Computing [x, p²] without the momentum operator definition

[syang9]

i've been trying to evaluate this commutator the 'easy' way--that is, without using the definition of the momentum operator. the farthest i got was trying to use this rule..

[A, BC] = [A, B]C + B[A, C]

so..
[x, p^2] = [x, p]p + p[x, p]

so i guess i get 2ihp. but that doesn't make sense, b/c there's an operator in that result. so i don't get what else I'm supposed to do. can anyone help me out?

 

[James R]

There's no reason you shouldn't have an operator in the result.

What you have done is fine.