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?

  • $\begingroup$ straightforward algebra and calculus? without needing any extra input from physics? $\endgroup$
    – arivero
    Commented Aug 24, 2015 at 3:45
  • $\begingroup$ Because here one works in Schrödinger rather than in Heisenberg representation, see, e.g., this answer. $\endgroup$
    – Roger V.
    Commented Oct 26, 2022 at 13:52

2 Answers 2


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 seeing 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.


The position $x$ is not a variable, it is an operator in quantum mechanics. Similarily, it's not always necessary to think of a quantum states as being given by a wavefunction $\psi(x)$, nor is it necessary to express the theory's time-dependence by write $\psi(x,t)$.

In abstract Dirac notation, the expectation value of position in a state $\lvert\psi\rangle$ is $$ \langle x \rangle = \langle\psi\rvert x \lvert\psi\rangle$$ and if you want to know its time-dependence, you have to decide whether to look at the time-dependence in the Schrödinger picture, where we write the time-dependence of the theory in the states $\lvert\psi(t)\rangle = \mathrm{e}^{-\mathrm{i}Ht}\lvert\psi(0)\rangle$ and the operators are constant in time, or in the Heisenberg picture, where the states are time-independent and the dependence is in the operators as $A(t) = \mathrm{e}^{\mathrm{i}Ht}A(0)\mathrm{e}^{-\mathrm{i}Ht}$.

Plugging in, we see that, in both cases, the time-dependence of the expectation value is $$ \langle x \rangle (t) = \langle \psi(0) \rvert \mathrm{e}^{\mathrm{i}Ht}A(0)\mathrm{e}^{-\mathrm{i}Ht}\lvert \psi(0)\rangle$$ and whether you say that it is the position operator that varies or the state (or some mixture) is a matter of taste.


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