5.73 Lecture #25
25 - 1
HSO + HZeeman
Coupled vs. Uncoupled Basis Sets
Last time:
2
matrices for J , J+, J–, Jz, Jx, Jy in jmj⟩ basis for J = 0, 1/2, 1
Pauli spin 1/2 matrices
r r
arbitrary 2 × 2
M = a0 I + a1 ⋅ σ
When M is ρ → visualization of fictitious vector in fictitious B-field
When M is a term in H → idea that arbitrary operator can be
decomposed as sum of Ji.
types of operators
aJ  e.g. magnetic moment (a is a known constant or a function of r)
r
q  how to evaluate matrix elements (e.g. Stark Effect)
J1 ⋅ J 2  e.g. Spin - Orbit
TODAY:
1.
HSO + HZeeman as illustrative
2.
Dimension of basis sets JLSMJ⟩ and LMLSMS⟩ is same
3.
matrix elements of HSO in both basis sets
4.
matrix elements of HZeeman in both basis sets
5.
ladders and orthogonality for transformation between basis
sets. Necessary to be able to evaluate matrix elements of
HZeeman in coupled basis. Why? Because coupled basis set
does not explicitly give effects of Lz or Sz.
revised 4 November, 2002
5.73 Lecture #25
25 - 2
Suppose we have 2 kinds of angular momenta, which can be coupled to each other to
form a total angular momentum.
r
L
orbital 
r

S
spin operate on different coordinates or in different vector spaces
r r r

J = L+ S
total 
The components of L,S, and J each follow the standard angular momentum
commutation rule, but
r r
[L,S] = 0
,
[Ji,L j] = ih Σ ε ijkLk
[J ,S ] = i
k
Σ ε ijk Sk .
k
These commutation rules specify that L and S act like vectors wrt J but as scalars wrt
to each other.
r
J → jm j
r
L → lm
r
S → sms
i
j
h
l
Coupled jlsm j vs. uncoupled lm sms representations.
l
*
*
*
matrix elements of certain operators are more convenient in one basis set than the
other
a unitary transformation between basis sets must exist
limiting cases for energy level patterns
l and s will
each give a
factor of h
ζn
 SO
H = ξ(r )l ⋅ s ≡ h l ⋅ s

will give a factor of
anomalous
g
value
of
e
1.
H Zeeman = − γB (l + 2s ) ≡ −(ω )(l + 2s )
z
z
z
z
z
0

l
h
−
(ζ
and ω 0 are in rad / s
* evaluate matrix elements in both basis sets
nl
)
* look at energy levels in high field γBz >> ζ n limit
l
* look at energy levels in low field γBz << ζ n limit
l
revised 4 November, 2002
5.73 Lecture #25
25 - 3
−

