left side of the diagram quantifies the Zeeman effect (weak field).
The top and bottom
lines on the extreme right show the energies of the states

1
,
1
)
+
)
and

1
,
−
1
)−)
, which
are eigenstates of the full Hamiltonian for all values of
B/A
.
The levels on the extreme right of Figure 9.2 show the energies described by
this formula in the case that
l
= 1 and
s
=
1
2
.
The fact that in a strong
magnetic field an atom’s energies depend on
m
l
and
m
s
in this way is known
as the
Paschen–Back effect
.
Zeeman effect
In a sufficiently weak magnetic field,
H
SO
affects the atom
more strongly than
H
Bs
. Then spinorbit coupling assigns different energies
to states that differ in
j
.
Consequently, when we use perturbation theory
to calculate the smaller effect of an imposed magnetic field, the degenerate
eigenspace in which we have to work is that spanned by the states that have
given values of
j,l
and
s
but differ in their eigenvalues
m
of
J
z
. Fortunately,
H
Bs
is already diagonal within this space because [
J
z
,S
z
] = 0. So the shift
in the energy of each state is simply
E
b
=
(
j,m,l,s

H
Bs

j,m,l,s
)
=
μ
B
(
m
+
(
j,m,l,s

S
z

j,m,l,s
)
)
.
(9
.
27)
As we saw in
§
7.5, our basis states do not have welldefined values of
S
z
– in general they are linear combinations of eigenstates of
L
z
and
S
z
:

j,m,l,s
)
=
s
summationdisplay
m
′
=
−
s
c
m
′

l,m
−
m
′
)
s,m
′
)
,
(9
.
28)
where the coefficients
c
m
′
are Clebsch–Gordan coefficients (eq. 7.146). In any
concrete case it is straightforward to calculate the required expectation value
of
S
z
from this expansion.
However, a different approach yields a general
formula that was important historically.
In the classical picture, spinorbit coupling causes the vector
S
to precess
around the invariant vector
J
. Hence, in this picture the expectation value of
S
is equal to the projection of
S
onto
J
.
4
The classical vector triple product
formula enables us to express
S
in terms of this projection:
J
×
(
S
×
J
) =
J
2
S
−
(
S
·
J
)
J
so
J
2
S
= (
S
·
J
)
J
+
J
×
(
S
×
L
)
.
(9
.
29)
In the classical picture, the expectation value of the vector triple product
on the right side vanishes. If its quantum expectation value were to vanish,
4
This heuristic argument is often referred to as the
vector model
.
9.2
Variational principle
189
the expectation value of the
z
component of the equation would relate
(
S
z
)
,
which we require, to the expectation values of operators that have the states

j,m,l,s
)
as eigenstates, so our problem would be solved. Motivated by these
classical considerations, let’s investigate the operator
G
≡
J
×
(
S
×
L
)
so
G
i
≡
summationdisplay
jklm
ǫ
ijk
J
j
ǫ
klm
S
l
L
m
.
(9
.
30)
It is straightforward to check that its components commute with the angular
momentum operators
J
i
in the way we expect the components of a vector to
do:
[
J
i
,G
j
] = i
summationdisplay
k
ǫ
ijk
G
k
.
(9
.
31)
From equation (9.30) it is also evident that
J
·
G
= 0.
In Problem 7.24
identical conditions on the operators
L
and
x
suffice to prove that
(
x
)
= 0 in
any state that is an eigenket of
L
2
. So the steps of that proof can be retraced
with
L
replaced by
J
and
x
replaced by
G
to show that for the states of
interest
(
G
)
= 0.
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 Spring '15
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 Physics, mechanics, The Land, probability amplitudes