What is spin in QFT, non-relativistic QM, and classical physics? When can we ignore spin? In section 4.1.1 of quantum field theory book by M. Schwartz, the author wants to calculate electron scattering by photons and writes the following interaction:
$$ V= \frac{1}{2}e\int dx \psi_e(x)\phi(x)\psi_e(x)$$
Where, obviously, $\psi_e$ is the electron field and $\phi$ is the photon field. The only difference between the real QED interaction term is that he treats the photon and electron fields as scalars. In fact he writes:

"the factor of 1/2 comes from ignoring spin and treating all fields as representing real scalar particles" 

It seems to me that the author says that this formula can be derived from QED in a particular limit. I tried to work out such calulation, but I failed to derive the 1/2 factor. 


*

*What does ignoring spin mean?

*There is a particular limit in wich it makes sense to derive this result from QED? 

*If this is the case, how can we obtain such result starting from real QED?
More generally, 


*What is the relation between spin in QFT, non-relativistic QM and classical physics?  

 A: You shouldn't ignore spin even in the nonrelativistic limit. Schwartz is only ignoring spin for pedagogical reasons. Calculating scattering amplitudes with scalars is easier to do, so he wants you to learn how to do that first before getting into all the complications of spin 1/2 and spin 1 particles (where you need to worry about fermions and gauge invariance, respectively). 
As an example of where spin shows up in a nonrelativistic problem, consider filling the electron orbitals of, say, helium. By the Pauli exclusion principle, if the electron had no spin you'd expect to have to fill the lowest two orbitals. But electrons do have spin, so in addition to the $n,l,m$ quantum numbers there are two internal polarization states associated with spin. So in fact we can fit two electrons into the lowest orbital, one with spin up and one with spin down. If you ignored the electron spin this would not be possible and chemistry would be completely different. 
I wouldn't worry too much about where the 1/2 comes from for now, it's a detail and really the only way to fully understand where it comes from is to understand the scattering calculation in the first place. But, just as a side note, different place where you see a similar factor of 1/2 is in comparing the kinetic term for a real scalar field:
\begin{equation}
-\frac{1}{2} \partial _\mu \phi \partial^\mu \phi,
\end{equation}
to a complex scalar field 
\begin{equation}
-\partial_\mu \Phi^* \partial_\mu \Phi. 
\end{equation}
You can see you need this for example by constructing the propagator. You will find in both cases the normalization guarantees that the propagator goes like $Z/p^2-i\epsilon$ with $Z=1$ (any other normalization would give you $Z\neq 1$). There's a similar normalization requirement underlying the $1/2$ in your cubic vertex. 
Spin is "really" about representations of the poincaire group. There is a classical analogue of spin at least for bosons (integer spin), it more or less amounts to dealing with tensor valued fields. from the perspective of quantum theory, a "classical" field is a coherent state with many quanta. For fermions there isn't really a classical analogue because the Pauli exclusion principle forbids you from making coherent states of fermions. 
