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The Hamiltonian of a non-relativistic charged particle in a magnetic field is

$$\hat{H}~=~\frac{1}{2m} \left[\frac{\hbar}{i}\vec\nabla - \frac{q}{c}\vec A\right]^2$$.

Under a gauge transformation of the magnetic potential:

$$\vec A ~\rightarrow~ \vec A + \vec\nabla \chi,$$

the wavefunction of the particle transforms as

$$\Psi~\rightarrow~ \Psi\exp(\frac{iq\chi}{\hbar c}).$$

When $\chi$ is real, the wavefunction simply gains an extra phase factor. However, when $\chi$ is imaginary, there is a measurable change to the wavefunction. This seems to contradict the fact that the magnetic field is invariant under the gauge transformation. How do I resolve this?

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A good explanation is given here – Murod Abdukhakimov Feb 29 '12 at 11:34

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$\chi$ is a real-valued function. This is part of the definition of the gauge transformation, since $U(1)$ is a one dimensional group. In general, when talking about gauge transformations in particle physics, group parameters are restricted to be real.

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So we cannot have an imaginary gauge even though the resulting magnetic field is the same? Is this related to how the magnetic potential is more fundamental than the magnetic field, like what the Aharanov-Bohm effect has shown? – leongz Mar 2 '12 at 15:15
The reason the gauge transformation is defined as it is is not because it's the most general transformation which allows the magnetic field to be the same, because obviously it's not. It's the simplest transformation which enables local gauge invariance. You only need one (real) degree of freedom to make the covariant derivative. In principle you could have multiple degrees of freedom in the gauge transformation, but it just hasn't been found to be necessary for electromagnetism. (That changes once you introduce the weak and strong forces.) – David Zaslavsky♦ Mar 2 '12 at 16:55
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$\chi$ can be any reasonable function, real-valued, imaginary-valued, whatever. No variable change can change physics although new wave function and its new equation may be different from the old ones ;-)

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Elaborate please. I have never seen an imaginary gauge. – C.R. Feb 29 '12 at 9:44
@KarsusRen: the gauge transformation is an introduction of new variables $\vec{A}^{\prime}$ and $\Psi^{\prime}$, isn't it? When $\chi$ is real, the new equations have the same form as the old ones, but the solutions are different numerically. The case of imaginary $\chi$ is not different in this respect from the case or a real one. – Vladimir Kalitvianski Feb 29 '12 at 12:25

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