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One way I've seen the Schrödinger equation expressed for the position wave function is

$$\frac{i\hbar\partial\Psi\left(\vec{r},t\right)}{\partial{t}}=-\frac{\hbar^2}{2m}\nabla^2\Psi\left(\vec{r},t\right)+V\left(\vec{r},t\right)\Psi\left(\vec{r},t\right)$$

with, $\hbar$ being the reduced plank constant $m$ being the mass, $t$ being the time, $\vec{r}=(x,y,z)$, $V\left(\vec{r},t\right)$ being the potential function, $\Psi\left(\vec{r},t\right)$ being the wave function and $\nabla^2\Psi\left(\vec{r},t\right)=\frac{\partial^2\Psi\left(\vec{r},t\right)}{\partial{x^2}}+\frac{\partial^2\Psi\left(\vec{r},t\right)}{\partial{y^2}}+\frac{\partial^2\Psi\left(\vec{r},t\right)}{\partial{z^2}}$

In the above equation, given $\hbar$, $V\left(\vec{r},t\right)$, $m$, and $\Psi\left(\vec{r},0\right)$, there is a unique $\Psi\left(\vec{r},t\right)$, however even if $\Psi\left(\vec{r},0\right)$, and $V\left(\vec{r},t\right)$ are the same in two inertial reference frames this does not mean that $\Psi\left(\vec{r},t\right)$ are the same in both inertial reference frames as the velocity of the position wave function could be different in one inertial reference frame from in another, and in non relativistic quantum mechanics Galilean relativity still applies. The way to fix this is to include a galilean transformation in the above equation. The galilean transformation is $\vec{r}'=\vec{r}-\vec{v}t$, so the above equation becomes

$$\frac{i\hbar\partial\Psi\left(\vec{r},t\right)}{\partial{t}}=-\frac{\hbar^2}{2m}\left(\frac{\partial^2\Psi\left(\vec{r},t\right)}{\partial{x'^2}}+\frac{\partial^2\Psi\left(\vec{r},t\right)}{\partial{y'^2}}+\frac{\partial^2\Psi\left(\vec{r},t\right)}{\partial{z'^2}}\right)+V\left(\vec{r},t\right)\Psi\left(\vec{r},t\right)$$

with $v_{xw}$, $v_{yw}$, and $v_{zw}$ being related to the velocity of the wave function.

I also saw in the answers to this question that the Schrödinger equation in momentum space is

$$i\hbar\frac{\partial}{\partial{t}}\Psi(\vec{p},t)=\frac{\vec{p}^2}{2m}\Psi( \vec{p},t)+V(\vec{p},t)\Psi(\vec{p},t)$$

meaning that in velocity space the Schrödinger equation would be

$$i\hbar\frac{\partial}{\partial{t}}\Psi(\vec{v},t)=\frac{\vec{v}^2m}{2}\Psi( \vec{v},t)+V(\vec{v},t)\Psi(\vec{v},t)$$

I don't know if in the general case the position velocity wave function would be separable and unless it is separable, it would be necessary to take the position velocity wave function $\Psi\left(\vec{r},\vec{v},t\right)$.

What I'm confused about is what the Schrödinger equation would be in position velocity space. I don't know how to rigorously derive the Schrödinger equation in position velocity space from the Schrödinger equation in position space and the Schrödinger equation in velocity space, meaning the best I know how to do is to make an educated but blind guess as to what the Schrödinger equation might be in position velocity space, and while in some cases a blind guess can be correct in most many cases it isn't.

I understand that when there are two particles the Schrödinger equation describing the combined position time wave function is

$$\frac{i\hbar\partial\Psi\left(\vec{r_1},\vec{r_2},t\right)}{\partial{t}}=-\frac{\hbar^2}{2}\left(\frac{\nabla_1^2\Psi\left(\vec{r_1},\vec{r_2},t\right)}{m_1}+\frac{\nabla_2^2\Psi\left(\vec{r_1},\vec{r_2},t\right)}{m_2}\right)+V\left(\vec{r_1},\vec{r_2},t\right)\Psi\left(\vec{r_1},\vec{r_2},t\right)$$

Also

$$\frac{\partial^2c}{\partial(a+b)^2}=\frac{\partial^2c}{\partial{a}^2}+\frac{\partial\left(\frac{\partial{c}}{\partial{a}}\right)}{\partial{b}}+\frac{\partial\left(\frac{\partial{c}}{\partial{b}}\right)}{\partial{a}}+\frac{\partial^2c}{\partial{b}^2}=\frac{\partial^2c}{\partial{a}^2}+\frac{2\partial^2c}{\partial{a}\partial{b}}+\frac{\partial^2c}{\partial{b}^2}$$

and

$$\frac{\partial{a}}{\partial(bc)}=\frac{\partial{a}}{c\partial{b}}$$

So I'm wondering if the Schrödinger equation in position velocity space would be

$$\frac{i\hbar\partial\Psi\left(\vec{r},\vec{v},t\right)}{\partial{t}}=-\frac{\hbar^2}{2m}\left(\frac{\partial^2\Psi\left(\vec{r},\vec{v},t\right)}{\partial{x^2}}+\frac{\partial^2\Psi\left(\vec{r},\vec{v},t\right)}{\partial{y^2}}+\frac{\partial^2\Psi\left(\vec{r},\vec{v},t\right)}{\partial{z^2}}+\frac{2\partial^2\Psi\left(\vec{r},\vec{v},t\right)}{t\partial{x}\partial{v_x}}+\frac{2\partial^2\Psi\left(\vec{r},\vec{v},t\right)}{t\partial{y}\partial{v_y}}+\frac{2\partial^2\Psi\left(\vec{r},\vec{v},t\right)}{t\partial{z}\partial{v_z}}+\frac{\partial^2\Psi\left(\vec{r},\vec{v},t\right)}{t^2\partial{{v_x}^2}}+\frac{\partial^2\Psi\left(\vec{r},\vec{v},t\right)}{t^2\partial{{v_y}^2}}+\frac{\partial^2\Psi\left(\vec{r},\vec{v},t\right)}{t^2\partial{{v_z}^2}}\right)+\frac{\vec{v}^2m\Psi\left(\vec{r},\vec{v},t\right)}{2}+V\left(\vec{r},\vec{v},t\right)\Psi\left(\vec{r},\vec{v},t\right)$$

or if it would be something else.

