zeeman

• Hydrogen Atom: 3D Spherical Coordinates
– Ψ = (spherical harmonics)(radial) and probability density P
– E, L 2 , Lz operators and resulting eigenvalues
• Angular momenta: Orbital L and Spin S
– Addition of angular momenta
– Magnetic moments and Zeeman effect
– Spin-orbit coupling and Stern-Gerlach (Proof of electron
spin s)
• Periodic table
Zeeman Effect
– Relationship to quantum numbers n, l, m
– Trends in radii and ionization energies

1D Cartesian
Schrödinger Equation: Coordinate Systems
Kinetic energy
2
ħ
2m
2
d ψ
dx
−
2
( x)
+ V
Potential
energy
Total energy
( x) ψ ( x) = Eψ
( x)
3D Cartesian Kinetic energy Potential energy Total energy
2
⎡
2 2 2
ħ ∂ ∂ ∂ ⎤
− ⎢ + + ⎥ψ ( x) + V ( x,
y,
z) ψ ( x) = Eψ
( x)
2 2 2
2m
⎣∂x
∂y
∂z
⎦
Convert to spherical using: x = r sinθ cos φ, y = r sinθ sin φ, z = r cosθ
3D Spherical
Kinetic energy
Potential
energy
2 2 2
ħ 1 ∂ ⎡ 2
1 1
sin
2 2 2 2
( )
2 r ∂ψ ⎤ ħ ⎡ ∂ ⎛ ∂ψ ⎞ ∂ ψ ⎤
− − ⎢ ⎜ θ ⎟ + ⎥ + V r ψ
µ r ∂r ⎢ r ⎥
⎣ ∂ ⎦ 2µ r ⎣sinθ ∂θ ⎝ ∂θ ⎠ sin θ ∂φ
⎦
= Eψ
Total energy

Hydrogen Atom: 3D Spherical Schrödinger Equation
“Rewritten” Schrodinger Eqn.:
ˆ 2
p
ψ
2µ
2 2 1 ∂ ⎡ 2 ∂ ⎤
where pˆ
= −ħ
+
eff n
,
∂ ⎢
r
2
r r ⎣ ∂r
⎥ ⎦
( r,
θ , Q) V ψ ( r,
θ , Q) = E ψ ( r,
θ Q)
Eigenfunctions:
( r θ , φ ) = R ( r ) Y ( θ φ )
,
n lm
ψ ,
nlm
Laguerre
Polynomials
Spherical
Harmonics
Eigenvalues:
E
n
2
2
2
−z
E0
1 ⎛ ke ⎞
= where E
2
0 ⎜ ⎟
n
= µ ≈ 13.6eV
2 ħ
⎝
⎠

Hydrogen Atom: 3D Spherical Schrödinger Equation
3 Quantum Numbers (3-dimensions)
n = energy level value (average radius of orbit)
n = 1, 2, 3 …
l = angular momentum value (shape of orbit)
l = 0, 1, 2, … (n – 1)
m = z component of l (orientation of orbit)
m = -l,(-l + 1) ..0,1,2,..+l
How many quantum states (n, l, m) exist for n = 3? Is there a general
formula?

Wave Functions : Formulas
n = 1
l = 0
n = 2
l = 1

Wave Functions: Angular Component
m = –2 m = –1 m = 0 m = 1 m = 2
l = 0
s-orbital
l = 1
p-orbitals
sinθ sinφ
cosθ
sinθ cosφ
l = 2
d-orbitals
sin 2 θ sin2φ
sinθcosθ
sinφ
3cos 2 θ–1
sinθcosθ
cosφ
sin 2 θ cos2φ

Wave Function: Radial
Component
http://cwx.prenhall.com/bookbind/pubbooks/giancoli3/chapter40/multiple3/deluxe-content.html
What is the relationship between the number of zero crossings for the radial
component of the wave function and the quantum numbers n and l?

l = 0 s-orbitals
Wave Functions: Angular & Radial Components
100
l = 1
p-orbitals
From: http://pcgate.thch.uni-bonn.de/tc/people/
hanrath.michael/hanrath/HAtomGifs.html
200 211 210 211
l = 2
d-orbitals
300 311 310 311 322 321 320 321 322
400 411 410 411 422 421 420 421 422
l = 3
f-orbitals
433
432 431 430 431 432 433

Probability Density: Formula
∗
P = ψ ψ ( r, θ , φ)
dV
=
∫
∞ π 2π
∫∫ ∫
0 0 0
∗
ψ ψ
r
2
sinθ dϕdθ
dr
Volume Element dV
P
=
4π
∞
∫
0
∗
ψ ψr
∞
∫
0
2
dr for spherically symmetricψ
2
( ) ( π )
P = P( r) dr where P r = 4 r
∗
ψ ψ
⇒For small ∆r, can use P = P(r) ∆ r (analogous to 1D case)

Probability Density: “Density” Plots
2
ψ
200 n = 2 ψ
210 n = 2 ψ
21 ± 1
2
2
s-orbital
l = 0
p-orbitals
l = 1
m = 0
m = 1
“Spherical Shells”
“Dumbbells”

Probability Density: Cross Sections
http://webphysics.davidson.edu/faculty/dmb/hydrogen/default.html
Can you draw the radial probability functions for the 2s to 3d wave functions?

