I have a problem understanding a specific bit of Dirac notation. Take, as an example this derivation:

enter image description here

I'm dubious about the step from line 3 to 4. When momentum operator acts on the momentum eigenstate, it drops an eigenvalue p. First question: shouldn't p be just a number? If we want to have an explicit space representation of the momentum operator shouldn't we write:

$\langle x| \hat{p} | p \rangle = \langle x|-i\hbar \frac{\partial}{\partial x}|p\rangle $

instead? But then are we allowed to just take the derivative out of the product? How to justify that we can?


The notation $$ \langle x \vert -\mathrm{i}\hbar\partial_x \vert p \rangle$$ would be non-sensical since $\lvert p \rangle$ does not depend on $x$.

Since $\psi_p(x) := \langle x \vert p \rangle$ is the position wavefunction representation of the abstract momentum eigenket $\lvert p \rangle$, the action of the momentum operator on this object is given by $-\mathrm{i}\hbar\partial_x$ while the action of the position operator is given by multiplication by the variable $x$.1

But the way your cited source seems to do it is indeed misleading. The $p$ in $p\langle x \vert p\rangle$ is indeed just a number. Writing it as $-\mathrm{i}\hbar\partial_x$ probably uses prior knowledge that $\langle p \vert x \rangle = \mathrm{e}^{\mathrm{i}xp/\hbar}$.

1That this is the correct and essentially unique way to implement the canonical commutation relation on the $L^2(\mathbb{R},\mathrm{d}x)$ space of position wavefunctions is the content of the Stone-von Neumann theorem.

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  • $\begingroup$ Thanks for your answer! So $\langle x| \hat{p} |p\rangle = -i\hbar \frac{\partial}{partial x} \langle x | p \rangle $ is correct, yes? How can this be shown? Or is it just a definition of an operator? If it is, I just find it confusing that the operator appears between a bra and a ket, rather than to the left; it looks to me as if it was supposed to act on p alone. And regarding the last part of your answer: if p should just be a number, is my source incorrect or is it just sloppy notation? $\endgroup$ – Glo Sep 17 '15 at 18:27
  • $\begingroup$ @Glo: One "physicists' way" to show it is outlined by me here. Also, if you already know that $\langle p \vert x \rangle = \exp(\mathrm{i}xp/\hbar)$, then the way your source shows it is indeed correct. Otherwise, I have no idea what they're doing. But you may also take this as a definition of the operator on wavefunction space and just check that it fulfills the canonical commutation relation. There's no "one true way" to proceed in this case. $\endgroup$ – ACuriousMind Sep 17 '15 at 18:51
  • $\begingroup$ The post you linked concludes that $\hat{p} = \int dx|x\rangle \frac{-i\hbar\partial}{\partial x} \langle x|$. So I could write: $\langle x | \hat{p} |p\rangle = \int dx \langle x | x \rangle \frac{-i\hbar\partial}{\partial x} \langle x | p \rangle = \int dx \frac{-i\hbar\partial}{\partial x} \langle x|p \rangle $. This looks *almost* like what I want, except that the partial is inside the integral. This can't be right since if e.g. $\langle x| p \rangle$ is constant then we get a zero. How does that work then? @ACuriousMind $\endgroup$ – Glo Sep 17 '15 at 19:11
  • $\begingroup$ Also what do we make of this matrix notation for operators: link ? If you take $Q = \hat{p}, |n\rangle = |p\rangle, |m\rangle = |x\rangle $ and, as you wrote in the post you linked, $ \hat{p} = \frac{-i\hbar\partial}{\partial x} $, don't we get exactly the same nonsensical thing from the top of your initial answer, $ \langle x \vert -\mathrm{i}\hbar\partial_x \vert p \rangle$? $\endgroup$ – Glo Sep 17 '15 at 19:18
  • $\begingroup$ @Glo: You have to be careful. When taking the result from the post I linked, you do not get $\langle x \vert x \rangle = 1$, but $\langle x \vert x' \rangle = \delta(x - x')$, which recovers $\langle x \vert \hat{p} \vert p \rangle = -\mathrm{i}\hbar\partial_x \langle x \vert p \rangle$ since you can just carry out the integral thanks to the delta function (at the physicists' level of rigor). For your second comment, you can't take $n = p$ and $m = x$, because $\lvert x \rangle$ is not an eigenstate of $\hat{p}$. $\endgroup$ – ACuriousMind Sep 17 '15 at 20:13

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