# Is this treatment of the momentum operator in the Dirac formalism allowed?

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

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.

• 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?
– Glo
Sep 17, 2015 at 18:27
• @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. Sep 17, 2015 at 18:51
• 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
– Glo
Sep 17, 2015 at 19:11
• 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, $\langle x \vert -\mathrm{i}\hbar\partial_x \vert p \rangle$?
– Glo
Sep 17, 2015 at 19:18
• @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 $\hat{p}$. Sep 17, 2015 at 20:13