Probability Density: Cross Sections
http://cwx.prenhall.com/bookbind/pubbooks/giancoli3/chapter40/multiple3/deluxe-content.html
Rank the states (1s to 3d) from smallest to largest for the electron’s most
PROBABLE radial position.
For which state(s) do(es) the most probable value(s) of the electron's position
agree with the Bohr model?

Probability Density: Problem
For the ground state n = 1, l = 0, m = 0 of hydrogen, calculate the
probability P(r)∆r of finding the electron in the range ∆r = 0.05a o
at r = a o
/2
[ ]
2
100 π 100
P ( r) ∆ r = (4 r ) Ψ ( r)
∆r
−r ao
−2r a
e
2 e
100 ( ) [ 100 ( )]
1.5 3
π ao
π ao
where Ψ r = and Ψ r =
2
o
⎛ −ao
ao
2 e ⎞
P100 ( ao
) ∆ r = ( π ao
) (0.05 a )
⎜
3
o
π a ⎟
⎝ o ⎠
after substitution of r, Ψ , and ∆r
P
100
−1
100
( a ) ∆ r = e ( 0.05) = 0.018
o

Orbital Angular Momentum L:
Related to Orbital “Shape”
• Magnitude of Orbital Angular Momentum L
ˆ 2 2 ⎡ 1 ∂ ⎛ ∂ ⎞ 1 ∂ ⎤
2
Lψ ( r, θ , φ ) = − ħ ⎢ ⎜sinθ ⎟ + ψ = l
2 2
( l + 1)
ψ
sinθ ∂θ ∂θ sin θ ∂φ
⎥ ħ
⎣ ⎝ ⎠
⎦
( )
Eigenvalues: L = l l + 1
ħ
• Z-component of L
ˆ ∂
Lzψ ( r, θ , φ ) = − iħ
ψ = mħψ
∂φ
Eigenvalues:
L
z
= mħ

Orbital Momentum L: Vector Diagram
For l = 2, find the magnitude of the angular momentum L and the possible m
values. Draw a vector diagram showing the orientations of L with the z axis.
l
=
2
z
L = √6h = 2.45h
L
L
= l( l+1) ħ = 2(2+1)
= 6 ħ or 2.45 ħ
ħ
2h
h
55º
24º
m = 2
m = 1
m = 0
m
L
z
= − l to l = 0, ± 1 ±
= m ħ = 0, ± 1 ħ, ± 2ħ
2
−h
−2h
m = −1
m = −2
Can you draw the vector diagram for l = 3? For j = 3/2?

Spin Angular Momentum S:
Property of Electron
• Magnitude of Spin Angular Momentum S
Eigenvalues:
S
= s s + 1
( )
ħ
1 ⎛ 1 ⎞ 3
For one electron: S = ⎜ + 1⎟
ħ =
2 ⎝ 2 ⎠ 4
ħ
• Z-component of S
Eigenvalues:
S
ħ
For one electron: S z =
2
z
=
m
s
ħ
Example for s = 1/2
z
h/2
-h/2
s = √0.75h
m = 1/2
m = −1/2

Angular Momentum: Link to Magnetic Moments
• Orbital angular momentum L and spin angular momentum s of
electrons result in magnetic moments µ l and µ s .
⎛ qv ⎞ q q
r ( vr )
⎛ L
→ ⎜ ⎟ = = ⎜
⎞
⎟ L = mvr
⎝ 2π
r ⎠ 2 2 ⎝ m ⎠
2
Remember that µ = i A π
where
Orbital l = 0, 1, 2, …
−gLµ
B
µ
l
= L = l ( l + 1)
gLµ
B
ħ
z-component
µ = −m g µ
lz l L
B
Spin: s = ½
−gs
µ
B
µ
s
= s = s ( s + 1)
gs
µ
B
ħ
z-component µ = −m g µ ≈ ± µ
sz s s B
B
where
µ
eħ
5 eV
= = 5.79 × 10
−
and g , g = gyromagnetic ratios
2m
T
B L s
e

Zeeman Effect: Splits m values
• Orbital magnetic moment µ L interacts with an external
magnetic field B and separates degenerate energy levels.
l = 1
NO B
Field
B Field
m = 1
m = 0
m = –1
l = 0
m = 0
U = −µ ⋅ B ⇒ U = −µ
B
ext
assume z
direction
lz
Different energies for
different m l
values!

“Anomalous” Zeeman Effect: More Lines??
Zeeman
????
l = 1
m l = 1
m l = 0
m l = –1
Add external
magnetic field
l = 0
m l = 0
Why are there more energy levels than expected from the Zeeman effect?

