TITLE: Is this treatment of the momentum operator in the Dirac formalism allowed?
QUESTION [1 upvotes]: 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?

REPLY [1 votes]: 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.
{"set_name": "stack_exchange", "score": 1, "question_id": 207545}