A. Eigenvectors and eigenvalues of the propagator

We will now find the solutions to the eigenvector equation

$${{\mathcal{D}} ^{-1}} ^{\mu \nu} ( k ) a _{\nu} = \lambda _{k} a ^{\mu} \,,$$ (34)

where ${{\mathcal{D}} ^{-1}} ^{\mu \nu} (k)$ is defined in equation (21).

Firstly, there is a ``trivial'' gauge solution $a _{\mu} = H (k) k _{\mu}$, since $k_\mu$ is orthogonal to both terms enclosed in round brackets in the definition (21) of ${\mathcal{D}} ^{-1}$; this can be seen from the relations

$$\left( k ^{2} g ^{\mu \nu} - k ^{\mu} k ^{\nu} \right) k _{\mu} = 0$$

and

$$\left( \bar{k} ^{2} \bar{g} ^{\mu \nu} - \bar{k} ^{\mu} \bar{k} ^{\nu} \right) k _{\mu} = 0$$

using the properties $\bar{g} ^{\mu \nu} k _{\nu} = \bar{g} ^{\mu \nu} \bar{k} _{\nu}$ and $\bar{k} ^{\nu} k _{\nu} = \bar{k} ^{\nu} \bar{k} _{\nu}$.

A second, non trivial polarization is $\tilde{k} ^{\mu} = \epsilon ^{\mu \alpha 0 1} k _{\alpha}$ in the electric case ( $\tilde{k} ^{\mu} = \epsilon ^{2 3 \mu \alpha} k _{\alpha}$ in the magnetic case), since $\tilde{k} ^{\mu} k _{\mu} = 0 = \tilde{k} ^{\mu} \bar{k} _{\mu}$ thanks to equations (19) and (20); $a _{\mu} = \tilde{k} _{\mu}$ is associated to the eigenvalue

$$k ^{2} - \frac{\kappa \bar{k} ^{2}}{k ^{2} - m _{\mathrm{A}} ^{2}}\,.$$ (35)

Then, remaining independent physically relevant polarizations must be orthogonal to both $k _{\mu}$ and $\tilde{k} _{\mu}$, hence they can be parametrized as

$$a ^{\mu} = \epsilon ^{\mu \nu \alpha \beta} d _{\nu} \tilde{k} _{\alpha} k _{\beta} \,;$$ (36)

since $a ^{\mu}$ given by (36) is not affected by the ``gauge transformation''

$$d _{\nu} \longrightarrow d _{\nu} + \lambda _{1} \tilde{k} _{\nu} + \lambda _{2} k _{\nu} \,,$$

it then follows that only two components of $d _{\nu}$ are physically relevant. One of these, which we will call $\tilde{k} ^{\perp} _{\mu}$, is obtained when $d _{\nu} = \bar{k} _{\nu}$: it is thus orthogonal to $k ^{\mu}$, $\bar{k} ^{\mu}$, $\tilde{k} _{\mu}$ and corresponds to the eigenvalue $k ^{2}$. This is the same eigenvalue of the last eigenvector, $E ^{\mu}$, which is given by

$$E ^{\mu} = \epsilon ^{\mu \nu \rho \sigma} k _{\nu} \tilde{k} _{\rho} \tilde{k} ^{\perp} _{\sigma} \,.$$

This can be verified by inserting it into (18) and observing that

$$\bar{k} ^{2} \bar{g} _{\mu \nu} - \bar{k} _{\mu} \bar{k} _{\nu} \propto \tilde{k} _{\mu} \tilde{k} _{\nu} \,.$$

The above equality holds because

$$\bar{k} ^{2} \bar{g} _{\mu \nu} - \bar{k} _{\mu} \bar{k} _{\nu}$$

is a projector onto the space orthogonal to $\bar{k} _{\mu}$ in the $2$-dimensional subspace, where the direction orthogonal to $\bar{k} _{\mu}$ is nothing but $\tilde{k} _{\nu}$.