The first time some kind of gauge fixing appears is during the Gupta-Bleuler procedure, which is used to be able to quantize the photon field:

The basic gauge invariant Lagrangian leads to $\Pi_0=0$ which is incompatible with the canonical commutator relations. Furthermore the Green Function, i.e. the propagator, for the corresponding equation of motion does not exists. Therefore one adds a term to the Lagrangian $\frac{1}{2} (\partial_\mu A^\mu)^2$, which isn't gauge invariant. Nevertheless, now $\Pi_0 \neq 0$ and the propagator can be derived. But in return unphysical degrees of freedom appear (longitudinal/timelike) photons appear, which are eliminated by the weak Lorenz condition, that guarantees we only pick physical states.

Instead of $\frac{1}{2} (\partial_\mu A^\mu)^2$ one can add $\frac{1}{2\zeta } (\partial_\mu A^\mu)^2$, which is called a gauge fixing term to the Lagrangian. The parameter $\zeta$ is the gauge parameter which determines which gauge we are working in. $\zeta =1 $ for the Feynman a.k.a. Lorenz gauge, $\zeta = \infty$ for the unitary gauge etc. The propagator is then $\zeta$ dependent, but all physical observables are of course gauge independent.

A similar problem appears for the gluon fields. Again, a gauge fixing term is introduced, but this time in order to secure unitarity of the S-Matrix ghosts fields are needed.

These problems seem to appear because we try to describe a massless spin-1 field, which has two physical degrees of freedom, in a covariant way, which means a four vector. In the unitary gauge, i.e. without a gauge fixing term, and imposing a gauge condition, for example the Coulomb gauge from the beginning, no timelike/longitudinal photons appear. But the Coulomb gauge isn't Lorentz invariant ($A_0=0$). For a covariant description we need a gauge fixing term.

I'm a little confused about these concepts and their connection:

  • How exactly does the gauge fixing term work? I understand that it is a term that destroys gauge invariance, but I do not understand how it fixes a gauge. (In this context often the term Lagrange multiplier is used, but can't make the connection. If someone could explain how this concept works in this context, it would help me a lot.)

  • Are the longitudinal/timelike photons, in some sense, ghosts, too? Nevertheless, for the photon case those unphysical degrees of freedom are eliminated by an extra condition, for the gluon field, unphysical degrees of freedom are introduced additionally (Ghost term in the Lagrangian), in order to secure unitarity. Is there some connection between these concepts? What happens to the longitudinal/timelike gluons? Are ghost fields only needed if we want to work in an arbitrary gauge

  • Whats the reason ghosts are needed? (Mathematically to make sense of the theory, i.e. make the S matrix unitary again, but) Is it because we want to work in an arbitrary gauge and with a non-covariant description with a fixed gauge from the beginning this problem wouldn't appear? Gluons carry charge themselves and therefore they can form loops. In those Gluon loops we have to add all contributions, including the unphysical ones (longitudinal/timelike) which makes the S-matrix non-unitary?! In contrast to the photon case, the contribution of these loops can't be cancelled by a weak-Lorentz condition (which defined what we understand as physical states), and therefore the ghosts are in some sense the equivalent to the weak-Lorentz condition?!

I'm trying to understand this using the canonical formulation of QFT, but unfortunately most books explain this using the path integral approach. Any idea or reading tip would be much appreciated!

  • 2
    $\begingroup$ We usually prefer that one post contains only one question, so it would be better if could create as many question as you have bullet points. And have a look at these lectures note, they might be what you're looking for : eduardo.physics.illinois.edu/phys582/582-chapter9.pdf $\endgroup$
    – Adam
    Commented Oct 9, 2014 at 7:26
  • 3
    $\begingroup$ I don't want to unilaterally reopen this thread since I answered it, but does it really deserve to be closed? Currently 7 users have bookmarked it as a favorite. $\endgroup$
    – Qmechanic
    Commented Feb 12, 2018 at 12:07

1 Answer 1


It seems OP's main question concerns the systematics of gauge-fixing. We interpret/reformulate OP's questions as essentially the following.

The original gauge-invariant action $S_0$ is unsuitable for quantization, so we add a non-gauge invariant gauge-fixing term to the action. Obviously we cannot add any non-gauge invariant term to the action.

  1. What is the principle that dictates which gauge-fixing terms are allowed and which are not?

  2. And how can the theory be independent of gauge-fixing?

Why we need gauge-fixing and Faddeev-Popov ghosts in the first place were explained in my Phys.SE answer here.

Now, to not get bogged down by technical details, it is actually more convenient to use the BRST formalism. Recall that the BRST formalism is a modern generalization of the Gupta-Bleuler formalism. The BRST transformation $\delta$ basically encodes the gauge transformations.

Moreover, recall that the BRST transformation $\delta$ is Grassmann-odd and nilpotent $\delta^2=0$, and that the original action $S_0$ is BRST-invariant $\delta S_0=0$. The total/gauge-fixed action

$$\tag{1} S_{\rm gf}~=~S_0+\delta\psi$$

is the original action $S_0$ plus a BRST-exact term $\delta\psi$ that depends on the so-called gauge-fixing fermion $\psi$, which encodes the gauge-fixing condition. In other words: different gauge-fixing means different $\psi$.

Note that while the gauge-fixed action $S_{\rm gf}$ is no longer gauge invariant, it is still BRST-invariant.

The BRST-exact term $\delta\psi$ in the action (1) contains both the Faddeev-Popov ghost term and the gauge-fixing terms. Not everything goes: There is an intricate balance between the various terms to ensure that we have only modified the original action $S_0$ with a BRST-exact amount $\delta\psi$, which cannot alter the BRST cohomology, and thereby, in turn, cannot alter the notion of physical states.

This basically answers question 1 and 2 at the conceptional level. For more details, see also e.g. my Phys.SE answer here.

Phrased differently, without the use of the BRST formalism: The usual Faddeev Popov trick (cf. e.g. Ref. 1) precisely codifies the balance between the Faddeev-Popov ghost term and the gauge-fixing terms. They go hand in hand. In the simplest situations, the Faddeev-Popov ghosts decouples, and can be integrated out.


  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, Section 9.4.

Not the answer you're looking for? Browse other questions tagged or ask your own question.