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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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  • $\begingroup$ A good explanation is given here $\endgroup$
    – Murod Abdukhakimov
    Commented Feb 29, 2012 at 11:34

4 Answers 4

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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 (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.

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  • $\begingroup$ 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? $\endgroup$
    – leongz
    Commented Mar 2, 2012 at 15:15
  • $\begingroup$ 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.) $\endgroup$
    – David Z
    Commented Mar 2, 2012 at 16:55
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    $\begingroup$ Suggestion to the answer (v1): Stress that the Lie group $U(1)$ is a manifold with the number of real dimensions equal to one. $\endgroup$
    – Qmechanic ♦
    Commented Oct 9, 2013 at 0:14
  • $\begingroup$ @Qmechanic I edited, see what you think. $\endgroup$
    – David Z
    Commented Oct 9, 2013 at 1:14
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    $\begingroup$ @TheQuantumMan That sounds like a good thing to post a separate question about. $\endgroup$
    – David Z
    Commented Sep 16, 2023 at 7:08
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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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    $\begingroup$ Elaborate please. I have never seen an imaginary gauge. $\endgroup$
    – Siyuan Ren
    Commented Feb 29, 2012 at 9:44
  • $\begingroup$ @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. $\endgroup$
    – Vladimir Kalitvianski
    Commented Feb 29, 2012 at 12:25
  • $\begingroup$ Would you please provide some examples where such an imaginary gauge is used? $\endgroup$
    – Jonathan Huang
    Commented Nov 30, 2023 at 6:43
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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, http://download.springer.com/static/pdf/480/art%253A10.1140%252Fepjc%252Fs10052-013-2371-4.pdf?auth66=1381456528_6b6a376576161b4f3d18182317776008&ext=.pdf ), 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 $\chi$.

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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.

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