In this document


on Landau levels, in section 4, page 19, "Transitionally invariant Gauge", they analyze the free electron in a uniform magnetic field problem, with the gauge $\vec{A}$ = $(0, Bx)$.

This leads to the wave function being a combination of harmonic oscillator in the $x$ dimension and a infinite strip in the $y$ direction, as shown in their Figure 6, page 20, with the center of the harmonic oscillator being bumped over by some amount which depends on the $y$-momentum(wave number) $k_y$.

On page 21 the probability current is said to be going up on one side of the harmonic oscillator (where $x > \dfrac{\hbar c}{eB} k_y$) and down on the other. By up and down I mean $+y$ direction and $-y$ direction.

How can a wave function have its right side be sending off probability, forever, upwards, and left side be sending off probability, forever, downwards? It would seem like this is saying as time progresses, the wave function eventually is splitting in two with its right side going off to $+y$ infinity, and its left side going off to $-y$ infinity, and after enough time passes its right and left sides will have separated their separate ways each arriving at the $+y$ vertical infinity location in the universe and the other at the $-y$ vertical infinity..


1 Answer 1


The state is an energy eigenstate and all the phase change is in the y direction in a periodic fashion, so there is an actual period $T$ where after time $T$ the wave function is exactly the same. So it isn't really changing in time, it is just changing its overall phase.

The wave in the paper is unnormalized in the y direction. And it was chosen to be a plane wave. A plane wave has a regular phase change, and hence a steady canonical momentum.

But the kinetic momentum differs from the canonical momentum by the vector potential. So the kinetic momentum is different in different places. And it is determined because we chose a phase and a gauge. You could add a constant to the vector potential and the kinetic momentum would be different, because you are fixing the canonical momentum to be nice because that made the wave mathematically easy to describe.

Since the wave is unnormalized in the y direction it extends forever in the $\pm \hat y$ directions and so isn't really changing. And we just gave it that current out of convenience. There are lots of currents consistent with the same probability. There are also lots of gauges, but normally if you adjust the gauge you also have to adjust the phase if you want the physics to be the same, but in this case we wanted the phase to be a nice plane wave.

If you look at the example where they add an electric field, effectively it is just adding a constant to the vector potential. So never read too much into a wave function, and know which parts relate to observations.

In this case nothing is really changing in time, since the global overall phase is unphysical. And we selected a certain canonical momentum from the plane wave phase and then selected a gauge and that is what made kinetic momentum be what it is. It isn't anything deeper than that.

  • $\begingroup$ After thinking on your words some more, and reading this line in lassp.cornell.edu/clh/p683/1.1DEF.pdf in the section Semiclassical Wavefunctions on page 5 of 8, "The .. standard spin wavefunction - having a completely definite $S_z$, and completely indefinite $S_x$ or $S_y$ - is similar to the ribbon-like electron eigenstates we found in either Landau or symmetric gauge." I believe what you are saying is that by pointing my finger in some arbitrary direction and calling it "$+y$" I am not suddenly making the wave function be truly ribbon-like in that orientation. $\endgroup$
    – a00
    Oct 5, 2015 at 15:04
  • $\begingroup$ It's analogous to spin, where when we have an electron known to be in $+z$ spin, it is totally indeterminate if we pick some arbitrary direction our finger points and call it $+y$, how much angular momentum the electron has along that axis. $\endgroup$
    – a00
    Oct 5, 2015 at 15:04
  • $\begingroup$ As for how the wave function can be sending off probability infinitely in two different directions, yet never "splitting apart" as time goes on, I note that for a plane wave even NOT in an electromagnetic field, the probability current turns out to be $\hbar k$. I attribute this to the idea that if you have an infinite waterfall, there will always be a point BEFORE the current point in the waterfall that can supply you with new probability "stuff"; likewise there is always a further downstream point into where your off-given probability stuff can "sink". $\endgroup$
    – a00
    Oct 5, 2015 at 15:05
  • $\begingroup$ Thank you for the answer. It was especially thought-provoking your statement that after time T the wave function returns to its original state. $\endgroup$
    – a00
    Oct 5, 2015 at 15:18
  • 1
    $\begingroup$ @a00 All energy eigenstates are like that. $\endgroup$
    – Timaeus
    Oct 5, 2015 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.