1. Non-degenerate Perturbation Theory
Problem :
can't solve exactly.
But
with
Unperturbed eigenvalue problem. Can solve exactly.
Therefore, know and
called perturbations
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2. complete, orthonormal set of ket vectors
Solutions of
complete, orthonormal set of ket vectors
with eigenvalues and
Kronecker delta
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3. Substitute these series into the original eigenvalue equation
Expand wavefunction
and
also have
Have series for
Substitute these series into the original eigenvalue equation
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4. Sum of infinite number of terms for all powers of l equals 0.
Coefficients of the individual powers of l must equal 0.
zeroth order - l0
first order - l1
second order - l2
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5. First order correction
also substituting
Substituting this result.
Want to find and
Expand
Then
After substitution
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6. Therefore, the left side is 0.
After substitution
Left multiply by
unless n = i, but then
Therefore, the left side is 0.
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7. The first order correction to the energy.
We have
The first order correction to the energy.
(Expectation value of in zeroth order state )
number, kets normalized, and transposing,
Then
Absorbing l into
The first order correction to the energy is the expectation value of
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8. First order correction to the wavefunction
Again using the equation obtained after substituting series expansions
Left multiply by
Equals zero unless i = j.
Coefficients in expansion of ket in terms of the zeroth order kets.
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9. is the bracket of with and .
Therefore
The prime on the sum mean j  n.
zeroth order ket
energy denominator
correction to zeroth order ket
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10. First order corrections
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11. Second Order Corrections
Using l2 coefficient
Expanding
Substituting and following same type of procedures yields
l2 coefficients have been absorbed.
Second order correction due to first order piece of H.
Second order correction due to an additional second order piece of H.
Second order correction due to first order piece of H.
Second order correction due to an additional second order piece of H.
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12. Energy and Ket Corrected to First and Second Order
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13. 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.
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14. cubic “force constant” quartic “force constant”
harmonic oscillator – know solutions
zeroth order eigenvalues
zeroth order eigenkets
perturbation c and q are expansion coefficients like l.
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15. In Dirac representation
First consider cubic term.
Multiply out. Many terms.
None of the terms have the same number of raising and lowering operators.
(At second order will not be zero.)
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16. has terms with same number of raising and lowering operators.
Therefore,
Using
Only terms with the same number of raising and lowering operators are non-zero. There are six terms.
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17. Sum of the six terms Therefore 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.
With
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18. Perturbation Theory for Degenerate States
and
normalize and orthogonal
and
Degenerate, same eigenvalue, E.
Any superposition of degenerate eigenstates is also an eigenstate with the same eigenvalue.
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19. n linearly independent states with same eigenvalue
n linearly independent states with same eigenvalue system n-fold degenerate
Can form an infinite number of sets of Nothing unique about any one set of n degenerate eigenkets.
Can form n orthonormal
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20. Want approximate solution to
zeroth order
Hamiltonian
perturbation
zeroth order eigenket
zeroth order
energy
But is m-fold degenerate. Call these m eigenkets belonging to the m-fold degenerate E10
orthonormal
With
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21. zeroth order ket having eigenvalue,
Here is the difficulty
perturbed ket
zeroth order ket having eigenvalue,
But, 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.
The correct zero order ket depends on the nature of the perturbation.
p states of the H atom in external
magnetic field – p1, p0, p-1
electric field – px, pz, py
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22. Substituting the expansions for E and into
To solve problem
Expand E and
Some superposition, but we don’t know the cj. Don’t know correct zeroth order function.
Substituting the expansions for E and into
and obtaining the coefficients of powers of l, gives
zeroth order
first order
want these
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23. Use projection operator
To solve
substitute
Need
Use projection operator
The projection operator gives the piece of that is
Then the sum over all k gives the expansion of in terms of the
Defining
Known – know perturbation piece of the Hamiltonian and the zeroth order kets.
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24. Substituting this and gives
this piece becomes
Substituting this and gives
Result of operating H0 on the zeroth order kets.
Left multiplying by
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25. 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,
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
We don’t know the c’s and the
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26. is a system of m of equations for the cj’s.
One equation for each index i of ci.
Besides trivial solution of only get solution if the determinant of the coefficients vanish.
We know the
Have mth degree equation for the
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27. Get system of equations for the coefficients, cj’s.
Solve mth degree equation – get the Now have the corrections to energies.
To find the correct zeroth order eigenvectors, one for each , substitute (one at a time) into system of equations.
Get system of equations for the coefficients, cj’s.
Know the
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.
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28. The solutions to the mth degree equation (expanding determinant) are
Therefore, to first order, the energies of the perturbed initially degenerate states are
Have m different (unless some still degenerate).
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29. Correction to wavefunctions
Again using equation found substituting the expansions into the first order equation
Left multiply by
Orthogonality makes other terms zero. Normalization gives 1 for non-zero brackets.
gives gives 0
Therefore
Normalization gives Aj = 0 for j  m. Already have part of wavefunction for j  m
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30. First order degenerate perturbation theory results
Correct zeroth order function. Coefficients ck determined from system of equations.
Correction to zeroth order function.
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