Let $Q$ be the BRST operator. Define physical state as those in $\mathrm{ker}\,Q$ (modulo its image): $$ Q|\psi_\mathrm{physical}\rangle\equiv 0\tag1 $$

It is often claimed1 that this condition becomes the Gupta-Bleuler condition if the gauge group is abelian (i.e., in QED): $$ (\partial\cdot A)^+|\psi_\mathrm{physical}\rangle\equiv 0\tag2 $$ but I've never seen an explicit proof of this claim. In fact, it appears to me that $(1)$ is local and non-linear in the fields, while $(2)$ is non-local and linear. Therefore, I don't really expect these conditions to be fully equivalent; or, at least, this equivalence is rather non-trivial. Am I being naïve? How can you prove that BRST theory is really equivalent to Gupta-Bleuler when the algebra is abelian?

1: See e.g. the last page on Timo Weigand's notes, or some remarks in the Scholarpedia article Becchi-Rouet-Stora-Tyutin symmetry (ctrl+f "gupta").


1 Answer 1


Condition (1) is not equivalent to condition (2) in the sense that the former would reduce to the latter by any algebraic manipulation. However, what is true is that on the ghost-free part of the intermediate space of states on which the BRST condition is imposed, it cuts out precisely the same subspace as the Gupta-Bleuler condition.

As so often, a good reference for this is "Quantization of Gauge Systems" (QoGS) by Hennaux and Teitelboim, in this case chapter 19, where the BRST quantization of a free Maxwell field is rather explicitly done.

First, a brief overview over the BRST part of the story, known in a more general form as the "quartet mechanism":

We quantize an extended system consisting of the dynamical fields $A^\mu$ and a ghost $C$ as well as an anti-ghost $\bar{C}$. After some conventional choices, the BRST charge in terms of Fourier modes is (eq. (19.36) in QoGS) $$ Q = \int \left( c^\ast (k) a(k) + c(k) a^\ast(k) \right)\mathrm{d}^3k, \tag{A}$$ where $a = a^3 + a^0$ (i.e. the creation/annihilation operator for a mixed longitudinal/timelike excitation) and $c$ is the Fourier mode of the ghost. With $b = a^3 - a^0$ and $\bar{c}$ as the Fourier mode of the anti-ghost, the number operator for the modes we want to turn out as unphysical is (eq. (14.62) in QoGS) $$ N = \int \left( a^\ast b + b^\ast a + \bar{c}^\ast c + c^\ast \bar{c} \right)\mathrm{d}^3 k, \tag{B}$$ and this operator commutes with the BRST charge and is BRST exact with $N=[Q,K]$ for the "fermion" $K = \int \left(b^\ast(k)\bar{c}(k) + \bar{c}^\ast(k) b(k) \right)\mathrm{d}^3 k$. This implies that BRST-closed eigenstates of $N$ with non-zero eigenvalue are BRST-exact, so the only contribution to non-zero BRST cohomology can come from zero eigenstates of $N$. This means the physical state space is the space generated by the longitudinal photon modes $a^1,a^2$, which agrees with the result of the Gupta-Bleuler formalism after quotienting out the spurious null states.

We now can also note that in the ghost-free part of the space, $Q\lvert \psi\rangle = 0$ holds for all states not involving $a$, i.e. the BRST-closed states include the unphysical states generated by the null mode $b$ as well as the physical states generated by $a^1,a^2$. This is precisely the same as the kernel of the Gupta-Bleuler condition $(\partial\cdot A)^+$, which is the space with $p^\mu \zeta_\mu = 0$ where $p$ is the momentum normalized to $(1,0,0,1)^T$ and $\zeta$ the polarization. Note that the spurious null modes correspond to the polarization $\zeta \propto p$ and that they are generated by $$ \sum_\lambda \alpha_\lambda a^{\ast\lambda},$$ where $\alpha$ are numbers determined by $$ \zeta^\mu = \sum_{\lambda,\lambda'} \alpha_\lambda \eta_{\lambda\lambda'} \epsilon^\mu(\lambda')$$ for standard basis 4-vectors $\epsilon(\lambda)$, so we have \begin{align} \zeta^\mu & = \sum_{\lambda,\lambda'} \alpha_\lambda \eta_{\lambda\lambda'} \epsilon^\mu(\lambda') = -\alpha_0 \epsilon^\mu(0) + \sum_i \alpha_i \epsilon^\mu(i) \implies \alpha_0 = -\alpha_3, \end{align} meaning the spurious states inlcuded in the Gupta-Bleuer condition in addition to the transverse state from $a^1,a^2$ are precisely the ones generated by $\alpha_3 (a^0 - a_3) \propto b$, in full agreement with the BRST-closed ghost-free space.


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