Tuesday, March 27, 2018

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 (Aμ,ϕ,ϕ). 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. (Aμ(x),ϕ(x),ϕ(x))  (Aμ(x)+μα(x),eiα(x)ϕ,eiα(x)ϕ(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 Aμ so that one picks out one of the solutions. His question was the following:



Instead of imposing a gauge condition on Aμ, why do we not impose a gauge condition on ϕ? 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 (Aμ,ϕ,ϕ) 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 (Aμ10, ϕ,ϕ0). 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 Aμ so that Aμ1. Now, Aμ 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 ϕ. For instance, one can require that ϕ=ϕ. We now have a real field corresponding to a spin-0 particle. However, Aμ 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 A0,A1,A2 and A3. A similar statement holds for the spin-0 d.o.f. in Aμ. 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 Aμ1, I mean that Aμ contains only a massless spin-1 d.o.f.



Answer



If ϕ is non-zero, fixing the phase of ϕ 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 ϕ 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 ϕ can vanish, the phase function isn't uniquely defined, which means the gauge transformation is not invertible. This gauge isn't quite a gauge.


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