Why Don't Angular Momentum and the Square of Momentum Commute?

[Legion81]

I have been told that L and P^2 do not commute, but I don't see why. It seems like the commutator should be zero.

$$<br /> \left[ \vec{L} , P^2 \right] = \left[ L^k , P_i P_i \right]<br /> = \left[ L_k , P_i \right] P_i - P_i \left[ L_k , P_i \right]<br /> = \left( - i \hbar \epsilon_{i}^{km} P_m \right) P_i - P_i \left( - i \hbar \epsilon_{i}^{km} P_m \right) <br /> = - i \hbar \epsilon_{i}^{km} \left( P_m P_i - P_i P_m \right) <br /> = - i \hbar \epsilon_{i}^{km} \left[ P_m , P_i \right] <br /> = 0<br />$$

What is wrong with this?

 

[Legion81]

Oh, I found one mistake but it still leads me to the same conclusion. The second equality should be plus, not minus: [A,BC] = [A,B]C+B[A,C]:

$$\left[ \vec{L} , P^2 \right] = \left[ L^k , P_i P_i \right] = \left[ L^k , P_i \right] P_i + P_i \left[ L^k , P_i \right] = \left( - i \hbar \epsilon_{i}^{k m} P_m \right) P_i + P_i \left( - i \hbar \epsilon_{i}^{k m} P_m \right) = - i \hbar \epsilon_{i}^{k m} \left( P_m P_i + P_i P_m \right)$$
and since $$P_m , P_i$$ commute,
$$\left[ \vec{L} , P^2 \right] = - 2 i \hbar \epsilon_{i}^{k m} P_m P_i = 2 \left[ L^k , P_i \right] P_i = 2 \left[ \vec{L} , P^2 \right]$$
but the only thing which satisfies
$$\left[ \vec{L} , P^2 \right] = 2 \left[ \vec{L} , P^2 \right]$$
is zero.
So I'm back to saying that L and P^2 commute. Any ideas?

 

[fzero]

Your calculations are correct. If $$\hat{L}_i$$ did not commute with $$\hat{P}^2$$, angular momentum would not be conserved for a free particle.

 

[Legion81]

fzero, I didn't even think of that! A free particle would have zero potential energy, so for L to be conserved it would still have to commute with the Hamiltonian, and that means L and P^2 must commute (2m factor is irrelevant). I thought for sure I was making a mistake somewhere since my mind is a notorious hangout for mathematical absurdities. Maybe I can get some points back that were taken off on my test. Thanks.