the detF factor is included by repeated determination of the calculation
of F −1 . This makes the inclusion of detF very costly indeed. It becomes
increasingly hard as the quark mass becomes smaller because F becomes illconditioned.
(The eigenvalues of F range between some fixed upper limit and
the quark mass, so this range increases as m q → 0). In the real world the u
and d quarks have very small mass of a few MeV and so in the past there has
not been sufficient computer power available to include them as sea quarks or,
if they have been included, their masses have been much heavier than their
254

eal values. Chiral symmetry is spontaneously broken in the real world giving
rise to a Goldstone boson, the π, whose mass consequently vanishes at zero
quark mass (m 2 π ∝ m q ). For small quark masses, where m π is small, we have a
well-developed chiral perturbation theory which tells us how hadron masses and
properties should depend on the u/d quark masses (or equivalently m 2 π) and we
can make use of this to extrapolate down to physical u/d quark masses from the
results of our lattice QCD simulation provided that we are able to work at small
enough u/d quark masses to be in the regime where chiral perturbation theory
works. In general m u/d < m s /2 is necessary for an accurate extrapolation.
Actually in earlier days the heavy mass of the up- and down quarks was
evenwelcome, in order to get a reasonable calculation for the heavier mesons.
E.g the Rho-meson would not be the outcome of a calculation but two pions,
whose quantum numbers can couple to the Rho. Similarly with the Delta-Isobar,
it exists in lattice codes only if the pion mass is heavier than the delta-nucleon
mass splitting.
Heavy quarks Heavy quarks (b and c) represent a rather different set of
issues to those for light quarks in lattice QCD. They have large quark masses,
m Q , and therefore large values of m Q a at any value of a at which we are able
to do QCD simulations. This means that we risk large discretisation errors
when handling these quarks if we use a formalism in which the errors are set by
the size of ma. If m Q a > 1 then no amount of improvement will give a good
discretisation for these quarks. This is particularly true for the b quark which
has a mass of around 5 GeV. To reduce m Q a below 1 requires a −1 > 5 GeV or
a lattice spacing < 0.04 fm which is incredibly expensive to simulate (as well as
being wasteful).
14 Linear chiral Sigma model (Gell-Mann–Levy)
The model must be understood in a special way. One cannot treat it as a fully
field theoretical model which includes the full Dirac sea because then instabilities
occur (See Ripka and Kahana etc. and Sieber). One must understand it as
a classical one with some mild quantum approximations. The best way to
understand it is described in the procedure, how to solve it for the properties
of a baryon. there one considers only the valence quarks and believes that the
sea quarks are absorbed into the mesonic fields, which then also have a kinetic
energy term.
14.0.4 Fock states and variational principle
The Lagrangean is given, the only unknown parameters are the value of the
coupling constant g, the vacuum value of the sigma field is taken to be f π .
Evaluate first the Hamiltonian density H(§)of the system of the system,
which is well defined in terms of the fields ψ,σ,π t . Write down the quantized
255

quark-, sigma- and pion field operators:
ψ(r) = ∑ {
}
ψ nljm (r)c nljm exp(−iɛ nlj t) + χ nljm (r)d † nljm exp(+iɛ nljt)
nljm
̂σ(r) = (2π) − 3 2
∫
d 3 k(2ω σ (k)) − 1 2 (a(k)exp(ikr) + a † (k)exp(−ikr))
̂π t (r) = (2π) − 3 2
∫
d 3 k(2ω π (k)) − 1 2 (bt (k)exp(ikr) + b † t(k)exp(−ikr))
Then you define coherent states as quantum states for the boson fields
|Σ〉 = N − 1 2 exp(
∫
d 3 kη(k)a † (k)) |0〉
with the properties
∑
∫
|Π〉 = N − 1 2 exp( d 3 kζ t (k)b † t(k)) |0〉
t
a(k) |Σ〉 = η(k) |Σ〉
and e.g.
b t (k) |Π〉 = ζ t (k) |Π〉
∫
〈Σ 1 |Σ 2 〉 = N − 1 2
1 N − 1 2
2 exp(
d 3 kη ∗ 1(k)η 2 (k))
and you define the quark state as
∣ q
3 〉 〉
∣
= ∣c † 1 c† 2 c† 3
with (for the lowest positive single plarticle state)
ψ nljm (r) = (r|c † nljm |0〉 = 1 ( )
u(r)
4π iv(r)σ a ̂r a |χ〉
with the hedgehog structure
|χ〉 = 1 √
2
(|u ↑〉 − |d ↓〉)
The hedgehog baryon has the Fock-state:
〉
∣
|Ψ〉 = ∣c † 1 c† 2 c† 3 |Π〉 |Σ〉
Here one realizes the approximations. The boson fields are quantized using
the creation operators of the free fields as basis and assuming coherent statess.
256

