# Converting Dirac Notation to Coordinate Space

I am trying to find how to explicitly calculate: $$\langle x|p\rangle$$. Intuitively, this means the momentum operator is acting on a state $$\phi$$ which is then projected into the coordinate space. Hence:

$$\langle x|p|\phi\rangle = \frac{\hbar}{i}\frac{\partial \phi}{\partial x}.$$

However, how do I show this explicitly using Dirac notation (if my conclusion is even correct)? My attempt:

$$\langle x|p|\phi\rangle = \int \langle x|p| x'\rangle\langle x'| \phi\rangle dx' = \int \langle x|\frac{h}{i} \partial_x| x'\rangle \phi(x') dx' .$$

I am stuck on how to handle the derivative here. I know this is an identity that is commonly used, but an explanation of why it is true would be greatly helpful.

• This question is closely related to the following physics.stackexchange.com/q/212964
– J.L.
Nov 3, 2022 at 18:31
• Also related: How does the momentum operator act on state kets? Nov 3, 2022 at 23:52
• What are you trying to calculate? $\langle x|p\rangle$, $\langle x|p|\phi\rangle$, or $\langle x|p$? You've written the first one, but it makes very little sense in the context of the rest of the question. Nov 3, 2022 at 23:56

There are typos in your expression: it should be $$\phi(x')$$, and there's no integration over $$p$$'s. You're then taking the derivative w/r to $$x$$ but integrating w/r to $$x'$$ so: \begin{align} \langle x\vert p\vert\phi\rangle &= \int dx' \langle x\vert -i\hbar \partial_x \vert x'\rangle \langle x'\vert \phi\rangle \tag{1} \end{align} by $$\langle x\vert \hat p=-i\hbar \partial_x \langle x\vert$$. Then pull out the derivative (you are integrating over $$dx'$$, not $$dx$$) to get \begin{align} \langle x\vert p\vert\phi\rangle= -i\hbar \partial_x \int dx'\langle x\vert x'\rangle\langle x'\vert \phi\rangle \end{align} and remove the identity $$\mathbb{I}=\int dx' \vert x'\rangle\langle x'\vert$$ to reach $$\langle x \vert p\vert\phi\rangle = -i\hbar\partial_x \langle x\vert \phi\rangle=-i\hbar \partial_x \phi(x)$$
Nota: To show $$\langle x\vert \hat p=-i\hbar \partial_x\langle x\vert$$, you need \begin{align} \langle x\vert p\rangle=\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar} \, ,\qquad \mathbb{I}&=\int dp \vert p\rangle\langle p\vert \end{align} so that \begin{align} \langle x\vert \hat p=\int dp'\langle x\vert \hat p \vert p'\rangle \langle p'\vert&= \int dp' p' \langle x\vert p'\rangle \langle p'\vert\, ,\\ &= \int dp' p' \frac{1}{\sqrt{2\pi\hbar}}e^{ip'x/\hbar} \langle p'\vert\, ,\\ &= -i \hbar \partial_x \int dp' \frac{1}{\sqrt{2\pi\hbar}}e^{ip'x/\hbar} \langle p'\vert\, ,\\ &= -i \hbar \partial_x \int dp' \langle x\vert p'\rangle \langle p'\vert\, ,\\ &= -i\hbar \partial_x \langle x\vert \, . \end{align} In fact that's really the only thing you need: you could completely bypass the $$\int dx' \vert x'\rangle\langle x'\vert$$ step of Eq.(1)
You used $$p = {1 \over i} \partial_{x}$$ too soon. That expression is already in the position basis. Really what you should write is that $$$$\langle x\prime|p| x\rangle = \delta(x-x\prime) {1 \over i}\partial_{x}$$$$