5.73 Lecture #14
(See CTDL 1095-1107, 1110-1119)
14 - 1
Last time: derivation of all matrix elements for Harmonic-Oscillator: x, p, H
“selection rules”
“scaling” x n ij x n ii
∝ i n / 2 i j n in steps of 2
e.g. x
3
dimensionless quantities x
~ p
~
=

 m
ω h


1/ 2 x
= ( h m
ω ) −
1/ 2 p
H
~
=
1 h ω
H
“ annihilation ”
“ creation ”
“ number ”
“commutator” a
= a
† =
(
2
2
− /
−
( ) not aa
( )
†
)
−
~
[ a,a † ] = + 1 a n
= n n
−
1 a
† n
= ( n
+
1
) n
+
1 n
= n n modified 9/30/02 10:13 AM

5.73 Lecture #14
14 - 2 a little more: a
01
=
1
1/ 2 a
=





0
0
0
0
0
1 0
0
2
0
0
0 0 0 0
0
0
3 0
O





0 0 0 0 0 a n = selection rule for a n ij








0
0
L L L
0
L
0
L
( )
L
L j - i
= n selection rule for a
† ij j - i
= − n n
= [ ] −
/
( ) n
0
[ ( ) m ( ) n jk
]
= δ selection rule


(
( ) k
− n
)
!
(
( ) j
− m
)
!




0
( n
+
1
)
1 !
L
!


L
(one step to right of main diagonal)
0
0
L


( n
+ q
)
!
 q !









(n steps to right)
Selection rules are obtained simply by counting the numbers of a † and a and taking the difference.
The actual value of the matrix element depends on the order in which individual a † and a factors are arranged, but the selection rule does not.
Lots of nice tricks and shortcuts using a , a † and a † a modified 9/30/02 10:13 AM

5.73 Lecture #14
14 - 3
One of the places where these tricks come in handy is perturbation theory.
We already have: 1. WKB: local solution, local k(x), stationary phase
2. Numerov–Cooley: exact solution - no restrictions
3. Discrete Variable Representation: exact solution,
ψ as linear combination of H-O.
Why perturbation theory?
• replace exact H which is usually of ∞ dimension by H eff which is of finite dimension. Truncate infinite matrix so that any eigenvalue and eigenfunction can be computed with error < some preset tolerance.
Fit model that is physical (because it makes localization and coupling mechanisms explicit) yet parametrically parsimonious
• derive explicit functional relationship between the n-dependent observable and n e.g.
E n = ω e
( n
+
1 / 2
) − ω e x e
( n
+
1 / 2
) 2 + ω e y e
( n
+
1 / 2
) 3 hc
• establish relationship between a molecular constant ( ω e
, ω e x e
, …) and the parameters that define V(x) e.g.
ω e x e
↔ ax
3
There are 2 kinds of garden variety perturbation theory:
1. Nondegenerate (Rayleigh-Schrödinger) P.T. → simple formulas
2. Quasi-Degenerate P.T. → matrix H eff finite H eff is corrected for “out-of-block” perturbers by “van Vleck” or “contact” transformation
~4 Lectures
Derive Perturbation Theory Formulas
* correct E n
and ψ n
directly for
“neglected” terms in exact H
* correct all other observables indirectly through corrected ψ modified 9/30/02 10:13 AM

5.73 Lecture #14
14 - 4
Formal treatment
E n
= λ 0
E
(0) n
+ λ 1
E
(1) n
+ λ 2
E
(2) n
λ 2
ψ n
= λ 0 ψ (0) n
+ λ 1 ψ (1) n
λ 1
ψ × ψ ′
H
= λ 0
H
(0) + λ 1
H
(1)
λ is an order-sorting parameter with no physical significance. Set λ = 1 after all is done. λ = 0 → 1 is like turning on the effect of H (1) . Equations must be valid for
0 ≤ λ ≤ 1.
Plug 3 equations into Schr. Equation H
ψ n
= E n
ψ n
and collect terms according to order of λ .
λ 0
terms
H
( 0 ) ψ ( 0 ) n
=
E
( 0 ) n
ψ ( 0 ) n left multiply by
ψ (0) m
H
(0) mn
= E
(0) n
δ mn requires that H
(0)
be diagonal in
ψ eigenvalues
↓
{ }
(0) n
↓ and eigenfunctions
CALLED ZERO ORDER MODEL
of H
(0)
CALLED BASIS
FUNCTIONS modified 9/30/02 10:13 AM

