Ehrenfest tells us that for $\hat{p}$

$$\partial_t \langle p \rangle = \langle -\partial_x V \rangle$$

I also understand the basic steps in deriving this result directly by taking the time derivative of $\langle p \rangle$, do integration by parts a couple of times, replace time derivatives of wave functions using Schrodinger, etc.

With that in mind, I can't find the gap in the logic of the following:

$$\partial_t \langle p \rangle = -i\hbar \partial_t \int \limits_{-\infty}^{\infty} dx \; \psi^* (\partial_x \psi) = -i \hbar \int \limits_{-\infty}^{\infty} dx \; \lbrace (\partial_t \psi^*)(\partial_x \psi) + \psi^* (\partial_t(\partial_x \psi)) \rbrace.$$

Then using integration by parts on the second term, with the boundary terms going to zero

$$-i \hbar \int \limits_{-\infty}^{\infty} dx \; \lbrace (\partial_t \psi^*)(\partial_x \psi) - (\partial_t \psi^*) (\partial_x \psi) \rbrace = 0$$

I must be missing something here. I feel like it has to do with a misunderstanding of the commutation relation between momentum and the Hamiltonian when the Hamiltonian is written as $\hat{H} = i \hbar \partial_t$ instead of $$\hat{H} = \dfrac{\hat{p}^2}{2m} + V(x)$$ (if that is correct to even say, I infer it from Schrodinger's Equation, $\hat{H} \psi = i \hbar \partial_t \psi$).

• You "integrated by parts" with the time derivative... but the integration is over the spacial derivative... That's not how integration by parts works...
– hft
May 1, 2015 at 5:47

Your integration by parts is incorrect. You are integrating over space, so you can only move the spacial derivative rather than time derivative onto the $\psi^*$ factor. Or put another way, what you are calling the boundary term and throwing away is actually the expression you started with. You are replacing $$\psi^*\partial_t\partial_x\psi=\partial_t(\psi^*\partial_x)-\partial_t\psi^*\partial_x\psi$$ so you are just undoing your inital use of the product rule and obviously the other terms will cancel.