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.

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

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.

 

[paranormal]

There are a few things that are wrong with this reasoning. First, the commutator of two operators is generally not equal to the product of the operators times the commutator of their components. In this case, the commutator of L and P^2 cannot be simplified to the commutator of L and P, as P^2 is a composite operator made up of P and itself.

Additionally, the commutator of two operators is not necessarily zero, even if one of the operators is the square of the other. In this case, while P^2 is the square of P, L and P do not have a simple relationship that would result in a zero commutator.

Finally, the commutator of L and P^2 is not equal to the commutator of P^2 and L. The order in which operators are written in the commutator matters and can change the result. In this case, the commutator of L and P^2 is not equal to zero, but the commutator of P^2 and L is. This is a fundamental property of commutators and is related to the non-commutative nature of operators in quantum mechanics.

In conclusion, the commutator of L and P^2 is not zero, and this does not contradict the fact that L and P^2 do not commute. It is important to carefully consider the properties of operators and their commutators when making statements about their relationships.