I'm trying to convince myself about a derivation of the two polarizations of the photon, but I'm stuck in last step.
Here is what I have done so far. I'm using as primary reference a post asked here here. I started using Maxwell's equations to arrive at the wave equation of the radiative/transverse part of the vector potential $\vec{A}$: $$\nabla^2 \vec{A}_T - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \vec{A}_T = -\frac{4\pi}{c}\vec{j}_T$$
Here I have argumented that in principle this equation have 4 degrees of freedom/polarizations in the vaccum, but since the potentials have a certain arbitrariness, we have a gauge freedom, so writing $A^{\mu}$ as $$A^{\mu} = \epsilon^\mu e^{i(\vec{k}\cdot\vec{x}-\omega t)}$$
and imposing the Lorenz Gauge $\partial_\mu A^\mu=0$, we get
$$k_\mu \epsilon^\mu=0$$
which tell us that we 1 of that initial 4 polarizations is not physical. Here $k_\mu \equiv (\omega/c, \vec{k})$ is the 4-momentum of the photon.
Since the photon have zero mass, $k^2 = 0$, and we can write $k_\mu$ as
$$k_\mu = |k|(1,0,0,1)$$
From this we can write a $\epsilon^\mu$ that satisfies the lorenz gauge as $$\epsilon^\mu = \alpha_1(0,1,0,0)+\alpha_2(0,0,1,0)+\alpha_3(1,0,0,1)$$
At this point I applied an additional gauge transformation give by: $$\epsilon^\mu \rightarrow \epsilon'^\mu = \epsilon^\mu + \partial_\mu \color{red}\Lambda = \epsilon^\mu + \partial_\mu \left(\color{red}{\lambda e^{ikx}}\right)$$
On the post I've attached, we read that: $$\epsilon^\mu \rightarrow \epsilon'^\mu = \alpha_1(0,1,0,0)+\alpha_2(0,0,1,0)+(\alpha_3+i\lambda |k|)(1,0,0,1)$$
which upon choosing $\lambda = -i|k|/\alpha_3$ reduces to just two polarizations.
My issues are:
- Why the 2nd gauge transformation acts only in the component $\alpha_3(1,0,0,1)$?
- Is not the term $(\alpha_3+i\lambda |k|)(1,0,0,1)$ missing an exponential, like $(\alpha_3+i\lambda |k|)(1,0,0,1)e^{ikx}$?
I appreciate any guiding thoughts on this two issues.
