# A quantum mechanical rigid roator

1. Jan 18, 2014

### hokhani

1. The problem statement, all variables and given/known data

Consider the Hamiltonian for a rigid rotator, constrained to rotate in xy plane and with moment of inertia I and electric dipole moment$\mu$ in the plane, as $H_0=\frac{L_z^2}{2I}$. Then suppose that a constant and weak external electric field,$E$, in the direction of x is applied so the perturbation is $H'=-\mu E cos (\phi)$.

2. Relevant equations

The eigenfunctions of $H_0$ are double degenerate and the degeneracy is not removed even in high orders of perturbation. How should I solve it? could anyone please help me?

3. The attempt at a solution

Last edited: Jan 18, 2014
2. Jan 18, 2014

### Staff: Mentor

The problem statement doesn't include a question.

Have you calculated $\langle m' | H' | m \rangle$? And found it to be always zero for $m'\neq m$?

3. Jan 18, 2014

### hokhani

The degenerate eigenfunctions are $|m\rangle=\frac{1}{\sqrt {2\pi}} e^(im\phi)$ and $|-m\rangle=\frac{1}{\sqrt {2\pi}} e^(-im\phi)$ and $\langle -m|cos(\phi)|m\rangle =0$ (ms are integer). Hence all the elements of perturbation matrix between degenerate kets would be zero.

Last edited: Jan 18, 2014
4. Jan 18, 2014

### Staff: Mentor

I made a quick calculation, and it seems indeed that the degeneracy is not lifted. How is that a problem? The field still shifts the levels.

5. Jan 18, 2014

### hokhani

Could you please give me your solution? Or at least, how much does the electric field move the levels?

6. Jan 18, 2014

### Staff: Mentor

You should calculate $\langle m' | H' | m \rangle$ yourself. Rewriting $\cos \phi$ in terms of exponentials helps in evaluating the integrals.

7. Jan 18, 2014

### hokhani

Ok, it is very easy to calculate it:
$\langle m' | H' | m \rangle=-\frac{\mu E}{2}(\delta(m',m+1)+\delta(m',m-1))$
Making the $H'$ matrix on the two degenerate kets, m and -m, we observe that all the four matrix elements are zero. Could you please tell me what is my mistake?

8. Jan 19, 2014

### Staff: Mentor

I'm sorry, but the I don't understand what question you are trying to answer.