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5.80 Small-Molecule Spectroscopy and Dynamics
Fall 2008
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5.80 Lecture #15
Fall, 2008
2
Page 1 of 6 pages
2
Lecture #15: ∏ and ∑ Matrices
Last Time:
2 , A
,A
 ± on |A α M 〉 basis set
effect of A
A
i
case (a) basis set
|n(L) Λ S ∑〉 |v〉 |ΩJM〉
z, L
 ± , selection rules
L-destroyed, but not Λ: L
2 matrix elements
 ROT = B(R)R
H
 ROT nΛS∑ v ΩJM = δ δ δ δ δ δ δ δ
n′Λ′S′ ∑′ v′ ΩJM H
n′n Λ′Λ S′S ∑′∑ v′v Ω′Ω J′J M′M
Diagonal:
× Bv[J(J + 1) – Ω2 + S(S + 1) – ∑2 + L2⊥ ]
∆Ω = ∆∑ = ±1 within Λ–S multiplet state (S-uncoupling):
 ROT nΛS∑ v ΩJM = –B [J(J + 1) – (Ω ± 1)Ω]1/2[S(S + 1) –
nΛS∑±1 v Ω ±1JM H
v
(∑ ± 1)∑]1/2
**** In some of my handouts I call J + 1/2 = x. Here, I'll call it y *****
Here x = J(J + 1), y = J + 1/2
For example: Start by listing all relevant basis states.
Λ S
Σ
2
n 1 1/ 2 1/ 2
∏ 3/2
2
∏
2
n 1 1 / 2 −1 / 2
∏1/2
2
n −1 1 / 2 −1 / 2
∏−3/2
2
n −1 1 / 2 1 / 2
∏−1/2
 ROT ( 2 ∏) =
H
∏ 3/2
2
∏1/2
Bvπ ×
2
∏ −1/2
2
∏−3/2
⎛
x − 7
4
⎜
⎜ sym
Bvπ ⎜
⎜ 0
⎜ 0
⎝
2
x − 94 + 43 − 14
sym
0
0
−[x − 43 ]1/2 [ 43 + 14 ]1/2
x − 14 + 43 − 14
0
0
0
0
∆Ω = ±1
∆Λ = ±2
x − 14 + 43 − 14 − [x − 43 ]1/2
sym
⎞
⎟
x + 41
0
0
⎟
⎟
1/2
0
x + 41 ( x − 43 ) ⎟
0
sym
x − 47 ⎟⎠
Two identical blocks for Ω > 0 and Ω < 0 - later we will consider parity basis.
What about 2∑+? Class should do this.
(x − 43 )
1/2
0
0
0
0
1
9
x − 4 + 43 − 14
5.80 Lecture #15
Fall, 2008
Page 2 of 6 pages
∆Ω = ∆Λ = ±1 between Λ-S multiplet states (L-uncoupling)
 ROT nΛS∑ v ΩJM = –B [J(J + 1) – (Ω ± 1)Ω]1/2 × n′Λ ±1 L nΛ
n′Λ +1S∑ v′ Ω ±1JM H
v′v
±

β
a perturbation parameter to be determined by a fit to the spectrum. ↑
Today:
spin-orbit
 SO , H
 SS , H
 SR ⎧⎨effective operators
H
⎩
matrix elements
Matrix elements of 2∏, 2∑ effective H.
∆ S=0
 SO =
 ·S
H
→ AL
∑ a(ri ) i ·si ⎯for⎯⎯⎯
only
(restricted validity operator replacement)
i
spin-spin
2 ⎡⎣  2  2 ⎤⎦
∆ S=0
 SS = ⎯for
H
⎯⎯⎯
→
λ 3Sz − S + another term ∆ ∑ = − ∆ Λ = ±2
only
3
spin-rotation
 SR = γR
 ·S
H
usually λ, γ are very small with respect to A and are dominated by second-order spin-orbit effects (thru
van Vleck transformation)—discussed later
 SO is very important
H
SO

