# Maxwell's equations as the particular case of massive vector field equation

I'll start from the beginning. Massive spin-1 representation $\left( \frac{1}{2}, \frac{1}{2}\right)$ is realizing by vector field $A_{\mu}$ and conditions $$(\partial^{2} + m^{2})A_{\mu} = 0, \quad \partial_{\mu}A^{\mu} = 0. \qquad (.0)$$ It's not hard to get the solution: $$A_{\mu} = \int \sum_{\lambda = 0}^{3}e_{\mu}^{\lambda }(\mathbf p )\left( a_{\lambda }(\mathbf p)e^{-ipx} + a^{*}_{\lambda }(\mathbf p)e^{ipx}\right) \frac{d^{3}\mathbf p}{\sqrt{2(2 \pi)^{3}}}, \qquad (.1)$$ where $e_{\mu}^{\lambda }$ are polarization 4-vectors, for which we can choose conditions $$e_{\mu}^{1} = (0, \mathbf e^{1} (\mathbf p)), \quad e_{\mu}^{2} = (0, \mathbf e_{2}(\mathbf p)), \quad e_{\mu}^{3} = \left( 0, \frac{\mathbf p}{\epsilon_{\mu}} \right), , \quad e_{0}^{\mu} = (1, 0), \quad (\mathbf e_{1,2} \cdot \mathbf p ) = 0.$$ and so $$a_{0} = \frac{\mathbf p^{2}}{\mathbf p^{2} + m^{2}}a_{3}. \qquad (.2)$$ So field has three independent polarizations.

Then let's have massless case. In my opinion, we may use solution $(.1)$ for it, but also we can analyze the conditions $(.0)$ in case $m = 0$, from which follows, that in some frame we may use $A_{0} = 0$. From this there is a result $e_{\mu}^{0} = 0$, and so $e_{\mu}^{3} = 0$. Massless spin-1 field has 2 independent polarization states, as it must be.

Also, equations $(.0)$ can be derived independently of Proca equation, so it's not necessary $m \neq 0$ condition.

So I have a few questions:

1. Are my statements correct?

2. If yes, can they be interpreted as transition from massive reps of Poincare group (with two non-trivial Casimir operators $W_{\mu}W^{\mu}, P_{\mu}P^{\mu}$) to massless rep (which is characterized by the helicity)?

This is a very intuitive way of arriving at helicity-1 representations, but it is not totally correct. Essentially, you are ignoring the continuous spin representations of the Poincare group. When you do induced representations of a semi-direct product like the Poincare group, you fix some momenta as a representative element of an orbit of $SO(3,1)$ and then look for the little group that stabilizes this momenta. In the case of massive particles for example you choose $(m,0,0,0)$ and find that the little group is SU(2) giving rise to the spin degrees of freedom of massive particles. Little group transformations then transform spin states into other spin states.
But in the case of a massless representation, choose for example $(E,0,0,E)$. If you play around for a while you will find that a much larger group, $ISO(2)$, the isometries of the plane, stabilizes this momenta. The representation theory of this group can be obtained as it is for the Poincare group ($ISO(3,1)$). But because of the two "momenta" parameters in $ISO(2)$ most of these representations will be infinite dimensional. This would mean that such a massless particle has infinitely many internal degrees of freedom. Since this is not observed, people choose the "representative momentum" (0,0) i.e. the vacuum state for $ISO(2)$. This corresponds to setting the $ISO(2)$ "momentum generators" to zero, and the little group here is then $SO(2)$. Irreducible representations of $SO(2)$ are one dimensional, but you have to include two such representations by CPT, giving rise to the two degrees of freedom of the photon.
So what I am saying is that the $m \to 0$ limit for massive particles isn't really well defined, you could land on the discrete helicity representations or the continuous spin representations. Because of this it is safer to just start with the correct representations of the Poincare group. Say that you want the helicity-1 representation. Try to put it in $A_\mu$ with polarizations $\epsilon_\mu$. Note that even if you choose only two polarizations to be nonzero when you do a little group transformation they shift by something proportional to the momenta. Interpret this as gauge invariance and make the identification on physical states.
Also, all representations of the Poincare group will be characterized by the Casimirs $P^2, W^2$. It is a straightforward calculation to show that $W^2=m^2J^2$ for massive states. Your idea basically amounts to just setting m=0 in this expression, which allows you to show $W^\mu =\lambda P^\mu$ with $\lambda$ the helicity that labels the states. However, you should actually go back through the calculation and plug $(E,0,0,E)$ into the expression for the Pauli-Lubanski vector. Then you will find that to show $W^2=0$ you need certain linear combinations of the Lorentz generators to vanish on the states. These are exactly the "$ISO(2)$ momenta operators" that you need to set to zero to get the helicity representations.