Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
A good explanation is given here – Murod Abdukhakimov Feb 29 '12 at 11:34
up vote 9 down vote accepted

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

In principle, I suppose you could perform a transformation on the wavefunction that looks just like a $U(1)$ gauge transformation except that the parameter can be complex. But the resulting group of transformations would not be $U(1)$, it would be some two-dimensional group, because a complex number parametrizes two dimensions.

share|cite|improve this answer
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 Z Mar 2 '12 at 16:55
Suggestion to the answer (v1): Stress that the Lie group $U(1)$ is a manifold with the number of real dimensions equal to one. – Qmechanic Oct 9 '13 at 0:14
@Qmechanic I edited, see what you think. – David Z Oct 9 '13 at 1:14

$\chi$ 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 ;-)

share|cite|improve this answer
Elaborate please. I have never seen an imaginary gauge. – Siyuan Ren 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

You can use electromagnetic gauge transforms with complex, rather than real $\chi$. However, I don't think they would be as useful as transforms with real $\chi$, because, if $\chi$ is not real, the equations of motion change, so there is no gauge invariance (see, e.g., Eqs. 20,21 of my article in the European Physical Journal C (free access, ), where the equations of motion after a gauge transform with a complex $\chi$ (which is equal to $\alpha$ of my article, up to a constant factor) are written out for the Dirac field interacting with electromagnetic field. To avoid confusion, please see the note between Eqs. 16 and 17).

Let me also note that magnetic field does not change under a gauge transform with a complex $\chi$.

share|cite|improve this answer

I think because the wavefunctions are required to be normalized so that $\psi^{*}\psi$ represents the probability or probability density of finding the particle, so their amplitude are not allowed to scale arbitrarily. That's why the gauge field can only be real.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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