Probability current in free quantum particle wavefunction

Question

Context:

In Griffith's book on quantum mechanics, the probability current formula (which indicates the rate of decrease in probability over time, at $x$) is given as: $$ J(x,t):=\frac{i\bar{h}}{2m}\left(\Psi\frac{\partial \Psi^{*}}{\partial x}-\Psi^{*}\frac{\partial \Psi}{\partial x}\right), $$

and the wavefunction of a stationary state for a free particle is given by: $$ \Psi_{k}(x,t)=Ae^{i\bigl(kx-\frac{\bar{h}k^{2}}{2m}t\bigr)}.$$

Then in one of the problems (2.19), the author asks the reader to calculate $J$ for such $\Psi_{k}(x,t)$, and find its direction of flow.

Applying the above equations quickly gives the correct value: $J=|A|^{2}\bar{h}k/m$, which is a positive value. The "Solutions manual" gives this answers and says that, therefore, $J$ points in the positive $x$ direction "as you would expect".

Question:

My question here is why should we have obviously "expected" this? The meaning here may be a bit subtle. I understand the correct mathematical result, but not the intuition that would make us readily "expect" this (my problem is not in the answer itself, but in the remark that such answer should have been "expected")... Yes, looking at the exponent in $\Psi$'s equation indicates that the wavefunction will move in the positive $x$ axis direction as time passes, but that doesn't necessarily mean that probability at a given point will subsequently "drop", as to give rise to a positive probability current there (i.e. we know that $\frac{\partial}{\partial t}|\Psi|^{2}=-\frac{\partial}{\partial x}J$). In fact, the given $\Psi$ above is a sinusoidal wave that keeps increasing and decreasing in any direction we flow, with constant $|\Psi|$. So why should it be obvious or "expected" to get a reduction in probability as the wavefunction shifts to the right in time?

[Update: even if the phase velocity here is (from the ratio of the coefficients of $x$ and $t$ in the exponential term) known to be positive as $\frac{\bar{h}k}{2m}$ and therefore the wave's phase appears to travel to the right side, why should we interchange the meaning of "phase velocity moving to the right" with the statistical "probability increase to the right (probabiltiy distribution function)" for such wave? The wave sinusoidal tails extend to $\pm \infty$ still, and we cannot perform any expectation calculations on such non-normalizable function (so we cannot statistically affirm that expectation of momentum is moving in either direction, for example, because we cannot perform the calculation). After all, we know that phase doesn't contribute much to statistical information, given a fixed energy wavefunction, since the modulus will remove it anyway.]

Answer 1

The expectation value of the momentum is in the positive direction. So you expect the particle is moving to the right.

---

This answer has gotten a little derailed with a discussion involving normalizable versus non-normalizable wavefunctions. I'm going to address this here, but I want to emphasize that I think this is a technical detail that Griffiths is trying to avoid. His comment that you should "expect" positive probability current was supposed to indicate exactly what I said above.

So: non-normalizable wavefunctions. There are probably dozens of ways to view non-normalizable wavefunctions, but ultimately they all come down to saying that non-normalizable wavefunctions are approximations to normalizable wavefunctions, and that you can treat them identically for most purposes. Here's one way to view a plane wave. Let's view our plane wave $\psi(x)=e^{ikx}$ as an approximation to an actual, normalizable wavefunction $\psi(x)=f(x)e^{ikx}$, where $f(x)$ is some $\mathcal{L}^2$-integrable function that varies much slower than $e^{ikx}$. For example, we might have $f(x)=e^{-{x^2}/{\sigma^2}}$, with $\sigma>>k$. Then if we calculate the expectation value of the momentum operator, we get

\begin{align} \langle \hat p\rangle &=\int dx\ f^*(x)e^{-ikx}(-i\hbar\frac{\partial}{\partial x})f(x)e^{ikx}\big/\int dx\ |f(x)|^2\\ &=\int dx\ f^*(x)e^{-ikx}(-i\hbar f'(x)e^{ikx}+\hbar k f(x)e^{ikx})\big/\int dx\ |f(x)|^2\\ &=\int dx\ (-i\hbar f'(x)f^*(x)+\hbar k |f(x)|^2)\big/\int dx\ |f(x)|^2\\ &= \hbar k -i\hbar \frac{\int dx\ f'(x)f^*(x)}{\int dx\ |f(x)|^2}\\ \end{align}

You can check for yourself that in the limit that $f(x)$ is very slowly varying, the second term vanishes. So we say that the plane wave has momentum $\hbar k$. This is one reason why plane waves are good approximations to real solutions. If you mostly look like a plane wave, you mostly have the properties of a plane wave.

If you like, you can also check that a wave of the form $f(x)e^{ikx}$ has approximately the same probability current as $e^{ikx}$, with the error vanishing in the limit where $f(x)$ is slowly varying. That means that a plane wave probability current must match the normalizable probability current, and that the plane wave momentum must match the normalizable momentum. Hopefully for a normalizable wave with positive momentum, you agree that you "expect" positive momentum and positive probability current to go hand-in-hand. Since the non-normalizable plane wave is a good approximation to both the momentum and the probability current of the normalizable wave, you should "expect" positive probability current and positive momentum to go hand-in-hand for plane waves as well.

Answer 2

Perhaps this is another way to look at it. I'll use the analogy of optics. There we know that light generally propagates in a direction perpendicular to the wavefront. So if I have an optical field (ignoring the amplitude) $$ \Psi = \exp[i\theta({\bf x})] , $$ Then I can get the phase as $$ \theta({\bf x}) = \frac{1}{i2} \ln\left( \frac{\Psi}{\Psi^*} \right) . $$ The propagation direction is then given by $$ {\bf k} = \nabla\theta({\bf x}) = \frac{1}{i2} \left( \Psi^*\nabla\Psi - \Psi\nabla\Psi^* \right) . $$ So you can see this gives the form of the current you are looking for apart from the replacement $\partial_x\to\nabla$.

Does this help?

Answer 3

Waves are such a common thing to see in physics that sometimes you'll miss out on a proper "introduction" to such an analysis. I think you're right in trying to figure out what is "obvious" about this (and I think I know what you're missing).

At t=0 we can look at what the wave profile looks like as a function of x. (It just looks like a cosine wave). Let's "lock-in" on a part of this cosine wave and keep track of that point when it evolves in time. Let's keep track of this point when $\Psi = A$, which is when $e^{i \theta}=1 \implies \theta = 0 \implies kx-\frac{hk^2}{2m}t=0$ Now if we solve for x, we will now at all times where this part of the wave is: $x(t) = \frac{hk^2}{2m}t$ Now we can see that as time increase, this point always moves forward! (And you can do this for any point along the wave). I believe this is the most "obvious" way of looking at your wave.