# Zeeman Effect - average shift is zero?

1. May 23, 2014

### unscientific

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

Part (a):Find the first order shift in energy
Part(b): What is the degeneracy after perturbation? Find Show average shift in energy is zero.

2. Relevant equations

3. The attempt at a solution

I've shown part (a), the troubling part is part (b).

Part (b)
With the perturbation, the degeneracy is lifted. Hence degeneracy = 0.

For $j=l+\frac{1}{2}$, shift is $\Delta E_+ = \frac{1}{4}mc^2 \alpha^4\frac{(l+\frac{1}{2})(l+\frac{3}{2}) - l(l+1) - \frac{3}{4}}{n^3 l(l+\frac{1}{2})(l+1)}$

For $j=l-\frac{1}{2}$, shift is $\Delta E_- = \frac{1}{4}mc^2 \alpha^4\frac{(l-\frac{1}{2})(l+\frac{1}{2}) - l(l+1) - \frac{3}{4}}{n^3 l(l+\frac{1}{2})(l+1)}$.

The total perturbation is given by the sum of positive perturbation and negative perturbation:

$$\delta E= \Delta E_+ + \Delta E_-$$
$$\delta E = \frac{mc^2 \alpha^4}{2n^3 l}$$

Clearly, the average perturbation for a given n and l is not zero. How does it even become zero then?

2. May 23, 2014

### dauto

j = l + 1/2 and j = l - 1/2 have different degeneracies. They contribute different weights to the average.

Last edited: May 23, 2014
3. May 23, 2014

### unscientific

Each has a weight of $2l-1$?

4. May 23, 2014

### dauto

No, they have weights 2j+1 for two different values of j

5. May 23, 2014

### unscientific

Oh yeah, because when two angular momenta $j_1$ and $j_2$ are combined, each state in $J = j_1 + j_2$ ranges from -J to +J in steps of 1.

I keep forgetting that this is a two-angular momentum problem, not a simple one with only spin.

$$\delta E = (2j_+ +1)\Delta E_+ + (2j_- +1)\Delta E_-$$

I worked it out, and it appears to be zero.

Last edited: May 23, 2014