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".


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.]

  • $\begingroup$ Your beef appears to be with plane waves used for illustration, which is routine in QM, as a first step to putting normalizable wavepackets together. Most wave treatments reassure the reader first on how to intuit a situation on the basis of the constituent waves involved, the useful exercise displayed here. $\endgroup$ Commented Dec 26, 2017 at 22:59
  • 2
    $\begingroup$ $|\psi|^2$ does not have peaks an troughs. $\psi$ does, as does $\psi^2$. $|\psi|^2=\psi^*\psi$ is constant. $\endgroup$
    – JEB
    Commented Dec 26, 2017 at 23:34
  • $\begingroup$ As you proceed to atoms, you will further see that the azimuthal component of the probability current of a static bound state of an m >0 atom is also non vanishing: a perpetual azimuthal probability swirl of a static configuration! $\endgroup$ Commented Dec 26, 2017 at 23:48
  • $\begingroup$ Related. $\endgroup$ Commented Dec 27, 2017 at 0:41
  • $\begingroup$ Also related. $\endgroup$ Commented Dec 27, 2017 at 0:55

3 Answers 3


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.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Dec 31, 2017 at 13:13
  • $\begingroup$ @JahanClaes Would you also consider addressing this point: physics.stackexchange.com/questions/380655/… , which also wasn't clear in Griffiths' book... Your insight into this would be much appreciated. $\endgroup$
    – user135626
    Commented Jan 29, 2018 at 14:47

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?

  • $\begingroup$ This is an elegant way to do it -- many thanks! But I just wonder whether this is what the author had in mind as he so casually remarked in the solution manual that we should "expect" the current direction to be to the right, without much explanation (maybe by imagining it to be the same as the travel direction of the wavefunction). Do you think he meant such notion or something else? I just fear that I am missing a subtle point. $\endgroup$
    – user135626
    Commented Dec 27, 2017 at 13:24
  • $\begingroup$ Not sure, but if the "expected" refers to what one would expect from a plane wave, then one can consider this approach as a generalization of that. $\endgroup$ Commented Dec 27, 2017 at 13:49
  • $\begingroup$ So, with this, one could say that if a plane wave represents a wavefunction, then its probability current will always point in the same direction as its direction of propagation (phase velocity vector direction). Is that what you meant? $\endgroup$
    – user135626
    Commented Jan 2, 2018 at 23:14
  • $\begingroup$ A single plane wave is a bit of an idealization, but if one wants to make that assumption then yes, that's what one will get. $\endgroup$ Commented Jan 7, 2018 at 3:30

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.

  • $\begingroup$ What you have derived is the phase velocity of the wave. However, this is different from the statistical profile of the signal (the probability distribution function), which is still spanning towards $\pm \infty$. Therefore, I am not sure why we should interchange the meaning of phase travelling to the right with probability increasing to the right, if such a wavefunction itself is infinitely long to the left and right. Even expectation calculations fail for such non-normalizable case. $\endgroup$
    – user135626
    Commented Dec 26, 2017 at 21:36
  • $\begingroup$ Okay so I wasn't correct that that was what you're stuck on. $\endgroup$ Commented Dec 26, 2017 at 21:46
  • $\begingroup$ No worries. BTW you forgot to remove the square power on $k$ in velocity equation. $\endgroup$
    – user135626
    Commented Dec 26, 2017 at 22:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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