Having defined the expectation value of position as follows

$$

\langle x \rangle = \int x {\lvert\Psi(x,t)\rvert}^2dx

$$

The time derivative of the expectation value is derived in my literature in the following way:

$$

\frac{d\langle x \rangle }{dt} = \int x\partial_t{\lvert\Psi\rvert}^2dx = \ldots

$$

From here it is straightforward algebra and calculus to the answer.

$$

\frac{d\langle x \rangle }{dt} = \frac{-i\hbar}{2m}\int\overline{\Psi}\partial_x\Psi dx

$$

What strikes me as odd is the fact that the author didn't write

$$

\frac{d\langle x \rangle }{dt} = \int \partial_t\left(x\lvert\Psi\rvert^2dx\right)

$$

Somehow, the position $x$ is treated as a constant. There is no explanation to why this is the case. From above, $x$ is somehow unchanging in time and the only thing that alters the expectation value is the wave function. This might be the case but then what is the interpretation of $x$?

**Why doesn't $x$ vary in time?**

## Best Answer

Because in QM, the wave function contains all the relevant information, and $x$ is the coordinate parameter.

You are expecting $x$ to be dependent on time because you are used to see it as the particle's coordinate, i.e. a function of time that says where the particle is for each time parameter value.

But here, $\psi$ has this information, and $x$ is just the parameter value for which you want to evaluate the odds of finding the particle. So $x$ is not the position of the particle in the sense that is a time function giving the particle's position; its just another parameter on which $\psi$ depends.

This is similar to wave mechanics, in the sense that you use $x$ as a variable on which the function of wave form depends, but is not a particular coordinate of the wave, is just a coordinate of the space where the wave exists.