Does anybody knows how to show that the position expectation value
$$\langle x\rangle = \int_{-\infty}^{\infty} x|\psi(x)|^2dx$$
can be expressed in terms of momentum?
$$\langle x\rangle = i\hbar\int_{-\infty}^{\infty} \tilde{\psi}^*(p)\frac{\partial \tilde{\psi}(p)}{\partial p}dp$$
My attempts:
\begin{eqnarray}
\langle x \rangle &=&\int \langle \psi |\hat{x}| \psi \rangle \int dp |p\rangle \langle p | \\
&=& \int \langle \psi |\hat{p}|p\rangle\langle p |\psi \rangle dp \\
&=& \int \psi^*(p) x \psi(p) dp
\end{eqnarray}
I am stuck at this part … any help will be appreciated.
Best Answer
You just need to heed the definitions of your book, $\langle p|\psi\rangle=\tilde{\psi}(p)$, $\langle \psi|x\rangle= \psi^* (x)$, $\langle x|p\rangle=e^{ixp/\hbar}/\sqrt{2\pi\hbar}$, insert two complete states and integrate by parts, $$\langle x\rangle = \langle \psi |\hat{x}| \psi \rangle =\int dx dp ~~ \langle \psi |\hat{x}|x\rangle \langle x |p\rangle \langle p |\psi \rangle \\= \int dx dp ~ \psi^* (x) ~x~ \frac{e^{ixp/\hbar}}{\sqrt{2\pi\hbar} } \tilde{\psi}(p) \\= \int dx dp ~ \psi^* (x)\frac{\hbar}{i} \frac{\partial_p e^{ixp/\hbar}}{\sqrt{2\pi\hbar} } \tilde{\psi}(p) \\= i\hbar \int dp ~\left (dx~ \psi^* (x) \frac{ e^{ixp/\hbar}}{\sqrt{2\pi\hbar} }\right )~ \partial_p\tilde{\psi}(p) \\ =i\hbar\int dp~ \tilde{\psi}^*(p)\frac{\partial \tilde{\psi}(p)}{\partial p} ~.$$
So, as often, a variable is gotten rid of in favor of its conjugate through the standard Fourier trick.