Why don't we consider the "most general" spin 1 Lagrangian, but only a special case? The most general Lorentz invariant, renormalizable Lagrangian for a spin 1 field $A_\mu$ reads
\begin{equation}\mathscr{L}_{\text{Proca}}=  C_1 \partial^\mu A^\nu \partial_\mu  A_\nu + C_2 \partial^\mu A^\nu \partial_\nu  A_\mu + C_3 A^\mu A_\mu + C_4 \partial^\mu A_\mu.\end{equation}
However, in the textbooks this Lagrangian only shows up as a very special case
\begin{equation}\mathscr{L}_{\text{Proca}}=  \frac{1}{2}(\partial^\mu A^\nu \partial_\mu  A_\nu - \partial^\mu A^\nu \partial_\nu  A_\mu ) + m^2 A^\mu A_\mu .\end{equation}


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*Why is the term linear in $A_\mu$ usually neglected, i.e. why is $C_4=0$ usually chosen? What would change for $C_4 \neq 0$?

*What would go wrong or what would change if we wouldn't consider the very special case $C_2/C_1=-\frac{1}{2}$?

 A: The equations of motion (and thus the free particle states) will not be changed if we add a term to the Lagrangian which is in the form of a total divergence of a vector $\partial_\mu W^\mu$. Consider the $C_2$ term and rewrite it, up to divergences, as:
$$\partial^\mu A^\nu \partial_\nu A_\mu = \partial^\mu(A^\nu \partial_\nu A_\mu) - A^\nu \partial^\mu \partial_\nu A_\mu = \partial^\mu(A^\nu \partial_\nu A_\mu) - \partial_\nu(A^\nu \partial^\mu  A_\mu) - (\partial_\mu A^\mu)^2$$
We can now redefine the quantities in your Lagrangian as $$\epsilon = sign (C_1), \, \eta = sign(C_3), \xi = \frac{|C_1|}{C_1 + C_2}, m^2 = |\frac{C_4}{C_1}|, V_\mu = 2 \sqrt{|C_1|} A_\mu, F_{\mu\nu} = V_{\mu,\nu} - V_{\nu,\mu} $$
to obtain
$$\mathcal{L}_\mathrm{Proca} = -\frac{1}{4}\epsilon F^{\mu\nu}F_{\mu\nu} + \frac{m^2}{2} \eta V^\mu V_\mu - \frac{1}{2 \xi} (\partial^\mu V_\mu)^2$$
As an exercise, you can compute the equations of motion of this Lagrangian, decompose the linear equations into a transversal solution $V^\mu_\mathrm{T}$ which has $\partial_\mu V^\mu = 0$, and longitudinal solutions $V^\mu_\mathrm{L}$ which have nonzero divergence, and you will obtain separate equations of motion in the form
$$(\Box + m^2) V^\mu_\mathrm{T} = 0$$
$$(\Box + \eta \xi m^2) V^\mu_\mathrm{L} = 0$$
I.e., the transversal solutions are a Proca field of mass $m$. Further analysis shows you that $V^\mu_\mathrm{L}$ is in fact a spin-0 field of mass $\sqrt{\eta \xi} m$, $V^\mu_\mathrm{L} \sim \partial^\mu \phi$. This also explains why we write the Proca kinetic term as $\sim F^{\mu\nu} F_{\mu\nu}$, because it includes only transversal kinetics and $F^{\mu\nu}_\mathrm{L} \sim \partial_\mu \partial_\nu \phi - \partial_\nu \partial_\mu \phi = 0$.
The limit $\xi \to \infty$ makes this scalar mode infinitely heavy and thus inactive, but in some approaches to quantization of Proca field, $\xi$ is left finite, canonical quantization is executed, and only after that you take the $\xi \to \infty$ limit.
A: I tend to believe that there is a mistake in the question. May you check ?
Since you have a factor 1/2 in front of the mass term, you are dealing with the complex case of field.
So in this case, how could you have a factor 1/2 in front of the kinematic term ?
Also, your sign seems in the wrong direction.
Have you really seen a book that contained your formula ? May you give the reference ?
Please find below my demonstration that there is a problem.
Do you see a problem in my derivation ? (I start from formula of wikipedia)

