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!

  • 3
    $\begingroup$ See the Ehrenfest theorem. $\endgroup$
    – Javier
    Commented Jan 6, 2017 at 21:55
  • 4
    $\begingroup$ To markup angle brackets for averages, expectation values and bra-kets, use either \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. $\endgroup$ Commented Jan 6, 2017 at 21:58
  • 6
    $\begingroup$ LaTeX is more difficult to master than QM. $\endgroup$ Commented Jan 6, 2017 at 22:09
  • $\begingroup$ @Javier consider writing that into an answer? $\endgroup$
    – Kyle Oman
    Commented Jan 6, 2017 at 22:51

1 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 $\left[\hat{x},\hat{p}\right]=i\hbar$

  • 1
    $\begingroup$ While this is correct, Ehrenfest's theorem is often introduced early (before students have the commutator form shown here), so it may not be as useful to the OP as it is to an experienced physicist. $\endgroup$ Commented Jan 6, 2017 at 22:08
  • $\begingroup$ $<\hat{x}> = \int dV \psi^*(x,t) x\psi(x,t)$. Differentiating this by $t$ and using Schroedinger equation would also give the result. $\endgroup$
    – kryomaxim
    Commented Jan 6, 2017 at 22:10
  • $\begingroup$ Right. That's the usual development early in the course, and the one that should be accessible to anyone ready to ask the question. $\endgroup$ Commented Jan 6, 2017 at 22:11

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