Question about the “wave function” on Relativistic Quantum Mechanics (RQM) and Quantum Field Theory (QFT)

I'm enrolled on a short and conceptual couse on RQM and QFT and the professor made a distinction about the Klein-Gordon (K-G) equation on RQM and the K-G equation on QFT. Roughly speaking, he said that RQM is about particles and QFT in about fields, and the object which captures that distinction is in fact the wave function. Then he said that the wave function on K-G of RQM is the common wave function from non-relativistic quantum mechanics, and the "wave function" from QFT is, also, a scalar field. He said that in RQM the wave function $$\psi$$ is for one electron and in QFT in for the field $$\Psi$$ of electrons, but since the wave function $$\psi$$ is already an scalar field, in the sense of pure mathematics, how can describe one electron and not a field?

• "the "wave function" from QFT" is not a spinor field, it is a scalar field. If your question is not about spinors, why mention them at all, especially when the statement is wrong? May I suggest to remove all references to spinors? – AccidentalFourierTransform Jul 27 at 17:32
• @AccidentalFourierTransform Dirac spinors still satisfy the Klein-Gordon equation, so I don’t understand your objection. – Bob Knighton Jul 27 at 17:44

• You start from the Lagrangian, then apply the principle of least action and get the equation of motion (Euler-Lagrange equation). The solutions are expressed as a superposition of waves with wave vector $\vec k$ and angular frequency $\omega$. Then you work out the conjugate momenta of the fields. To go from classical to quantum mechanics you apply the commutation relations to the fields and their conjugate momenta. So you get creation and annihilation operators in the momentum space, that is $a^\dagger(\vec k)$ and $a(\vec k)$. A field is a superposition of both the operators. – Michele Grosso Jul 29 at 21:42