H =
∑
a(ri )̂ i ·ŝ i
i
electron (not
component)
⎛
⎞
⎜⎜ not
∑ a̂ i ·ŝ j
⎟⎟
⎝
⎠
i, j
L
S
i ,ŝ i are vectors with respect to J → i ·ŝ i is scalar (∆J = ∆M = ∆Ω = 0) with respect to J .
ŝ i is vector with respect to S → i ·ŝ i is vector with respect to S → ∆S = 0, ±1, ∆∑ = 0, ±1
i·si does not operate on |ΩJM〉, only on |nΛS∑〉; it is therefore NOT INDEPENDENT of Ω because, as
vector with respect to L and S, its matrix elements are not independent of Λ and ∑.
5.80 Lecture #15
Fall, 2008
Selection rules (ASSERTED)
Page 3 of 6 pages
∆J = 0
∆Ω = 0
+↔
/ – (LAB INVERSION I ) (parity)
g↔
/ u (body inversion î )
+
∑ ↔ ∑– (σv)
∆S = 0, ±1
∆∑ = –∆Λ = 0, ±1
 SO is a one-electron operator, so it has non-zero matrix elements only between electronic
H
configurations differing by a single spin-orbital. (e.g. π orbital = 1α, 1β, –1α, –1β spin-orbitals)
 is vector with respect to
Special simplification (due to simple form of Wigner-Eckart Theorem). If B
 ) with respect to angular momentum ( A
 , then ∆B = 0 matrix elements of a vector operator ( B
 ) may
A
 by b A
 (where b is a constant, often called a reduced matrix element)!
be evaluated by replacing B

a ( ri )  i is vector with respect to L
ŝ is vector with respect to S
i
For ∆L = 0, ∆S = 0 matrix elements
∑ a (r ) ·ŝ
i
i
i
 ·S
→ AL
(limited validity operator replacement)
i
SO
⎡
⎤
1

H = A ⎢L zSz + ( L +S− + L −S+ )⎥
⎣
⎦
2
E.g., for 2∏
⎛ ⎞ A
 SO 2 ∏
( ) 1
∏±3/2 H
±3/2 = A ±1 ⎜± ⎟ =
⎝ 2⎠ 2
⎛ ⎞
A
2
 SO 2 ∏
( ) 1
∏±1/2 H
±1/2 = A ±1 ⎜  ⎟ = −
⎝ 2⎠
2
all ∆Ω ≠ 0 matrix elements are = 0.
2
SS
2 ⎡⎣  2  2 ⎤⎦ 2 ⎡ 2

H ⎯⎯⎯
→ λ 3Sz − S = λ ⎣3∑ −S(S +1)⎤⎦ + additional term
∆ S=0
3
3
5.80 Lecture #15
Fall, 2008
Selection rules
∆S = 0
∆Ω = 0
∆S = 0
∆∑ = 0
+↔
/ –
g↔
/ u
+
∑ ↔
/ ∑–
Page 4 of 6 pages
[also ±1, ±2 neglected here]
[also ∆∑ = –∆Λ = ±2 (Λ-doubling in 3∏0 neglected here)]
SR

 ·S = γ ( J – L
 – S )·S = γ[J·S − L·S − S2 ]
H = γR



we already know how to
deal with all three of these!
Now we are ready to set up full 2∏, 2∑+ matrix. Start with all matrix elements of 2∏3/2 and then 2∏1/2 and
then 2∑1/2 etc.
v,n, ∏ 3/2
2
elect
vib
ROT
SO
SS
SR







H = H + H + H + H + H + H n, 2 ∏ 3/2 , v =
Λ∑
3∑2 – S(S + 1)
Ω ∑
Λ ∑ S(S+1)
2
2 ⎛ ⎛ 1⎞
3⎞
1
⎛31
2
Te ( n ∏ ) + G (v∏ ) + A ∏ (1·1 / 2 ) + λ ⎜ 3·⎜ ⎟ − ⎟ + γ ∏ ⎜ · − 1· −
⎝2 2
3 ⎝ ⎝ 2⎠
4⎠
2
2
–Ω
2
2
+S – ∑
3⎞
⎟
4⎠
always = 0 for
S = 1/2 states!
9 3 1
1
1
⎛
⎡
⎤
+Bv∏ ⎢ J(J + 1) − + − + L2⊥ ⎥ = E v∏ + A ∏ − γ ∏ + Bv∏ ⎜ J(J + 1) −
⎝
⎣
4 4 4
⎦
2
2
include with Te
+ G in E v∏
y ≡ J + 1/2, thus y is an integer since J is half-integer for 2∏ and 2∑.
Get same results for
2
 2∏
∏−3/2 H
−3/2 .
2
1
1
 2∏
+1)
+1 /4 ⎤
∏1/2 H
A∏ − γ∏ + Bv∏ ⎡J(J
1/2 = E v∏ −


