Vectors of polarizations from vector boson field solution

Let's have the solution for vector boson Lagrangian in form of 4-vector field:

$$A_{\mu } (x) = \int \sum_{n = 1}^{3} e^{n}_{\mu}(\mathbf p) \left( a_{n}(\mathbf {p})e^{-ipx} + b_{n}^{+} (\mathbf p )e^{ipx}\right)\frac{d^{3}\mathbf {p }}{\sqrt{(2 \pi )^{3}2 \epsilon_{\mathbf p}} },$$ $$A^{+}_{\mu } (x) = \int \sum_{n = 1}^{3}e^{n}_{\mu}(\mathbf p) \left( a^{+}_{n}(\mathbf {p})e^{ipx} + b_{n} (\mathbf p )e^{-ipx}\right)\frac{d^{3}\mathbf {p }}{\sqrt{(2 \pi )^{3}2 \epsilon_{\mathbf p}} },$$ where $e^{n}_{\mu}$ are the components of 3 4-vectors, which are called polarization vectors. There are some properties of these vectors: $$e_{\mu}^{n}e_{\mu}^{l} = -\delta^{nl}, \quad \partial^{\mu}e^{n}_{\mu} = 0, \quad e_{\mu}^{n}e_{\nu}^{n} = -\left(\delta_{\mu \nu} - \frac{\partial_{\mu}\partial_{\nu}}{m^{2}}\right).$$ The first two are obviously, but I have the question about the third. It is equivalent to the transverse projection operator $(\partial^{\perp })^{\mu \nu}$ relative to $\partial_{\mu}$ space. So it realizes Lorentz gauge in form of $$\partial^{\mu}A^{\perp}_{\mu} = 0, \quad A^{\perp}_{\mu} = \left(\delta_{\mu \nu} - \frac{\partial_{\mu}\partial_{\nu}}{m^{2}}\right)A^{\nu},$$ which is need for decreasing the number of components $A_{\mu}$ as vector form of representation $\left(\frac{1}{2}, \frac{1}{2} \right)$ of the Lorentz group by one (according to the number of components of spin-1 field).

I don't understand how to interprete this property. Can it be interpreted as matrice of dot product of 4 3-vectors $e_{\mu}$, which makes one of component of $A_{\mu}$ dependent of anothers three?

Answer

In a sum on polarizations, like $\sum_\lambda~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)$, there is a fundamental difference if you are considering all the polarizations, or only the physical polarizations .

If you take all polarizations, the sum is equals to $g_{\mu\nu}$, and it is in fact a normalizations of the $e_{\mu}^{\lambda}(k)$. This sum is non-physical.

$$\sum_{all ~polarizations~~ \lambda}~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)=-g_{\mu\nu} \tag{1}$$

If you consider only the physical polarizations, this sum is physical, and you will get the pole of the propagator, which is a physical quantity too :

$$\sum_{physical ~polarizations~~ \lambda}~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)=-(g_{\mu\nu} - \frac{k^\mu k^\nu}{m^2})\tag{2}$$

The propagator here is :

$$D_{\mu\nu}(k) = \frac {-(g_{\mu\nu} - \frac{k^\mu k^\nu}{m^2})}{k^2-m^2} \tag{3}$$