My notes claim that $\langle \hat p \rangle = m \frac{d}{dt} \langle \hat x \rangle$. Is this true? I don't see why it should be.
Ok, so I tried to derive it using the Schrödinger equation, as kryomaxim suggested:
$$\frac{d}{dt} \langle x \rangle =\frac{d}{dt} \int \psi^\ast x \psi dx = \int \partial _t (\psi^\ast x \psi) dx = \int (\psi^\ast_t x \psi + \psi^\ast x \psi_t )dx \\ = \int x \left( \left( \frac{i \hbar}{2m} \frac{\partial^2}{\partial x^2}\psi + \frac{1}{i\hbar}V(x) \psi \right) ^\ast \psi + \psi^\ast \left(\frac{i \hbar}{2m} \frac{\partial^2}{\partial x^2}\psi + \frac{1}{i\hbar}V(x) \psi \right) \right) dx \\ = \int x \left( \left(- \frac{i \hbar}{2m} \frac{\partial^2}{\partial x^2}\psi^\ast - \frac{1}{i\hbar}V(x) \psi^\ast \right) \psi + \psi^\ast \left(\frac{i \hbar}{2m} \frac{\partial^2}{\partial x^2}\psi + \frac{1}{i\hbar}V(x) \psi \right) \right) dx \\ =\int x \left( \left( -\frac{i \hbar}{2m} \frac{\partial^2}{\partial x^2}\psi^\ast \right) \psi + \psi^\ast \frac{i \hbar}{2m} \frac{\partial^2}{\partial x^2}\psi \right)dx \\ = \frac{i \hbar}{2m} \int x \left( \left( -\frac{\partial^2}{\partial x^2}\psi^\ast \right)\psi + \psi^\ast \frac{\partial^2}{\partial x^2} \psi dx \right) \\ $$
Now I'm not sure how to make progress. Is this correct so far? And does the time derivative necessarily commute with the integral?
Thanks for the help!
\langle
and\rangle
or\left<
and\right>
(the latter will scale to the contents, the former will not). Both the plain characters<
and>
and the macros\lt
and\gt
are typeset as relational operators and have too much space around them for use as brackets. $\endgroup$ – dmckee --- ex-moderator kitten Jan 6 '17 at 21:58