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In QFT, electromagnetism is represented by the quantum field $\hat{A}_\mu$, and fermions (matter) by the quantum field $\hat\psi$. The same kind of formalism is used for both phenomena, even if the methods of quantization might be somewhat different.

On the other hand - classically we use the classical version of $A_\mu$ to describe the electromagnetic interaction, but to describe classical electrons, one "dispenses with" the Dirac field $\psi$ altogether and uses classical/relativistic mechanics, in which the electron is represented by a trajectory $x^\mu(\tau)$.

If we accept the quantum theory as fundamental, and independend of the existence of any classical approximation - this distinction feels odd.

Why is it that one kind of field remains a field classically, while another type of field becomes a trajectory (or multiple ones)?

So my questions essentially:

  • Is there any kind of clear reason for that?

  • Is there any situation where treating the Dirac equation as a classical field equation allows one to derive useful/physically true results about classical electrons? Maybe a statistical description?

It is especially the latter in which I am interested in.

Bence Racskó
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1 Answers1

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This is a well-formulated question. You are using the "reverse logic" in which any classical theory is a limit of the correct quantum theory, but this logic has limitations, and you ask these questions about one of them:

  1. At classical level, the electron is a point particle (elementary charge) and no more than that. As of such, according to the tenets of classical electromagnetism in flat or curved spacetime, it could be described by a point of mass m and charge (-e) located at a certain point P of a Lorentzian manifold which is locally parametrized by a set of 4 functions $x^{\mu} (\tau)$. There is no proper field interpretation to give to a massive particle because classical field theory has a clear dichotomy field-particle. In classical field theory, the electron is a source term for the electromagnetic field, thus it enters the Maxwell equations by the 1-form called electric current density $j$. You can say that this $j$ is a "field" (it is mathematically a 4-vector field), for it allows the interaction electron-em field to be written as a Lagrangian density (i.e. a 4-D integration, not a 1-D one as in the case of $x^{\mu}$). This story of classical electromagnetism is neatly described by PAM Dirac in his GR booklet of about 80 pages from 1975.

  2. A classical ("dequantized") Dirac field exists only in introductory QFT texts to illustrate the concept of field quantization starting from the "Dirac Lagrangian density". There is a very clear reason why a Dirac field cannot exist classically. The number 4 of fields at each point in spacetime is a consequence of the existence of spin 1/2 (thus a quantum effect) and parity invariance (this is also quantum-based). Indeed, the $j$ in point 1. is $\bar{\psi}\gamma\psi$, but the physical interpretation of each of the 3 terms requires quantum mechanics.

DanielC
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  • The Dirac field/equation is however used as a relativistic wave equation to describe a spin 1/2 particle for some applications: for example, fine structure of the energy levels in the hydrogen atom, scattering by black holes... So one can extract physics from it. – Rudyard Jul 22 '19 at 11:45
  • I don't know the answer to OP's question, but this answer is not correct.
    1. The path integral for QED is well defined in the limit of $\hbar \to 0$ which gives a classical Dirac field coupled to a $U(1)$ gauge field. You can represent the spinor part as a current but it isn't necessary
    2. The existence of spin 1/2 particles is relativistic but not quantum--it comes from the fact that that SO(3+1) has a $1/2 \otimes 1/2$ representation which shows up in the spin current (see Belinfante–Rosenfeld stress–energy tensor on Wikipedia). Parity is a well-defined classical symmetry, too.
     – Elijah_Lew-Smith Apr 03 '23 at 21:53