The fermion fields do not show any Dirac sea. This is achieved technically by
considering only positive energy states and
Since the quarks are the sources for the boson fields those have the structure
〈Σ| ̂σ(r) |Σ〉 = σ(r) ::::::::: σ(r) ⇔ η(k)
corresponding to
⎛
〈Π| ̂π t (r) |Π〉 = π t (r) = ⎝
⎛
ζ t (k) = ⎝
x
r
y
r
z
r
⎞
⎠ Φ(r)
⎞
ik x
ik y
⎠ A(k) :::::::::: A(k) ⇔ Φ(r)
ik z
Thus the informations in η(k),ζ t (k) and σ(r),Φ(r) are equivalent.
The energy of the system is given by
E = 〈Ψ |HΨ〉
in terms of the functions σ(r),Φ(r),u(r),v(r). The fields are assumed timeindependent.
One determines the functions σ(r),Φ(r),u(r),v(r) by the variational
procedure:
δE
δu(r) = δE
δv(r) = δE
δσ(r) = δE
δΦ(r) = 0
This results in a set of four differential equations with 4 unknown functions.
The equations are non-linear. Their result corresponds to a soliton. In the end
the full Fock-state is known. There are, however, no quantum numbers yet.
14.0.5 Projection techniques
The quantum numbers are determined in the following way using Peierls-Yoccoz
perojection techniques:
One defines the projection operator:
P J MK = 2J + 1
8π 2
P T M T K T
= 2T + 1
8π 2
∫
∫
dΩD J∗
MK(Ω)R(Ω)
dΩ T D T ∗
M T K T
(Ω T )R T (Ω T )
where R and R T are rotation operators in spin and isospin SU(2)
R(α,β,γ) = exp(−iαJ 3 )exp(−iβJ 2 )exp(−iγJ 3 )
257

and analogously for R T . The projection operators have the properties
(
P
J
M,K
) †
= P
J
K,M
P J ′
K ′ ,M ′P J M,K = δ J,J ′δ M,M ′P J K ′ ,K
One can show that the most general state with given angular momentum and
isospin , obtained by rotating Ψ in space and isospin space, can be written as
|J,T,M,α,M T 〉 = ∑
PM,KP J M T T K T
|Ψ〉
with
∑
g (J,T,α)
K,K T
K,K T
J,T,M,α,M T
|J,T,M,α,M T 〉 〈J,T,M,α,M T | = I
which means, that Ψ is a linear combination of the projected states, that is, it
contains the states with good angular momentum and isospin as components.
∑
|Ψ〉 = |J,T,M,α,M T 〉 〈J,T,M,α,M T | Ψ〉
J,T,M,α,M T
The coefficients g (J,T,α)
K,K T
are the results of a non-orthogonal diagonalization procedure:
∑ (
)
h (J,T)
K,K T ,K ′ ,K
− E (J,T,α) n (J,T)
T
′ K,K T ,K ′ ,K
g (J,T,α)
T
′ K ′ ,K
= 0
T
′
with the kernels
h (J,T)
K,K T ,K ′ ,K ′ T
= 〈Ψ|HPKK J ′P K T T ,K |Ψ〉
T
′
= 〈Ψ|PKK J ′P K T T ,K |Ψ〉
T
′
The normalization of the coefficients is done by
∑
(J,T,α g ′ ) ∗
g (J,T,α)
K ′ ,K
n (J,T)
T
′ K,K T ,K ′ ,K
= δ
T
′ α′ α
n (J,T)
K,K T ,K ′ ,K ′ T
K ′ ,K ′ T
Apparently the eigenvalues E (J,T,α) of these equations are the energies (masses)
of the baryon with the corresponding quantum numbers. One has some simple
sum rules:
∣ ∣
∑ ∣∣∣∣∣
∑
∣∣∣∣∣
2
g (J,T,α)
K,K T
n (J,T)
M,K,M T K T
= 1
JTαMM T K,K T
This sum rule tells, how much of the state JTαMM T is contained in the state
Ψ. And the next sum rule tells, how much of the mean field energy 〈Ψ|H |Ψ〉
comes from the stateJTαMM T .
∣ ∣∣∣∣∣
∑ ∑
E (J,T,α)
JTαMM T
g (J,T,α)
K,K T
K,K T
n (J,T)
M,K,M T K T
∣ ∣∣∣∣∣
2
= 〈Ψ|H |Ψ〉
258

