QED lagrangian: gauge fixing term

Question

I have a question about the structure of the QED lagrangian, in particular the free photon lagrangian which is contained in it. My premise is: I only know how to exploit canonical quantization in order to quantize a theory; I don't know how to use the path integral formulation.

The QED lagrangian is: $$ \mathcal{L}=-\frac{1}{4}F^{\mu \nu}F_{\mu \nu}+\bar{\psi}(i\gamma^{\mu}D_{\mu}-m)\psi, $$ so I assume that the free photon theory exploited here is $$ \mathcal{L_{free}}=-\frac{1}{4}F^{\mu \nu}F_{\mu \nu}. $$ However, I also learnt that $\mathcal{L_{free}}$ jointed with Lorenz's gauge cannot give us a covariant quantization for the electromagnetic field (by means of the canonical quantization, at least). In fact, we introduce the following lagrangian: $$ \mathcal{L_{feyn}}=-\frac{1}{4}F^{\mu \nu}F_{\mu \nu}-\frac{1}{2 \xi}(\partial_{\mu}A^{\mu})^2 $$ with Feynman gauge choice $\xi=1$. This, jointed with Gupta-Bleuer constraint, gives us the physical states of the electromagnetism.

So: why do we adopt $\mathcal{L_{free}}$ instead of $\mathcal{L_{feyn}}$? I know that the latter is not gauge-invariant, but the covariant quantization of the theory is achieved through that, so this point is not clear to me.

Answer 1

1. Your gauge-fixing Lagrangian $\mathcal{L}_\text{feyn}$ only fixes the gauge if the Lagrange multiplier $1/\xi$ is dynamical, i.e. the Lagrangian is thought of as a functional of both $A$ and $1/\xi$. Then the equations of motion for $1/\xi$ fix the gauge. When you say "with $\xi = 1$", then you're effectively integrating out the Lagrange multiplier and back to using the original Lagrangian $\mathcal{L}_\text{free}$ and imposing its equations of motion - the gauge - as a constraint by hand. That is, you're not actually using $\mathcal{L}_\text{feyn}$ as anything other than a handwavy excuse to make your gauge choice not seem quite so arbitrary. If you were really quantizing $\mathcal{L}_\text{feyn}$, then you should also treat the Lagrange multiplier as a new quantum field - do you?

2. This isn't your fault, this is how Gupta-Bleuler quantization works. It's a very well-working hack, but it's a hack. However, it's an important hack in that it is the precursor of BRST quantization of general gauge theories, which stll retain the intermediate step of constructing such a gauge-fixed Lagrangian, quantizing it (with all the auxiliary fields!) and then throwing all the states they didn't actually want to get away (according to a well-defined criterion).

3. The gauge-fixed Lagrangian makes a bad starting point for the theory because it does not lend itself well to coupling to other fields charged under the symmetry - because you'd need to add new gauge fixing terms for every charged field you add. $\mathcal{L}_\text{free}$ is the true free Lagrangian because you can easily couple it to currents made from different fields.