Consider a charged particle (electron) moves in xy plane under a magnetic field pointing along the z direction, i.e., $\vec{B}=B\hat{z}$. As a consequence, we can write down three different gauges-

  1. symmetric gauge: $\vec{A}=(A_x,A_y)=\frac{B}{2}(-y,x)$,
  2. translational invariant gauge along x: $\vec{A}=B(-y,0)$, and
  3. translational invariant gauge along y: $\vec{A}=B(0,x)$.

As previously addressed and asked in another post Charged quantum particle in a magnetic field, the answers were the three different ground state wavefunctions resulting from the above three different gauge choices are connected by linear transformations, and giving rise to same energy spectrum. Hence, everything is consistent. Working through the detailed derivation, for instance Hitoshi Murayama's Landau levels note, we see that

  1. the symmetric gauge gives rise to ground state wavefuction: $\psi_n (z,\bar{z})=N_n z^n \exp\left(\frac{-eB\bar{z}z}{4\hbar c} \right)$, where $z=x+iy$ and $\bar{z}=x-iy$, while
  2. $\vec{A}=B(0,x)$ gives rise to different ground state wavefunction (unnormalized): $\psi_0 (x,y)=\exp\left(ik_y y -\frac{eB}{2\hbar c}\left(x-\frac{\hbar c}{eB}k_y \right)^2 \right)$.

It is obvious that the two ground states are different, and thus $|\psi|^2$. My question is how our different choice of gauge, coming from the same physical magnetic field, results in different $|\psi|^2$, even though it leads to same energy spectrum?

  • $\begingroup$ I think you must be careful when trying to compare these two wave-functions. As you know, the GS w-f are highly degenerate, and so you might be able to construct one w-f of one gauge as the superposition of w-f of the other. $\endgroup$
    – Adam
    Mar 29, 2016 at 5:32
  • $\begingroup$ @AccidentalFourierTransform: I think the point of the OP is that physical quantities are gauge independent, and you may ask why $\langle \hat x\rangle$ would change with the gauge, since the w-f changes. $\endgroup$
    – Adam
    Mar 29, 2016 at 8:50
  • $\begingroup$ @AccidentalFourierTransform I know that. But I think that the answer is a bit more subtle than just saying "it has to be gauge invariant, so it is". $\endgroup$
    – Adam
    Mar 29, 2016 at 9:19

2 Answers 2


Of course, observables such as $|\psi|^2$ have to be gauge invariant, and the two-wave functions of gives by the OP, in two different gauge, gives obviously two different probability distributions.

The resolution of the paradox comes from the fact that the two wave-functions describe in fact two different ground-states, as it can be seen as follows. $\vec A_1=\frac B2(-y,x)$ and $\vec A_2=B(0,x)$ are related by a gradient of the function $\chi=\frac B2 xy$, $\vec A_2=\vec A_1+\nabla \chi$, which implies that for a given wave-function $\psi_2$ translational invariant gauge along $y$, $\psi_2 =N\exp\left(ik_y y -\frac{eB}{2\hbar c}\left(x-\frac{\hbar c}{eB}k_y \right)^2 \right)$, corresponds a wave-function $\psi_1$ in the other gauge, $\psi_1=e^{i e \chi}\psi_2$. It is easy to show that $$ \psi_1=N' e^{k_y z-\frac{eB}{2}z^2}e^{-\frac{eB}{4}z\bar z} , $$ which can be rewritten as $$ \psi_1=f(z)e^{-\frac{eB}{4}z\bar z} , $$ and is indeed a ground-state wave-function in the symmetric gauge.

One has to remember that the wave-functions given by the OP are just two elements of two basis (corresponding to two gauges) that can describe the massively degenerate states of one particle in a magnetic field.

Edit: To clarify a little bit. The two wave-functions in the OP's question do not describe the same physical state. For a given $n$ or $k_y$, they are all valid ground-state wave-functions, and form a basis to describe the massively degenerated ground-state. But when changing gauge, one will not generally map from one basis state of one gauge onto one basis state of the new gauge, but it will generally be a superposition, as can be seen in the above example.

  • $\begingroup$ thank you for the vivid explanation. It is clear to me now that the two wf's are related by a gauge transformation. However, what do we expect to see experimentally? As you agreed, the two wf's give rise to two different prob. dist. Or, do we see superposition of both? $\endgroup$ Mar 30, 2016 at 6:55
  • 1
    $\begingroup$ @thihakyaw: experimentally, it depends in what state you prepare the system. Say you prepare it in some way in the state $\psi_2$. Then you could say equivalently that you prepared it in the state $\psi_1$, since 1- you don't know the gauge since it is not observable, 2- all the properties of these two states are the same. However, of course, a state $\psi_1'$ (with its companion $\psi_2'$) will be different and have different properties. $\endgroup$
    – Adam
    Mar 30, 2016 at 7:02
  • $\begingroup$ thanks again for the explanation. I guess we have no control over of which state to prepare, except we might cool down the system to one of its many degenerate ground states. Or, do we really have a control to which state to begin with? sorry for my persistent question, I am new in this. thanks. $\endgroup$ Mar 30, 2016 at 11:19
  • 1
    $\begingroup$ @thihakyaw: We could imagine measuring the position of the particle at $t=0$. Then $\psi(x)=\delta(x-x_0)$. You can then decompose the delta function onto the basis of your choice (for example, that of the symmetric gauge). You can then also change gauge, and rewrite it in terms of the basis of the other gauges. It's all the same physics. $\endgroup$
    – Adam
    Mar 30, 2016 at 11:40
  • $\begingroup$ @thihakyaw: see also my edit, maybe it clarifies things a bit. $\endgroup$
    – Adam
    Mar 30, 2016 at 11:45

See old paper of Swenson in American Journal of Physics (magnifying on exactly the above issues) and a new resolution that will appear in a forthcoming paper by G. Konstantinou & K. Moulopoulos..

  • 2
    $\begingroup$ Referencing forthcoming papers is bad style - it simply doesn't answer anything and it looks like you are one of the authors (are you? Self-referencing is ok, but you should say so). In addition, your citation is more than incomplete (I can find five articles fitting the description) and we have the policy that you should, at least in a few words, sketch the content of the paper that gives an answer to the question. $\endgroup$
    – Martin
    Jul 17, 2016 at 12:29
  • $\begingroup$ Yes, I completely agree with Martin. Would you, @user123823, specify the old paper of Swenson? DOI or url to the paper? Do you also have the arXiv link to the paper by G. Konstantinou et. al.? Thanks. $\endgroup$ Jul 18, 2016 at 13:12

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