To calculate a nucleon property one has to write an observable Ô in terms of
̂ψ, ̂σ, ̂π t and has then to calculate
O (α)
J,T,M,M T
= 〈J,T,α,M,M T |Ô |J,T,α,M,M T 〉
which is now well defined. The calculations of the form factors are done in this
way. The calculations are succesfull in several respects. If one adjusts the coupling
constant g of the fermion-pion coupling such that the mass of the nucleon
is reproduced (M = 938MeV ), then the electric form factor of the proton is
reasonably well reproduced. The absolute values of the magnetic moment are
correct to 30%. One has also disadvantages: One obtains g A = 1.8 and the
nucleon-delta-splitting comes out about about 150 MeV. The experimental values
are g A = 1.25 and 300 MeV. The nucleon mass is reproduced because one
automatically obtains some sort of correlation energy such that in the limit of
strong deformation in flavour space we have approximately
〈Ĵ2〉
E J,T = E 0 +
J(J + 1)
2Θ
One can also ask for the component a certain state|J,T,α,M,M T 〉 is contained
1 1
2 2
20%
3 3
in the soliton. One obtains for the lowest state for a given J,T,%:
2 2
50%
5 5
2 2
25%
One sees at the numbers that higher components are only contained in Ψin
a negligible way. This is interesting, compared to the outcome of semiclassical
methods, where we have rigid rotations in the coordinate and isospin space and
hence all quantum nunbers are present.
In the above procedure one performs first the variation and after that the
projection (variation before projection). However, one obtains only good results
if one applies the variation after projection:
−
2Θ
δ 〈J,T,α,M,M T |Ĥ |J,T,α,M,M T 〉 = 0
This means, one varies the already projeted states. In this case one has different
functions σ(r),Φ(r),u(r),v(r) for each state |J,T,M,α,M T 〉 . This is an
interesting result, which has not fully been exploited yet. Show form
factors of
gen hh with
15 Restbestände
projections
Current algebra, Schwinger terms A simple argument shows, that the
Schwinger terms do not vanish. Asume for a moment that they are absent. Then
we have from the above formulae (because of antisymmetry of the structure
coefficients) [
j
a
0 (x,t),j a i (y,t) ] = 0
which implies [
j
a
0 (x,t),∂ i j b i (y,t) ] = 0
259

Assuming current conservation we obtain
[
j
a
0 (x,t),∂ 0 j b 0(y,t) ] = 0
Taking the vacuum expectation value and inserting a complete set of energy
eigenstates we have
〈0| [ j a 0(x,t),∂ 0 j b 0(y,t) ] |0〉 =
= ∑ n
(〈0| J 0 (x,t) |n〉 〈n| ∂ 0 J 0 (x,t) |0〉 − 〈0| ∂ 0 J 0 (x,t) |n〉 〈n| J 0 (x,t) |0〉
= i ∑ n
(
exp(ipn (x − y)) + exp(−ip n (x − y)) ) E n |〈0| J 0 (0) |n〉| 2
because the time derivative, applied to the states, cancels the terms from the
translation operators (used to change vom J(x) to J(0). In the limit x → y this
means that ∑
E n |〈0| J 0 (x,t) |n〉| 2 = 0
n
>From the fact, that all E n ≥ 0 follows that 〈0|J 0 (x,t) |n〉 = 0 for all n. Thus
we would have J 0 = 0 identically. Thus the Schwinger terms dissappear only if
the current j 0 = 0, which is the trivial case.
Since M N has the dimension of an energy and since a is the only dimensionful
quantity we must have the relation
M N =
1
aλ(g(a))
and hence there must be a critical value g cr such that
lim λ(g) = ∞
g−→g cr
We can see the effect immediately at the harmonic oscillator. The eigenfunctions
are known
( mω
) (√ ) ]
1/4 1 mω
< x|n >= √
πħ 2n n! H n
ħ x exp
[− mωx2
2ħ
260