5.73 Lecture #26
26 - 1
LMLSMS⟩
HSO + HZeeman in JLSM

J⟩ and 
Last time:
HSO = ζl ⋅ s
(
HZeeman = −γBz LZ + 2S Z
)
→ JLSM J
→ LM L SM S
OK to set up H in either basis
problem about HZeeman in Coupled Basis → need to work out explicit transformation
between basis sets to evaluate matrix elements of HZeeman in coupled basis.
Today:
1.
Ladders and Orthogonality method for JLSMJ⟩ ↔ LMLSMS⟩
(coupled↔uncoupled) transformation, term by term.
2.
evaluate HZeeman in coupled basis for 2P state.
3.
Correlation Diagram, Noncrossing Rule
* simple patterns without calculations
* guidance for “intermediate case”
War between two limits
* one term creates ∆Eij( 0 ) ≠ 0
* other term causes Hij(1) ≠ 0
The two terms play opposite roles in the two basis sets.
4.
Stepwise picture of level structure working out from 2 opposite limits
* strong spin-orbit, weak Zeeman
* strong Zeeman, weak spin-orbit
Distortions from limiting patterns (via 2nd-order nondegenerate perturbation
theory) give the “other” (pattern distorting) parameter.
How does a zero-order picture identify the “picture defining” and the “picture
destroying parameters.
updated November 4, 2002
5.73 Lecture #26
26 - 2
(
)
(
HZeeman = −γBz Lz + 2S z = −γBz J z + S z
)
OK Not OK
for coupled basis
to evaluate matrix elements of Lz or S z in JLSM J
need to work out JLSM J =
∑ a m LM L SM S = M J − M L
L
ML
ladders and orthogonality
J ± = L± + S±
begin with " extreme" J = L + S M J = J ; M L = L, M S = S
basis states where there is a 1:1 correspondence
JLSM J = J = LM L = L SM S = S
=L+S
(
)
J − L + S L S L + S = L− + S − LL SS
[( L + S)( L + S + 1) − ( L + S)( L + S − 1)]1/ 2 L + S L S L + S − 1 =
[ L( L + 1) − L( L − 1)]1/ 2 LL − 1 SS + [S( S + 1) − S( S − 1)]1/ 2 LL SS − 1
L + S L S L + S −1 =
[2L ]1/2 LL − 1
c
LL SS − 1
1 /2
for 2P L = 1, S = 1/2
3 2 1 12 12
1 /2
[2(L + S )]
 L 
L + S L S L + S −1 = 

 L + S
 2
=
 3
SS + [2S ]
1/2
1 /2
 S 
LL − 1 SS + 

 L + S
1 1
1 0
2 2
u
 1
+
 3
1/2
11
1 1
−
2 2
1 /2
LL SS − 1
u
orthogonality: there are only 2 MJ = 1/2 possibilities, J = 3/2 and J = 1/2 for specified
L,S
updated November 4, 2002
5.73 Lecture #26
1 11
1
1
= − 
 3
2 22
1/ 2
26 - 3
11
2
10
+ 
 3
22
1/ 2
11 1 2 − 1 2
(I always choose – sign in front of smaller coefficient - arbitrary phase choice)
Nonlecture: Summary for 2 P
(coupled → uncoupled)
11
3 13
1
= 11
2 22
22
3 11
 2
1
=
 3
2 22
1/2
1 11
 1
1
=−
 3
2 22
11
 1
10
+
 3
22
1/2
1/2
11
 2
10
+
 3
22
 (2 /3)1/2 
1 1
11 −
=
1/2 
2 2
 (1 /3)  u
1/2
 −(1 /3)1/2 
1 1
11 −
=
1/2 
2 2
 (2 /3)  u
continue down to MJ = –1/2, MJ = –3/2 (or start at MJ = –3/2 and ladder up.
Now work in Coupled Representation for HSO + HZeeman
ζ
HSO + HZeeman = n l ⋅ s − γBzJ z − γBzS z
l
h
diagonal, easy
2

h
s
J (J + 1) − L( L + 1) − S(S + 1)
l
⋅
=

recall 
2
J z = hM J
[
]
off diagonal in J
(can’t be off
diagonal in
L,S,MJ - WHY?)
For 2P, there are 2 2 × 2 MJ = 1/2 and MJ = –1/2 blocks.
J
Evaluate
L S
MJ
3 /2 1 1 /2 1 /2 S z 3 /2 1 1 /2 1 /2
by inserting the above transformation into the uncoupled basis set.
L ML S MS
2

1
 1 0 1 / 2 1 / 2 S z 1 0 1 / 2 1 / 2 + 1 1 1 / 2 −1 / 2 S z 1 1 1 / 2 −1 / 2  =
3

3
1
 2   1   1   1  

h
 3   2  +  3   − 2   = h  6 
THERE ARE NO OFF-DIAGONAL MATRIX ELEMENTS OF HZeeman IN
THE UNCOUPLED BASIS SET.
updated November 4, 2002
5.73 Lecture #26
2
3 / 2 1 1 / 2 1 / 2 Sz 1 / 2 1 1 / 2 1 / 2 = −
3
Sz =h
1 /2
 1 2
 +
 2 3
1 /2
26 - 4
 1
2
−  = −
 2
3
etc. for Sz (2P)
Basis State

1 / 2
1/ 2


1 / 6 −2 / 3


sym −1 / 6


1/ 2


−1 / 6 +2 / 3


sym
1/6




−1 / 2

3/2 1
1 /2
1/2 3 /2
3/2 1
1/2 1
1/2 1/2
1/2 1/2
3 / 2 1 1 / 2 −1 / 2
1 / 2 1 1 / 2 −1 / 2
3 / 2 1 1 / 2 −3 / 2
(trace Sz = 0)
2 2 × 2’s and 2 1 × 1’s
If we had worked in uncoupled representation, HZeeman would have been fully diagonal but HSO

