# BRST cohomology and Gupta-Bleuler$.$

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").

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.