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

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The discussion focuses on evaluating the commutator [x, p^2] without directly using the momentum operator's definition. The user applies the commutation rule [A, BC] = [A, B]C + B[A, C] and arrives at the expression [x, p^2] = [x, p]p + p[x, p], ultimately yielding 2ihp. The conclusion emphasizes that having an operator in the result is acceptable and expected in quantum mechanics.

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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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