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

## Homework Statement

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.

## Homework Equations

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

## 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.

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

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?

vela
Staff Emeritus
Homework Helper
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##.