Lecture 14 - 1 5.73 #14

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5.73

Lecture

#14

Perturbation Theory I

(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

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

0

0

0

1

0

0

0

0

2

0

a

n

= selection rule for a n

ij

0

0

L

0

L L

0

L

j - i

selection rule for a

ij j - i

= n

L

L

L

= − n n

[

=

[ ] −

/

( ) n

( ) m ( ) n

] jk

0

=

δ selection rule

( k

( )

− n

)

!

( j

( )

− j !

m

)

!

0

0

0

3

0

0

0

0

O

0



( n

0

+

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

usually stop at

λ

2

ψ n

= λ

0

ψ

(0) n

+ λ

1

ψ

(1) n

usually stop at

λ

1

(because all observables involve

ψ × ψ ′

) order-sorting is MURKY

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)

V(x)

=

2

[ kx

2 bx

4 e

− ax

2

]

V(x)

=

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) n

1st-order correction to E is just 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

=

∑ k

ψ

(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

ψ

n

H

ψ

k n

=

E n

H

(1) n,k

Σ ′ k

E

(

0

) n

H

(

1

) k,n

E

(

0

) k

E

(

2

) n

=

Σ

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)

H

(1)

=

1

2 kx

2

= ax

3 p

2

+

2m 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

need matrix elements of

x

3 two ways to do this

* matrix multiplication x

3 i l

= ∑ j,k x ij x jk x k l

*

a

,

a

tricks

x

3 = 

 h m

ω



= 

 h

2 m

ω



x

~

3 = 

 h m

ω



[

a

3 +

(

+

[

2

/

(

a

+

a

]

)

3

+

aaa

)

+

(

+ † +

14 - 8

)

+

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

=

=

[

[

( n

−1

)

( n

+ 3

) ( n

− 2

+ 2

) (

) n

]

1

/

2

+1

)

]

1

/

2 square root of larger q.n.

(

+ +

aaa

)

= because

=

3

aa a

+

( )

aaa

n

1 n

† =

= n

+

[ ]

=

= −

a

+

a

modified 9/30/02 10:13 AM

5.73

Lecture

#14

14 - 9

(

aa

a

[

3

a

+

a a

aa a

+

+ 3

a a

a

]

n

+1 n

a

)

= 3

a

a

a

= 3

(

+ 3

a

+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 n n

2 for H - O 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

E

( )

n

0

H

1

nk

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 n

Allowed

2nd Indices n

+ n

+

,

4 ,

+

1 n n

2 ,

, n

2 , n

− n

+ n

2

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

+ n!

7

)

!

1/ 2



n

7/2 other less extreme

∆ n transitions go as lower powers of

1

ω

and n

modified 9/30/02 10:13 AM

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