non-degenerate perturbation theory problem : can't solve exactly. but with unperturbed...

Non-degenerate Perturbation Theory

Problem :

n n nH E

can't solve exactly.

But0 2' "H H H H

0 0 0 0Lim 0 n n nH E

with

Unperturbed eigenvalue problem.Can solve exactly.

0nE 0

n

2' "H H

Therefore, know and .

called perturbationsCopyright – Michael D. Fayer, 2007

Solutions of

0n 0 0 0

0 1 2, ,

0 0n m mn

0 0 0 0n n nH E

complete, orthonormal set of states

with eigenvaluesand

0 0 00 1 2, , ,E E E

Kronecker delta

1

0nm

n m

n m

Copyright – Michael D. Fayer, 2007

Expand wavefunction

0 2n n n n

0 2n n n nE E E E

and

H nnE

Have series for

Substitute these series into the original eigenvalue equation

n n nH E


0 2' "H H H H

also have

0 00 0 0 0 00 ' 0'n n n n n n n n nH E H H E E

0 0 0 20' "n n n n n n n n nH H H E E E

0

Sum of infinite number of terms for all powers of equals 0.

Coefficients of the individual powers of must equal 0.

0 0 0 0 0n n nH E

0 0 0 0' " 0n n n n n n n n nH H H E E E

0 0 0 0' 0n n n n n nH H E E

zerothorder - 0

firstorder - 1

secondorder - 2


0 0 0'n nn n nH E E H

nE n

First order correction

Want to find and .

n

0i i

in c

Expand

0 0 0 0 0n i i i i i

i i

H c H c E

Then

Substituting this result.

also substituting

After substitution

0 0 0 0'i i n i n ni

c E E E H


After substitution

0 0 0 0'i i n i n ni

c E E E H

0n

0 0 0 0 0 0'n i i n i n n ni

c E E E H

Left multiply by

0 0 0 0 0 0'i i n n i n n ni

c E E E H

0 0 0n i

0 0 0n nE E

unless n = i,but then

Therefore, the left side is 0.


0 0' 0n n nE H

We have

0 0 0 0' 0n n n n nE H

0 0'n n nE H The first order correction to the energy.

Then0

n n nE E E

0n n nE E E

0 0'n n n nnE H H

Absorbing into and nn nnH E

The first order correction to the energy is the expectation value of . 'H


0 0 0 0'i i n i n ni

c E E E H

First order correction to the wavefunction

Again using the equation obtained after substituting series expansions

0 0 0 0 0 0'j i i n i j n ni

c E E E H

0 0 0 0'j j n j n nc E E E H

0 0 0 0'j j n j nc E E H

0 0

0 0

'j n

j

n j

Hc j n

E E

0jLeft multiply by

Equals zero unless i = j.

Coefficients in expansion of ket in terms of the zeroth order kets.Copyright – Michael D. Fayer, 2007

0 0

0 0

'j n

j

n j

Hc j n

E E

0 0

'jnj

n j

Hc

E E

'jnH 'H 0j 0

nis the bracket of with and .

0 00 0

'( )

jnn n j

j n j

H

E E

Therefore

The prime on the sum meanj n.

zeroth order ket correction to zeroth order ket

energy denominator


First order corrections

0 0 0 00 0

'( )

jnn n j jn j n

j n j

HH H

E E

0 0 0' 'n n nn nn n nE E H H H


Second Order Corrections

Using 2 coefficient

Expanding nn

0 0' ni in

n nni n i

H HE H

E E

Substituting and following same type of procedures yields

2 coefficients have been absorbed.

Second order correction dueto first order piece of H.

Second order correction due to anadditional second order piece of H.

0

20 0 0 0 0 0

' ' km mn nn knn k

k m n k n m n k

H H H H

E E E E E E

00 0

'( )

knk

k n k

H

E E

Second order correction dueto first order piece of H.

