The expectation value of momentum is given by:
$$
\langle p\rangle = \int_{-\infty}^{\infty}\psi^{*}(x)\left(-i\hbar\frac{\partial}{\partial x}\right)\psi(x)dx
$$
How can I show that the above expression is equivalent to this?
$$
\langle p\rangle = \int_{-\infty}^{\infty}p|\tilde\psi(p)|^{2}dp
$$
I have tried to use that
$$\psi(x)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\tilde\psi(p)e^{ipx / \hbar}dp$$
and
$$\psi^{*}(x)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\tilde\psi^{*}(p)e^{-ipx / \hbar}dp$$
Then
$$
\langle p \rangle = \int_{-\infty}^{\infty} \left[\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\tilde\psi^{*}(p)e^{-ipx / \hbar}dp \right
]\left(-i\hbar\frac{\partial}{\partial x}\right
)\left[\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\tilde\psi(p)e^{ipx / \hbar}dp \right]dx$$
$$=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\left
[ \left (\int_{-\infty}^{\infty} \tilde\psi^{*}(p)e^{-ipx / \hbar}dp \right)
(-i\hbar) \left (\int_{-\infty}^{\infty} \frac{\partial}{\partial x}\tilde\psi(p)e^{ipx / \hbar}dp \right)\right]dx$$
$$=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\left
[ \left (\int_{-\infty}^{\infty} \tilde\psi^{*}(p)e^{-ipx / \hbar}dp \right)
(-i\hbar) \left (\int_{-\infty}^{\infty} \frac{ip}{\hbar}\tilde\psi(p)e^{ipx / \hbar}dp \right)\right]dx
$$
But I don't know if this is the right approach or if I'm doing the right thing.
Best Answer
If you represent the wave function $\psi(x)$ with it's fourier transform,
\begin{eqnarray*} \psi(x) &=& \frac{1}{\sqrt{2\pi \hbar}}\int \tilde{\psi}(p)e^{\frac{ipx}{\hbar}}dp\\ \psi(x)^\star &=& \frac{1}{\sqrt{2\pi \hbar}} \int \tilde{\psi}^\star(q)e^{\frac{-iqx}{\hbar}}dq \end{eqnarray*}
(where p and q are almost like "dummy" momenta), then you can rewrite the expectation value of momentum as follows:
\begin{eqnarray} \langle p \rangle &=& \int \psi^\star \left(-i\hbar \frac{\partial}{\partial x}\right)\psi dx\\ &=& \frac{1}{2\pi \hbar} \int \tilde{\psi}^\star(q)e^{\frac{-iqx}{\hbar}}\left(-i\hbar \frac{\partial}{\partial x}\right) \tilde{\psi}^\star(p)e^{\frac{ipx}{\hbar}} dpdqdx \end{eqnarray}
Now if you apply the derivative with respect to $x$, you'll spit out a $p$ in the integrand
\begin{eqnarray} &=& \frac{1}{2\pi \hbar} \int \tilde{\psi}^\star(p) \tilde{\psi}^\star(q)e^{\frac{i(q-p)x}{\hbar}} \left(p\right) dpdqdx \end{eqnarray}
and exchanging integration order to integrate over $x$ first -- since we know these functions to be $L^2$ integrable --yields the (scaled) dirac delta function:
\begin{eqnarray} &=& \frac{1}{2\pi \hbar} \int \tilde{\psi}^\star(p) \tilde{\psi}^\star(q)\hbar \delta(q-p) \left(p\right) dpdq \\ \langle p \rangle &=&\frac{1}{2\pi } \int \vert\tilde{\psi}(p)\vert^2 p dp \end{eqnarray}
There's a missing factor of $2\pi$ in there, but I trust you'll find it if you do it carefully by hand.