- second order corrections, and
- some problems with degeneracy.

In first order perturbation theory the small shift in energy is given by the following formula:

That is that the first order shift in the energy of the
*n ^{th}* energy eigenfunction
(

Calculating the second order shift in energy requires a bit more work:

The (unnormalized) perturbed wavefunction () is given by:

where |*k*>=_{k} are the unperturbed wavefunctions.

Consider then the (fairly bogus) example of a small additional (1-d) spring force:

*V*(*x*)=½*kx ^{2}*(1+)

*V'*(*x*)=½*kx ^{2}*

It's "fairly bogus" because we can exactly solve this problem (since it's just a slightly stronger spring problem and we've solved the spring problem), but thats the point: we can then compare the perturbation theory approximation to the exact answer.

The exact answer just amounts to a shift in our energy unit *e*.

*E _{n}*=(2

*e*=½(*k/m*)^{½}

(1+)^{½}=1 + ½
- ^{2}/8

Before we can calculate the second order shift we should calculate the first order shift. Thus we need to find the following integral:

<*n*|*x'*^{2}|*n*>

The virial theorem would give us that integral, but instead lets do it using the recursion relations.

Thus the first order shift in energy is itself proportional to the unperturbed energy:

Calculating a second order shift looks to be quite difficult: an infinite number of integrals must be performed and then we must sum an infinite number of terms. Luckily all of the integrals can be calculated via the recursion relation and via orthogonality and almost all of them are zero.

This results exactly reproduces what we expected from our Taylors
expansion of (1+)^{½} shown above.

The solution to the problem is to find a new basis for the
degenerate subspace such that <*k*|*V'*|*n*>
is zero. For example, in the 3-d oscillator at *E'*=7
there are 6 degenerate states (5 *d* [*l*=2] states
and 1 *s* [*l*=0] state). Any linear combination of these
states also solves Schrödinger's equation with *E'*=7.
With work, we can find six new wavefunctions (just linear combinations of the old
|*n _{r}*

The new basis will be found by finding the eigenvectors of
the matrix for *V'* on the degenerate subspace. So we need
to begin by finding all 36 3-d integrals:

<*n _{r}*

Our example perturbing potential:
*V'*=(*x*^{2}-*y*^{2})
can be usefully expressed in terms of the spherical harmonics: *Y*_{2 ±2}

The *Y _{lm}*(,)
part of the needed 36 integrals is easily found via the Wigner-Eckart
Theorem (Schiff p.223, many details in Edmonds

are called Wigner's 3-j symbols
and are essentially symmetric versions of the Clebsch-Gordan coefficients
(<*j*_{1}*m*_{1},*j*_{2}*m*_{2}|*j*_{3}*m*_{3}> a.k.a., vector coupling coefficients). (3-js should not be confused
with Racah 6-j symbols which are involved
in "recoupling" of angular momentum.) *Mathematica*
knows about all of the above functions; Abramowitz and Stegun (p.1006)
have formulas for some low *j* Clebsch-Gordan coefficients;
the particle data handbook has a nice table of
*j*<=2 Clebsch-Gordan coefficients (click here
for a printable postscript version of their table). Here are some basic properties
of 3-j symbols:

The key point in these formulas for the *Y _{lm}* integrals is that
<

In this example, our basis states are *l*=2 (5 |*lm*> states: |22>, |21>, |20>, |2-1>, |2-2>)
and *l*=0 (|00>) and the perturbing potential *V'* is essentially
*Y*_{2 2}+*Y*_{2 -2}. Using the above result for
*Y _{lm}* integrals only the following matrix elements are non-zero
(i.e., the potential "connects" only the listed states)

<22|*V'*|20> & <22|*V'*|00>

<20|*V'*|2-2> & <00|*V'*|2-2>

<21|*V'*|2-1>

A moment's thought should convince you that the first two rows of matrix elements
are related as *Y ^{*}_{lm}*=(-1)

Note that all diagonal matrix elements (i.e., expectation values) are zero as
*Y ^{*}_{lm}*

Note that the even-*m* basis states are totally unconnected to the
odd-*m* basis states, so our 6×6 matrix for *V'* has two blocks
placed along the diagonal (with zero connection):
2×2 matrix of odd-*m* basis states (which we order: |21>,|2-1>) and
4×4 matrix of even-*m* basis states (which we order: |22>, |20>, |00>, |2-2>)

The radial part of the integral must now be tackled. The angular part of the
perturbation *x*^{2}-*y*^{2} ended up in
*Y*_{2 2}+*Y*_{2 -2}, the radial part is
*r*^{2}, so we need to evaluate integrals like:

In the case *l*=*l'* we have exactly the "weighting function"
*x*^{l+½}*e*^{-x} needed
for the orthogonality of the Laguerre polynomials (i.e.,
=*l*+½). The recursion relation for
the Laguerre polynomials can then be used to convert
*x L _{n}*(

Note that we could have also derived this result via the Virial Theorem.

The other needed case is *l*=*l'*+2. The strategy
here is to steal one *x* from *r*^{2} and another
from the weight function and to re-express
*x*^{2}*L*^{l+½} in terms
of *L*^{l'+½} through a double use
of the formula (Abramowitz & Stegun 22.7.31 or Szegö 5.1.14):

Multiplying our *r* integral result by our ,
integral result gives us the required matrix elements. Here are the results:

*Mathematica's* `Eigensystem[m]` command gives us the eigenvectors
and eigenvalues of the matrix `m`. Here are the results.

Thus the 6 degenerate *E'*=7 states "split" into states with the following
energies: *E'*=7+2, *E'*=7+,
*E'*=7 (doubly degenerate), *E'*=7-, and
*E'*=7-2.

This result can be compared to the exact result since the problem is still exactly
the sum of three oscillators: *k _{x}*=(1+)

In the unperturbed system there are 6 combinations that produce
*E'*=7: (*n _{x}*,

For the four states that became non-degenerate in first-order, we can go on to find their second order correction in the usual way and compare the calculation to the exact results shown above. For the remaining degenerate pair more work is needed to find the "right" linear combination, see Schiff p.250.

Let us focus on the second order correction of a particular state: the
*E'*^{(1)}=+ state (|021>-|02-1>)/2^{½}=|*B*>.
In order to calculate the second order shift we need to find an infinite number
of integrals: <*k*|*V'*|*B*> where *k*
represents any *n _{r}lm* state. We don't have to worry
about states in the

<*B*|*V'*|*B*> =

The angular integral part of <*B*|*V'*^{2}|*B*> gives
4/21; the radial part 63/4, so the result is 3^{2}.

Putting the results together we find the same second order correction calculated
(much easier) via the *xyz* separation: ½^{2}.