Second order correction due to anadditional second order piece of H.

0 0 0 0ni in n i i nH H H H


0 0' ni in

nni n i

H HH

E E

0 'nnE E H

0 00 0

'( )

jnn n j

j n j

H

E E

0

20 0 0 0 0 0

' ' km mn nn knk

k m n k n m n k

H H H H

E E E E E E

00 0

'( )

knk

k n k

H

E E

Energy and Ket Corrected to First and Second Order


x

E

Example: x3 and x4 perturbation of the Harmonic Oscillator

Vibrational potential of molecules not harmonic.

Approximately harmonic near potential minimum.

Expand potential in power series.

First additional terms in potential after x2 term are x3 and x4.


22 3 41

2 2

pH k x c x q x

m

cubic “force constant”

quartic “force constant”

20 21

2 2

pH k x

m

0 1

2H aa a a

00

1

2E n

harmonic oscillator – know solutions

n

zeroth order eigenvalues

zeroth order eigenkets

3 4'H c x q x perturbationc and q are expansion coefficients like .

0When and 0,c q H H Copyright – Michael D. Fayer, 2007

'nnH n H n

3 4n c x q x n

3 4c n x n q n x n

1

20

2x a a

k

In Dirac representation

33x a a

First consider cubic term.

3 2 3, , , .a a a aa a a

3 0n x n

Multiply out. Many terms.

None of the terms have the same number of raising and lowering operators.


2 2 44 0

24n x n n a a n

k

4

a a

4 0n x n

has terms with same number of raising and lowering operators.

Therefore,

1 2n aaa a n n n

1n a a aa n n n

21n aa aa n n

2n a aa a n n

1n aa a a n n n

1n a aaa n n n

Using1/ 2 1/ 21 and ( 1) 1a n n n a n n n

Only terms with the same number ofraising and lowering operators arenon-zero.There are six terms.


426( 1 2)n a a n n n

2 220

2

3 1

2 2nn

qH n n

k

Sum of the six terms

Therefore

0 k m 2 4 20k m

22

0 2 20

1 3 1

2 2 2nE n q n nm

With

Energy levels not equally spaced.

Real molecules, levels get closer together – q is negative.

Correction grows with n faster than zeroth order term

decrease in level spacing.


1 1H E

Perturbation Theory for Degenerate States

2 2H E

1 2and

normalize and orthogonal

1 2and Degenerate, same eigenvalue, E.

1 1 2 2

1 1 2 2 1

If

with

c c

c c c c

H E

Any superposition of degenerate eigenstates is also an eigenstatewith the same eigenvalue.


n linearly independent states with same eigenvaluesystem n-fold degenerate

Can form an infinite number of sets of .

Nothing unique about any one set of n degenerate eigenkets.

i

Can form n orthonormal from the degerate .i nn


0 ' j j jH H E

0 0 0 0j j jH E

Want approximate solution to

zeroth orderHamiltonian

perturbation

0

zeroth ordereigenket

zeroth orderenergy

0 0 01 2, m

0iEBut is m-fold degenerate.

Call these m eigenkets belonging to the m-fold degenerate E0

0 0 0 01 2 1mE E E E

orthonormal

With


0i i

Here is the difficulty

0

perturbed ket zeroth order ket having eigenvalue, 01E

0i

0 0 0 01 1 2 2i m mc c c

0i 0 .iBut, is a linear combination of the

We don’t know which particular linear combination it is.

is the correct zeroth order ket, but we don’t know the ci.


To solve problem

i 01E E E

0

1

m

i j j ij

c

Expand E and

Some superposition, but we don’t know the cj.Don’t know correct zeroth order function.

