**Context:** 

In Griffith's [book](https://rads.stackoverflow.com/amzn/click/1107179866) 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.]