For a free particle of mass m, with Hamiltonian
ˆH=ˆP22m,
where ˆP=−iℏ∂∂x.
The commutative relation is given by
[ˆX,ˆH]=iℏmˆP
In the common eigenstate of ˆH and ˆP, |e,p⟩, can we do the following?
⟨e,p|[ˆX,ˆH]|e,p⟩=⟨e,p|ˆX(ˆH|e,p⟩)−(⟨e,p|ˆH)ˆX|e,p⟩=⟨e,p|ˆX(e|e,p⟩)−(⟨e,p|e)ˆX|e,p⟩=e(⟨e,p|ˆX|e,p⟩−⟨e,p|ˆX|e,p⟩)=0
Since the ˆH is Hermitian, the above derivation doesn't seem to show any flaw. Given the commutative relation, Eq (1), we know the result is wrong. What's wrong with the above derivation?
[EDIT]
Following the comment by Luboš Motl, I have worked out the solution and would like to share it here. The link provided by Qmechanic had the solution closely related to this question.
⟨e′,p′|[ˆX,ˆH]|e,p⟩=⟨e′,p′|ˆX(ˆH|e,p⟩)−(⟨e′,p′|ˆH)ˆX|e,p⟩=(e−e′)⟨e′,p′|ˆX|e,p⟩
Note that:
e−e′=p22m−p′22m=(p+p′)(p−p′)2m
⟨e′,p′|ˆX|e,p⟩=−iℏδ′(p−p′)
where δ′(⋅) is the derivative of the Dirac function, with respect to p.
Then we get
(e−e′)⟨e′,p′|ˆX|e,p⟩=−iℏ(p+p′)2m⋅(p−p′)δ′(p−p′)=−iℏ(p+p′)2m⋅(−δ(p−p′))=iℏ(p+p′)2mδ(p−p′)
As we take the limit p→p′:
limp→p′iℏ(p+p′)2mδ(p−p′)→iℏmpδ(p−p′)
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