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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.

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A wave function is the projection of a state in the coordinate basis, where the coordinate basis is defined as the basis that diagonalizes the position operator ${\hat x}$. In QFT, the analog of this is the fundamental field operator ${\hat \Psi}$ itself. Eigenvalues of this are defined as $$ {\hat \Psi} (t,\vec{x}) | \phi \rangle = \phi(\vec{x}) | \phi \rangle $$ Then, given a state $| \alpha \rangle$, its wavefunction is defined $$ \alpha [ \phi (\vec{x})] = \langle \phi | \alpha \rangle~. $$ Remember that in QFT, the wave function is really a wave-functional.

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Weinberg's final chapter isn't ad hoc. It's the answer to your question.

In ordinary quantum mechanics, the wave function corresponding to some single particle state $|\psi\rangle$, is $$\psi(x)=\langle x|\psi\rangle$$ where $|x\rangle$ is an eigenstate of the position operator with position $x$.

Now in relativistic QFT, with quantum field $\Psi$, it is problematic to define a position operator, but $$\Psi^\dagger (x)|0\rangle$$ creates a single particle which acts in some sense as if it is localized at $x$. It is a Fourier transform of momentum states, and it has the right transformation properties under the Poincaré group.

So if we have a single particle state $|\psi\rangle$ in QFT, $$\psi(x)=\langle 0|\Psi(x)|\psi\rangle$$ is an object that is defined just like the ordinary QM wave function. And from the operator equation of motion for $\Psi$ we get the wave function equation of motion for $\psi$. So the solutions of the Dirac equation thought of in the wave function sense really are applicable to these states $|\psi\rangle$ in the genuine QFT.

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