quantum field theory - Gauge fixing and degrees of freedom

Today, my friend (@Will) posed a very intriguing question -

Consider a complex scalar field theory with a $U(1)$ gauge field $(A_\mu, \phi, \phi^*)$. The idea of gauge freedom is that two solutions related by a gauge transformation are identified (unlike a global transformation where the solutions are different but give rise to the exact same observables), i.e. $$(A_\mu(x), \phi(x), \phi^*(x)) ~\sim~ (A_\mu(x) + \partial_\mu \alpha(x), e^{i \alpha(x)}\phi, e^{-i \alpha(x)}\phi^*(x)).$$ The process of "gauge fixing" is to pick one out of the many equivalent solutions related via gauge transformation. The usual procedure of gauge fixing is to impose a condition on $A_\mu$ so that one picks out one of the solutions. His question was the following:

Instead of imposing a gauge condition on $A_\mu$, why do we not impose a gauge condition on $\phi$? Wouldn't this also pick out one the many equivalent solutions? Shouldn't this also give us the same observables? If so, why do we not do this in practice?

After a bit of discussion, we came to the following conclusion:

The idea of gauge symmetry comes from the requirement that a quantum theory involving fields $(A_\mu, \phi, \phi^*)$ have a particle interpretation in terms of a massless spin-1 particles and 2 spin-0 particles. However, prior to gauge fixing, the on-shell degrees of freedom include those of a massless spin-1 particle and 3 spin-0 fields ($A_\mu \equiv 1 \otimes 0,~\phi,\phi^* \equiv 0$). We would now like to impose a gauge condition to get rid of one scalar degree of freedom. There are two ways to do this -

1. 
   

   Impose gauge condition on $A_\mu$ so that $A_\mu \equiv 1$. Now, $A_\mu$ corresponds to a massless spin-1 particle and the complex scalar corresponds to two spin-0 particles. This is what is usually done.

   
   


2. 
   

   Impose a gauge condition on $\phi$. For instance, one can require that $\phi = \phi^*$. We now have a real field corresponding to a spin-0 particle. However, $A_\mu$ still contains the degrees of freedom of both a massless spin-1 and a spin-0 particle.

   
   

I claimed that the second gauge fixing procedure is completely EQUIVALENT to the first one. However, the operator that now creates a massless spin-1 particle is some nasty, possibly non-Lorentz Invariant combination of $A_0, A_1, A_2$ and $A_3$. A similar statement holds for the spin-0 d.o.f. in $A_\mu$. Thus, the operators on the Hilbert space corresponding to the particles of interest are not nice. It is therefore, not pleasant to work with such a gauge fixing procedure.

In summary, both gauge fixing procedures work. The first one is "nice". The second is not.

Is this conclusion correct?

NOTE: By the statement $A_\mu \equiv 1$, I mean that $A_\mu$ contains only a massless spin-1 d.o.f.

Answer

If $\phi$ is non-zero, fixing the phase of $\phi$ is a perfectly valid gauge condition. It's used frequently in Standard Model calcuations involving the Higgs field, where it goes by the name unitarity gauge. This is a nice gauge in some ways, because it makes manifest the fact that there's a massive vector field in the system.

Edit: Some caution is required with unitary gauge. It's a complete gauge when you can reasonably treat $\phi$ as non zero, because it uses every degree of freedom in the gauge transformation. This means for example that it's ok to use in perturbative calculations around a Higgs condensate. But when $\phi$ can vanish, the phase function isn't uniquely defined, which means the gauge transformation is not invertible. This gauge isn't quite a gauge.