For one dimensional quantum mechanics [ˆx,ˆp]=iℏ.
Does this fix univocally the form of the ˆp operator? My bet is no because ˆp actually depends if we are on coordinate or momentum representation, but I don't know if that statement constitutes a proof. Moreover if we choose ˆxψ=xψ is the answer of the following question yes?
For the second one
(ˆxˆp−ˆpˆx)ψ=xˆpψ−ˆpxψ=iℏψ,
but I don't see how can I say that ˆp must be proportional to ∂∂x. I don't know if trying to see that ˆp must satisfy the Leibniz rule and thus it should be proportional to the x derivative could help. Or using the fact that ˆx and ˆp must be hermitian
Any hint will be appreciated.
Answer
No. You can add an arbitrary constant shift (or an arbitrary operator commuting with x) without affecting the CCR.
For 1-dimensional QM, the general solution of the CCR with ˆx represented as multiplication by x on wave functions with argument x is ˆp=ˆp0−A(ˆx) , where ˆp0 is the canonical momentum operator , and A(x) is an arbitrary function of x.
Proof. The difference ˆA:=ˆp0−ˆp commutes with ˆx, hence is a function of ˆx.
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