2
2
⎢⎣
⎥⎦
y2
y2 – 2
7⎞
⎟
4⎠
5.80 Lecture #15
Fall, 2008
2
Get same results for
2
Page 5 of 6 pages
 2∏
∏−1/2 H
−1/2 .
 2∑
∑1/2 H
1/2 =
2
1 1
 2∑
∑−1/2 H
γ∑ + Bv∑ ⎡J(J
+1) −1 / 4 + 3 / 4 −1 
/ 4⎤
−1/2 = E v∑ − A ∑ 0· −
2 2
⎢⎣
⎥⎦
2
y
↑ always for ∑-states
[ASIDE: we have two explicit cases where, by evaluation of matrix elements, we see that
 Λ ∑Ω = −Λ − ∑−Ω H
 − Λ − ∑−Ω . But be careful, this is not true for
Λ ∑Ω H
 Λ = −Λ′ H
 − Λ ! Non-automatically-evaluable matrix elements.]
Λ′ H
Off-Diagonal Matrix Elements
Always ask what operator do we need to get non-zero matrix element between specified basis states?
2
∏ 3/2
 2∏
H
1/2 =
2
∏ −3/2
 2∏
H
−1/2
1
( J + S− + J − S + )
2
1/2
⎡1 ⎛
3⎞ ⎤
= −0A∏ + ⎢ ⎜ x − ⎟ ⎥ γ ∏ − Bv∏
4⎠ ⎦
⎣2 ⎝
∆Ω = 0
2
 2∏
∏ 3/2 H
−1/2 = 0
∆Ω = 2
2
 2∏
∏ 3/2 H
−3/2 = 0
∆Ω = 3
–BJ–L+
2
 2 ∑+
∏ 3/2 H
1/2
J+
⎡
3 1⎤
= − v∏ B(R) v∑ ⎢ J(J +1) −
⎣
2 2 ⎥⎦
B
1/2
v∏ v∑
= −βv∏ v∑ ⎡⎣y −1⎤⎦
2
2
 2∑
∏ 3/2 H
−1/2 = 0
∆Ω = 2
all done with 2∏3/2
2
 2∏
∏1/2 H
−3/2 = 0
1/2
1/2
1 3⎤ ⎡ 3 1 ⎛ 1⎞ ⎤
⎡
⎢⎣ x − 2 2 ⎥⎦ ⎢⎣ 4 − 2 ⎜⎝ − 2 ⎟⎠ ⎥⎦
[y2 – 1]1/2
1
∆Ω = 2
1/ 2
n∏ L + n′∑
β(∏,∑)
5.80 Lecture #15
Fall, 2008
from AL·S
2
Page 6 of 6 pages
from 2BL·S
⎡1
⎤
2
2
+
 2∑
∏1/2 H
1/2 = v∏ ∏1/2 ⎢ A + B(R)⎥ L +S− v∑ ∑1/2
⎣2
⎦
1/ 2 ⎡
= ⎣⎡S(S +1) − ∑∏ ∑∑ ⎤⎦ ⎢ v∏ v∑
⎣
= 1⎡⎣α v∏ v∑ + β v∏ v∑ ⎤⎦
–BJ–L+
2
 2∑
∏1/ 2 H
−1/ 2
n∏
⎤
A
L +
n′∑ + Bv∏ v∑ n∏ L + n′∑ ⎥
2
⎦
β
α
⎡
1 ⎛ 1 ⎞⎤
= −Bv∏ v∑ ⎢J(J +1) − ⎜− ⎟⎥
⎣
2 ⎝ 2 ⎠⎦
= −βv∏ v∑ y
y
1/2
n ∏ L + n′∑
2
all done with 2∏1/2
–BJ–S+
2
 2∑
∑1/2 H
−1/2
J+
1/2
1 ⎛ 1⎞ ⎤
⎡
= −Bv∑ ⎢ J(J + 1) − ⎜ − ⎟ ⎥
2 ⎝ 2⎠



⎢
⎥
2
y
⎣
⎦
= −Bv∑ y
S+
⎡3
⎢4 −
⎣
1/2
1 ⎛ 1⎞ ⎤
⎜− ⎟
2 ⎝ 2 ⎠ ⎥⎦
all done with 2∑1/2.
Are we done? Not quite. Must worry about 2∑+ ~ 2∏–3/2 and 2∑+ ~ 2∏–1/2 matrix elements. What
happens to the 〈∏|L+|∑〉 unevaluable factor? Need to consider effects of σv(xz) reflections and ∑+, ∑–
symmetry in order to get the correct relative signs of off-diagonal matrix elements.