# Can an Electromagnetic Gauge Transformation be Imaginary?

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

$\chi$ 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.
$\chi$ 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 ;-)
@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