# Why eigenvalues of L_x^2 and L_z^2 identical?

1. Oct 6, 2013

### fandango92

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

Calculate the eigenvalues of the $L_x^2$ matrix.
Calculate the eigenvalues of the $L_z^2$ matrix.
Compare these and comment on the result.

2. Relevant equations

$L_x=\frac{1}{2}(L_+ + L_- )$

3. The attempt at a solution

I have derived eigenvalues for each: $0$ and $\hbar^2$ for both $L_x^2$ and $L_z^2$. But why are they identical? I'm finding it difficult in qualitatively explaining why the eigenvalues are the same for both.

2. Oct 6, 2013

### Oxvillian

Because the God of Physics does not care which direction you call the $x$-direction and which direction you call the $z$-direction.

3. Oct 6, 2013

### fandango92

Sorry I forgot to mention this is for $l=1$.

Okay, but I used $L_z$ eigenvalues of $m\hbar$, where $m=-1,0,1$ in this case, and used $L_x=\frac{1}{2}(L_+ + L_- )$. I have called the z component the one in which is certain, so how can the x component squared in this case have the same eigenvalues as the z component squared?

4. Oct 6, 2013

### vela

Staff Emeritus
The operators will have the same eigenvalues for the reason Oxvillian said, but that's not saying anything about the state a particle is in. If a particle is in an eigenstate of $\hat{L}_z$, it's not in an eigenstate of $\hat{L}_x$.