I am trying to prove that the position operator in momentum space is $i\hbar \partial/\partial p$ but my derivation is missing one sign. Can someone spot the error?
Start with
$$<\hat x> = \int\limits_{-\infty}^{\infty} \psi^*(x) x \psi(x) \mathrm{d}x$$
Taking the fourier transform of both $\psi(x)$
$$ \frac{1}{2\pi} \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \psi^*(\bar k_x) e^{i\bar k_xx} \, \mathrm{d} \bar k_x x \int\limits_{-\infty}^{\infty} \psi(k_x) e^{-i k_xx} \mathrm{d} k_x \mathrm{d} x $$
Taking integration by parts on the right most integral yields
$$ \frac{1}{2\pi} \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \psi^*(\bar k_x) e^{i\bar k_xx} \, \mathrm{d} \bar k_x x * \frac{1}{ix} \int\limits_{-\infty}^{\infty} e^{-i k_xx} \frac {\partial \psi(k_x)}{\partial k_x} \mathrm{d} k_x \mathrm{d} x $$
We see the middle term gets canceled out.
Now using the following identity
$$ \delta(k_x-\bar k_x) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} e^{-ix(k_x-\bar k_x)} \, \mathrm{d} x $$
we arrive at
$$ \frac{1}{i} \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \psi^*(\bar k_x) \delta(k_x-\bar k_x) \mathrm{d} \bar k_x \frac {\partial \psi(k_x)}{\partial k_x} \mathrm{d} k_x $$
$$ \frac{1}{i}\int\limits_{-\infty}^{\infty} \psi^*(k_x) \frac {\partial \psi(k_x)}{\partial k_x} \mathrm{d} k_x $$
Very lastly, we use the following identity
$$\frac{\partial}{\partial k_x} = \frac{\hbar \partial}{\partial p_x}$$
We prove
$$ \frac{\hbar}{i} \int\limits_{-\infty}^{\infty} \psi^*(k_x) \frac {\partial \psi(k_x)}{\partial p_x} \mathrm{d} k_x $$
So
$$ x = \frac{\hbar \partial}{i\partial p_x}$$
So close, not quite right. Because the actual answer is $$ x = \frac{-\hbar \partial}{i\partial p_x}$$ or $$ x = \frac{i\hbar \partial}{\partial p_x}$$
Can someone spot the mistake. Thanks!
