Evaluating Commutator [x, p^2]: Need Help!

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The discussion focuses on evaluating the commutator [x, p^2] without using the momentum operator's definition. The user applies the rule [A, BC] = [A, B]C + B[A, C] and arrives at the expression [x, p^2] = [x, p]p + p[x, p], leading to a result of 2ihp. However, confusion arises regarding the presence of an operator in the outcome. Participants clarify that having an operator in the result is acceptable, confirming that the user's approach is correct. The conversation emphasizes the validity of the method used for evaluating the commutator.
syang9
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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?
 
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There's no reason you shouldn't have an operator in the result.

What you have done is fine.
 

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