How is polarization vector in QFT related to polarization in classical electrodynamics?

As i know in classical electrodynamics polarization shows the orientation of the electric vector in a plane perpendicular to the direction of propagation of light.

But in quantum field theory polarization vector is defined differently from that in classical electrodynamics and also using the gauge transformations we have some degrees of freedom in choosing it.

Now my question is how are these two related? is there any relation between them?

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 Maxwell 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 Maxwell 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. 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 Hooft and Feynman. 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. In addition, the states with this weird temporal-longitudinal polarization have zero norm, zero energy, zero momentum, etc. Therefore you can just discard them without any physical consequence, and effectively eliminate another degree of freedom. At the end, you only retain the good old transverse polarizations.

tl;dr: In any gauge choice, there are two polarizations, perpendicular to the direction of propagation.

Polarization of the electric field

Once you have the four-potential, it's easy to obtain the electric field. For the four-potential above $$E^i = - F^{0i} = \partial^i A^0 - \partial^0 A^i = i \omega_k \left( -\epsilon^i(k, \lambda) a_{k,\lambda} e^{-ikx} + {\epsilon^i}^*(k, \lambda) a^*_{k,\lambda} e^{ikx} \right) = 2 \omega_k \textrm{Im}[\epsilon^i(k,\lambda) a_{k,\lambda}]\sin(\omega_k t - \vec{k}\cdot \vec{x})$$ So we have recovered a typical plane wave of amplitude $$E_0 = 2\omega_k \textrm{Im} (a_{k,\lambda})$$ and polarization vector $$\epsilon^i$$.

Of course you can have the supperposition of several plane waves with different polarizations, and in this way obtain circular polarizations and elliptical polarizations. In QFT, circular polarized photons are specially important becuase they are helicity eigenstates.

• In addition, the states with this weird temporal-longitudinal polarization have zero norm, zero energy, zero momentum, etc. Can you please explain this statement or refer to a document that explains this in detail? Standard QFT lecture notes make the same statement as yours with much of the details left out. – Nanashi No Gombe Jul 30 '17 at 8:28