Finding the position operator in momentum space

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Homework Statement

Given $\hat{x} =i \hbar \partial_p$, find the position operator in the position space. Calculate $\int_{-\infty}^{\infty} \phi^*(p) \hat{x} \phi(p) dp$ by expanding the momentum wave functions through Fourier transforms. Use $\delta(z) = \int_{\infty}^{\infty}\exp(izy) dz$.

Homework Equations

I'll explicitly state them in my attempt

The Attempt at a Solution

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First we focus on the integrand. We move over to position space through Fourier transformations $$\phi^*(p) \hat{x} \phi(p) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{2 \pi \hbar} \exp(-ipx/\hbar) \psi(x) i \hbar \frac{\partial}{\partial p} \left(\exp(ipy/\hbar) \psi^*(y) \right) dx dy $$ my reasoning is that I can move over conjugate of the spatial wave function as it does not explicitly depend on momentum. Upon simplifying a little we get $$ \phi^*(p) \hat{x} \phi(p) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{-y}{2 \pi \hbar} \exp\left(\frac{ip(y-x)}{\hbar}\right) \psi(x)\psi^*(y)dx dy $$ which we integrate with respect to $p$ to get
$$ \int_{-\infty}^{\infty} \phi^*(p) \hat{x} \phi(p) =\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{-y}{2 \pi \hbar} \exp\left(\frac{ip(y-x)}{\hbar}\right) \psi(x)\psi^*(y)dx dy dp$$ utilizing the dirac delta function relation from earlier, we find that $$ \int_{-\infty}^{\infty} \exp\left(\frac{ip(y-x)}{\hbar}\right) dp = \delta\left(\frac{y-x}{\hbar}\right) $$ which, when put back into our integral and integrated with respect to y, simply substitutes $y \rightarrow x$ in the integrand (by the definition of the dirac delta function $$ \int_{-\infty}^{\infty} \phi^*(p) \hat{x} \phi(p) =\int_{-\infty}^{\infty} \frac{-x}{2 \pi \hbar} \psi(x)\psi^*(x)dx$$ which is close, I think? I may be off by a factor of $-1/2π\hbar$? Ideally, since $\hat{x} = x$ in position space, I should get
$$ \int_{-\infty}^{\infty} x \psi(x)\psi^*(x)dx = \int_{-\infty}^{\infty} \psi(x) \hat{x} \psi^*(x)dx $$
no? I don't know, i feel like I've gone so much through the weeds that either everything is wrong or my brain is fried and I can't see that I've actually gotten the right answer (unlikely). A bit of guidance would be much appreciated!

EDIT: I published this on safari and the formatting seemed off (some sentences were not being displayed). It looks fine on chrome, though. Weird!

 

[TSny]

Use $\delta(z) = \int_{\infty}^{\infty}\exp(izy) dz$.

There is a missing factor of $2 \pi$ somewhere in this equation.

First we focus on the integrand. We move over to position space through Fourier transformations $$\phi^*(p) \hat{x} \phi(p) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{2 \pi \hbar} \exp(-ipx/\hbar) \psi(x) i \hbar \frac{\partial}{\partial p} \left(\exp(ipy/\hbar) \psi^*(y) \right) dx dy $$

I think the signs in the arguments of the two exponential functions here need to be switched.

See for example https://quantummechanics.ucsd.edu/ph130a/130_notes/node82.html

utilizing the dirac delta function relation from earlier, we find that $$ \int_{-\infty}^{\infty} \exp\left(\frac{ip(y-x)}{\hbar}\right) dp = \delta\left(\frac{y-x}{\hbar}\right) $$

Again, a missing factor of $2 \pi$. Also, you need to handle the $\hbar$ correctly when using this. See the second equation here
https://en.wikipedia.org/wiki/Dirac_delta_function#Scaling_and_symmetry