How do you derive $x = ihbar d/dp$? Why does this equation hold true?

[Danielk010]

TL;DR

I am trying to study for an upcoming exam, and one of the topics is about the relation between position and momentum in qm. I want to understand why the equation holds true and the proof for the equation.

The equation comes from the solution of a homework equation we were given from A Modern Approach to Quantum Mechanics: 2nd edition by Townsend:
$$
\textbf{Show}

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

and

$$
\langle \varphi | \hat{x} | \psi \rangle
= \int dp \, \langle p | \varphi \rangle^* \, i \hbar \frac{\partial}{\partial p} \langle p | \psi \rangle
$$

The solution just states that we observe the above equation. Is there a proof someone can show or an explanation on why this formula holds true?

 

[dextercioby]

Hint: insert the completion relation of the eigenbasis of $\hat{x}$, then cheekily write the $x$ as a p- derivative.

 

[Danielk010]

Hi, thank you response. This is from an homework problem, in which the TA for my class has already provided us the solution. They don't explain how:

$$xe^{\frac{-ipx}{\hbar}} = i\hbar \frac{d}{dp} e^{\frac{-ipx}{\hbar}}$$
If I insert the completion relation of the eigenbasis of x (assuming this is what you mean as the completion relation)
$$\int_{-\infty}^{\infty} \psi(x) |x\rangle \,dx$$
then I would just get this (I think):
$$x\int_{-\infty}^{\infty} \psi(x) |x\rangle \,dx$$
Is this the correct way to prove the first equation or the equation below?

$$x = i\hbar \frac{d}{dp}$$​

When you say completion relation of the eigenbasis of x, do you mean the commutator relation between momentum and positon?