Magnetism and Magnetic Materials
DTU (10313) – 10 ECTS
KU – 7.5 ECTS
Module 3
08/02/2001
Crystal fields
Sub-atomic – pm-nm
But with some surrounding
environment
Intended Learning Outcomes (ILO)
(for today’s module)
1.
2.
3.
4.
5.
6.
7.
Explain why paramagnetism is T-dependent whereas diamagnetism is not
Estimate the value of the Curie constant for a given paramagnetic substance
Predict the ground state of ions by applying Hund’s rules
Explain the origin of the spin-orbit interaction, and describe its main effects
Compare Hund’s rule predictions with data on 4f and 3d elements
Describe how crystal fields arise
Explain phenomena such as crystal field splitting, Jahn-Teller distortions, low/high spin states
Flashback
Einstein de Haas:
-measure g-factor

N a  0 e2
2
 
0
r

i 0
Vm 6me i
Diamagnetism:
-small
-T-independent
-Orbital size
M  ngJ  B JBJ (y)  M S BJ (y)
2J 1  1
 y 
2J 1
BJ (y) 
coth
y coth 
 2J  2J
2J 
2J
g  JB
y J B
k BT
Paramagnetism:
-small
-T-dependent ---> Curie law
-Total angular momentum J
Van Vleck paramagnetism
If J=0, in principle there is no paramagnetic term.
However, if we go second-order, and consider the
possibility of excited states (off-diagonal matrix
terms) with nonzero J, then we have:
E0  
n
  20 B2
0  B (L  gS)  B n
E0  En
N

V n
Another contribution to the
paramagnetic susceptibility
(there’s one more…mobile
electrons – Pauli)
2
0 (Lz  gSz ) n
En  E0
2
Which is positive (para), and T-independent.
Why is it T-indepenent?? And
why was the Langevin term Tdependent instead?
John H. van Vleck, Nobel prize lecture
Lande’ g-value and effective moment
J 1
BJ (y) 
y
3J
J=1/2
J=5
2
M n0eff


MS
3k BT
eff  gJ  B J(J 1)
3 S(S 1)  L(L 1)
gJ  
2
2J(J 1)
Estimate the Curie constant for
a paramagnetic ionic salt with
a=0.3 nm, J=S=3/2
J=3/2
Curie law: =CC/T
Check: where are we?
All atoms and ions are diamagnetic
diamagnetism arises from a perturbation of the ground state
diamagnetism is small and T-independent
Whenever J differs from zero, we observe a paramagnetic response
J can be either from OAM or from Spin or both
paramagnetism is larger than diamagnetism
but still small at room T
The question now is:
What gives angular momentum to an atom?
Why are some atoms “more magnetic” than others?
That’s what we focus on today.
The multi-electron atom and the Hund’s rules
2 
2
 pi2
Ze
Ze
Hˆ   
 
2m 4  0 ri  i j 4  0 | ri  ri |
i 
With many electrons, it gets
messy. How do electrons
“choose” which state to occupy?
(1) Arrange the electronic wave function so as to maximize S. In this
way, the Coulomb energy is minimized because of the Pauli
exclusion principle, which prevents electrons with parallel spins
being in the same place, and this reduces Coulomb repulsion.
(2) The next step is to maximize L. This also minimizes the energy
and can be understood by imagining that electrons in orbits
rotating in the same direction can avoid each other more
effectively.
(3) Finally, the value of J is found using J=|L-S| if the shell is less
than half-filled, J=L+S is the shell is more than half-filled, J=S
(L=0) if the shell is exactly half-filled (obviously). This third rule
arises from an attempt to minimize the spin-orbit energy.
2S+1L
J
Find the electronic structure of Fe3+,
Ni2+, Nd3+, Dy3+, and determine
their spin configuration
Spin-orbit and the fine structure
2
e
2
Hˆ  Hˆ 0   B (L  gS)  B 
(B

r
)
 S  L

i
8me i
Bso 
0 I
2r

For the atomic Hamiltonian
we’ve considered so far, L and S
were good quantum numbers.
Problem is: they are not…
0 Zev 0 Zeh

2
4 r
4 mer 3
For multi-electron atoms:
1
H so   (g B ms S)  B so  S  L
2
Z 4e2 h2

2  0 a03n 3l(2l 1)(l 1)
This is an opportunity to put QFT in
action! Try to re-derive spin-orbit in a
fully relativistic framework.
H so  S  L
Where the sign of Lambda
depends on the shell
occupancy.

Spin-orbit in the multi-electron atom
For multi-electron atoms:
H so  S  L  

2S
SL
Where the sign of the
energy depends on the shell
occupancy (see table).
This justifies Hund’s third rule,
whenever spin-orbit is a
significant perturbation.
If spin-orbit dominates (large
atomic number, as it goes as Z4),
the L-S coupling scheme fails.
Alternative: j-j coupling.
Composition of angular momentum
J  LS
Possibilities:
J=L+S, L+S-1…|L-S|

How many?
LS
 (2J 1)  (2L 1)(2S 1)
J|LS|
Without spin-orbit, L and S are
good quantum numbers (i.e. L and

S are conserved), and J is not
useful.
With spin-orbit, L and S are not
good quantum numbers (i.e. L and
S are not conserved, although L2, S2
and J2 are), and J becomes
important. States are |L,S,J,MJ>
Summary and example
Fine structure of the Co2+ ion:
3d7: S=3/2, L=3, J=9/2, gJ=5/3
Data and comparison (4f and 3d)
Hund’s rules seem to work well for
4f ions. Not so for many 3d ions.
Why?
How do we measure the effective
moment?
Origin of crystal fields
When an ion is part of a
crystal, the surroundings (the
crystal field) play a role in
establishing the actual
electronic structure (energy
levels, degeneracy lifting,
orbital “shapes” etc.).
Hˆ  Hˆ 0  Hˆ so  Hˆ cf  Hˆ Z
Vcf 
1
4  0
Hˆ cf 
 (r)
 r  r' d r'
3
3

(r)
V
(r)d
r
 0 cf
Not good any longer!
A new set of orbitals
Octahedral
Tetrahedral
Crystal field splitting; low/high spin states
The crystal field results in a new set of orbitals where to
distribute electrons. Occupancy, as usual, from the lowest to the
highest energy. But, crystal field acts in competition with the
remaining contributions to the Hamiltonian. This drives
occupancy and may result in low-spin or high-spin states.
Orbital quenching
Examine again the 3d ions. We
notice a peculiar trend: the
measured effective moment
seems to be S-only. L is
“quenched”. This is a
consequence of the crystal
field and its symmetry.
Vcf 
1
4  0
 (r)
 r  r'
d 3r'
p x  l  1, ml  1  l  1, ml  1
d xy  l  2, ml  2  l  2, ml  2
d x 2 y 2  l  2, ml  2  l  2, ml  2 
dz 2  l  2, ml  0
Examples
Is real. No differential (momentumrelated) operators. Hence, we need
real eigenfunctions. Therefore, we
need to combine ml states to yield
real functions. This means,
combining plus or minus ml, which
gives zero net angular momentum.
Jahn-Teller effect
In some cases, it may be
energetically favorable to
shuffle things around than to
squeeze electrons within
degenerate levels.
EJT  A  B 2

Sneak peek
1
2
Interactions
Ferromagnetism (Weiss)
Wrapping up
•Temperature dependencies
•Curie Law
•Van Vleck paramagnetism
•Hund’s rules
•Spin-orbit
•Crystal field
•Orbital quenching
•Jahn-Teller distortions
Next lecture: Friday February 11, 8:15, KU
Interactions (MB)