Which gauge fixing conditions are allowed to choose for this approach?

For example (following the steps of Peskin in 9.4) I can choose a "modified" lorenz gauge ( for a abelian theory):

$$ (\partial_\mu \hat{A}^\mu)^² = 0 $$

With $ \hat{A}^\mu = A^\mu + \partial_\mu \alpha $ is the gauge transformed of $ A $

For the ghosts I end up with: $$ L= c(\partial_\mu \hat{A}^\mu) \partial^2 \overline{c}$$ or $$ L= c(\partial_\mu A^\mu + \partial^2 \alpha) \partial^2 \overline{c}$$

I assumed to find an equal result as for the very similar Lorenz condition. Or if not I thought the formalism would introduces ghost's to keep the theory stable. But even then I can not interpret this Lagrangian. The ghosts have no kinetic term and the interaction vertex with the gauge field depends on moment of the gauge field and the $ p^2 $ of the ghosts. But the ghost are massless here so $ p^2 = m^2 =0$. What is wrong with this gauge fixing? What says this Lagrangian about the interactions and "propagator" of the ghost fields?

  • $\begingroup$ I don't really see the issue there. If you choose a weird gauge-fixing condition, you end up with weird-looking gauge-fixed actions. Additionally, why do you claim there's no kinetic term? The kinetic term arises generically from rewriting the Faddeev-Popov determinant as an integral over the ghost fields. Also, $\bar c (\partial^2\alpha)\partial^2 c$ becomes $(\partial^2\alpha)\partial c \partial c$ upon partial integration, which is a kinetic term. $\endgroup$
    – ACuriousMind
    Oct 28, 2015 at 14:45
  • $\begingroup$ It is about how this approach reacts to unconventional choices. I am not sure since $\alpha (x) $ is a scalar function. So shouldn't it be seen as a scalar field interacting with the ghosts rather than consider it to be a prefatory ? $\endgroup$
    – LOQ
    Oct 28, 2015 at 15:22
  • $\begingroup$ Oh, yes. Wait, I only now noticed the square on your condition. That square is not an allowed gauge condition, since it does not uniquely pick a configuration (for any configuration $\hat{A}^\mu$ that fulfills it, $-\hat{A}^\mu$ does, too). $\endgroup$
    – ACuriousMind
    Oct 28, 2015 at 15:34
  • $\begingroup$ But the Lorenz gauge doesn't choose a unique configuration neither.The minus sign is a different solution like there are a lot of them for the Lorenz gauge, why is this a problem? Nevertheless I did the calculations with $(\partial_\mu A^\mu )^3$ and got nearly the same "problematic" result $$ c (\partial_\mu A^\mu )^2 \partial^2 \overline {c} $$ maybe the pure virtual ghost fields not necessarily need a propagator and the interaction just modifies the gauge propagator? $\endgroup$
    – LOQ
    Oct 29, 2015 at 15:28

1 Answer 1


There is nothing wrong here. Choosing non-linear gauge conditions leads to uncanonical forms of the action.

  1. Generically, nothing ensures that for arbitary gauge conditions $F(A) = 0$ we will obtain some sort special form of the gauge fixed action. If one picks "odd" forms for the gauge fixing condition, then the resulting ghost action will contain unusual terms, too. In particular, in electromagnetism, for gauges that cannot be written as $$ F(A) = k\dot{A}_0 + \chi(A_1,A_2,A_3),k\in\mathbb{R} \tag{1}$$ where $\chi$ is some function of the $A_i$, the Faddeev-Popov procedure does not lead to the usual kinetic term for the ghosts. This constraint on the gauge choice comes from the formal way the gauge is fixed in the Hamiltonian formalism by ways of a gauge fixing fermion, and from the fact that $\dot{A}_0$ is a Lagrange multiplier rather than a proper dynamical variable. One can choose other forms for the gauge-fixing fermion to implement other conditions, but then there is no general argument that the resulting action will contain canonical terms for the ghosts.

  2. The term in the action is not a "kinetic term", anyway because $c$ and $\bar c$ are not conjugates of each other. $c,\bar c$ are independent Grassmann variables, and while one is a ghost and the other an antighost, there is no relation between ghosts and antighosts. They are independent variables. Treating this as a kinetic term works since ultimately the ghosts decouple in any case, but it is not a kinetic term as it would be if $c$ was a complex scalar field and $\bar c$ was its conjugate.

What must be stressed in particular is that, in full generality, just picking some gauge condition $F(A)$ and trying to implement it may fail for reasons that are not obvious from the usual treatment of the quantization of the electromagnetic field. This is because the proper quantization method is either by the BRST formalism in the Hamiltonian picture, or by the BV-antifield formalism for the Lagrangian formalism. In both approaches, the gauge is fixed by a gauge-fixing fermion, which does not straightforwardly related to a gauge-fixing condition $F(A) = 0$ in all cases, but only when it is assumed to take special forms, such as being linear in certain ghost variables, which then leads to constraints on the gauge-fixing condition such as $(1)$.

  • $\begingroup$ Minor comment to the answer (v1) about terminology: BV-antifield formalism is usually also considered a BRST formulation. $\endgroup$
    – Qmechanic
    Oct 29, 2015 at 16:51

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