Quantum Mechanics: Degenerate Perturbation Theory on square well

1. Jun 23, 2009

Evil Harry

1. The problem statement, all variables and given/known data
Hi I am trying to apply degenerate perturbation theory to a three dimensional square well v= 0 for x, y,z interval 0 to a, perturbed by H' = xyz (product) from 0 to a, otherwise infinite. I need to find the correction to energy of the first excited state which I know is triply degenerate. I am using Griffith's text book.

2. Relevant equations

3. The attempt at a solution

The problem I am having is in constructing the matrix. the one that looks like
Waa Wab
Wba Wbb

Or H11 H12
H21 H22
as some other text books calls it. I know for this problem I need to use a three dimensional matrix the above is just to clarify. Specifically my problem arises when i need to solve the values of the non diagonal elements. say Wab because this leads to an integral that I can't seem to solve the integral of say:
integrate x sin(Ax) sin(Bx) between 0 and a, where A and B are different because the wave functions are different say that n=1 for the argument in A and n=2 for the argument in B. so that the wave functions are $$\psi$$ 112 and $$\psi$$211. Now I am trying to avoid solving the integral having been unsucsesful in numerous attempts and instead use another approach of finding an operator that commutes with H' and H (the unpeturbed system). So my new problem is an operator that I can use. I have never before used this method and there isn't really a worked example in the book. So any help with this integral (see bottom) or in using this method will be much appreciated. How do I know which operator to use and since they have simultaneous eigenfunctions wouldn't the diagonal elements be the same as using non degenerate perturbation theory <ψ | H'|ψ > or am I now using different eigenfunctions than those of the square well?
E=∫_0^a▒〖xsin(xπ/a) sin⁡(2πx/a)dx〗
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 23, 2009

Avodyne

Your integral is of the form x sin(ax) sin(bx). There are many ways to do this. You could first use sin(ax)sin(bx) = (1/2)cos((a-b)x) - (1/2)cos((a+b)x), then you have to integrate x cos(cx), which can be done by integration by parts.

3. Jun 23, 2009

Evil Harry

Thanx for the help, I forgot about that identity, haven't used it in a while.