# 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 $$t$$.