On counting the DOF of EM wave polarization

Question

I've got a few questions:

1. In a direvation I've seen, it's 4->3->2: in Lorenz gauge, by solving d'Alembert equation we get $A^\mu=\mathcal{A}\varepsilon^\mu e^{i(kz-\omega t)}$, and $\varepsilon^\mu$ now has 4 DOF, by using two gauge transformations it is reduced to two.

   but I think it should be 6->4->2, since $\varepsilon^\mu$ is a complex unit vector and those differ by a phase factor are regarded as the same polarization

   (for an $n$ dimensional polarization vector, there are $2n-2$ dofs -- $2n$ for every component being a complex number, 2 got dropped because it has unit length and global phase doesn't matter).

2. The statement that EM field has 2 degrees of freedom at one point seems confusing to me.

   First, every $A^\mu$ is actually a function, and the constraints are about the derivatives, so 'at one point' seems confusing to me.

   Second, my understanding is that for every frequency, there are 3 degrees of freedom for the direction of the wave and 2 degrees of freedom for the polarization, so there should be 5 DOF in total for every frequency.

Answer 1

I guess DOF means "degrees of freedom". When saying that the number of DOF of the electromagnetic field is 2, we should remember that we are talking about the gauge field $A_\mu$. This four-vector usually has 4 independent components, that satisfies $\partial_{\mu}F^{\mu\nu}=0$ in the vacuum. Applying the Lorenz condition, $\partial_{\mu}A^{\mu}=0$, it simplifies to $$\partial_{\nu}\partial^{\nu}A^{\mu}=0$$ Taking $A^{\mu}=\epsilon^{\mu} e^{i k.x}$, where $k.x=k_{\mu}x^{\mu}=- \omega t + \boldsymbol{k}.\boldsymbol{x}$, and substituting this into the above equation we find $$k^2=0$$ which is another way of saying that the mass of photon is zero. Since $k^{\mu}$ is a null vector, using the spatial $SO(3)$ symmetry, it can be parametrized as $$k^{\mu}=|k|(1,0,0,1)$$ where we take $\eta_{\mu\nu}=(-1,1,1,1)$. Now let'e get back to the Lorenz condition. For the above $A^{\mu}$ and $k^{\mu}$, it gives $$k_{\mu}\epsilon^{\mu}=0$$ The most general four-vector $\epsilon^{\mu}$ satisfying this relation with the above $k^{\mu}$ is parametrized with three parameters $\alpha_{0,1,2}$: $$\epsilon^{\mu}= \alpha_0 (1,0,0,1)+ \alpha_1 (0,1,0,0)+ \alpha_2 (0,0,1,0) $$ So far we have used "Maxwell equations + Lorenz gauge condition"; the result is that photon seems to have 3 dof: $\alpha_{0,1,2}$.

However, there does exist some residual gauge redundancy which has not fully fixed by the Lorenz condition. It is easy to see that if we make a gauge transformation $$A_{\mu}\rightarrow A'_{\mu}=A_{\mu}+\partial_{\mu}\Lambda$$ with $$\partial_{\nu}\partial^{\nu}\Lambda=0\,,$$ the new gauge field $A'_{\mu}$ will satisfy the Lorenz gauge, too. In other words, all gauge transformation satisfying the above condition are still to be fixed.

To fix this gauge freedom, let's choose $\Lambda = \lambda e^{i k . x}$. Then $$\epsilon^{\mu}\rightarrow \epsilon'^{\mu}= \big(\alpha_0+i\lambda|k|\big) (1,0,0,1)+ \alpha_1 (0,1,0,0)+ \alpha_2 (0,0,1,0) $$

Choosing $\lambda=- i |k|/\alpha_0$ we arrive at $$\epsilon'^{\mu}= \alpha_1 (0,1,0,0)+ \alpha_2 (0,0,1,0) $$ This is the final result: a photon with two dof: $\alpha_{1,2}$. The above choice of $\lambda$ is the famous $$\text{Coulomb gauge:}\,\,\, \boldsymbol{\epsilon}.\boldsymbol{k}=0$$

In summary, combining "Maxwell equations + Lorenz gauge+ Coulomb gauge", motion of photon along the $z$ direction can be described in terms of only two polarization vectors $$\epsilon^{\mu}_1=(0,1,0,0)\,,\,\,\,\,\, \epsilon^{\mu}_1= (0,0,1,0)$$ that are both transverse to $\boldsymbol{k}$.

Hope this helps.

See also Counting degrees of freedom of gauge bosons