Time-Independent Perturbation Theory

Prepared by: James Salveo L. Olarve Graduate Student

January 27, 2010

IntroductionThe presentation is about how to solve the approximate new energy levels and wave functions to the perturbed problems by building on the known exact solutions to the unperturbed case.

The intended reader of this presentation were physics students. The author already assumed that the reader knows Dirac braket notation.

This presentation was made to facilitate learning in quantum mechanics.

The Perturb Hamiltonian

The Hamiltonian of a quantum mechanical system is written

Here, is a simple Hamiltonian whose eigenvalues and eigenstates are known exactly.

We shall deal only with nondegenerate systems; thus to each discrete eigenvalue there corresponds one and only one eigenfunction

And will be the additional term (can be due to external field)

EE

n

nE

Task:

To find how these eigenkets and eigenenergies change if a small term (an external field, for example) is added to the Hamiltonian, so:

So on adding

Assumption:

In perturbation theory we assume that is sufficiently small that the leading corrections are the same order of magnitude as itself, and the true energies can be better and better approximated by a successive series of corrections, each of order compared with the previous one.

Strategy:

Expand the true wave function and corresponding eigenenergy as series in

It is more convenient to introduce dimensionless parameter λ

The series expansion

match the two sides term by term in powers of λ (taking λ=1).

Zeroth Term:

First Order Correction:

Second Order Correction: nEnEnEnHnH nnn2''2''2

Eq. 1

Eq. 2

Matching the terms linear in on both λ sides (taking λ =1)

Taking the inner product of both sides with

Since it is normalized

The Hamiltonian is a Hermittian Operator

Eigenvalue Equation

1 oooo nn

*T

HH

nEnHn

First Order Correction

Eq. 1

First Order CorrectionSo,

We find first order correction

for energy

1D continuous spectrum

nEnnEnnHnnEn nnn''''

nnEnnEnHnnnE nnn''''

dzHEn

'*'

'''' nnEnnEnnEnHn nnn

nHnEn''

First Order Correction nEnEnHnH nn

''''''''

nHEnEH nn'''

mcn nm

nm'

nHEmcEH nnmn

nm

''

mEmH m

nHEmcEE nnmnm

nm

''

Solving for the 1st order change in the wave function

Since form a complete set thenn

The eigenvalue equation for unperturbed state m

nHlnlEmlcEE nnmnm

nm

''

Taking inner product with l

First Order Correction

Cases: I.

)( nl

nHlnlEmlcEE nnmnm

nm

''

1 nlnl So,

0 mlml

nHnnHlEn'''

Now,

Cases: II.

)( nl So, 0 nlnl

1 mlml

nHlcEE nlnl

'

mn

nm

lnnl

nl EE

nHmc

EE

nHl

EE

nHlc

'''

Therefore the wave function correction to first order is:

mEE

nHmn

mnnm

'

'

Second Order Correction

Taking the inner product with yieldsn

nEnnEnnEnnHnnHn nnn2''2''2

nEnEnEnHnH nnn2''2''2

nnEnnEnnEnHnnnE nnnn2''2''2

mnnm

n EE

nHmmHnnHnE

'

'''2

Now, '''2'2 nnEnHnE nn

But, 0)(' mn

nm

nm

nm

nmcmncnn

mnnm

n EE

nHmE

2'

2Second order correction for energy

Eq. 2

Finally,The Eigenenergy

The Wave Function

...2'0 nnnn EEEE

...

2'

'0

mnnm

nn EE

nHmnHnEE

...' nnn

...'

m

EE

nHmnn

mnnm

Twofold Degeneracy

When the unperturbed states are degenerate then two or more distinct states share the same energy. As a consequence of that the ordinary perturbation theory fails.

Suppose that:

Note that any linear combination of these states, ba 0

0 baba

is still an eigenstate of , with the same eigenvalue H E

aaH E bbH E

Typically the perturbation will break the degeneracy

We can’t even calculate the first-order energy because we don’t know what unperturbed states to use.

'H

Twofold DegeneracyThe “good” unperturbed states in the general form EH

with'HHH

...22' EEEE

...22'

Plugging these and collecting like powers of ...... '''' EEEHHH

'''' EEHH

Taking the inner product with a

'''' EEHH aaaa

'''' aaaa EEEH

Note: ba 0

Twofold Degeneracy

Then, baabaa EH ''

baaabaaa EEHH '''' ''' EHH baaa

0 baba

Let: aaaa HW ' baab HW '

'EWW abaa

where: jiij HW ' baji ,,

Twofold DegeneracySimilarly, the inner product with yields

b

'EWW bbba

Multiplying at the right hand sideabW abbbba WEWW '

ababbbabba WEWWWW ' ababbbabba WEWWWW '

from'EWW abaa aaab WEW '

aaaabbabba WEEWEWWW '''

0''' 2

aaaabbbbabba WEEWWEWWW

Twofold Degeneracy 0''' 2

aaaabbbbabba WEEWWEWWW

0''2 baabbbaabbaa WWWWWWEE

a) Suppose 0Then, baabbbaabbaa WWWWWWEE ''2

2

42' baabbbaabbaabbaa WWWWWWWWE

2

42' baabbbaabbaa WWWWWWE

here,

baabbbaabbbbaaaabaabbbaabbaa WWWWWWWWWWWWWW 424 222

*22 42 ababbbbbaaaa WWWWWW

Twofold Degeneracyb) Suppose 0

ba 0 1

From:'EWW abaa 00 abW'EWW bbba '0 EWbb

Which is consistent with'E

aabbaabbaa WWWWWE 2'

2

1

bbbbaabbaa WWWWWE 2'

2

1

The states and were already the “correct” linear combinations.

The answers for are precisely what we would have obtained using nondegenerate perturbation theory.

IMPLICATION: a

b

'E

In matrix form:

'E

WW

WW

aaba

abaa

Evidently the are nothing but the eigenvalues of the -matrix.

And the “good” linear combinations of the unperturbed states are the eigenvectors of W.

'E W

Reference:

Retrieved from http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/Time_Ind_PT.htm, January 19 2010, Michael Fowler.

Introduction to Quantum Mechanics. David J. Griffiths. 1994