Polarization Vectors

Turning now from classical mechanics to quantum mechanics, a free photon can be represented by a wave function $A^\mu$, which satisfies the equation

$$\Box A^\mu = 0 .$$ (7.97)

A plane-wave solution to this equation is

$$A^\mu(x;k) = \frac{\varepsilon^\mu(\vec{k},\lambda)}{\sqrt{2\omega V}} (e^{-ik\cdot x} + e^{ik\cdot x}) ,$$ (7.98)

where $k^\mu = (\omega,\vec{k})$ is the four-momentum and $\varepsilon^\mu$ is the unit polarization four-vector of the photon. We will discuss the polarization and normalization presently.

On substituting into the equation, we fine $k$ satisfies

$$k^2 = 0, \quad\textrm{that is}\quad m = 0 .$$ (7.99)

The polarization vector has four components and yet it describes a spin-1 particle. The Lorentz condition $\partial_\mu A^\mu=0$, gives

$$k \cdot \varepsilon = 0,$$ (7.100)

and this reduces the number of independent components of $\varepsilon^\mu$ to three. Moreover, we have to explore the consequences of the additional gauge freedom. Choose a gauge parameter

$$\Lambda = ia e^{-ik\cdot x}$$ (7.101)

with constant $a$ so that $\Box\Lambda$ is satisfied. This, together with the solution for $A^\mu$ shows that the physics is unchanged by the replacement

$$\varepsilon_\mu \rightarrow \varepsilon_\mu^\prime = \varepsilon_\mu + a k_\mu .$$ (7.102)

In other words, two polarization vectors $(\varepsilon_\mu,\varepsilon_\mu^\prime)$ which differ by a multiple of $k_\mu$ describe the same photon. We may use this freedom to ensure that the time component of $\varepsilon^\mu$ vanishes, $\varepsilon^0\equiv 0$; and then the Lorentz condition reduces to

$$\vec{\varepsilon}\cdot\vec{k} = 0 .$$ (7.103)

This (noncovariant) choice of gauge is known as the Coulomb gauge.

We see that there are only two independent polarization vectors and that they are both transverse to the three-momentum of the photon. For example, for a photon traveling along the $z$-axis, we may take

$$\vec{\varepsilon}_1 = (1,0,0), \quad \vec{\varepsilon}_2 = (0,1,0) .$$ (7.104)

These are referred to as linear polarization vectors. The linear combinations

$$\vec{\varepsilon}_R = -\frac{\varepsilon_1+i\varepsilon_2}{\sqrt{2}} \quad +1\ \textrm{helicity} ,$$ (7.105)

$$\vec{\varepsilon}_L = +\frac{\varepsilon_1-i\varepsilon_2}{\sqrt{2}} \quad -1\ \textrm{helicity} ,$$ (7.106)

are called circular polarization vectors. A free photon is thus described by its momentum $\vec{k}$ and a polarization vector $\vec{\varepsilon}_i$. Since $\vec{\varepsilon}_i$ transforms as a vector, we anticipate that it is associated with a particle of spin-1. If $\vec{\varepsilon}$ were along $\vec{k}$, it would be associated with a helicity-zero photon. This state is missing because of the transversability condition $\vec{k}\cdot\vec{\varepsilon}=0$. It can only be absent because the photon is massless.

In a special Lorentz frame $\epsilon^\mu$ is pure space-like, $\epsilon^\mu=(0,\vec{\epsilon})$, with $\vec{\epsilon}\cdot\vec{\epsilon}=1$. In an arbitrary Lorentz frame $\epsilon^\mu$ is space-like and normalized to $\epsilon^2=-1$. The normalization constant of the plane wave for a photon is chosen such that the energy in the wave $A^\mu$ is just $\omega=k_0=\vert\vec{k}\vert$ (ie. $E = \hbar \omega$ for a single photon). To verify this, we compute

$$E_\textrm{photon} = \frac{1}{2} \int d^3x \langle \vec{E}^2+\vec{B}^2 \rangle.$$ (7.107)

