electromagnetism - Transverse polarizations of a massless spin 1 particle

Physical polarization vectors are transverse, $p\cdot{\epsilon}=0$, where $p$ is the momentum of a photon and $\epsilon$ is a polarization vector.

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Physical polarization vectors are unchanged under a gauge transformation $\epsilon + a\cdot{p}=\epsilon$, where $a$ is some arbitrary constant.

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For a massless spin $1$ particle, in the Coulomb gauge,

one common basis for the transverse polarizations of light are

$$\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)\qquad \epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0).$$

This describes circularly polarized light and are called helicity states.

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In the centre of mass frame, in the positive $z$-direction, the polarization vectors of a photon are

$$(\epsilon_{\mu}^{\pm})^{1}=\frac{1}{\sqrt{2}}(0,1,\pm i,0)\qquad (\epsilon_{\mu}^{\pm})^{L}=\frac{1}{\sqrt{2}}(0,1,\mp i,0).$$

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1. Why are physical polarization vectors transverse?



2. How is $\epsilon + a\cdot{p}=\epsilon$ a gauge transformation? The gauge transformations I know are of the form $A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\Lambda$.


3. How do you define transverse polarization of light? For example, why are $\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)$ and $\epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0)$ the transverse polarizations of light?


4. Why do the polarization vectors $\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)$ and $\epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0)$ correspond to transverse polarizations of light?


5. Why are these polarization vectors $\epsilon_{\mu}^{1}=\frac{1}{\sqrt{2}}(0,1,i,0)$ and $\epsilon_{\mu}^{L}=\frac{1}{\sqrt{2}}(0,1,-i,0)$ called helicity states?


6. In the centre of mass frame, in the positive $z$-direction, why are the polarization vectors of a photon given by $(\epsilon_{\mu}^{\pm})^{1}=\frac{1}{\sqrt{2}}(0,1,\pm i,0)$ and $(\epsilon_{\mu}^{\pm})^{L}=\frac{1}{\sqrt{2}}(0,1,\mp i,0)?$

Answer

I will answer the first two questions and leave the rest to others.

First question - We describe the photon with $A_{\mu}$ but this has 4 degrees of freedom.But we ultimately need just 2. To reduce the number we use the transverse condition and say the momentum is perpendicular to the polarization vector. So any polarization vector will be transverse to the direction of motion. This gives us 3 degrees of freedom. To get to 2, we set up an equivalence class of polarization vectors. This leads to second question

Second question - $ A_{\mu} \rightarrow A_{\mu }+\partial_{\mu} \lambda $ in momentum space becomes $ \epsilon \rightarrow \epsilon + \alpha p $ . This is gauge transformation because notice $ \epsilon \cdot p \rightarrow e \cdot p + \alpha p \cdot p $ but $ p \cdot p = 0 $ so in fact $ \epsilon\text{ and } \epsilon + \alpha p $ represent the same physical state in the same way that $ A_{\mu} $ and $A_{\mu }+\partial_{\mu} \lambda $ represent the same physical state.