• Electron’s spin magnetic moment µ s interacts with internal B field caused by its
orbital magnetic moment µ l and separates energy levels.
Spin-Orbit Coupling: Splits j values
Spin up:
High Energy
U
S L, B l
e-
µ
s
⋅ B = −µ
szB
= ± µ
BB
=
int
µ s
p+
v
l = 1
s = 1/2
j = 3/2
∆U = 2µ BB
j = 1/2
Spin down:
Low Energy
µ s
L, B l
e-
S
p+
v
l = 0
s = 1/2
j = 1/2

• Special Case:
Vectors
J = L
J =
Angular Momentum Addition: L + S gives J
+ S
j( j +1)ħ
L
+
S
Quantum Numbers
j = l + s,
l − s
m = − j, − j + 1,... j −1,
j
j
Example: l = 1, s = ½
j
= 1+
1
2
3
2
=
3
2
m j = − , − , , and m j
1
2
1
2
3
2
and
j
=
1−
= −
1
2
1
2
=
,
1
2
1
2
j = 3/2 j = 1/2

l = 1
“Anomalous” Zeeman Effect: Spin-Orbit + Zeeman
s = 1/2
Spin-Orbit
j = 3/2
Zeeman
mj = 3/2
mj = 1/2
mj = –1/2
mj = –3/2
j = 1/2
mj = 1/2
mj = –1/2
l = 0 s = 1/2
j = 1/2
mj = 1/2
mj = –1/2
• Quantum numbers m j (j-j coupling) for HIGHER Z elements.

• General Case:
Angular Momentum Addition: General Rules
Vectors
Jtot
= J1
+ J 2
J = j( j + 1)ħ
tot
J 1 + J 2
Quantum Numbers
j = ( j + j ) ( j + j − )
1
2
, 1 2 1 ,... j1
− j2
m j = − j,
− j + 1,... j −1,
j
Example: j 1 = 3/2, j 2 = 3/2
j
max
3 and j
0
3 3 3 3
2 2 min 2 2
j = 3, 2, 1, 0
m
j
= + = = − =
= − − − =
3, 2, 1, 0,1, 2, 3 for j 3

Stern-Gerlach Experiment: ALSO splits m values
⎛ dB ⎞
⎜ Fz = µ
z ⎟
• A magnetic force ⎝ dz ⎠ deflects atoms up or down by an
amount that depends on its magnet moment and the B field gradient.
• For hydrogen (m lz = 0), two lines are observed (spin up, spin down).
– Since l = 0, this experiment gave direct evidence for the
existence of spin.
Case of m l = -1, 0, 1
What is the “story” of this experiment? Did Stern & Gerlach know about

Stern-Gerlach Experiment: Problem
The angular momentum of the yttrium atom in the ground state is
characterized by the quantum number j = 5/2. How many lines would
you expect to see if you could do a Stern-Gerlach experiment with
yttrium atoms?
Remember that in the Stern-Gerlach experiment all of the atoms with
different m j values are separated when passing through an
inhomogeneous magnetic field, resulting in the presence of distinct
lines.
How many lines would you expect to see if the beam consisted of
atoms with l = 1 and s = 1/2?

Multi-electron Atoms
• Schrödinger equation cannot be solved exactly for multi-electron
atoms because Coulombic repulsion “mixes” variables.
– Estimate energies using single-electron wave functions and
“correcting” energies with 1 st -order perturbation theory.
V
int
=
2
2
ke
r − r
1
• Orbitals “fill” in table as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p . .
– Only ONE electron per state (n, l, m l , m s ) - Pauli Exclusion Rule!
– Why is 4s filled before 3d? ⇒ 4s orbital has a small bump near
origin and “penetrates” shielding of core electrons better than 3d
orbital, resulting in a larger effective nuclear charge and lower
energy.

Noble Gas
Halogen
Group VI
Group V
n
1
Periodic Table
l = 1 (p)
2
3
4
5
6
7
l = 0 (s)
l = 2 (d)
l = 3 (f)
Alkali
Group III
Group IV

Periodic Table: Trends for Radii and Ionization Energies
Maxima = alkali metals
• Effective atomic radii decrease
across each row of table.
– Why? Effective nuclear charge
increases and more strongly
attracts outer electrons,
decreasing their radius.
• Ionization energies increase
across each row of table until the
complete “shell” is filled.
– Alkali atoms easily give up s-
orbital electrons.
– Halogens have strong affinity
for outer electrons.
Maxima = noble gases

APPENDIX: Wave Functions Formulas
Spherical Component Y l,m Complete Wave Function ψ n,l,m
s-orbital
n = 1 l = 0
l = 0
m = ±1
p-orbital
l = 1
n = 2
l = 0,1
d-orbital
l = 2
f-orbital
l = 3
l = 1 m = ±1
n = 3
l = 0,1,2