The Lagrangian for a massive vector field (without sources) has the form
$$ L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{1}{2}M^{2}A_{\mu}A^{\mu}$$
and is not gauge invariant due to the mass term. The eq.s of motion are
$$\partial _{\mu}F^{\mu\nu} + M^{2}A^{\nu} = 0$$ and by deriving a second time
$$\partial_{\mu} A^{\mu} = 0$$
that is not a gauge-fixing. Instead it arises dynamically and reduces the d.o.f. from 4 to 3. In the electromagnetic case (massless vector fields) once you fix the gauge you have an additional residual gauge condition that reduces the d.o.f. 298
from 3 to 2. The e.o.m. can be solved in the momentum space $$A^{\mu}(x) = \frac{1}{4\pi^{2}} \int \frac{d^{3}k}{2\omega}(e^{ik\dot x}\epsilon^{\mu}(\mathbf{k}) + c.c.)$$
What I think <PERSON> wants to say is that in the massive case you can always choose a reference frame on the particle and so the time component of the polarization vector $\epsilon^{\mu}$ became redundant.
When I think of three possibilities for spin-1, I think {+1,0,−1}. When I think >of a spin-1 "vector state," I think the three positions in the vector represent {+1,0,−1}and not {,,}
Maybe you are talking about the $z$-component of the spin. For $\frac{1}{2}$-spin it can also point in 3 directions, in fact there is the quantum number $helicity$ that is the projection on the direction of motion of the spin.
Here I understand that due to special relativity, is a function of , but since >I don't see the connection to the polarization states, I am missing the >relationship to the amplitude. I believe <PERSON> when he cites this dependence of the amplitude, but where does it come from?
In an analogous way to the scalar field, from the expression of the $A^{\mu}(x)$ you can see that the creation and absorption amplitude are proportional to the polarization vectors.
How <PERSON> is able to conclude immediately that it is "fixed proportional to ${}−\frac{{}_{}}{^{2}}$?
For the Lorentz invariance $$\sum \epsilon_{\nu}^{(a)}(k) \epsilon_{\lambda}^{(a)}(k) = -g_{\nu\lambda}A(k^{2})+k_{\nu}k_{\lambda}B(k^{2}) $$ You can multiply the previous for $k^{\nu}$ and applying $k^{\mu}\epsilon_{\mu}^{(a)} = 0$ you find $$0 = -k_{\lambda}(-A(k^{2})+k^{2}B(k^{2})) \iff B(k^{2}) = \frac{A(k^{2})}{m^{2}}$$ Also in the case $\nu = \lambda$ for the completeness relation $$\sum \epsilon_{\nu}^{(a)}(k) \epsilon_{\nu}^{(a)}(k) = -1 = A(k^{2})$$ 346