# Homework Help: Commutator of L^2 with x_i

1. Feb 6, 2010

### mat1z

Task: The task is to compute the commutator of L^2 with all components of the r-vector. It seems to be an unusual task for I was unable to find it in any book.

Known stuff: I know that $$[L_i,x_j]=i \hbar \epsilon_{ijk} x_k$$ ($$\epsilon_{ijk}$$ being the Levi-Civita symbol). Now I would go about as follows (summation implied):

My attempt: $$[L^2,x_j]=[L_iL_i,x_j]=L_i[L_i,x_j]+[L_i,x_j]L_i=i \hbar \epsilon_{ijk} (L_i x_k + x_k L_i)$$

Question: Is this correct? Is there a physical meaning to this result other than that they do not commute (with the usual implications)?

Regards, Matti from Germany

Last edited: Feb 6, 2010
2. Feb 6, 2010

### Maxim Zh

Yes, it's correct.

When two operators commute their eigenvectors coincide. In quantum mechanics it means that the corresponding physical magnitudes can be measured simultaneously.

3. Feb 6, 2010

### mat1z

So, does that mean that the only real information contained in commutators as such (and not their application, as in constructing angular momentum eigenstates) is in it being zero or not? It's clear to me that zero commutators imply instantaneous measurability but I thought maybe there was something to be seen from the "value" that the commutator takes...

4. Feb 6, 2010

### jdwood983

Not instantaneous measurements, but simultaneous diagonalization--you will be able to find eigenstates of both A and B if [A,B]=0.
Having a non-zero commutators means the observables cannot be simultaneously known (cf. Heisenberg's uncertainty principle).

5. Feb 6, 2010

### mat1z

Sorry, simultaneous measurements is what I was trying to say, of course. Thanks.