Transverse polarizations of a massless spin 1 particle

Question

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 1

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.

Answer 2

First off, a transverse vector is a spatial vector that is perpendicular to the direction of propagation, i.e. perpendicular to the momentum. A photon does not have a "center of mass" frame because it has no mass, but your vectors make sense if we assume you meant the frame in which the momentum of the photon is purely in the temporal and spatial z-directions. This answers the question that you confusingly labeled by 3.,4. and 6. - the $\epsilon^{\pm}$ span the two-dimensional space of spatial vectors perpendicular to the momentum $p = (p_0,0,0,-p_0)$, and are hence an allowed basis for the transverse vectors.

Now, as for why there are only transverse polarizations of light allowed, this requires carrying out the full quantum field theoretic quantization of the massless spin-1 field that is the electromagnetic vector potential. In particular, the transverse condition arises rather clearly in the Gupta-Bleuler quantization using the Lorenz gauge $\partial_\mu A^\mu = 0$. Carrying out the procedure, one finds that implementing the Lorenz gauge as $\partial_\mu A^{+\mu}\lvert \psi\rangle = 0$ for $A^+$ the annihilation operator part of $A$ leads to the gauge condition being realized within expectation values, and eliminates all negative norm states from the naive Fock space.

In the remaining space, we have Fock states associated to the two transverse polarization, and to a timelike/longitudinal polarization, and the latter are called the spurious states. They have zero norm, and we quotient them out to obtain an actual Hilbert space. This is allowed because one can show that the spurious states decouple from all physical processes in the sense that not only is their norm zero, but also the expectation value of any operator vanishes for them and they have no overlap with any transverse states. In the full quantum theory, the decoupling is ensured by the Ward identities.

These spurious states and the invariance of physical processes under adding/subtracting them is what your "Physical polarizations are invariant under $\epsilon\mapsto \epsilon + a\cdot p$." is supposed to mean. A given polarization $\zeta$ decomposes as $\zeta = \zeta_t + \zeta_s$ where $\zeta_s = a\cdot p, \zeta_s^2 = 0$ and $\vec \zeta_t \cdot \vec p = 0,\zeta_t^2 < 0$, and we quotient the relation $$ \lvert \vec p,\zeta\rangle \sim \lvert \vec p,\eta\rangle \iff \lvert \vec p,\zeta\rangle = \lvert \vec p,\eta\rangle + \lvert \vec p,\sigma\rangle $$ for $\sigma^2 = 0$ out of the Hilbert space, that is, we identify states that only differ by a spurious polarization state. Adding such a spurious Hilbert space corresponds to the classical residual gauge symmetry of harmonic functions we have in the Lorenz gauge. After quotienting this relation out, the remaining Fock space has only the transverse polarization states left (or, more precisely, every equivalence class in the quotient has exactly one representant that is purely transverse).

The states with circular polarization in your basis are called helicity because they are eigenstates of the helicity operator. For more on helicity for photons, see e.g. this answer or this answer of mine.