Difference between classical fields and wavefunctions? Relation between QM and Classical field theory

Question

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.

Answer 1

Classical field theory (CFT) is not an appropiate framework to study wavefunctions. It is a hard lesson to learn, I know. The notation and math might be similar to the quantum theory which comes before second quantization, but the interpretation is all over the place. What happens upon measurement is completely disconnected from the CFT.

If you were to check for time invariance in the lagrangian, you'd obtain a locally conserved Noether charge corresponding to energy. But we know from experiment that a particle can exchange a quantum of energy at any place, anytime. So, how come that the CFT said that the energy is spread all over space? The answer is that it isn't. The CFT is an incorrect description of particle dynamics.

Answer 2

Unfortunately, the historical naming of the main conceptual elements of quantum mechanics (QM), relativistic QM, and Quantum Field Theory (QFT) is confusing.

If we indicate as classical fields the elements of any theory based on fields of scalar or vector functions with real or complex values, thus commuting objects, the solutions of Maxwell equations of electromagnetism, the solutions of Schrödinger equation of non-relativistic QM, and the solutions of Dirac's equation are all classical fields.

The main difference, in the context of the present question, between classic electromagnetic fields like ${\bf E}({\bf r},t)$ and (quantum) non-relativistic wavefunctions $\psi({\bf r},t)$ is that the latter are used as elements of the domain of an algebra of non-commuting operators, representing the observable elements of the theory. Instead, the former are directly the observables. Therefore, from this point of view, there is no essential difference between non-relativistic wavefunctions and four-component Dirac spinors. They are all classical fields that can be obtained by solving partial differential equations or systems of equations. The quantum character of such fields is entirely in their interpretation as elements of the domain of non-commuting operators.

The situation is different in the case of QFT. There, we deal with fields of operators, not all pairwise commuting.

In a way, it would be possible to say that the adjective quantum should be reserved for the element of the theory where the non-commutation of the product appears. This point of view can also help to make sense of the traditional name of second quantization to indicate the passage from scalar-valued to operator-valued wavefunctions.

Answer 3

It seems that the source of confusion here is the distinction between the first quantization and the second quantization. For "particles", like electrons, first quantization means that we describe them by a wave field (rather than by Newton equations or something like that) - the particles thus acquire "wave properties". Second quantization is then just a mathematical procedure, which allows to treat systems with many particles more conveniently.

Electromagnetic field is already a field classically. Thus, the procedure called second quantization is actually the first quantization for this field, which then acquires particle properties - e.g., we can count photons, attribute to them spin, etc.

Uniform treatment of particles and electromagnetic fields suggests using the same formalism for them (although there is no obligation to do so.)

Related: How does quantization arise in quantum mechanics?