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It recently occurred to me I could exploit the fact the operators on the wave function are partial differentials and use an identity. To make explicit:

$$ i \hbar d \psi(x,y,z,t) = (i \hbar \frac{\partial \psi }{\partial t}dt -(-i \hbar \nabla \psi \cdot dr)) $$

Writing in terms of momentum and energy:

$$ i \hbar d \psi(x,y,z,t) = (\hat E \psi dt - \hat p \psi . dr ) $$

Dividing both sides by $dt$:

$$ i \hbar \frac{d \psi(x,y,z,t)}{dt} = (\hat E \psi - \hat p \psi . \frac{dr}{dt} ) $$

Questions

What is the physical meaning of $\frac{dr}{dt}$ in this context? It seems to be the velocity of something but I'm not sure what.

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In brief, no, it does not have any physical meaning. In a non-relativistic context the wavefunction $\psi$ is defined as a function on the space of physical space coordinates, with an extra dependence on time. As the coordinate space itself does not depend on time, in this context $dr/dt$ does not have any physical (nor mathematical, for that matter) meaning. There is no total differential of $\psi$ with respect to $t$.

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  • $\begingroup$ What if I had a moving coordinate system? $\endgroup$ – drewdles Jul 20 '17 at 18:48
  • $\begingroup$ Then you would have to redefine partial derivatives. You would set $\vec{r}'(\vec{r},t)$ and $t'(\vec{r},t)$, and differentiate to obtain $\partial/\partial r^{i}=(\partial r'^{j}/\partial r^{i})\,\partial/\partial r'^{j}+(\partial t'/\partial r^{i})\,\partial/\partial t'$ and $\partial/\partial t=(\partial r'^{j}/\partial t)\,\partial/\partial r'^{j}+(\partial t'/\partial t)\,\partial/\partial t'$. (cont'd) $\endgroup$ – Giorgio Comitini Jul 20 '17 at 19:01
  • $\begingroup$ Galileian time needs $t'=t$ so that $\partial/\partial r^{i}=(\partial r'^{j}/\partial r^{i})\,\partial/\partial r'^{j}$ and $\partial/\partial t=(\partial r'^{j}/\partial t)\,\partial/\partial r'^{j}+\partial/\partial t'$. In the last one (\partial r'^{j}/\partial t) is in fact a velocity, but still, no total derivative. In this case the velocity must be regarded as a parameter with respect to which the dynamics of the quantum system is defined. It is no more ambiguous than the concept of a coordinate system: one chooses a classical system (i.e. the lab) with respect to which (cont'd) $\endgroup$ – Giorgio Comitini Jul 20 '17 at 19:08
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    $\begingroup$ things as positions etc. are well defined classically. In this case the extra velocity parameter shows up in the average of $\vec{r}'$ so that indeed, if you define the velocity to be the time derivative of the average position, that extra velocity is taken into account. However in order to ensure that the change of coordinates itself makes sense a more accurate analysis is needed. $\endgroup$ – Giorgio Comitini Jul 20 '17 at 19:15
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What you calculate is how the observed value of $\psi$ changes for an observer who moves around with velocity $\frac{dr}{dt}$. As others have said, it is difficult to come up with a situation where something like that might be physically relevant - as we know, observers can not actually observe the value of $\psi$. It might be possible that terms like this appear in pilot wave theory where a point particle moves in the "matter wave" that is the wave function.

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