Think about an electron been accelerated from rest in a static electric field. If we treat the problem classically, in which the electron is just a point charge. The velocity of the electron would increase linearly with respect to time.

Now if we treat this problem quantum mechanically, in which the electron is represented by a wave packet (the field still classical). Assume we have solved the Schrodinger equation and have the time-dependent wave function $\psi(x,t)$, what should we calculate to represent the "velocity" of the electron?

One possibility is to calculate the group velocity of the wave packet $v_g$. The other way is to calculate the probability current $$ j=\frac{\hbar}{2mi}\left[\psi^*\nabla\psi - \psi \nabla\psi^*\right]. $$

Are the group velocity of the wave packet and the probability current essentially have the same physical meaning? If they do, how to prove they are the same, and if they are different, what are the differences?


1 Answer 1


The probability current by itself is actually composed of the momentum operator defined as

$$\hat {\vec p}\psi(\vec r)=\frac{\hbar}i\nabla \psi(\vec r).$$

Integrate the current $j$ over all space, and you'll obtain the expected value of the group velocity. It's easy to see that it's not phase velocity: the latter depends on the zero point of potential energy, which is arbitrary, while the momentum operator doesn't depend on the potential.

So, yes, both approaches—group velocity via dispersion relation, and probability current—yield the same result.


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