# The meaning of gauge-fixing in covariant quantization of the electromagnetic field

I am having trouble wrapping my head around the idea behind the covariant quantization for the electromagnetic field that is usually done in textbooks (I'm currently following Mandl & Shaw and David Tong's notes in my QFT course). The starting point for quantizing the electromagnetic field is to talk about gauge freedom and how it can be useful for treating electromagnetism, and then you conveniently choose the Lorentz condition as an auxiliary constraint that you want the 4-potential $$A_\mu$$ to satisfy; after quantization, however, this condition becomes a constraint on the space of acceptable physical states, and not on the field operators themselves: "physical states" $$\left|\Psi\right\rangle$$ are those such that $$\partial_{\mu}A^{\mu+}\left|\Psi\right\rangle = 0$$ (Gupta-Bleuler condition). Now, my question is twofold:

1. Is there any quantization procedure that does not rely on gauge fixing? I am still not very comfortable with the idea of you having to fix a gauge in order to consistently quantize a theory that you formulated explicitly as gauge invariant from the get-go, and it seems to me that this is done at a high cost -- gauge fixing manifests itself as a constraint on physical states!
2. (Somewhat related to the 1st question) How does the Gupta-Bleuler condition exactly fixes the gauge, since it is not a condition on the operators, but on the physical states? Am I supposed to think of these two things as being totally equivalent?

Firstly, you have to fix the gauge since the 4-potential $$A_\mu$$ would have 4 degrees of freedom instead of the 2 polarizaton of the photon.

In general there are 3 different ways to quantize a theory which presents a first class constraints (which is the case for the Electromagnetic field). To whom is not used to first class constraints, they are those constraints $$C_\alpha$$ which generates gauge transformations as $$\delta z^A=\{z^A,\epsilon^\alpha(z)C_\alpha(z)\}$$ where $$z^A$$ are the coordinates in the phase space and $$\epsilon^\alpha$$ are the gauge parameters.

Now the methods are:

1. Reduced Phase-Space method;
2. Dirac-Gupta-Bleuer method;
3. BRST method.

## Reduced Phase Space:

The first method fix a single representative in the gauge orbit through a gauge fixing condition $$F^\alpha(z)=0$$ with $$det\{C^\alpha,F^\beta\}|_{C=F=0}\neq0$$ (this leads to have a new set of constraints which are now of the II class, and there can be defined Dirac parenthesis as new poisson parenthesis in order to quantize in the canonical way).

## Dirac-Gupta-Bleuer

The second method consist in quantize all the phase-space, leaving in all the non-physical variables. Since we have quantize everything, now we must require that our constraints, when acting on physical states, give zero: $$\hat{C}_\alpha(\hat{z})|\psi_{ph}\rangle=0$$ This the prefered method to quantize the electromagnetic field, in this way there can be removed all the non-physical degrees of freedom after the quantization.

## BRST Lagrangian

This method is the more complicated one, but in my opinion is the most fascinating one. The idea, here, is to add new degrees of freedom, called ghosts (with grassman parity opposite to the constraints' one) and their momenta (with same parity as the ghosts) (In order to understand this procedure, you will have to be familiar with some basic notions of supersymmetry).

Now, it can be proved that there exists a BRST charge $$Q$$ with some properties, in particular it is nilpotent ($$Q^2=0$$), which generates the BRST transformation.

In this contex, a physical state is defined if it is BRST-invariant, i.e. $$\hat{Q}|\psi_{ph}\rangle=0$$ But thanks to the nilpotency of $$Q$$, one can consider the Cohomology class of the physical state, which means that two sates are equivalent if they differ by an exact term, i.e. $$(|\phi_{ph}\rangle\sim|\psi_{ph}\rangle)\iff(|\phi_{ph}\rangle=|\psi_{ph}\rangle+\hat{Q}|\chi\rangle)$$ for an arbitrary $$|\chi\rangle$$.

This seem just an overcomplicated procedure, but when one consider the functional integral with all these new terms, one has the freedom to add a new term to the action which is an exact term in order to make a lot of simplifications and to arrive at the right result in a more sofisticated way.

I leave here a reference (textbook) which cover the quantization of gauge theories: Marc Henneaux and Claudio Teitelboim "Quantization of gauge systems".

1. Why do we gauge-fix the path integral in the first place? If we were doing lattice gauge theory, we didn't need to gauge-fix. But in the continuum case, (the Hessian of) the action for a gauge theory has zero-directions that lead to infinite factors when performing the path integral over gauge orbits. To avoid this, we gauge-fix.

2. The Gupta–Bleuler condition is not a gauge-fixing condition per se, but rather a condition to determine physical states. In general there are also conditions on an observable $$\hat{\cal O}$$. This is perhaps easiest to see in the BRST formulation. A physical state $$|\Psi \rangle$$ and an observable $$\hat{\cal O}$$ should both be BRST-invariant $$\hat{Q}|\Psi \rangle~=~0 \qquad\text{and}\qquad [\hat{Q}, \hat{\cal O}]~=~\hat{0}.$$

• I'll try to check BRST formulation, I had never heard of it before. So regarding your first point, are you saying one motivation to gauge-fix is to not "overcount" physical states more than once when doing the path-integral? And so there is no formal reason for you to do that in canonical quantization except if you reach out to the path integral formulation? – Bruno De Souza Leão Oct 9 '18 at 12:36