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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.

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  • $\begingroup$ $\psi$ in the original Dirac equation is a "classical field" in the sense its values depend on position and time and are complex-valued and values at arbitrary positions in space commute with each other. Mathematically, $\psi$ is complex-valued four-component field. $\endgroup$ Sep 20, 2023 at 15:25
  • $\begingroup$ Ok, so $\psi$ is just a wavefunction with four components right? Sorry for these basic questions $\endgroup$
    – Filippo
    Sep 20, 2023 at 15:40
  • $\begingroup$ I'm no expert on this, but I don't think Dirac spinor is usually called wave function. I think this is because it does not have a clear interpretation in terms of physics measurements, like $\psi$ in non-relativistic Schr. equation has. $\endgroup$ Sep 20, 2023 at 15:53
  • $\begingroup$ But if it's not even a wavefunction, the classical Dirac field has no physical meaning? $\endgroup$
    – Filippo
    Sep 20, 2023 at 16:08
  • $\begingroup$ Dirac proposed an interpretation in terms of charge density (hole theory), but it has problems. $\endgroup$ Sep 20, 2023 at 16:36

3 Answers 3

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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.

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  • $\begingroup$ Ok, so CFT can be used to describe classical fields like EM and Gravitation but also as a "necessary tool" for QFT (in the sense that we always start with a classical $\mathcal{L}$ before second quantize etc...)? The fact that we can find the Dirac equation from a CFT is thus a coincidence? If we would like to describe a (quantum) relativistic particle of Spin 1 what should we use ?? In the RQM page of Wikipedia I see that CFT is more used for QFT rather than RQM as I thought. There is also a generalization to a higher spins, would be interesting to know where it comes from. $\endgroup$
    – Filippo
    Sep 21, 2023 at 10:39
  • $\begingroup$ "So, how come that the CFT said that the energy is spread all over space? The answer is that it isn't." OK, then maybe you can also answer this question physics.stackexchange.com/questions/700934/… $\endgroup$
    – hyportnex
    Sep 21, 2023 at 13:00
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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.

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  • $\begingroup$ Ok, so relativistic QM is just the covariant version of QM which is different to QFT right?Is it correct to say that QFT is just a generalization to many particles of relativistic QM? I think that this is one thing I'm confused about. Also you said that spinors can be regarded as wavefunctions, so they can be used as in standard QM to compute probabilities etc... right? $\endgroup$
    – Filippo
    Sep 21, 2023 at 9:08
  • $\begingroup$ @Filippo A part of their transformation properties, spinorial wavefunctions are only multi-component wavefunctions. Yes, relativistic QM is the version of QM consistent with the relativity principle. However, such consistency introduces some inconsistency that can be cured only by going to QFT. I would not say that relativistic QM cannot deal with many particles. Instead, it can deal with a varying number of particles, which is the new physical fact in the relativistic regime. $\endgroup$ Sep 21, 2023 at 13:49
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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?

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  • $\begingroup$ Thank you, the main point I was trying to understand is if we can use the Dirac solutions to compute physical measurements like we do in QM (in that sense it would be a RQM formulation). But I've read that there would be some issues like negative probabilities... So can we say that to describe a single relativistic electron we have to use QFT (and thus promoting the Dirac field to operators)? $\endgroup$
    – Filippo
    Sep 21, 2023 at 9:51
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    $\begingroup$ @Filippo To my knowledge Dirac does not produce negative probabilities (which is a problem if we use Klein-Gordon). For a single electron we can use Dirac without second quantization - e.g., the derivation of spin-orbit coupling is done this way. However for a variable number of particles (i.e., when we actually consider interactions where they can be created or annihilated) second quantization is more handy. Dirac electron is both relativistic and quantum. $\endgroup$
    – Roger V.
    Sep 21, 2023 at 9:56
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    $\begingroup$ @Filippo Sure, we can, and we use Dirac solutions to compute physical measurements like in nonrelativistic QM. All the properties of heavy elements we can simulate with ab-initio methods are related to the solutions of Dirac equations for atomic systems. See, for instance, Bachelet, Hamann & Schluter "Pseudopotentials that work: From H to Pu" giovannibachelet.it/curriculum/ppthatwork.pdf $\endgroup$ Sep 21, 2023 at 13:55
  • $\begingroup$ Thank you, I haven't seen this point stressed enough in any book. If you want to edit and add those things in your answer I'll accept and take the question as solved :) $\endgroup$
    – Filippo
    Sep 21, 2023 at 14:06

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