Probability current and Travelling waves Consider a travelling wave $\Psi(x,t) = A e^{i(kx-\frac{\hbar k^2}{2 m})}$
According to the formula for probability current $J = \frac{\hbar}{m} Im( \Psi^*\frac{\partial\Psi}{\partial x})$, the probability per unit time flowing along the $x$ direction is $\frac{\hbar k}{m}|A|^2$.
But we know the probability density corresponding to $\Psi$ to be $|\Psi|^2 = |A|^2$ . Because a length  $v =\frac{\hbar k}{2 m}$ of the wavefunction $\Psi$ passes any point $x$ in unit time, the probability current is $J = v 
 |\Psi|^2 = \frac{\hbar k}{2m}|A|^2$.
But this is exactly half the probability current obtained by formula.
Where did I go wrong?
 A: Very interesting question! Whenever dealing with a wave's velocity, it's a good idea to ask yourself if you should be dealing with a phase velocity or group velocity. In most introductory examples of wave mechanics they are both the same since the waves are not dispersive. However, for quantum mechanical waves this is not true, since $\omega$ is not proportional to $k$, but indeed $$\omega = \frac{\hbar k^2}{2m}.$$
In this case, the phase velocity and group velocity differ by a factor of 2:
\begin{aligned}
\text{Phase velocity:  }\quad v_p & = \frac{\omega}{k} = \frac{\hbar k}{2m},\\
\text{Group velocity:  }\quad v_g & = \frac{\partial \omega}{\partial k} =\frac{\hbar k}{m}.
\end{aligned}
Your analysis is correct, it's just that you need to use the group velocity in it. It is much easier to see why this is the case for a superposition of waves, but essentially a good rule of thumb is to say that for any wavepacket, the velocity through space is the group velocity. It is the group velocity that behaves (in a very crude sense) as we would expect the "velocity" of the classical particle to behave. Feynman deals with this a little in Chapter 7 of Volume 3.
Also related: LubosMotl's answer to this question: Speed of a particle in quantum mechanics: phase velocity vs. group velocity.
