Degenerate Perturbation Theory Question

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  • #1
T-7
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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)=[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
 

Answers and Replies

  • #2
pam
458
1
I don't see how your <12|V|21> integral could be zero.
 
  • #3
kdv
348
6
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:
  • #4
kdv
348
6
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)=[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

It does seem to me that you are right. Here the degeneracy is not lifted.
 
  • #5
pam
458
1
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.
 
  • #6
T-7
64
0
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.
 
  • #7
olgranpappy
Homework Helper
1,271
3
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.
 
  • #8
pam
458
1
|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.
 
  • #9
kdv
348
6
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...
 
  • #10
T-7
64
0
Thanks for your thoughts folks. Much appreciated. :-)
 

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