# Specifying the state of polarization of a photon, a classical polarized beam of light, and connection with $\textbf{E}\pm i\textbf{B}$

1. In Classical Electrodynamics, the state of polarization of a monochromatic electromagnetic wave is specified by the direction of the electric field. For example, $\textbf{E}=\textbf{E}_0\cos(\textbf{k}\cdot\textbf{r}-\omega t)$ where $\textbf{E}_0=E_0\hat{\textbf{x}}$, represents a linearly polarized wave, polarized along $x-$axis. Similarly, $$\textbf{E}=E_{0x}\hat{\textbf{x}}\cos(\textbf{k}\cdot\textbf{r}-\omega t)+E_{0y}\hat{\textbf{y}}\sin(\textbf{k}\cdot\textbf{r}-\omega t)$$ represents a elliptically polarized wave where direction of the field is given by $$\tan\theta=\frac{E_y}{E_x}=\frac{E_{0y}}{E_{0x}}\tan(\textbf{k}\cdot\textbf{r}-\omega t).$$

-So here the polarization is specified in terms of the direction of the electric field.

1. On the other hand, the state of polarization of single photons are specified in a different manner. One writes $A^\mu(x)$ in terms of the Fourier modes as $$A^\mu(x)=\sum\limits_{\lambda=0}^{3}\int\frac{d^3\textbf{k}}{(2\pi)^{3/2}\sqrt{2\omega_k}}[\epsilon_{\lambda}^{\mu}a_{\lambda}(k) e^{-ik\cdot x}+\epsilon_{\lambda}^{\mu *}a_{\lambda}^{\dagger}(k)e^{ik\cdot x}]$$ where $\epsilon^\mu$ is called polarization four-vectors.

From this, using the Lorentz gauge condition, one can show that for given direction of propagation $\textbf{k}$, $\textbf{k}\cdot\boldsymbol{\epsilon}=0$, only two components of $\epsilon^\mu$ are independent. Thus a photon (quanta of quantized $A^\mu$ field) is said to have two independent states of polarization.

-Here the polarization is specified in terms of $\boldsymbol{\epsilon}$.

1. The $(0,1)$ and $(1,0)$ representation of the Lorentz group, is said to correspond to $\textbf{E}\pm i\textbf{B}$. These are also referred to as left circularly polarized photon and right circularly polarized photon. Which now gives yet another different definition of polarization.

-Here the polarization is specified in terms of both the electric and magnetic field.

Note that I have given two different and apparently unrelated definitions of polarization in 1 (where the state of polarization is specified by the direction of the electric field) and 2 (where the state of polarization is specified by $\boldsymbol{\epsilon}$).

Why are the definitions of polarization so different in classical electrodynamics and quantum field theory? I believe, these two definitions are related but I'm unable to see the connection.

How does the definition 3, fit into and reconcile with definition of left and right circularly polarized light, as encountered in 1?

Note that in the QFT approach given in Your question we build the free photon field operator from massless irreducible representations of the Poincare group with helicities $\pm 1$, which is the projection of the full angular momentum on the direction of momentum $\mathbf{k}$. I.e., electric field of single photon has the given helicity. In classical electrodynamics (and, in fact, always when many-photon interactions are important) we typically talk about the polarization of electric and magnetic fields being constructed from many photons, which is not the same as the single photon helicity.
The true conditions for the massless field of the helicity $+1$ is $$\tag 1 F_{\mu\nu} = +i\tilde{F}_{\mu\nu}, \quad \partial_{\mu}F^{\mu\nu} = 0$$ while for the massless field of the helicity $-1$ is $$\tag 2 F_{\mu\nu} = -i\tilde{F}_{\mu\nu}, \quad \partial_{\mu}F^{\mu\nu} = 0,$$ where $\tilde{F}_{\mu\nu} =\frac{1}{2}\epsilon_{\mu\nu\alpha\beta}F^{\alpha\beta}$ (the Bianchi identity is satisfied automatically).
It just turns out that the helicity coincides with the circular polarization. You can prove it by using $(1),(2)$. Let's introduce the expansion $F_{\mu\nu} = (\mathbf{E},\mathbf{B})$, $\tilde{F}_{\mu\nu} = (-\mathbf B, \mathbf E)$. We obtain $$\tag 3\mathbf E_{\pm} = \mp i\mathbf{B}_{\pm}, \quad \nabla \cdot \mathbf E_{\pm} = 0, \quad \nabla \times \mathbf B_{\pm} = \partial_{t}\mathbf E_{\pm}$$ The second equation specifies the expansion in terms of divergenceless eigenvectors in the momentum space: $$\mathbf{E}_{\pm}(\mathbf k,t) = \mathbf{e}_{\pm}(\mathbf k)E_{\pm}(t), \quad \mathbf k \cdot \mathbf{e}_{\pm} = 0, \quad \mathbf e_{\pm}^{*}(\mathbf k) \cdot \mathbf e_{\mp}(\mathbf k) = 0, \quad |\mathbf{e}_{\pm}(\mathbf k)|^{2} = 1$$ Next, the first equation of $(3)$ being inserted in the third one gives (we've taken into account the dispersion relation $\omega = |\mathbf{k}|$) $$[\mathbf k \times \mathbf{e}_{\pm}(\mathbf k)]E_{\pm}(t) = \mp ik\mathbf{e}_{\pm}(\mathbf{k})E_{\pm}(t)$$ Let's choose the direction $\mathbf k = (0,0,k)$, and write $\mathbf e_{+} = (a,b,0)$, $\mathbf e_{-} = (c,d,0)$. Then You'll obtain $$(b,-a,0)E_{+}(t)= - i(a,b,0)E_{+}(t), \quad (d, -c,0)E_{-}(t) = +i(c,d,0)E_{-}(t)$$ For arbitrary functions $E_{\pm}$ You obtain $$b = -ia, \quad d = +ic$$ Together with the relation $|\mathbf e_{\pm}|^{2} = 1$ You obtain (up to imaginary unit) $$\mathbf{e}_{+} = \frac{1}{\sqrt{2}}(1, -i, 0), \quad \mathbf{e}_{-} = \frac{1}{\sqrt{2}}(1,i,0)$$ So, the helicity expansion of the one-photon state coincides with circular polarizations expansion.
• What do you mean in eq. (1) and eq. (2)? The standard electromagnetic tensor $F_{\mu\nu}$ is a real 2-form, and the Hodge duals of real forms are real, writing $F_{\mu\nu} = \mathrm{i}\tilde{F}_{\mu\nu}$ is nonsensical (as is obvious from the definition of $\tilde{F}$ you give - there is no way for an imaginary unit to enter). – ACuriousMind Jan 17 '17 at 13:58
• @ACuriousMind : I construct the massless irreducible representation of the Poincare group with helicity $\lambda = \pm 1$ by starting from the spinors $A_{aa}, A_{\dot{a}\dot{a}}$ satisfying equations $$\partial^{a\dot{a}}A_{ab} = 0, \quad A_{ab} = A_{ba}, \quad \lambda = +1,$$ $$\quad \partial^{a\dot{b}}A_{\dot{a}\dot{b}} = 0, \quad A_{\dot{a}\dot{b}} = A_{\dot{b}\dot{a}}, \quad \lambda = -1$$ The direct sum of helicities $\pm 1$ reps $(1, 0) \oplus (0,1)$ is equivalent to the real tensor $F_{\mu\nu}$. But this is not the case when we have only one helicity. – Name YYY Jan 17 '17 at 14:17