# 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?

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.

• Hi @Anu3082, the "speed of the wave " is an ambiguous statement and isn't as well defined as the speed of a particle. The example you have given deals with a complex exponential function and it's easy to get confused with this, but keep in mind that such a function doesn't actually move anywhere with any velocity, it is defined over all space. A more physical example would be a gaussian wavepacket, for example. Now imagine something like that -- localised in space -- moving rightward. You can do the calculations and see that what you would call the "speed" of this "clump" is indeed $v_g$. Nov 29, 2020 at 17:17
• Check out this nice answer on the difference between phase and group velocities, and what they might represent physically. Remember, the quantity you call the velocity (actually the phase velocity) is often unphysical, even in classical physics, since it can give rise to speeds that are greater than $c$ and so on. So one must tread carefully in interpreting it. Nov 29, 2020 at 17:19
• I'll modify my answer in a little while to include a general wave packet. But one additional point I'd like to add is this: if you're interested in the speed of the transfer of information, this is usually the group velocity; the phase velocity is only defined for waves of infinite extent like the one you've described, and since these waves are described over all space, what you're actually measuring with $v_p$ is how fast the phase changes, rather than how fast information is transferred. i.e. for waves of the given form, $v_p$ isn't what you'd conventionally call the wave's speed. Nov 29, 2020 at 17:26
• I will tell you where I think I went wrong. But before, I see that using $v = v_g$ gives the correct result, it is also the speed to which the probability current is proportional to. Before I had thought, because $\Psi$ is sweeping past any point at speed $v_p$, the area underneath it squared $|\Psi^2|$ (at this point I almost pictured $\Psi$ like a sinusoidal wave passing that point, obviously I assumed that the area under a sinusoidal wave must move along with the sine), should also pass the point at speed $v_p$, therefore the probability current is its density times $v_p$. Nov 29, 2020 at 17:51
• I understand that $v_g$ is often the most useful speed, I never thought that would be useful in this question in any way because the wave has an infinite extent, and the idea of the "envelope" seemed impossible. But I would better lean on to the probability formula, although why it spits out the probability current to be proportional to $v_g$, in a place where there is no envelope (but $v_g$ still defined), I don't know. Thanks. Nov 29, 2020 at 18:00