5.73 Lecture #14
So we choose H (0) to be the part of H for which:
* it is easy to write a complete set of eigenfunctions and eigenvalues
* it is easy to evaluate matrix elements of common
“perturbation” terms in this basis set
* sometimes choice of basis set is based on convenience rather than “goodness” — doesn’t matter as long as the basis is complete.
examples: Harmonic Oscillator
Morse Oscillator
Quartic Oscillatr n-fold hindered rotor
V(x)
= 1
V(x)
=
[ kx
−
2 e
− ax
2 ]
V(x)
= bx
4
V n
(
φ
)
=
( )
(
1
− cos n
φ )
Now return to the Schr. Eq. and examine the λ 1 and λ 2 terms.
14 - 5
λ 1
terms
H
(
1
) ψ (
0
) n
+
H
(
0
) ψ (
1
) n
=
E
(
1
) n
ψ (
0
) n
+
E
(
0
) n
ψ (
1
) n multiply by
ψ (0) n from H operating to left
H
(1) nn
+
E
(0) n
ψ (0) n
ψ (1) n
=
E
(1) n
+
E
(0) n
ψ (0) n
ψ (1) n same
( could also require
ψ get rid of them
(0) n
ψ (1) n we do require this later
=
0
) modified 9/30/02 10:13 AM

5.73 Lecture #14
14 - 6
H
(1) nn
=
E
(1) 1st-order correction to E is just n expectation value of perturbation term in
H : H (1) .
return to
λ 1
equation and this time multiply by
ψ (0) m
H
(1) mn
H
(1) mn
ψ (0) m
+
E
(0) m
= ψ (0) m
ψ (0) m
ψ (1) n
ψ (1) n
(
E
(0) n
=
0
+
E
(0) n
−
E
(0) m
)
ψ (0) m
ψ (1) n
=
E n
H
(0)
(1) mn
−
E
(0) m completeness of
{ }
:
∑ k
ψ (0) k
ψ (1) n
ψ (0) k
ψ (1) n
= ∑ k
ψ (0) k
ψ (0) k
ψ (1) n but we know this
ψ (1) n
= ∑ ψ k
(0) k
E
(0) n
H
(1) kn
−
E
(0) k
* index of
ψ (1) n
matches 1st index in denominator
* n = k is problematic. Insist
Σ ′ k exclude k = n.
* we could have demanded
ψ (0) n
ψ ( ) n
1 =
0 modified 9/30/02 10:13 AM

5.73 Lecture #14
λ 2
terms
14 - 7 most important in real problems although excluded from many text books.
H
(
1
) ψ (
1
) n
=
E
(
1
) n
ψ (
1
) n
+
E
(
2
) n
ψ (
0
) n multiply by
ψ (0) n
ψ (0) n
ψ (1) n
=
0
ψ ( 0 ) n
H
( 1 ) ψ ( 1 ) n
↑
= 0 +
E
( 2 ) n completeness
k
( ) n
H
1
k
0
n
H
(1) n,k
Σ ′ k E
(
0
) n
H
(
1
) k,n
−
E
(
0
) k
E n
E
(
2
) n
=
k
E
(
0
) n
↑
H
(
1
) k,n
2
−
E
(
0
) k always first we have derived all needed formulas matrix element squared over energy difference in “energy denominator”
E
(0) n
, E
(1) n
, E
(2) n
;
ψ (0) n
,
ψ (1) n
Examples
V(x)
=
1
2 kx
2 + ax
3 (a < 0)
V(x)
H
(0) =
1
2 kx
2 p
2
+
2m
H
(1) = ax
3 x
(actually ax 3 term with a < 0 makes all potentials unbound. How can we pretend that this catastrophe does not affect the results from perturbation theory?) modified 9/30/02 10:13 AM

