Gauge transformation vs field excitation

Question

I think I'm fundamentally misunderstanding something.

Say I have a gauged Lagrangian for a complex scalar field $\phi$ with no SSB:

$$\begin{equation} \mathcal{L} = (D_{\mu}\phi)(D^{\mu}\phi)^{\dagger} - m^2 \phi \phi^{\dagger} - \frac{1}{4} F_{\mu \nu} F^{\mu \nu}\tag{1} \end{equation}$$

with $D_{\mu} = \partial_{\mu} + i A_{\mu}$.

And now suppose I parameterise my complex scalar field as $\phi(x) = r(x) e^{i \theta(x)}$ -- two real degrees of freedom excited around the vacuum at $\langle \phi\rangle = 0$. If I now plug this into the Lagrangian I get

$$\mathcal{L} = (\partial_{\mu} r + i \partial_{\mu} \theta r + i A_{\mu} r )(\partial_{\mu} r - i \partial_{\mu} \theta r - i A_{\mu} r ) - m^2 r^2 - \frac{1}{4} F_{\mu \nu} F^{\mu \nu}.$$

But by gauge invariance $A_{\mu}$ and $A_{\mu} + \partial_{\mu} \theta$ are exactly the same field (this may be the place where I'm doing something wrong), so

$$\mathcal{L} = (D_{\mu}r)(D^{\mu}r)^{\dagger} - m^2 r^2 - \frac{1}{4} F_{\mu \nu} F^{\mu \nu},\tag{2}$$

which only depends on the real excitation!

I'm very confused as to where the angular excitation has gone. Was it just never real in the first place?

If we had SSB, we'd happily eliminate the goldstones that corresponded to the gauge degrees of freedom by letting them get eaten by the gauge field in just such a way. The only difference here is there's no vev to give the $A$s a mass.

In this post, TwoBs' answer seems to do the same as I do, but it seems to me their argument rests on the fact that they have neglected to package up (where $h$ corresponds to my $r$) $\partial_{\mu} h$ and $A_{\mu} h$ into a covariant derivative again, and they claim this makes $\mathcal{L}$ non-gauge-invariant. I don't understand the argument. I don't feel like I have fixed a gauge anywhere, I've just expressed the $\phi$ field in a certain form, and since $A_{\mu}$ was a general field, $A_{\mu} + \partial_{\mu} \theta$ is surely also general.

Is there a difference between a gauge transformation and an angular excitation of the $\phi$ field? Do they only look the same at the level of the Lagrangian and in reality they're truly different?

Answer 1

Indeed you're doing something wrong, or at least confusing, when you say "...by gauge invariance $A_\mu$ and $A_\mu+\partial_\mu \theta$ are exactly the same field..." and then discard the $\partial_\mu \theta$: You're implictly imposing a gauge transformation with parameter $\theta$, which absorbs the phase into the gaueg field.

Also, your decomposition is only valid for nonzero fields, which includes extra confusion. For example, the modulus $r$ should be gauge invariant, you should have two massive scalars etc.

When there is a nonzero VEV, all you have writen goes through, and it's essentially the standard Higgs mechanism, where you can and routinely do shuffle "angular excitaion of $\phi$" and "part of gauge field" around by treating $\theta$ as parameter of a gauge transformation.

Answer 2

The issue is that the scalar has to be transformed as well when you perform a gauge transformation. Namely if you want to perform the change $$A_\mu + \partial_\mu \theta(x)$$ it has to come along with a gauge transformation of the scalar field by the same parameter, i.e. $$\phi \rightarrow e^{-i\theta(x)}\phi$$ I would recommend writing the complete Lagrangian first with whatever parameterization for $\phi$ you want, for example $\phi = \rho(x)e^{i\sigma(x)}$, (notice that so far the parameters have nothing to do with a gauge transformation). One does not need to speak about VEV even, to perform a gauge transformation. This way you will explicitly see if you can pick some $\theta(x)$ such that it eliminates the $\sigma(x)$ field.

If I am not mistaken what will happen is that $\sigma(x)$ will decouple for an appropriate $\theta(x)$ meaning it will not have interaction terms, but the $\sigma(x)$ kinetic term should remain and the total number of degrees of freedom conserved.