The Dirac field $\Psi(x)$ satisfies the Dirac equation $$(i\gamma^\mu\partial_\mu-m)\Psi(x)=0$$
When we quantize, each of the four components of the Dirac field becomes an operator that creates or destroys electrons or positrons with various spin states.
However, the Dirac field equation in an EM field reduces to the Pauli equation in the non-relativistic limit, which is an equation for a two component object, each component corresponding to the spin state of the electron.
But this is not how the Pauli equation is usually represented. It's components are usually said to be wave functions, with the usual probability amplitude, and this agrees with experiment.
My question is, given a quantum field operator, how does one obtain the corresponding wave function of usual QM?
I have looked at countless textbooks on QFT, and this is never discussed, apart from the final chapter of Weinberg QFT Vol1, but what happens there seems very arbitrary and ad-hoc.