lower case for 1e atom angular momenta
Notation: 
−

upper case for many - e angular momenta
two different CSCOs


(can©t be factored)  recall tensor product

uncoupled basis  states and “ entanglement ”
(explicitly factored) 
a) H elect , J 2 , J z , L2 , S2
coupled basis
nJLSM J
b) H elect ,
2
L
, L z , S2 , S z
nLM L SM S
2. Coupled and Uncoupled Basis Sets Have Same Dimension
COUPLED
r r r
J = L+S
L−S ≤ J ≤ L+S
each J has 2 J + 1 M J ’s
J
S
L
2( L + S ) + 1
2( L + S − 1) + 1
J = L+S
L + S −1
2( L + S − 2 ) + 1
LLLL
L+S−2
2( L − S ) + 1
every J contributes 2L + 1 to sum
If L > S , there are
2S + 1 terms in sum
(2S + 1)(2L + 1) + 2[S + (S − 1) + L (−S)] = (2S + 1)(2L + 1)
=0
⇑
total dimension
of basis set for
specified L,S
UNCOUPLED
LM L SM S
123123
2 L +1
2 S +1
total dimension (2 L + 1)(2 S + 1) again
term for term correspondence between 2 basis sets
∴ a transformation must exist:
Coupled basis state in terms of uncoupled basis states:
JLSM J = Σ a ML LM L SMS = M J − M L
ML
144
42444
3
constraint
Trade J, MJ for ML, MS, but MJ = ML + MS.
revised 4 November, 2002
5.73 Lecture #25
25 - 4
Uncoupled basis state in terms of coupled basis states:
L +S
OR
∑
LM J SMS =
b J JLSM J = M L + MS
1442443
J = L −S
3.
Matrix elements of HSO =
ζn
l
constraint
l
h
⋅s
A. Coupled Representation
r r r
J=L+S
J 2 = L2 + S2 + 2L ⋅ S
(useful trick!)
J 2 − L2 − S2
L⋅S =
2
J ′L′S′ M ′J L ⋅ S JLSM J =
(
h
2
)[
]
2 J (J + 1) − L( L + 1) − S(S + 1) δ J ′J δ L ′Lδ S ′Sδ M J′ M J
a purely diagonal matrix.
B. Uncoupled Representation
L ⋅ S = L zS z +
diagonal
(
1
L+ S − + L− S +
2
)
off-diagonal
L ′M′LS′ MS′ L ⋅ S LM LSMS = h 2δ L ′LδS ′S ×
can’t change L
can’t change S
1
1/ 2

[M LMSδ ML′ ML δ MS′ MS ] + [L (L + 1) − M′LM L ] ×

2
[S(S + 1) − M′ M ]
1/ 2
S
S
}
δ ML′ ML ±1 × δ MS′ MS m1
∆M L = −∆MS = 0, ±1
Nonlecture notes for evaluated matrices
S = 1 / 2,
L = 0,1, 2
2
S, 2P, 2D states
revised 4 November, 2002
5.73 Lecture #25
2 S +1
2
2
LJ
D
NONLECTURE for HSO : COUPLED BASIS
S1/ 2
HSO
COUPLED
=
P
HSO
COUPLED
=
L
( 2S1/ 2 ) 0
( 2 P1/ 2 ) 1
( 2 P3 / 2 ) 1
( 2 D3 / 2 ) 2
( 2 D5 / 2 ) 2
2
25 - 5
h
2
h
2
ζ ns (0)
ζ np
 −2
0

0

0
0

0
0
−2
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0

0
0

1
J = 1/2
J = 3/2
J
J(J+1) – L(L+1) – S(S+1) =
1/2 3 / 4 0 3 / 4 0
1 / 2 3 / 4 2 3 / 4 −2
3 / 2 15 / 4 2 3 / 4 +1
3 / 2 15 / 4 6 3 / 4 −3
5 / 2 35 / 4 6 3 / 4 +2
J =3/2
HSO
COUPLED
=
h
2
ζ nd
 −3 0 0 0
 0 −3 0 0

 0 0 −3 0
 0 0 0 −3

0 0 0 0
0 0 0 0

0 0 0 0
0 0 0 0

0 0 0 0

0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

0 0 0 0 0 0
0 0 0 0 0 0

2 0 0 0 0 0
0 2 0 0 0 0

0 0 2 0 0 0
0 0 0 2 0 0
0 0 0 0 2 0

0 0 0 0 0 2
J =5/2
center of gravity rule: trace of matrix = 0
(obeyed for all scalar terms in H)
revised 4 November, 2002
5.73 Lecture #25
2S +1
25 - 6
NONLECTURE for HSO : UNCOUPLED BASIS
L
2
S
HSO
UNCOUPLED
= hζ ns (1 2 ⋅ 0) = (0)
2
P
HSO
UNCOUPLED
= hζ np ×
ML
MS
1 / 2 1 / 2
−1 / 2  0
1/2  0

−1 / 2  0
1/2  0

−1 / 2  0
1
1
0
0
−1
−1
0
−1 / 2
0
2−1/2
2−1/2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2−1/2
−1 / 2
0
2−1/2
0





