# Time derivative of expectation value of position

It is necessary to distinguish between the *position*, *operator of position*, and *mean value of position*/*average position*. Here one works in *Schrödinger representation*, which means that all the time dependence is carried by the wave function, whereas the operators are time-independent. Moreover, in the *position representation* the operator of position is $\hat{x}=x$ - a time-independent number that should be integrated with the wave function.

In other words: the average position $\langle x\rangle$ is time-dependent, but its operator $x$ is time-independent.

You may also want to consult this answer.

An analogy might be useful.

Suppose you want to compute the time-dependence of the average weight of a population. The average weight is just \begin{align} \langle w\rangle = \int dw w N(w) \tag{1} \end{align} where $N(w)$ is the probability of people having weight $w$. Now, what changes with time is not the weight $w$: $1$kg today is the same as $1$kg tomorrow, but what changes in time is the probability $N(w)$ of having persons of a certain weight: some people will gain weight over time, some will loose weight so a better expression for the average time would be \begin{align} \langle w(t)\rangle = \int dw w N(w,t) \end{align} and of course the rate of change in this average is \begin{align} \frac{d\langle w(t)\rangle}{dt}= \int dw w \frac{N(w,t)}{dt}\tag{2} \end{align} Thus, in (2), what changes is the probability distribution. This $N(w,t)$ is in fact nothing but the probability distribution $\vert \psi(x,t)\vert^2$ in your problem.

$x$ is just a position variable or operator, if you prefer. It is not the position of the particle, which instead is $$\langle x\rangle = \int dx\, x \left|\Psi\right|^2~.$$ $\langle x\rangle$ may depend on $t$, but $x$ does not depend on $t$.