# QED lagrangian: gauge fixing term

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.

• I think that my question is slightly different: I understand the benefits of using $\mathcal{L}_{feyn}$ instead of $\mathcal{L}_{free}$. What I don't understand is: why do we adopt $\mathcal{L}_{free}$ as the free photon lagrangian in qed? – E. Marc. Jan 2 at 21:30
• @InitialObserver I don't understand your statement. Theoretical physics should not be a hack. The question is, what is the point of choosing a gauge invariant lagrangian if you are not going to use it anyway when push comes to shove? – my2cts Jan 2 at 22:15
• @Qmechanic The question that you refer to as a duplicate is different. Unless this question has been asked and answered adequately it should be reopened. – my2cts Jan 2 at 23:18
• What do you mean "we adapt $\mathcal{L}_{free}$"? The Lagrangian $\mathcal{L}_{free}$ is just an incomplete formulation before gauge-fixing. The Lagrangian should ultimately be gauge-fixed. Who are "we"? "Adapt" in which context? – Qmechanic Jan 2 at 23:59

## 1 Answer

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.

• What exactly do you mean by true feee lagrangian? Isn’t the gauge fixed lagrangian the one that gives the physical equations of motion? I guess I’m just slightly confused on point 3 – InertialObserver Jan 5 at 11:44
• @InertialObserver The physical equations of motion are Maxwell's equations. Half of them is an off-shell inherent property of the gauge field ($\mathrm{d}F = 0$), the other half ($\mathrm{d}{\star}F = 0$) are the equations of motion for both the free and the gauge-fixed Lagrangian. (Why do you think we chose the free Lagrangian to begin with if it didn't reproduce the equations of electromagnetism?!) – ACuriousMind Jan 5 at 11:47
• But don’t we need to fix a gauge to get rid of the last unphysical degree of freedom on the polarization ? – InertialObserver Jan 5 at 11:53
• It may be because I don't know more rigorous formalisms such as BRST, but I don't see how what you wrote answers the question. When trying to quantize the EM field in the obvious way using $\mathcal{L}_\mathrm{free}$ we quickly run into a problem and lose covariance, so how come we don't lose covariance when quantizing the QED lagrangian? I think this question is independent of the method we use in the end to make sense of the quantization of the EM fields, the important point is it doesn't work in the obvious way. Although I may have misinterpreted the question. – user2723984 Jan 5 at 11:54