This if for the Heaviside-Lorentz system of units. In the Gaussian system of units the constant in front of the integral would be $1/8\pi$. Since ($\phi=A^0=0$ in the Lorentz gauge)

$$\vec{E} = -\vec{\nabla}\phi -\frac{\partial\vec{A}}{\partial t} = 0 + i\sqrt{\frac{\omega}{2V}} \ \vec{\varepsilon} \ (e^{-ik\cdot x} - e^{ik\cdot x}) ,$$

$$= \sqrt{\frac{2\omega}{V}} \vec{\varepsilon} \sin k\cdot x .$$ (7.108)

$$\vec{B} = \vec{\nabla} \times \vec{A} = i\sqrt{\frac{\omega}{2V}} \ \hat{k} \times \vec{\varepsilon} \ (e^{-ik\cdot x} - e^{ik\cdot x}) ,$$

$$= \sqrt{\frac{2\omega}{V}} \ \hat{k} \times \vec{\epsilon} \ \sin k\cdot x$$ (7.109)

and

$$(\hat{k}\times\vec{\epsilon})\cdot(\hat{k}\times\vec{\epsilon}) = \epsilon_{ijk}\hat{k}_j\epsilon_k \epsilon_{inm}\hat{k}_n\epsilon_m ,$$

$$= (\delta_{jn} \delta_{km} - \delta_{jm} \delta_{kn}) \hat{k}_j \hat{k}_n \epsilon_k \epsilon_m ,$$

$$= \hat{k}_j \hat{k}_j \epsilon_k \epsilon_k - \hat{k}_j \hat{k}_k \epsilon_k \epsilon_j ,$$

$$= \hat{k}\cdot\hat{k} \vec{\epsilon}\cdot\vec{\epsilon} - \hat{k}\cdot\vec{\epsilon} \hat{k}\cdot\vec{\epsilon} ,$$

$$= \vec{\epsilon}\cdot\vec{\epsilon} - (\hat{k}\cdot\vec{\epsilon})^2 ,$$

$$= -1.$$ (7.110)

Thus $\vec{E}^2=\vec{B}^2$ and

$$E_\textrm{photon} = \frac{2\omega}{V} \int d^3x \sin^2(\omega t - \vec{k}\cdot\vec{x}) ,$$

$$= \frac{2\omega}{V} \int dx_1dx_2dx_3 \left( \frac{e^{i(\omega t - \vec{k}\cdot\vec{x})} - e^{-i(\omega t - \vec{k}\cdot\vec{x})}}{2i} \right)^2 ,$$

$$= -\frac{\omega}{2V} \left( e^{2i\omega t} \int dx_1 e^{-2ik_1x_1} \int dx_2 e^{-2ik_2x_2} \int dx_3 e^{-2ik_3x_3} \right.$$

$$+ \left. e^{-2i\omega t} \int dx_1 e^{2ik_1x_1} \int dx_2 e^{2ik_2x_2} \int dx_3 e^{2ik_3x_3} - 2\int dx_1 \int dx_2 \int dx_3 \right) ,$$

$$= -\frac{\omega}{2V} \left( \frac{e^{2i(\omega t -\vec{k}\cdot\vec{... ...3} - \frac{e^{-2i(\omega t -\vec{k}\cdot\vec{x})}} {8ik_1k_2k_3} - 2V \right) ,$$

$$= -\frac{\omega}{2V} \left( \frac{\sin 2(\omega t -\vec{k}\cdot\vec{x})} {4k_1k_2k_3} - 2V \right) ,$$

$$= \omega ,$$ (7.111)

where we have taken the time average.

The completeness relationship for polarization vectors is

$$\sum_{\lambda=1,2} \varepsilon_\mu(\vec{k},\lambda) \vareps... ...(\vec{k},\lambda) = -g_{\mu\nu} - A \hat{k}_\mu \hat{k}_\nu .$$ (7.112)