This question is related to the quantization of a massless vector field using the Gupta-Bleuler quantization.
Consider a vector field$A^\mu(x)$,$x\equiv x^\mu$. The field$A^\mu$ transforms according to a gauge transformation:$A_\mu\rightarrow A_\mu^{'}\equiv A_\mu-\partial_\mu\alpha$, where$\alpha$ is some function. Consider the following Lagrangian
$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu},\quad F_{\mu\nu}\equiv \partial_\mu A_\nu-\partial_\nu A_\mu.$$
This Lagrangian is in particular invariant under the gauge transformation of the field$A^\mu$. The quantization à la Gupta-Bleuler of the field$A^\mu$ consists in adding a term in the Lagrangian that breaks the gauge invariance; for a free parameter$\xi$ one has
$$\mathcal{L}_{GB}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{\xi}{2}(\partial_\mu A^\mu)^2.$$
The equations of motion of this second Lagrangian do not correspond to the ones coming from the first Lagrangian, which are the Maxwell equations. Given the Fock space$\mathcal{F}$ arrising from the Gupta-Bleuler quantization, one defines the physical states$|\psi\rangle$ as those satisfing$\langle\psi|\partial_\mu A^\mu|\psi\rangle=0$. Equivalently, this condition can be rewritten as
$$\partial_\mu A^\mu=\partial_\mu A_+^\mu+\partial_\mu A_-^\mu,\quad \partial_\mu A_-^\mu=\int \mathrm{d}\Omega_{\vec{k}}(-ik_\mu)a^\mu(\vec{k})e^{-ikx}\Rightarrow \partial_\mu A^\mu_-|\psi\rangle=0.$$
The original Lagrangian is such that the vector field$A^\mu$ has only two independent polarizations;$A^\mu$ has only two degrees of freedom. For the Gupta-Bleuler quantization, I would expect this to hold as well, otherwise we would be describing a different theory.
My question: is it correct that the Gupta-Bleuler quantization also leads to only two independent polarizations of the field$A^\mu$, and how can one show it?
1 Answer1
Consider the vector field$A^\mu(\vec{x},t)$, which is quantized à la Gupta-Bleuler. It can be expanded in plane waves as
$$A^\mu(\vec{x},t)=\int \mathrm{d}\Omega_\vec{k}\left(a^\mu(\vec{k})e^{-ikx}+a^{\mu\dagger}(\vec{k})e^{ikx}\right),\quad \mathrm{d}\Omega_\vec{k}=\frac{\mathrm{d}^3 k}{(2\pi)^3 2k_0}.$$
The condition$\partial_\mu A^\mu_-|\Psi\rangle=0$ for a physical state can be rewritten as follows; define first$L(\vec{k})=k_\mu a^\mu(\vec{k})$. A physical state of the Fock space can be defined as
$$\partial_\mu A^\mu_-|\Psi\rangle=0\quad\Leftrightarrow\quad L(\vec{k})|\Psi\rangle=0.$$
Consider now a single particle state given by
$$|\Psi\rangle=\int\mathrm{d}\Omega_\vec{k}f(\vec{k})\varepsilon(\vec{k})^\mu a_\mu^\dagger(\vec{k})|0\rangle,$$
where$f(\vec{k})$ is some function, and$\varepsilon^\mu(\vec{k})$ the polarization of the massless vector field. Using the relation$[a^\mu(\vec{k}),a^{\nu\dagger}(\vec{q})]=-\eta^{\mu\nu}\delta^3(\vec{k}-\vec{q})(2\pi)^3 2q_0$ between the ladder operators of the quantized massless vector field and the definition of a single particle state as above, we obtain that the relation$L(\vec{k})|\Psi\rangle=0$ implies
$$L(\vec{k})|\Psi\rangle=\int \mathrm{d}\Omega_\vec{q}f(\vec{q})k_\mu a^\mu(\vec{k})\varepsilon_\nu(\vec{q}) a^{\nu\dagger}(\vec{q})|0\rangle=-\int\mathrm{d}\Omega_\vec{q}f(\vec{q})2q_0(2\pi)^3\delta^3(\vec{k}-\vec{q})k_\mu \varepsilon^\mu(\vec{q})|0\rangle=0.$$
This last relation implies the condition$k_\mu\varepsilon^\mu(\vec{k})=0$ on the polarization, which reduces its degrees of freedom to$3$ (instead of the four of a generic four-vector object$O^\mu)$. We can further reduces the degrees of freedom of the polarization vector by noting that the polarization vector, as it satisifes$k_\mu\varepsilon^\mu(\vec{k})$, can be rewritten as$\varepsilon_\mu=\varepsilon_\mu^\perp+a_Lk_\mu$, where$\varepsilon_\mu^\perp=c_1\varepsilon_\mu^1+c_2\varepsilon^2_\mu$,$1,2$ are the directions transverse w.r.t.$\vec{k}$. Note for this to hold we require that$k_\mu k^\mu=0$, which is the case for a massless vector field. This relation implies that$k^\mu\varepsilon_\mu^\perp=0$, and that$\varepsilon_\mu^\perp$ has zero time component (by its definition of being transvere w.r.t.$\vec{k}$). Observe how$\varepsilon^\perp_\mu$ only has two degrees of freedom.
Using the above relations, we can compute the norm of a one particle state
$$\langle\Psi|\Psi\rangle=\int \mathrm{d}\Omega_\vec{p} \mathrm{d}\Omega_\vec{k}f^*(\vec{p})f(\vec{k})\varepsilon^{\mu *}(\vec{p})\varepsilon(\vec{k})\langle0 |a_\mu(\vec{p})a_\nu^\dagger(\vec{k})|0\rangle=\int \mathrm{d}\Omega_\vec{p}|\vec{\varepsilon}^\perp(\vec{p})|^2\geq 0,$$
where I used the commutation relation between the ladder operators, and that$\varepsilon^\perp_\mu$ has zero time component. This shows that only the element$\varepsilon^\perp_\mu$ affect the norm. As such, only two degrees of freedom are relevant when computing the norm of a physical state, which is the result one obtains from the quantization of the Lagrangian$\mathcal{L}=-\frac{1}{4} F_{\mu\nu}F^{\mu\nu}$, as wanted.
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