Does $j=\rho v$ hold in quantum mechanics?

Question

Let's consider the current of probablity $\vec{J}(\vec{x},t)$ associated to a particle of mass $m$ with wave function $\psi(\vec{x},t)$, given by

$$\vec{J}(\vec{x},t)=\frac{i\hbar}{2m}(\psi \nabla\psi^*-\psi^* \nabla\psi ).$$

The probability density is given by $\rho=\psi\psi^*$.

If we define the average speed of the particle as $\left<\vec{v}(t)\right>=\frac{d\left<\vec{x}(t)\right>}{dt}$, do we have that $\vec{J}(\vec{x},t)=\rho\left<\vec{v}(t)\right>$?

I'm not even sure if this is true in general (I know it's true for a plane wave) but I tried to do the calculation however I ran into some integral expressions and I'm not sure how to go forwards.

Answer 1

It sort of holds for cases with slowly varying potential. In the semi-classical(WKB) approximation in which we have a slowly varying potential $V(x)$, it's customary to express the wavefunction as $$\Psi(x,t)= \sqrt{\rho(x,t)}\exp[\dfrac{i}{h}S(x,t)]$$ where $\rho=|\Psi|^2$ by definition and $S$ is the action. Applying the definition for current $$\vec{J}=\dfrac{\hbar}{m}\Im(\Psi^*\nabla\Psi)$$

where with $\Im$ denotes imaginary part. It's easy to verify yourself that applying this on $\Psi(x,t)$ will give $$\vec{J}(x,t)=\rho(x,t)\frac{\nabla S(x,t)}{m}$$

For slowly varying potential and stationary solutions (i.e., $S(x,t)=S(x)-Et$ ), a valid approximation for the action is $$S(x)=\int_{x_1}^{x} p(x)dx$$ where $p(x)$ is the often dubbed classical/local momentum and given by $$p(x)=\sqrt{2m(E-V(x))}$$ and $x_1$ is arbitrary (often it's set to satisfy $V(x_1)=E$ ) or in other words $\nabla S=p(x)$ hence we have $$J(x)=\rho(x)\frac{p(x)}{m}=\rho(x) \ v(x)$$

Where $v(x)=p(x)/m$ (local velocity) by definition.

Answer 2

You could probably successfully work this out for the coherent states of a quantum harmonic oscillator, as they are fixed 2D Gaussians orbiting phase space in a circle. You can trivially work it out for eigenstates of some Hamiltonian as both $J$ and $\langle v \rangle = 0.$ However, if those Gaussians started expanding in size uniformly, one might see a radial-expansion component to $J$ which is not reflected in $\langle v \rangle.$

This sort of radial expansion is actually in some sense forbidden by quantum mechanics, interestingly enough. Like, you could create it with a slowly-time-varying potential for a Hydrogen atom, say: then it would seem to be the case that in the $s$-state $\langle v \rangle = 0$ but because the orbital is widening its spatial distribution you have nontrivial $\vec J$ as probability flows out and then back in. But naturally QM satisfies the Moyal equation, which is analogous to the Liouville equation, which expresses this idea of constant density while you follow trajectories and hence $\vec J$ pointing along a trajectory.

So there are possible analogues and limiting cases where this makes sense, but it could not possibly be provable as a fully general result of quantum mechanics the way you have expressed it. You might be able to get a velocity field out of quantum field theory and then use that instead of $\langle \vec v\rangle$, or you might be able to do something with the Moyal equation and Wigner functions that looks analogous. You could even be bold and just define $J/\rho$ as $v$.

Answer 3

Not in general.

The plane wave has a fixed velocity, which is then the average, but the components of a standard wavepacket will not move in lockstep: they will disperse, in general, and the conjectured relation will fail.

To avoid needless clutter and defocussing, stick to just one dimension, and non-dimensionalize to $m=1=\hbar$. Further assume that probability densities at infinity vanish.

By integration by parts, it follows that $$ \langle v(t)\rangle = \langle p \rangle= -i \int_{-\infty}^\infty \!\! \! dy ~~\psi^*(y)\partial_y \psi(y)=\int\!\! dy ~ J(y)~. $$ So you must contrast $\rho(x) \int dy J(y)$ to $~J(x)$.

In 1D, run to mama free right-moving Gaussian wavepacket, as always: $$ \psi(x,t) = \frac{(2/\pi)^{1/4}}{\sqrt{1 + 2it}} e^{-\frac{1}{4}\bar{v}^2} ~ e^{-\frac{1}{1 + 2it}\left(x - \frac{i\bar{v}}{2}\right)^2}\\ = \frac{(2/\pi)^{1/4}}{\sqrt{1 + 2it}} e^{-\frac{1}{1 + 4t^2}(x - \bar{v}t)^2}~ e^{\frac{i}{1 + 4t^2}\left((\bar{v} + 2tx)x - \frac{1}{2}t\bar{v}^2\right)},\\ \rho(x,t) = \sqrt{\frac{2/\pi}{{1+4t^2}}}~e^{-\frac{2(x-\bar{v}t)^2}{1+4t^2}}~. $$

$\langle p\rangle=\langle v \rangle=\bar{v}$, the group velocity of the right-moving wavepacket, of course, but it is spreading. For large t, its width increases as 2 t .

Now compute $$ J(x)= \frac{-i}{2}(\psi^*\partial_x \psi - \psi \partial_x \psi^*)= \frac{\bar{v} + 4tx}{1+4t^2} \rho (x) ~. $$ Indeed, at $t=0$, you had $J=\rho \bar{v}$ — but this is the last time you did:

At all subsequent times, while the center (and mean) of the wavepacket moves like $\langle x \rangle = \bar{v} t$ and so satisfies your conjectural rigid flow relation $\rho (\langle x\rangle) ~~\bar{v}=J (\langle x\rangle) $, any other x in it does not! This is not a surprise, since the Gaussian preserves its shape, but is spreading, so there is probability flow in it.

Your relation, however, does hold, also predictably, for non-dispersive states like coherent states (Schroedinger's wavepacket) which oscillate semi-classically without change in shape. Like plane waves, they move in lockstep.

Answer 4

Let $\vec{J}=\frac{i\hbar}{2m}(\Psi\frac{\partial \Psi^*}{\partial x}-\Psi^*\frac{\partial\Psi}{\partial x})$

Let $\Psi=Ae^{ikx}$.

Then $\Psi\frac{\partial \Psi*}{\partial x}=-ikAA^*$

If you subtract the complex conjugate, then

$\vec{J}=\frac{i\hbar}{2m}(-2ikAA^*)=\frac{\hbar kAA^*}{m}=vAA^*$

And $AA^*$=probability of being in state k.

So $J=\rho v$ holds here and a finite linear combination of discrete states in general.

How general can you get? You'd need to consider a full spectrum of both possible discrete and continuous eigenstates.