A Transverse polarizations of a massless spin 1 particle

[spaghetti3451]

Physical polarization vectors are transverse, $p\cdot{\epsilon}=0$, where $p$ is the momentum of a photon and $\epsilon$ is a polarization vector.
Physical polarization vectors are unchanged under a gauge transformation $\epsilon + a\cdot{p}=\epsilon$, where $a$ is some arbitrary constant.
1. Why are physical polarization vectors transverse?

2. Why is $p\cdot{\epsilon}=0$ the condition for the transverseness of $\epsilon$?

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

 

[dextercioby]

Your text is ambiguous (if not plain wrong) to use both classical and quantum terminology. The polarization vector is assigned to a classical em wave, while p is the photon's momentum (3 or 4-momentum, it's not clear from what you wrote). The quantum analogue to the classical wave's polarization vector is called the helicity operator. Leaving this aside, for 3. I can say that the $\epsilon$ is not unique which can be (comparing to the standard gauge field theory) thought of a gauge transformation. For 2. the answer should be obvious (hint: why are the E and B fields called transverse?). The answer to 2. automatically sheds light (pun intended!) on 1.

 

[spaghetti3451]

Thank you for the excellent answer.

I have another question.

Assume that, in the centre of mass frame, photon 1 moves in the positive $z$-direction so that photon 2 moves in the negative $z$-direction.

Then, the helicity states of the two photons $1$ and $2$ are given by $(\epsilon_{\mu}^{\pm})^{1}=\frac{1}{\sqrt{2}}(0,1,\pm i,0)$ and $(\epsilon_{\mu}^{\pm})^{2}=\frac{1}{\sqrt{2}}(0,1,\mp i,0)$.

1. Why is there a relative negative sign for the helicity state of each photon?
2. I know that these are so-called circular helicity states. But, how are these the components of these helicity states determined?
3. Is it a common terminology to call $\epsilon^{+}$ the positive helicity state and $\epsilon^{-}$ the negative helicity state?

 

[vanhees71]

There are two helicity states $\epsilon^{\pm}$. What you've written down is the standard basis for the radiation gauge $A^0=0$, $\vec{\nabla} \cdot \vec{A}=0$ for a photon with momentum in $3$-direction. Of course, only either the upper or the lower choice of signs gives a complete set. The one basis is just interchanging the basis vectors of the other.