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

fandango92

## 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.

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

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?

Staff Emeritus