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?

 

[vela]

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 $\hat{L}_x$.