How do I prove Ehrenfest's theorems using bra-ket notation?

Question

Townsend's A Modern Approach to Quantum Mechanics gives the following example:

Calculate the time rate of change of the expectation value of the position of a particle moving in one dimension using the following:

$$\frac{d}{dt} \langle A \rangle = \frac{i}{\hbar}\langle\psi(t)|[\hat{H},\hat{A}]|\psi(t)\rangle + \langle\psi(t)|\frac{\partial \hat{A}}{\partial t}|\psi(t)\rangle \space \tag{1}$$

They then show

\begin{align} \rightarrow \frac{d\langle x \rangle}{dt} &= \frac{i}{h}\langle\psi|\left[\frac{\hat{p}_x^2}{2m},\hat{x}\right]|\psi\rangle \tag{2}\\ &= \frac{i}{2m\hbar}\langle\psi|\left(\hat{p_x}\left[\hat{p_x},\hat{x}\right]+\left[\hat{p_x},\hat{x}\right]\hat{p_x}\right)|\psi\rangle\\ &=\frac{\langle\psi|\hat{p_x}|\psi\rangle}{m}=\langle p_x \rangle \end{align}

First of all, why is the right hand side of (1) immediately 0? Why is that left out of the subsequent calculation? Secondly, why in (2) is the potential term completely left out? That is, why doesn't (2) read: $$\rightarrow \frac{d\langle x \rangle}{dt} = \frac{i}{h}\langle\psi|\left[\frac{\hat{p}_x^2}{2m} + V(\hat{x}),\hat{x}\right]|\psi\rangle \tag{alternative 2} $$

I'm assuming it's because any function of $\hat{x}$ commutes with $\hat{x}$. Is that correct?

Answer 1

First I presume you mean the $\partial A/\partial t$ term. This is $0$ because $x$ as an observable does not depend explicitly on time.

Second, if the potential $V$ is a function $x$ (or more properly of $\hat x$), it will commute with $\hat x$ itself as you can verify by expanding the potential in a power series in $\hat x$. (There might be exceptions in some pathological cases.)