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This is a question on perturbation theory - which I am trying to apply to the following example.

## Homework Statement

The two-dimensional infinitely deep square well (with sides at x=0,a; y=0,a) is perturbed by the potential V(x)=[tex]\alpha(x^{2}+y^{2})[/tex]. What is the first-order correction to the energy of the first excited states (that is, those with quantum numbers (1,2), (2,1))?

## The Attempt at a Solution

I have calculated [tex]\langle\left\right 1,2^{(0)}|\widehat{V}|2,1^{(0)} \rangle[/tex]. In Maple this is:

> value( Doubleint( (2/a)*sin(Pi*x/a)*sin(2*Pi*y/a)*(x^2+y^2)*(2/a)*sin(2*Pi*x/a)*sin(Pi*y/a) ,x=0...a,y=0...a) );

which is zero.

I have calculated

[tex]\langle\left\right 1,2^{(0)}|\widehat{V}|1,2^{(0)} \rangle[/tex]. In Maple this is:

value( alpha*Doubleint(4*sin(Pi*x/a)*sin(2*Pi*y/a)*(x^2+y^2)*sin(Pi*x/a)*sin(2*Pi*y/a),x=0..a,y=0..a) )

which is

[tex]\alpha a^{2}\left( \frac{16\pi^{2}-15}{24 \pi^{2}} \right)[/tex]

Am I to conclude from this that the perturbation, in *this* case, does not lift the degeneracy, and that the first order correction to the energy is the same for both states, and is given above?

I had expected to form a matrix, work out its eigenvalues and eigenvectors, use that to construct a new basis etc. But the matrix I would form from the above would already be diagonalised, with a repeated eigenvalue of 1.

I'd appreciate some speedy words of wisdom :-)

Thanks folks