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For commutator $\langle p|[\hat{x},\hat{p}]|p\rangle$, we can compute it as

$$\langle p|[\hat{x},\hat{p}]|p\rangle = \langle p|\hat{x}\hat{p}|p\rangle-\langle p|\hat{p}\hat{x}|p\rangle=p\langle p|\hat{x}|p\rangle-p\langle p|\hat{x}|p\rangle=0,$$

but also $$\langle p|[\hat{x},\hat{p}]|p\rangle = \langle p|i\hbar|p\rangle=i\hbar,$$

where is the problem?

J.-H.
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  • The issue is p or x eigenstates are not directly normalisable states. Thus if you actually compute $\langle p|\hat{x}|p\rangle$, it would formally be infinite (very non-rigorous). You can do a semi-formal back of the envelope calculation by replacing one of the p eigenstate with p' to see that the full expression is actually neither 0 not i.hbar but infinite. One can make this rigorous using the so called Gelfand triple formalism. – Arnab Apr 09 '21 at 05:03
  • @Arnab Thank you, I was also confused of $\langle p | \hat{x} | p \rangle$, it is helpful to know Gelfand triple formalism. – J.-H. Apr 09 '21 at 05:15
  • Related, if not a duplicate Quantum expressions involving Dirac delta function. – Philip Apr 09 '21 at 10:03

1 Answers1

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Well, as we were taught in kindergarten, it is wrong to assume that $\infty - \infty = 0$.

\begin{align} p\langle p\vert\hat{x}\vert p\rangle - p\langle p\vert\hat{x}\vert p\rangle &= p\partial_p\delta (p-p)-p\partial_p\delta (p-p)\\ &= p\delta' (0)-p\delta'(0)\\ &= \infty - \infty \end{align}

But, importantly, your second calculation is also mistaken. Notice that $\langle p\vert q\rangle=\delta(p-q)$ and thus, $\langle p\vert i\hbar\vert p\rangle=i\hbar\delta (0)\neq i\hbar$. Notice that this is perfectly expected because the identity operator is $\delta (p-q)$ not $\delta_{pq}$.

All of this "craziness" arises because the eigenstates of the position and the momentum operators are not normalizable. They are not in the Hilbert space, they only provide a useful basis for the states in the Hilbert space. A famous argument that closely relates to your first calculation is that $[\hat{x},\hat{p}]$ cannot be $i\hbar$ because we know that the trace of a commutator must vanish. This argument does not work because we cannot define trace for infinite-dimensional Hilbert spaces, or, as the learned like to say, the position and the momentum operators are not "trace-class". See Trace of a commutator is zero - but what about the commutator of $x$ and $p$?.

  • "the identity matrix is (−) not ". What does that even mean? You can't have a matrix representation of the p or x eigenstates. The identity matrix, if it could have been defined in this basis, would have to be a kronecker delta just because it is a matrix. – Arnab Apr 09 '21 at 05:24
  • @Arnab I mean the component-representation of the identity operator, for example, $\langle p \vert \int dx \vert x\rangle\langle x\vert q\rangle = \delta (p-q)$. –  Apr 09 '21 at 05:26
  • I have updated the language from "identity matrix" to "identity operator" to be perfectly clear about what I mean. Thanks. –  Apr 09 '21 at 05:28