Is $\langle \hat p \rangle = m \frac{d}{dt}\langle \hat x \rangle$?

Question

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!

Answer 1

Use the equation of motion $\frac{d\langle\hat{x}\rangle}{dt} = -i\left<\left[\hat{H},\hat{x}\right]\right>$ and the Hamilton operator $\hat{H}$ of the Schroedinger equation. Moreover use commutator relation $\left[\hat{x},\hat{p}\right]=i\hbar$