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.

  • $\begingroup$ It doesn't make any sense to speak of massless or massive particles created by $A$. You have to specify an action to know which Lorentz irreps occur. Without an action, you can only count the number of d.o.f.'s. $\endgroup$
    – user1504
    Jul 24, 2013 at 23:08
  • 1
    $\begingroup$ Moreover, since $A$ and $\phi$ are mixed by the gauge symmetry, it doesn't really make sense to think about their particles separately. You want to consider gauge invariant observables like Wilson lines connecting $\phi$ and $\bar{\phi}$. $\endgroup$
    – user1504
    Jul 24, 2013 at 23:09
  • $\begingroup$ I definitely see that just counting degrees of freedom was too naive. I should think about how they transform under the Lorentz transformation as well. $\endgroup$
    – Prahar
    Jul 26, 2013 at 21:26

1 Answer 1


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.

  • $\begingroup$ I don't think that's what I'm talking about though. In that case, we are not fixing any gauge. We are explicitly breaking a global internal symmetry. Also, in that case, a d.o.f. from the Higgs field now becomes the longitudinal mode of a photon to transform it into a massive vector boson, thus changing the particle content of the theory. However, the "gauge fixing" I am talking about does not change the particle content of the field theory. It simply embeds a massless spin-1 and a spin-0 d.o.f. into one $A_\mu$. $\endgroup$
    – Prahar
    Jul 24, 2013 at 21:38
  • 3
    $\begingroup$ Eh? No. The unitary gauge is actually a gauge. It's the one you described in 2, fixing the phase of the scalar field to 0. (This uses a function's worth of d.o.f., much more than nailing down a global symmetry.) $\endgroup$
    – user1504
    Jul 24, 2013 at 22:54
  • $\begingroup$ OK. Sorry. I think I agree with you now. Here is my current understanding - Gauge fixing the phase of $\phi$, naturally gives you the d.o.f. of a massive spin-1 particle and a scalar field. My question to you is then - Can we decompose a massive spin-1 representation of the Poincare group into a massless spin-1 representation and a spin-0 representation? $\endgroup$
    – Prahar
    Jul 26, 2013 at 21:22
  • $\begingroup$ If so, then in principle, we should be able to find some combination of the $A_\mu$'s that transforms like a massless spin-1 field. We then have an operator (call it $O$) that creates a gauge boson. Then the theory where $A_\mu$ is gauge fixed with $A_\mu$ representing the massless spin-1 particle is EQUIVALENT to the theory where $\phi$ is gauge fixed and $O$ represents the massless spin-1 particle (By equivalent I mean, that they will give the same observables under the operator identification mentioned above). Is that right? $\endgroup$
    – Prahar
    Jul 26, 2013 at 21:25
  • $\begingroup$ You can decompose the underlying vector space, but the representations won't decompose. The mismatch will vanish with the mass, though. $\endgroup$
    – user1504
    Jul 26, 2013 at 22:35

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