We all know that Dirac equation comes up naturally by trying to build a relativistic generalization of the Schrödinger equation $$(i\gamma^{\nu}∂_{\nu}-m)\psi=0$$ and that the associated solutions are given by the usual decomposition $$\psi(x,t)=\sum_{s}∫ \frac{d^{3}p}{(2\pi)^{3}}\sqrt{\frac{m}{ω_{p}}}\bigg(b_{s}(p)u_{s}(p)e^{-ip\cdot x}+d^{*}_{s}(p)v_{s}(p)e^{i p\cdot{x}} \bigg)|_{p_{0}=ω_{p}}$$ $$\bar{\psi}(x,t)=\sum_{s}∫ \frac{d^{3}p}{(2\pi)^{3}}\sqrt{\frac{m}{ω_{p}}}\bigg(b^{*}_{s}(p)\bar{u}_{s}(p)e^{+ip\cdot x}+d_{s}(p)\bar{v}_{s}(p)e^{-i p\cdot{x}} \bigg)|_{p_{0}=ω_{p}}$$ As far as I know, here we are dealing with classical fields before promoting $b_{s}(p),d_{s}(p)$ to ladder operators, so I interpret them as simple functions over spacetime. At this point I've seen that is usual to define the "helicity operator" $$h:=\frac{1}{2|\stackrel{\rightarrow}{p}|} \begin{pmatrix} \stackrel{\rightarrow}{\sigma}\cdot{\stackrel{\rightarrow}{p}} & \mathbb{0}\\ \mathbb{0} & \stackrel{\rightarrow}{\sigma}\cdot{\stackrel{\rightarrow}{p}}\\ \end{pmatrix}$$ One could show that $[H,h]=0$ so helicity is a conserved quantity. Here in my mind I get confused, I've said that we are treating the Dirac field as a classical field, and NOT as a wavefunction... so why do we use QM here? It seems thus that the Dirac field is actually a wavefunction (would make sense since we start from the Schrödinger eq. to derive the Dirac one), is it the case?
If the answer is YES, then is it just a coincidence that we can "recover" QM from a classical field theory? Or it's more general? For example, if we take the Proca Lagrangian $$\mathcal{L}(x):=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-j_{\mu}A^{\mu}+\frac{m^{2}}{2}A_{\mu}A^{\mu}$$ would we get the wavefunction's EoM associated to a Spin-1 massive particle just by imposing the stationary condition as usual? In other words I'm trying to ouline the difference between QM, Classical FT and non relativistic QFT.
