# Degenerate Perturbation Theory Question

• T-7
In summary, the conversation discusses a question on perturbation theory and the application of it to a specific example of a two-dimensional infinitely deep square well. The potential V(x)=\alpha(x^{2}+y^{2}) is perturbed and the first-order correction to the energy of the first excited states is calculated. It is found that the perturbation does not lift the degeneracy and the first-order correction to the energy is the same for both states. The conversation also includes a discussion on the calculation of integrals and the resulting conclusion is that the potential does not lift the degeneracy due to the symmetry of the unperturbed potential and the perturbation treating x and y on equal footing.
T-7
Hello,

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)=$$\alpha(x^{2}+y^{2})$$. 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 $$\langle\left\right 1,2^{(0)}|\widehat{V}|2,1^{(0)} \rangle$$. 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

$$\langle\left\right 1,2^{(0)}|\widehat{V}|1,2^{(0)} \rangle$$. 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

$$\alpha a^{2}\left( \frac{16\pi^{2}-15}{24 \pi^{2}} \right)$$

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

I don't see how your <12|V|21> integral could be zero.

pam said:
I don't see how your <12|V|21> integral could be zero.

Why not?

The integral of sin(x) sin(2x) x^2 is zero since it's odd with respect to the center of the well and the integral sin(x) sin(2x) y^2 is obviously zero.

EDIT: Oops, I was wrong on this since I was thinking about a well going from -a/2 to a/2,

Last edited:
T-7 said:
Hello,

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)=$$\alpha(x^{2}+y^{2})$$. 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 $$\langle\left\right 1,2^{(0)}|\widehat{V}|2,1^{(0)} \rangle$$. 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

$$\langle\left\right 1,2^{(0)}|\widehat{V}|1,2^{(0)} \rangle$$. 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

$$\alpha a^{2}\left( \frac{16\pi^{2}-15}{24 \pi^{2}} \right)$$

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

It does seem to me that you are right. Here the degeneracy is not lifted.

Did you leave out the x^2+y^2 in the integrals from 0 to a?
If you are integrating from -a/2 to +a/2, then change your question.

pam said:
Did you leave out the x^2+y^2 in the integrals from 0 to a?
If you are integrating from -a/2 to +a/2, then change your question.

Hi,

I have done the integral on paper and on Maple. On Maple I put:

> 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) );

The (x^2+y^2) is there, as you can see. The result is zero (which is what I also get using paper and pen).

Since the well is defined from 0 to a, I have used those as the limits of the integrals.

pam said:
I don't see how your <12|V|21> integral could be zero.

|1> and |2> are orthogonal. So,

<1,2|x^2 + y^2 |2,1>

=<1|x^2|2><2|1> + <1|2><2|y^2|1>

= 0 + 0

= 0

OP: It's quite reasonable that the potential doesn't lift the degeneracy since the unperturbed potential (the length of the box) is the same in x and y and the perturbation also treats x and y on equal footing.

olgranpappy said:
|1> and |2> are orthogonal. So,

<1,2|x^2 + y^2 |2,1>

=<1|x^2|2><2|1> + <1|2><2|y^2|1>

= 0 + 0

= 0

OP: It's quite reasonable that the potential doesn't lift the degeneracy since the unperturbed potential (the length of the box) is the same in x and y and the perturbation also treats x and y on equal footing.
That's right! Sorry to have said the opposite. I didn't notice the x^2 did not enter the y integral.

pam said:
That's right! Sorry to have said the opposite. I didn't notice the x^2 did not enter the y integral.

Ah yes..Ok, my mistake too. Of course, the x integral does not give zero but the y part will give zero so the whole thing is zero. My mistake too...

Thanks for your thoughts folks. Much appreciated. :-)

## What is Degenerate Perturbation Theory?

Degenerate Perturbation Theory is a mathematical approach used in quantum mechanics to calculate the energy and properties of systems in which there are multiple states with the same energy level.

## When is Degenerate Perturbation Theory used?

Degenerate Perturbation Theory is used when the system being studied has degenerate energy levels, meaning two or more states have the same energy.

## How does Degenerate Perturbation Theory work?

Degenerate Perturbation Theory involves treating the degenerate states as a single system and then applying perturbation theory to calculate the energy and properties of the system.

## What are the limitations of Degenerate Perturbation Theory?

Degenerate Perturbation Theory is only applicable to systems with degenerate energy levels and is not suitable for systems with non-degenerate energy levels.

## What are the applications of Degenerate Perturbation Theory?

Degenerate Perturbation Theory is commonly used in quantum mechanics to study systems such as atoms, molecules, and nuclei. It is also used in other areas of physics, such as solid state physics and condensed matter physics.

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