0 
0 

1 2
0
0
0
Each box is for one value of MJ = ML + MS.
2
HSO
UNCOUPLED = hζ nd ×
D
ML
2
2
1
1
0
0
−1
−1
−2
−2
MS
1 / 2 1

−1 / 2  0
1 / 2 0

−1 / 2  0
1 / 2 0

−1 / 2  0

1 / 2 0
−1 / 2  0

1 / 2 0
−1 / 2  0
0
−1
0
1
0
0
0
0
1
1/2
0
0
0
0
0
0
0
0
−1 / 2
(3 / 2)
1 /2
0
(3 / 2)
1 /2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
−1 / 2
0
0
1/2
0
1
0
0
0
0
1
0
−1
0
(3 / 2)
1 /2
0
(3 / 2)
1 /2
0
0
0

0
0

0

0
0

0

1
revised 4 November, 2002
5.73 Lecture #25
25 - 7
4. Matrix Elements of H Zeeman = − γBz (L z + 2Sz )
A. very easy in uncoupled representation
H Zeeman
′ L z + 2Sz LM LSMS
uncoupled = − γBz L ′M ′LS′ M S
= − γBz h(M L + 2MS )δ L ′LδS ′Sδ M′L ML δ MS′ MS
strictly diagonal
B. coupled representation
Lz + 2S z = J z + S z
easy hard — no clue!
can’t evaluate matrix elements in coupled
representation without a new trick
5. If we wish to work in coupled representation, because it diagonalizes HSO,
need to find transformation
JLSM J = Σ a ML LM LSMS = M J − M L
ML
lengthy procedure:
J ± = L ± + S±
and orthogonality
Always start with an extreme ML, MS basis state, where we are
assured of a trivial correspondence between basis sets:
M L = L,
M S = S,
M J = M L + M S = L + S, J = L + S
J = L + S LSM J = L + S = LM L = L SM S = S
coupled
uncoupled
revised 4 November, 2002
5.73 Lecture #25
25 - 8
MJ
J8
67
67
8
J − | L + S LS L + S⟩ = (L − + S− ) LM L = L SMS = S
( L + S )( L + S + 1) 
 (

− L + S )( L + S − 1)
1/ 2
[
]
+[S(S + 1) − S(S − 1)]
L + S LS L + S − 1 = L( L + 1) − L( L − 1)
1 /2
1 /2
LL − 1SS
LLSS − 1
Thus we have derived a specific linear combination of 2 uncoupled basis
states.
There is only one other orthogonal linear combination belonging to the
same value of ML + MS = MJ: it must belong to the L + S − 1
LS
L + S −1
basis state.
lower J
NONLECTURE
Work this out for 2P
JLSM J = 3 / 2 1 1 / 2 3 / 2 = LM LSMS = 1 1 1 / 2 1 / 2 \
21/ 2 1 0 1 / 2 1 / 2 + 1 1 1 / 2 −1 / 2
JLSM J − 1 =
31/ 2
now use orthogonality:
J − 1LSM J − 1 = 1 / 2 1 1 / 2 1 / 2 =
− 1 0 1 / 2 1 / 2 + 21/ 2 1 1 1 / 2 −1 / 2
.
31/ 2
Continue laddering down to get all 4 J = 3/2 and all 2 J = 1/2 basis states.
 2
3 / 2 1 1 / 2 −1 / 2 =  
 3
1 /2
 1
1 0 1 / 2 −1 / 2 +  
 3
1 /2
1 −1 1 / 2 1 / 2
3 / 2 1 1 / 2 −3 / 2 = 1 −1 1 / 2 −1 / 2
 1
1 / 2 1 1 / 2 1 / 2 = − 
 3
1 /2
 2
1 0 1 / 2 −1 / 2 +  
 3
1 /2
1 −1 1 / 2 1 / 2
You work out the transformation for 2D!
Next step will be to evaluate HSO + HZeeman in both coupled and
uncoupled basis sets and look for limiting behavior.
revised 4 November, 2002