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Consider:

$$\frac{d\langle x\rangle}{dt}=\frac{d}{dt}\left(\int_{-\infty}^{\infty}x\psi^*\psi \ dx \right)=\frac{d}{dt}\left(\int_{-\infty}^{\infty}xP(x,t) \ dx \right)=\int_{-\infty}^{\infty}\frac{\partial}{\partial t}\left(xP(x,t) \right) \ dx.$$

If you now assume that $x$ is independent of $t$, it is not hard to show that Ehrenfest's theorem follows (using the continuity equation, boundary conditions, integration by parts, and the equation for the expected value of momentum). But why is this assumption necessary? Why can't you let $x=x(t)$?

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$x$ in that equation is just the coordinate in the wave function $\psi(x)$. It is not the position of a particle along any particular trajectory.

You may have seen this before in classical mechanics where you can integrate (for example) a function like the potential $V(x)$, which does not depend on the trajectory of the particle, even when that trajectory is defined.

For that reason there is no dependence of $x$ on anything. It is just a parameter that enters into the wave function.

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  • $\begingroup$ Oh, okay. Alright, that makes more sense, it's just the position parameter for the wavefunction. Thanks. $\endgroup$
    – agaminon
    Nov 8, 2022 at 17:36

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