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.]
Best Answer
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.