2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Note that $\frac{J}{\rho}=\frac{i\hbar}{2m}\left(\frac{\nabla\psi^\ast}{\psi^\ast}-\frac{\nabla\psi}{\psi}\right)$ becomes $\frac{hk}{m}=\frac{p}{m}=v$ for a plane wave $\psi\propto\exp i(k\cdot x-\omega t)$. $\endgroup$ – J.G. May 13 at 16:08
  • $\begingroup$ It is even more interesting in my naive understanding. In quantum mechanics of electromagnetism, the canonical momentum and the "kinetic momentum" are different due to the presence of the vector potential (just like in classical electrodynamics). The operator whose expectation value gives the correct density current is the one constructed using kinetic momentum operator and not the canonical one. So, in some sense, it just seems amazing that it is really the velocity operator (kinetic momentum operator) that enters the description of density currents--and not the momentum operator per se. $\endgroup$ – Dvij Mankad May 13 at 16:11
  • 1
    $\begingroup$ Your uncertainty is warranted. You can see how this is approximately true for a 1D wavepacket. But are you seeking a conterexample of the exact statement? $\endgroup$ – Cosmas Zachos May 13 at 16:20
  • $\begingroup$ Conceivably related. $\endgroup$ – Cosmas Zachos May 16 at 0:48
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $v$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.