# Eigenvalues and -functions

1. Mar 8, 2005

### broegger

I having trouble finding the eigenvalues and eigenfunctions for the operator

$$\hat{Q} = \frac{d^2}{d\phi^2},$$ ​

where $$\phi$$ is the azimuthal angle. The eigenfunctions are periodical,

$$f(\phi) = f(\phi + 2\pi),$$ ​

which I think should put some restrictions on the eigenvalues.

I think that the eigenfunctions are complex exponentials, and that the eigenvalues are 0,-1,-2,..., but I am not sure if this is correct. Also I have to determine if the spectrum is degenerate, that is, if two (or more) distinct eigenfunctions correspond to the same eigenvalue.

2. Mar 8, 2005

### Kane O'Donnell

You're right about the complex exponentials and wrong about the eigenvalues, but finding the e-values isn't hard, just substitute a complex exponential into the equation and see what happens.

Those numbers ...-1, 0, 1, 2, ... do parametrize the equation - they turn out to be quantum numbers, not eigenvalues. Make sure you understand why they have to be integers.

Degeneracy should follow from what the eigenvalues look like, so calculate them first.

Kane

3. Mar 8, 2005

### marlon

$$\frac{d^2 f}{d\phi^2} = Af$$

Where A is an eigenvalue. Now solve this diff equation and make a clear distinction between A positive and negative. Also keep in mind that the eigenfunctions need to be periodical.

marlon

4. Mar 8, 2005

### dextercioby

Okay.Better solve the ODE and then impose the periodicity condition.

Daniel.

5. Mar 8, 2005

### broegger

This is my solution-attempt:

For negative eigenvalues we have:

$$f(\phi) = e^{-iq\phi}$$​

This correspond to eigenvalues $$\lambda = -q^2$$. Since $$f(\phi)$$ is $$2\pi$$-periodical we have:

$$1 = e^{-iq2\pi}.$$​

This implies that $$q = 0, \pm1, \pm2, \ldots$$, which again implies that the eigenvalues are $$\lambda = 0, -1, -4, -9, \ldots$$. There are no positive eigenvalues.

Last edited: Mar 8, 2005
6. Mar 8, 2005

### dextercioby

Are u sure...?
$$\frac{d^{2}f(\varphi)}{d\varphi^{2}}=\lambda f(\varphi)$$

I'm getting some square roots...

Daniel.

7. Mar 8, 2005

### broegger

No, I'm not sure, not at all

From where do you get the square roots? If you differentiate twice you get $$(-iq)^2 = -q^2$$?

8. Mar 8, 2005

### dextercioby

??Well,the characteristic equation is
$$r^{2}=\lambda$$

,with the solutions:
$$r_{1,2}=\pm \sqrt\lambda$$

Daniel.

9. Mar 8, 2005

### broegger

I don't get it :( What do you get for the eigenvalues exactly?

10. Mar 8, 2005

### Galileo

The eigenvalues need not be real a a priori, but the periodicity implies it.

Since at first the eigenvalue lambda can be anything at all, best let $\lambda = -m^2$, where $m$ can be any complex number.

11. Mar 8, 2005

### broegger

$$\hat{Q}$$ is hermitian, so the eigenvalues must be real - isn't that correct? If we apply the condition $$\lambda = -m^2$$, we get the eigenvalues 0, 1, 4, 9,... since the periodicity implies that $$m = 0, \pm1, \pm2,...$$ Right?

12. Mar 8, 2005

### Galileo

Yeah, I overlooked the Hermiticity.

The eigenvalues are nonnegative as you said, so you get 0,-1,-4 etc.

13. Mar 8, 2005

### broegger

Thank you, Galileo. And the rest of you too :)

Last edited: Mar 8, 2005