The Nucleon-Nucleon Interaction in a Chiral Effective Field Theory

54 3. The derivation of nuclear forces from chiral Lagrangians
of those points, say, to (0" = A, 7r = 0). 5 Thus, the vacuum is characterized by some non-zero
value of the field 0".
Let us now co me back to the Goldstone theorem and consider the vacuum expectation value of the
operator 'l/Ji (x) . U sing the translational properties of field operators and assuming translational
invariance of the vacuum, we obtain
(3.47)
where TJi is some (non-zero) constant. Assuming furt her that -tfjTJj = K,f i- 0 if some of the 7]'S
do not vanish, we obtain using eq. (3.31)
K,f (OI[Qa, 'l/Ji(O)]IO) (3.48)
I d3 x (OI[Jö(x,
t), 'l/Ji(O)]IO)
I: I d3 x {(OIJö(O)e-iP,xln)(nl'l/Ji(O)IO) - (OI'l/Ji(O)ln)(nleiP.X Jö(O)IO)}
n
I:(27r)38(Pn) {(OIJö(O)ln)(nl'l/Ji(O)IO)e-iEnt - (OI'l/Ji(O)ln)(nIJö(O)IO)eiEnt}
n
i- 0,
where we have inserted a complete set of intermediate energy eigenstates Ln In)(nl and used the
translational invariance of the vacuum. Here Pn and En denote the momentum and energy of the
state In). Note that Qa should correspond to a broken generator because otherwise the right-hand
side of eq. (3.48) vanishes. Since K,f does not depend on time, differentiating the equation (3.48)
with respect to time leads to
I:(27r)38(Pn)En {(OIJö(O)ln)(nl'l/Ji(O)IO)e-iEnt + (0 I 'l/Ji (0) In) (nIJö(O)IO)eiEnt} = 0 . (3.49)
n
The positive and negative frequency parts can not mutually cancel. Therefore each of both terms
in the square bracket in eq. (3.49) must vanish for all states In) except those where En = 0
for Pn ---t O. Furthermore, the latter ones must exist since otherwise the equation (3.48) will
not be satisfied. Thus, the physical spectrum should contain massless states In) of spin zero
and the same parity and internal quantum numbers as Jö, whieh are called Goldstone bosons.
For a more rigorous and mathematically correct proof of the Goldstone theorem the reader is
referred to the original publications [61], [62] as well as to the references [155], [156] and [157] .
The plausible physical interpretation of the Goldstone theorem can be obtained from the simple
classieal example considered above and illustrated in fig. 3.1. There, the vacuum state corresponds
to the point (0- = A, 7r = 0), whieh is clearly not invariant under rotations in the 0"-7r surface. The
potential V (0", 7r) takes its minimum value for the points lying on the ring (j2 + 7r2 = A 2• Thus, no
additional energy is needed to move from one such point to another. The modes corresponding
to such transformations can be identified with the Goldstone bosons.6
Let us now co me back to the QCD Lagrangian. To decide whether the physieal vacuum is chiral
symmetrie or not we first note that the charges of SU(Nf)A carry negative parity. This is because
the ')'5 matrix enters the expression (3.21) for the axial-vector current Jc;t. Thus, if the chiral
symmetry would be realized in the Wigner mode, i. e. if the vacuum would be symmetrie, one
5In principle, the vacuum could also be given by a linear combination of those points. This possibility can,
however, be excluded for quantum fields in a space of infinite volume [133].
GIn this particular example such a mode can be defined using the polar angle in the surface CJ-7r.