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
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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.
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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.
Copyright – Michael D. Fayer, 2007
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.
Copyright – Michael D. Fayer, 2007
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.
Copyright – Michael D. Fayer, 2007
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.
Copyright – Michael D. Fayer, 2007
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.
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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.
Copyright – Michael D. Fayer, 2007
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.
Copyright – Michael D. Fayer, 2007
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.
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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 .
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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
Copyright – Michael D. Fayer, 2007
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.
Copyright – Michael D. Fayer, 2007
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