5.73 Lecture #14
14 - 8 need matrix elements of x 3 two ways to do this
* matrix multiplication x
3 i l
= ∑ x j,k ij x jk x k l
* a , a † tricks x
3 = 
 h m
ω 

= 
 h
2 m
ω 
 x
~
3 = 
 h m
ω 
 [
2
−
/
( a
+ a
†
) ]
3
[ a
3 +
( a aa
+ aa a
+ aaa
†
)
+
( aa a
+ a aa
† + a a a
)
+ a
† 3
] each group in ( ) has their own ∆ v selection rule (see pages 13-8 and 9): simplify using [ a , a † ] = 1
Goal is to manipulate each mixed a , a † term so that “the number operator” appears at the far right and exploit a
† a n
= n n
Only nonzero elements: a a
3 n
−3 n
†
3 n
+3 n
= [
[
( −1 ) ( n
− 2 ) ] 1
/
2
= ( n
+ 3 ) ( n
+ 2 ) ( n
+1 ) ] 1 / 2 square root of larger q.n.
( a aa
+ aa a
+ aaa
†
)
=
3 because a aa
= aa a
+
+
[ ]
[ ]
=
= aa a
−
+ a
,
† aa a a
( ) aaa
† n
−
1 n
=
= n modified 9/30/02 10:13 AM

5.73 Lecture #14
14 - 9
( aa
† a
†
3 a
+ a
† a
†
† aa
† + a a
+ 3 a
†
† a
† a n
+1 n
)
= 3 a
† a
† a
+ 3 a
†
= 3 ( +1 ) 1
/
2
+ 3 ( n
+1 ) 1
/
2
= 3 ( n
+1 ) 3
/
2
So we have worked out all x 3 matrix elements — leave the rest to P.S. #5.
Property other than E n
?
Use
ψ n
= ψ (
0
) n
+ ψ (
1
) n e.g. transition probability (electric dipole allowed vibrational transitions)
P n n ′
∝ x for H - O n n ′
2 x n n
′
2
(only
=
∆

 h
2(km)
1/ 2 m ω

 n >
δ n
>
,n
< n = ±1 transitions)
+ 1 for a perturbed H–O, e.g. H (1) = a x 3
ψ n
= ψ 0 n
ψ n
= ψ 0 n
+
Σ k
H
1 nk
E
( ) n
0 − E
( ) k
0
H
−
3
1 nn + 3 h ω
ψ ( ) n
0
+ 3
ψ ( ) k
0
+
H
1 nn + 1
− h ω
ψ 0 n + 1
+
H
+
1 nn − 1 h ω
ψ ( ) n
0
− 1
+
H
+
3
1 nn h
−
ω
3 ψ ( ) n
0
− 3 modified 9/30/02 10:13 AM

5.73 Lecture #14
14 - 10





1st index n n n n
+
+ n
−
−
3
1
1
3




 X




 n
Allowed
2nd Indices
+
4 , n
+
2 n n n
+
1 n n
−
,
+
2 ,
,
2 n
− n
, n
−
2
−
1
4





For matrix elements of X .
cubic anharmonicity of V(x) can give rise to ∆ n = ±7, ±5, ±4, ±3, ±2, ±1, 0 transition n x n
+
7
= x nn
+
7
2
≈
 h
 m
2 km m ω a
4
1/ 2
7
( )
ω 11 n
7


7/ 2 a
2
( −
3 h ω ) 2 
 ( n
+
7
)
!

1/ 2 n!

≈ n 7/2 other less extreme ∆ n transitions go as lower powers of
1
ω
and n modified 9/30/02 10:13 AM