Consider the Gupta-Bleuler quantization of a vector field. In this theory, the Fock space of states is restricted to those satsifying the condition$L(k)|\psi\rangle\equiv k^\rho a_\rho(\vec{k})|\psi\rangle=0$ (this condition comes from$\langle\psi|\partial_\mu A^\mu|\psi\rangle=0$ after some manipulations); such states satisfiying this condition are denoted physical states. An operator$O$ is associated to a physical observable if its action on a physical state gives back a physical states. Therefore, one gets$L(k)O|\psi\rangle=0$. I am trying to show that the helicity operator defined by
$$h = i \, \epsilon_{ijk} \, \int d\Omega_{\vec{p}} \, a_i^\dagger(\vec{p}) \, a_j(\vec{p}) \, p_k,$$
where a summation over$i,j,k$ is intended,$\epsilon_{ijk}$ the Levi-Civita symbol, is a physical operator. As such, I computed the following. I dropped the vector notation for the ladder operators for ease of notation.
$$L(k)h|\psi\rangle=i\epsilon_{ijk}\int d\Omega_{\vec{p}}d\Omega_{\vec{q}}k^\rho\varepsilon^\mu(q) a_\rho(k)a_i^\dagger(p)a_j(p)a_\mu^\dagger(q)|0\rangle,$$
where I used that$|\psi\rangle=\int d\Omega_{\vec{q}}\varepsilon^\rho(q)a_\rho^\dagger(q)|0\rangle$,$\varepsilon^\mu(q)$ a polarization vector. Now using the equality
$$[ a_\mu(k), \, a_\nu^{\dagger}(p)] = -\eta_{\mu\nu} \, \delta^{3}(\vec{k} - \vec{p}) \, (2\pi)^3 \, 2\omega_{k},$$
I could obtain the following
\begin{equation}L(k)h|\psi\rangle=-i\epsilon_{ijk}\eta_{j\mu} \, \int d\Omega_{p}k^\rho \varepsilon^\mu(p)a_\rho(k)a_i^\dagger(p)|0\rangle = i\epsilon_{ijk}\eta_{j\mu}\eta_{\rho i} k^\rho \varepsilon^\mu(k)|0\rangle = i\epsilon_{ijk}k_i\varepsilon_j(k)|0\rangle.\end{equation}
This last expression doesn't seem to vanish as we would want for the helicity operator to be a physical observable.My question: is there an error in this derivation? My guess is that yes, because the helicity operator being unphysical seems strange to me. However, I do not manage to spot such a mistake.
