Polarization and gauge symmetry
In QFT, the dynamical varible is the four-potential $A_\mu$. The electromagnetic field is defined by $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$, an antysymmetric tensor wich six independent components: 3 for the electric field and 3 for the magnetic field $E^i = - F^{0i} $, $B^i = -\frac{1}{2}\varepsilon^{ijk}F^{jk}$. The <PERSON> equations are $$\partial_\mu F^{\mu\nu} = \partial_\mu\partial^\mu A^\nu -\partial^\nu \partial_\mu A^\mu = 0$$ which is derived from the following Lagrangian $$\mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$ In addition, the four-potential has a gauge symmetry, meaning that observables don't change with the transformation $$A_\mu \to A_\mu - \partial_\mu \chi$$ This allows us to fix different gauges that can simplify (or not) the calculations. For example, one can choose $A$ so that $A^0 = 0$ and $\vec{\nabla}\cdot\vec{A}=0$ (called the radiation gauge). With these election, the <PERSON> equation reduces to a massless Klein-Gordon equation $$\partial_\mu \partial^\mu A^i = 0$$ A particular solution to this equation is $$A^\mu(k, \lambda) = \epsilon^\mu(k, \lambda) a_{k,\lambda} e^{-ikx} + {\epsilon^\mu}^(k, \lambda) a_{k,\lambda}^e^{ikx} \qquad \text{with}\quad k_\mu k^\mu =0$$Here $a_k$ is the amplitude of the mode (wich will become the annihilation operator in canonical quantization) and $\epsilon^\mu$ the polarization vector, and $\lambda$ is an index for the basis of polarization vectors. In the general solution, you have to sum over all possible momenta and polarizations. Using the radiation gauge, $A^0 = 0$ implies that $\epsilon^0 =0$ and $\vec{\nabla}\cdot\vec{A}=0$ implies $k_i\epsilon^i = 0$, this is, the usual polarization in classical electrodynamics, perpendicular to the direction of propagation. 818
As you can see, of the 4 possibles degrees of freedom for the polarization, one is eliminated via gauge freedom and the other with the equations of motion.
Another posible gauge choice is that of 't <PERSON> and <PERSON>. This time we don't impose anything on the four-potential directly, rather we break by hand the gauge symmetry at the Lagrangian $$\mathcal{L}' = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2$$ The point in this approach is that you need to have the canonical momenta conjugate to every component in $A_\mu$ for canonical quantization. As the Maxwell Lagrangian sets that $\partial_0 A^0 = 0$, you run in problems. You can eliminate the $A^0$ altogether (as above) or include an ad-hoc term to make $\partial_0 A^0 \neq 0$ and discard it later. With this new Lagrangian, you get again a massless Klein-Gordon equation $$\partial_\mu \partial^\mu A^\nu = 0$$ Now you have the four degrees of freedom for the polarization vector, or at least apparently. Another problem with this gauge is that a four-potential with polarization $\epsilon^\mu(k,0) =( \epsilon, 0, 0, 0)$ has negative norm. The solution is the Gupta-Bleuler quantization: we require any physical state $|\psi\rangle$, $|\phi\rangle$ to verify $$\langle \psi | \partial_\mu A^\mu |\phi \rangle = 0$$ When enforcing this condition, it is found that the temporal and longitudinal polarization must appear together in any physical state, thus reducing the degrees of freedom by one. 418