Zeeman
produces ∆E(ij0 ) ≠ 0 across which HSO has H(ij1) ≠ 0.
would be 2 2 × 2's and 2 1 × 1's. H
If we work in coupled representation, HSO is diagonal but H Zeeman is off - diagonal.
 SO
Zeeman
(0)
has Hij(1) ≠ 0.
H produces ∆Eij ≠ 0 across which H
WAR:
HSO tries to force system to coupled representation
HZeeman tries to force system to uncoupled representation
2 Limiting Cases: weak and strong spin-orbit.
updated November 4, 2002
5.73 Lecture #26
26 - 5
extreme level patterns
Bz → 0,ζ n >> 0
l
COUPLED
(connect same — MJ states)
Bz >> 0,ζ n → 0
l
UNCOUPLED
(M
2
1 / 2ζ np
P3 / 2
)
(–1, –1/2)
Bzγ
(0, –1/2)
0
P1/ 2
,MS
2Bzγ
0
2
L
−ζ np
(1, –1/2), (–1, 1/2)
− Bzγ
(0, 1/2)
−2Bzγ
(1, 1/2)
correlation diagram:
noncrossing rule for MJ: why?
states of same MJ do not cross
In coupled (strong spin - orbit) picture HZeeman is H(1)
In uncoupled picture (strong field) HSO is H(1)
2
P matrices for HSO + HZeeman
coupled
MJ = 3 / 2
MJ = 1 / 2
uncoupled
ζ / 2 − 2γBz

2
j = 3 / 2 ζ / 2 − γBz
3


sym
j = 1 / 2 

2
γBz 
3


1
−ζ − γBz 

3
ζ 2 − 2γBz
1 /2
ML,MS =
(1,−1 / 2)  −ζ 2 2 ζ 
(0,1 / 2)  sym − γB 
−1/2
z
CONTINUED AT BOTTOM OF PAGE 26-6
updated November 4, 2002
5.73 Lecture #26
26 - 6
Spin orbit only
Βz♦ 0
(J, MJ)
MJ
Zeeman only (ML, MS)
2Bγ
(–1, –1/2)
–3/2
Bγ
–1/2
(3/2, –3/2)
(0, –1/2)
ζ/2
(3/2, –1/2)
(3/2, +1/2)
(3/2, +3/2)
0
1/2
0•
+3/2
(1, –1/2)
(–1, 1/2)
–1/2
–Bγ
–ζ
(1/2, –1/2)
(1/2, +1/2)
(0, 1/2)
1/2
–2Βγ
(1, 1/2)
There are only repulsions (shown by vertical arrows) between same-MJ pairs.
coupled
uncoupled
1/ 2
E
±
M J =1 / 2
( γBz )2 − γBzζ 
 ζ γB   9
= − − z ±  ζ2 +

 4
2  16
4
4 
M J = −1 / 2 j = 3 / 2  ζ / 2 + 2γBz

3

sym
j = 1 / 2 
E
±
M J = −1/2
2 1/ 2

γBz

3
1 
−ζ + γBz 
3 
−
 ζ γB   9
1
=  − + z  ±  ζ 2 + γBz
 4
2  16
4
E MJ = −3 / 2
E± = same as coupled (even
though matrices are different)
( )
ζ 2 + 2γBz
2
γB ζ 
− z 
4 
 γB z 2 −1/2 ζ


 sym − ζ 2 
1 /2
E ± = same as coupled
ζ 2 + 2γBz
updated November 4, 2002
5.73 Lecture #26
26 - 7
Energy eigenvalues come out to be identical (as they must) in both
representations.
Coupled picture is good for 2nd order perturbation theory in the weak field
(|γΒ| << ζnl) limit.
Zeeman splitting (and tuning rates,
H
coupled
first order
picture
correction
( 0)
=H
E (1) = H Zee
SO
dE
, as B z is varied)
dB z
second order
correction
E
( 2)
=
2γΒz
H Zeeman
ij
2
E (i 0 ) − E (j0 )
MJ = –3/2
+2/3 γΒz
+
4
27
( γBz )2
ζ
MJ = –1/2
ζ 2
}
MJ = 1/2
+ 274
( γBz )2
ζ
–2/3 γΒz
MJ = 3/2
–2γΒz
∆MJ = 0
matrix elements
1/3 γΒz
MJ = –1/2
–ζ
}
− 274
( γBz )2
ζ
MJ = 1/2
–1/3 γΒz
– 274
( γBz )2
ζ
updated November 4, 2002
5.73 Lecture #26
26 - 8
r r
Uncoupled picture is good for strong field limit where Bz causes L + S to uncouple
r
from J and into the laboratory (Paschen - Back limit).
ZERO-ORDER
FIRST-ORDER
SECOND-ORDER
(ML, MS)
(–1, –1/2)
2γΒz
MJ
(–3/2)
+ζ/2
(–1/2)
γΒz
0
(0,–1/2)
2
+ζ /2γΒz
(1,–1/2)
(–1,1/2)
(1/2)
+ζ /2γΒz
–ζ/2
2
–ζ2/2γΒz
(–1/2)
–γΒz
(0,1/2)
–ζ2/2γΒz
(1/2)
–2γΒz
(1,1/2)
Regular
Zeeman
Pattern.
(3/2)
+ζ/2
Small distortions
from equal
intervals, from
which ζ may be
determined.
Extra distortions.
Repulsions
between SameMJ components
This is a regular Zeeman pattern, but with small distortions
(shown as vertical arrors on expanded scale) from equal
intervals, from which ζ may be determined.
updated November 4, 2002