0 0 0 01

1 1

m m

j j j jj j

H c E c

0 0 01

1

'm

i j jj

H E c E H

Substituting the expansions for E and into

0 ' i i iH H E

i

and obtaining the coefficients of powers of , gives

zerothorder

firstorder

want theseCopyright – Michael D. Fayer, 2007

0i k k

k

A

0 0 01

1

'm

i j jj

H E c E H

To solve

substitute

0' jH

0 0 0 0' 'j k k jk

H H

0 0k k

0' jH 0k

Need

Use projection operator

The projection operator gives the piece of that is .0iThen the sum over all k gives the expansion of in terms of the .

0' jH 0 0' 'kj k jH H

0 0' j kj kk

H H

Defining Known – know perturbation piece of theHamiltonian and the zeroth order kets.


0 0 01

1

'm

i j jj

H E c E H

0 0

1 1

m m

j j j kj kj j k

c H c H

this piece becomes

0i

0 0 0 0 0 0 0 01

1 1

m m

k k i k j i j j kj i kk j k j

E E A E c c H

Left multiplying by

Substituting this and gives 0

i k kk

A

0 0 0 0 01

1 1

m m

k k k j j j kj kk j k j

E E A E c c H

Result of operating H0 on the zeroth order kets.


0 0 0 0 0 0 0 01

1 1

m m

k k i k j i j j kj i kk j k j

E E A E c c H

Correction to the Energies

Two cases: i m (the degenerate states) and i > m.

i m

Left hand side – sum over k equals zero unless k = i.But with i m,

0 0 0 01 1 0 Therefore, i iE E E E

The left hand side of the equation = 0.

Right hand side, first term non-zero when j = i. Bracket = 1, normalization.Second term non-zero when k = i. Bracket = 1, normalization.

The result is

1

0m

ij j ij

H c E c

We don’t know the c’s and the . E sCopyright – Michael D. Fayer, 2007

1

0m

ij j ij

H c E c

is a system of m of equations for the cj’s.

11 1 12 2 1 0m mH E c H c H c

21 1 22 2 2 0m mH c H E c H c

1 1 2 2 0m m mm mH c H c H E c

One equation for each index i of ci.

1 2 0mc c c Besides trivial solution ofonly get solution if the determinant of the coefficients vanish.

11 12 1

22 2

1 2

0

m

m

m m mm

H E H H

H E H

H H H E

0 0jk j kH H

We know the

Have mth degreeequation for the .E s


Solve mth degree equation – get the . Now have the corrections to energies.

11 1 12 2 1 0m mH E c H c H c

21 1 22 2 2 0m mH c H E c H c

1 1 2 2 0m m mm mH c H c H E c

iE s

iETo find the correct zeroth order eigenvectors, one for each , substitute (one at a time) into system of equations.

iE

Get system of equations for the coefficients, cj’s.

1 1 2 2 1* * *, , m mc c c c c c

There are only m – 1 conditions because can multiply everything by constant.Use normalization for mth condition.

Now we have the correct zeroth order functions.

ijH Know the .


1, 2, mE E E

01 1i iE E E i m

The solutions to the mth degree equation (expanding determinant) are

Therefore, to first order, the energies of the perturbed initially degeneratestates are

Have m different (unless some still degenerate).iE s

01With

as 0iE E


0 01

1

m

k k j kjj

E E A c H

Correction to wavefunctions

Again using equation found substituting the expansions intothe first order equation

0 0 0 0 01

1 1

m m

k k k j j j kj kk j k j

E E A E c c H

0i i k m

Left multiply by

Orthogonality makes other terms zero. Normalization gives 1 for non-zero brackets.

gives 1 gives 0

1

0 01

m

j kjj

k

k

c H

A k mE E

Therefore

Normalization gives Aj = 0 for j m.Already have part of wavefunction for j m


First order degenerate perturbation theory results

10 0

0 01

m

j kjj

i i kk m k

c H

E E

01i iE E E

Correct zeroth order function.Coefficients ck determined fromsystem of equations.

Correction tozeroth order function.


non-degenerate perturbation theory problem : can't solve exactly. but with unperturbed...

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