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?


1 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.)

  • $\begingroup$ This is probably a stupid question but I'm just beginning QM. Why wouldnt $\partial \hat{x} / \partial t$ = $\hat{\nu}$ where $\hat{\nu}$ is some velocity operator. In fact I don't even know what it means to take the derivative of an operator. $\endgroup$ Mar 24, 2017 at 3:56
  • $\begingroup$ @johnmorrison $v=dx/dt$, not $\partial x/\partial t$, i.e. velocity is a total - not partial - derivative of the position. The partial derivative would "pick out" any explicit factors of $t$ in an operator. $\endgroup$ Mar 24, 2017 at 3:59
  • $\begingroup$ Yeah still not sure I understand. Wouldn't x only depend on t, thereby forcing $\partial x / \partial t$ = dx/dt? The example problem states that the particle may be moving. $\endgroup$ Mar 24, 2017 at 4:01
  • $\begingroup$ You need to review the distinction between total and partial derivatives. Irrespective, $\hat x$ here is an operator which does NOT depend on $t$. $\endgroup$ Mar 24, 2017 at 4:03
  • $\begingroup$ How is what I've stated about differentiation wrong? Is it not true that $\partial f(\eta) / \partial \eta$ = $df(\eta)/d \eta$? My confusion lies with the parameters along which an operator depends. How is $\hat{x}$ independent of time? Are you saying the act of measuring the position of an object is time independent? Because certainly the position of the particle isn't. $\endgroup$ Mar 24, 2017 at 4:12

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