In the Schrodinger equation, the statement of electromagnetic gauge invariance is that observables don't depend on the electromagnetic gauge. That is, if we let:

$$\partial_t \psi(\mathbf x,t) = \hat H \psi(\mathbf x,t) $$ $$\mathbf A' \equiv \mathbf A +\nabla \lambda$$ $$V' \equiv V - \frac{\partial \lambda}{\partial t}$$


$$\partial_t \psi(\mathbf x,t) = \hat H' \psi(\mathbf x,t) $$

will also give the same expectation values, where $\hat H'$ that depends on the modified potentials.

Often, however, I see authors drawing more attention to the fact that you can rewrite the Schrodinger equation in the same form as the original, provided the wavefunction also transforms by a local phase factor:

$$\psi'(\mathbf x,t) \equiv \exp(-i\lambda)\psi(\mathbf x,t).$$

Why exactly is this 'form invariance' significant, and how is it related to the fact that observables (e.g. electric and magnetic fields, expectation values) are not changed by the choice of gauge? I feel it is sometimes implied that the ability to write a transformed equation in the same form as the original implies that it is gauge invariant. Naively, one might assumed that you've merely 'defined away' observable differences by transforming your wavefunction.

If this is question is still poorly defined and semantic I'll just close it out.

  • $\begingroup$ Can you give an example of how the phrase 'form invariance on gauge transformation' is used? $\endgroup$
    – knzhou
    Aug 17, 2016 at 3:00
  • $\begingroup$ This appears to be mostly a semantic question which is difficult to answer because the context is lacking. I've never heard "form invariance on gauge transformation", and the usage of the Schrödinger equation and "local phase" implies you want to talk about a specific gauge invariance, since the notion of gauge invariance is more general than "local phases" or, indeed, quantum physics. Finally, I'm not sure what implications you're asking about in the second point - how bad a broken gauge symmetry is again depends on the context, and what it is broken by. $\endgroup$
    – ACuriousMind
    Aug 17, 2016 at 13:13

1 Answer 1


Gauge invariance means that something is invariant with respect to a gauge transformation. Gauge transformations are transformations of quantities, typically the gauge field $A^{\mu}$, according to $$ A^{\mu} \rightarrow A^{\mu}+\partial^{\mu} f $$ where $f$ is a scalar field. Here I show the simplest case for an Abelian symmetry. The non-Abelian case is a little more involved, but the idea behind it is similar.

Gauge transformations are associated with a gauge symmetry of a theory (usually expressed in terms of a Lagrangian). Such symmetry is represented by continuous group (Lie group), such as the U(1) Lie group for the Abelian case or SU(2), etc. for the non-Abelian cases.

Although gauge invariance is the generic term that people tend to use, one can differentiate between invariance or covariance. In the former case it usually refers to a quantity that remains unchanged after a gauge transformation. One can also say that it transforms as a singlet under the gauge symmetry. Covariance (not to be confused with the use for convariant as opposed to contravariant indices) is usually used for an expression such as an equation, where it states that the expression retained its form after the gauge transformation. This is the same as form invariance.

Gauge symmetries are local symmetries, which means that the transformation could vary from point to point (the scalar $f$ is a function of space and time). In is not a space-time symmetry such as translation or rotation invariance. As a result gauge transformations do not transform position and momentum vectors.

All physically measurable quantities are gauge invariant. That is why the electric field $\mathbf{E}$ and the magnetic field $\mathbf{H}$ are gauge invariant and directly measurable, while the magnetic vector potential (gauge field) $\mathbf{A}$, which is not gauge invariant, cannot be measured directly.


Firstly, let me just point out that if $\psi$ interacts with the gauge field then $\psi$ would also have to transform under the gauge transformation. Otherwise the interaction term would not be invariant.

The form invariance has a slightly different meaning compared to the invariance of observable quantities, but there is some relationship. Form invariance is saying that the formal expression of a theory does not depend on the transformation. This comes from the reason why these transformations represent a symmetry of the theory in the first place.

I am more familiar with the expression of a theory in terms of the Lagrangian. When the Lagrangian has a particular symmetry, which means that it is form invariant with respect to a particular set of transformations for all the fields that it contains, then it really tells me that nature behaves in a way that is not affected by these transformations.

One can now relate this with the invariance of observable quantities. If nature is not affected by these transformations, then the observable quantities in this theory should also not be affect by these transformations. Hence, the form invariance of the theory leads to (or implies) the gauge invariance of the observable quantities.

It may seem that if nature is not affected by these transformation, then these symmetry transformations are just an artifact of our poor ability to come up with a better way to formulate the theory. However, symmetries have significant implications for the properties of a theory. To give one example, it is because of the gauge symmetry that the mass of the photon is zero. Without this symmetry, there would have been no reason for the mass to be zero and in general it would not have been zero.

Hope this answers your question.

  • 1
    $\begingroup$ There is a lot of truth here, but it doesn't really answer my question I'm afraid; I will revise it as I think its unclear as stands $\endgroup$
    – anon01
    Aug 17, 2016 at 13:51
  • $\begingroup$ I tried to capture the essence of what you seemed to ask, but after writing the answer, I felt that what I wrote is probably the same as many other answers that one can find here. So, indeed, I'm not sure I know exactly what you are asking. $\endgroup$ Aug 18, 2016 at 4:21
  • $\begingroup$ Too busy to make the revision? $\endgroup$ Aug 24, 2016 at 4:30
  • $\begingroup$ question rewritten/revised. $\endgroup$
    – anon01
    Aug 24, 2016 at 5:59
  • $\begingroup$ That's a good neat answer though :). $\endgroup$
    – gented
    Aug 24, 2016 at 7:54

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