What would be the Schrödinger equation in position velocity space?

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    $\begingroup$ I imagine that it is downvoted because there is no "velocity space" verison of of Schroedinger, and your formulae with $x-vt$ are meaningless. Do you not know the chain rule for partial derivatives? (I have not downvoted you though) $\endgroup$
    – mike stone
    Aug 10, 2021 at 21:57
  • $\begingroup$ @mikestone I know the chain rule for ordinary derivates. $\endgroup$ Aug 10, 2021 at 23:27
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    $\begingroup$ What is "position velocity space"? $\endgroup$
    – J. Murray
    Aug 11, 2021 at 11:21
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    $\begingroup$ I suggest you learn how to change variables in partial derivatives. $\endgroup$
    – mike stone
    Aug 11, 2021 at 13:00
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    $\begingroup$ there's no obvious physical meaning to your proposed "position-velocity space". Moreover, Schrödinger equation is usually written either in position space, or in momentum space, not in a mixture of both. $\endgroup$ Aug 12, 2021 at 15:19

1 Answer 1

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In the general case the position wave function and the velocity wave function would not be independent implying that it would not be independent implying that it would be necessary to take the position velocity wave function $Ψ(r⃗ ,v⃗ ,t)$.

What is "the general case" you are referring to, if not experimental confirmation? The wave functions of position and velocity (i.e. momentum) indeed depend upon each other in mainstream quantum mechanics. You get from one to the other by a simple Fourier transform, period. No need to introduce any new equations or functions, unless this simplifies the Fourier math (which I don't know of).

Position and velocity are independent when it comes to their observables, because you cannot just derive observed position with respect to time in order to obtain observed velocity. That is actually the very point about quantum mechanics: that velocity/momentum is related to wavelength (according to the de-Broglie relations) rather than change of position. If you try to prepare a particle with a point-like wave-function (i.e. with a specific position), you will see that you cannot at all attribute a velocity to it, because velocity/momentum becomes infinitely uncertain for that configuration. I.e. if you observe position very precisely, you will know nothing about what you would observe when measuring momentum after that. Hence, observed momentum is something completely independent of observed position. Vice-versa, if you prepare a particle with a sharp velocity, it will have an infinitely uncertain position.

If you are inclined to describe the particle by a "position-velocity space wave function", you introduce additional information and you would have to model the corresponding differential equation so that it reflects the very well-explored Schrödinger equation accurately. Moreover you would have to specify under what circumstances your "richer" wave function can be described by the Schrödinger one. To stay with the mentioned point-like wave function: you are of course free to combine $\psi(r,t)=\delta(r-r_0)$ with the corresponding Fourier transform $\psi(p,t)=1$ (unnormalized) in order to obtain a function $\psi(r,p,t)=\psi(r,t)\cdot \psi(p,t)$. But what benefit does this offer? If there is no additional information, you can simply drop the "p-part".

Possibly you are thinking about "completing" the Schrödinger equation so that it becomes deterministic (hidden variables, in your case the wave amplitudes in position-velocity space), but again, you would have to construct your hidden variables so that they don't contradict the current state of affairs. And I don't know of any successful attempt to introduce hidden variables. This does not mean that I (personally) can rule out the existence of hidden variables, but that thousands of brilliant physicists would have missed the point. Be sure you know your tools well enough before you claim that you are smarter than them.

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  • $\begingroup$ velocity/momentum is related to wavelength (according to the de-Broglie relations) rather than change of position is not quite true since by Ehrenfest’s theorem $\langle p\rangle=m\frac{d\langle x(t)\rangle}{dt}$. $\endgroup$ Aug 15, 2021 at 12:19
  • $\begingroup$ I found from your answer that I used the wrong terminology in the quoted part of my question, for describing what I actually meant, and have edited it to more accurately reflect what I really meant. I noticed however that I had also implicitly made the assumption that position velocity wavefunctions can be not separable when I don't know whether or not they are allowed to be not separable so edited that part to reflect that. $\endgroup$ Aug 15, 2021 at 20:30
  • $\begingroup$ In your answer you implied that position and velocity being independent simply means that you cannot determine the velocity from the position with respect to time, however as I understand it A and B being completely independent means that the probability distribution for B does not depend on the value for A and you don't make it clear whether or not you also mean that the probability distribution for velocity is independent of observed position. $\endgroup$ Aug 15, 2021 at 21:27
  • $\begingroup$ The probability distribution for velocity (though more often momentum is used) is restricted by the probability distribution for position (and vice versa). Therefore they are related $\endgroup$ Aug 17, 2021 at 17:47
  • $\begingroup$ @NAMcMahon Hein? How can this be? Certainly there’s no classical relation between the two…. indeed just shift you coordinate axes by a fixed amount to get a shifted distribution of positions without affecting velocities at all. $\endgroup$ Aug 22, 2021 at 14:41

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