This question is probably answered somewhere in textbooks, but I haven't got there yet, sorry for my ignorance in advance.

There is a famous transversality condition in E&M and QED

$$\vec{k}\cdot \vec{A}=0$$

After reading a bit of Cohen-Tannoudji I got an impression that this condition is a consequence of us choosing a Coulomb gauge. However, if only $A_{\perp}$ is gauge invariant (btw, how can I see it?), doesn't it impose this choice $$\vec{\nabla} \cdot \vec{A}=0~?$$ Shouldn't the vector field be gauge invariant for a theory to even make sense? And if this is the case, can't we say that transversality condition is imposed by gauge invariance?


The vector potential is not gauge invariant, and there are many gauge potentials that correspond to the same physical electromagnetic field. Thus when we take the path integral $$\mathcal{Z} =\int D[A^\mu] \exp(-i S[A^\mu])$$ we are in effect over-counting field configurations.

The way to solve this is to add a gauge fixing condition so that each physical electromagnetic field configuration is counted only once. That is done by adding a Lagrange multiplier, say $\lambda \partial_\mu A^\mu$ in covariant form for the Lorenz gauge, or $\lambda \nabla \cdot\mathbf A$ for Coulomb gauge. There are of course other choices for other gauges.

The above generalizes to all gauge theories. The keywords to look for here is constrained Hamiltonian systems and BRST formalism. Here are some lecture notes using electrodynamics as an example, showing how to deal with gauge symmetries.

  • $\begingroup$ Strictly speaking, in quantum field theory you're not adding a Lagrange multiplier to the action. You're doing the Fadeev-Popov procedure, or adding a BRST exact term to the action to make your kinetic term invertible. Simply adding a Lagrange multiplier is what you do to gauge fix in classical field theory. Seeing your profile, I think you probably know all this already. :P $\endgroup$ – Arturo don Juan Apr 2 '19 at 16:40

First, $\mathbf{k}\cdot\mathbf{A}=0$ and $\nabla\cdot\mathbf{A}=0$ are the same condition, one in Fourier space and the other in regular space.

The theory by itself is gauge invariant. In other words, the equations for the potentials as they follow from Maxwell's equations are gauge invariant:

$$\partial^\mu \partial_\nu A^\nu - \partial^2 A^\mu = 0$$

You can check that this is invariant under $A^\mu \to A^\mu + \partial^\mu \chi$.

Now, what you can do is choose one particular gauge. That is, you impose some condition on the potential. When you do this you lose gauge invariance, for the simple reason that now you're working in a particular gauge. Two common examples of gauge conditions are Lorenz gauge ($\partial_\mu A^\mu=0$; this one is nice because it's Lorentz invariant) and Coulomb or transverse gauge: $\nabla \cdot \mathbf{A}=0$.

You seem to have the words a bit mixed up, too. The vector field (assuming you mean $\mathbf{A}$ or $A^\mu$, that is, the potentials) is not gauge invariant, because the vector field is precisely the thing that changes when you do a gauge transformation. In particular, $A_\perp$ is not gauge invariant: the reason that it's important is that in transverse gauge $\mathbf{A}(\mathbf{k})$ doesn't have a component parallel to $\mathbf{k}$ (hence the name). What is gauge invariant is the electromagnetic field itself, that is, $F_{\mu\nu}$ or the electric and magnetic fields if you prefer. They are constructed from the potentials in such a way that they remain unaffected if you do a gauge transformation.

And no, there's no need for anything to be gauge invariant for the theory to make sense. You postulate your equations and either they're gauge invariant or they're not. In QFT it's usually highly desirable for a theory to be based on a gauge symmetry (as the Standard Model is), but it's not necessary. The weak interaction as we see it at low energies is not gauge invariant, for example: the symmetry is broken.


A simple quick reaction that does not require long paragraphs is that the first equation is in momentum space and the other equation is the same equation but in position space thought I think strictly speaking in momentum space the condtion is $ k \cdot\epsilon =0 $ where $ \epsilon $ is the polarization vector.

To answer your question directly, how fundamental is the transversality condition? First note that $ A _{\mu}$ has 4 degress of freedom while we actually need 2. So before we move on we have to reduce the number of degrees of freedom. So as long as you are using a gauge theory (in this case an abelian gauge theory) you will need to impose 2 conditions to reduce the number of degrees of freedom. But if one finds a way of doing QED without using the four vector potential then presumably one would already have the right number of degrees of freedom making the imposition of transversality needless.


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