To recover classical electromagnetic fields from the quantum electromagnetic field, we consider coherent states of the form $$\exp \left(\int d\vec{r}\, \vec{A}(\vec{r}) \vec{a}^\dagger(\vec{r}) \right) |0\rangle$$ as described in this blog post. This state corresponds to a classical EM field with the vector potential $\vec{A}(\vec{r})$, containing an indeterminate number of photons.

I'm confused about how this works for spin 1/2 fields such as the electron. For some reason, the classical limit of the electron quantum field that we care about is a particle, not a field, so it looks like this procedure does not apply. Even if we did it, I have no idea what a "classical electron field" would look like, since we never talk about such a thing.

  • If I apply the procedure anyway, to get 'electron coherent states', what is the interpretation of these states? They have indeterminate electron number, which is strange.
  • What's the resulting classical electron field like?
  • How can we recover quantum electron particles from the quantum electron field?
  • $\begingroup$ Sorry if the wording is clunky, but I want to be really careful to distinguish between a classical particle, a classical field, a quantum particle, and a quantum field. $\endgroup$ – knzhou Mar 24 '16 at 3:21
  • $\begingroup$ In order to recover a "classical particle" from a quantum mechanical object you need a constant weak measurement of the position. Mott described this in 1929, already, and today you would need use a decoherence/density matrix approach to achieve the same thing in modern language. It doesn't "just follow" from free field theory. Why would it? The free field is always the behavior of a quantum mechanical object. To be very honest with you... Motl's article is not a very good place to start with this... he may know what he is talking about, but the presentation is atrociously handwaving... $\endgroup$ – CuriousOne Mar 24 '16 at 4:23
  • $\begingroup$ I would think one can use the procedure to get classical electric and magnetic fields from the electron field, not particles as the answer of Virgo points out $\endgroup$ – anna v Mar 24 '16 at 5:11

The classical limit of bosonic quantum mechanical systems with both finite and infinite degrees of freedom is pretty well understood from a mathematical standpoint (with complete rigour, and for quite general quantum states; see the references in the end).

With fermions on the other hand, the situation is more involved. The point is essentially that hinted by @Virgo 's answer, i.e. that by exclusion principle it is quite tricky to treat the limit of a large number of fermionic particles.

As far as I know, the question may be tackled in two separate steps. 1) A first step is to pass from the Fock space and quantum field dynamics to a sort of classical field fermion dynamics (you can see that as a mean field approximation $N\to\infty$). While for bosons the result of such a procedure is an effective dynamics on the one-particle space, for fermions you still have a dynamics involving all the different components with different number of particles. However, it may be possible (e.g. in the case of non-self-interacting-fermions) that the dynamics can be actually restricted to each $n$-particle sector. 2) Now suppose that you have a mean field dynamics that restricts to each sector. On the one particle sector, you would have let's say a Dirac equation for one electron in an external potential. Now you can there use the standard tools of (bosonic/quantum mechanic) semiclassical analysis to obtain the classical particle dynamics, taking also into account spin, in the limit $\hbar\to 0$.

The step (2) has been done e.g. here, for relativistic fermions. In the case of non-relativstic self-interacting fermions, this two-step procedure has been done with complete rigour recently, obtaining Hartree-Fock for the mean-field $N\to\infty$ approximation, and Vlasov equation then in the $\hbar\to 0$ limit (because for self-interacting fermions you can't just restrict to a fixed-particle sector because the dynamics involves other sectors, so you have to consider all particles at once, getting therefore a Vlasov equation). The references are for the step (1) this and for the step (2) this.

I remark that also for non-relativstic bosons it is possible to do both a mean filed and classical limit, and it is easier than with fermions. It has been done e.g. here. However for relativstic bosonic fields and quantum mechanical systems it is possible to simply do the classical limit $\hbar\to 0$ to obtain either classical field or classical particle dynamics. The latter has been done rigorously for essentially any "conceivable" system, using the results originating in the Weyl-Hörmander pseudodifferential calculus and the so-called Wigner measures. The former is much more complicated, but some (few) results have been recently obtained, essentially here, here, and here.

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That formula is for bosons. In the large particle limit the boson quantum field can be approximated by the classical field. Electrons are fermions and by the Pauli exclusion principle you can have only one particle per state. So you cannot go to the large particle limit in the same way as you can for a boson field which can have many particles in each state. For this reason it is not possible to get a classical field limit for the electron field. For non-relativistic electrons, quantum field theory reduces to multi-particle quantum mechanics. This limit does not apply to photons as they are always relativistic.

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  • $\begingroup$ Thanks for the answer! To make sure I understand, does this imply that for a massive bosonic field, there are both quantum particle (nonrelatistivic) and classical field (high $N$) limits? $\endgroup$ – knzhou Mar 24 '16 at 5:12
  • $\begingroup$ Yes, that's right. QFT reduces to QM in the non-relativistic limit. $\endgroup$ – Virgo Mar 24 '16 at 18:02
  • $\begingroup$ Also, bosons (massive and massless) can be described by a classical field in the large N-limit. $\endgroup$ – Virgo Mar 24 '16 at 20:35
  • $\begingroup$ If we have that units of E are just Newtons per coulomb what would the unit of electron field be? $\endgroup$ – Žarko Tomičić May 13 '19 at 8:26

Fermionic coherent states with a truly classical behavior are constructed in the survey

Zhang, W. M., & Gilmore, R. (1990). Coherent states: theory and some applications. Reviews of Modern Physics, 62(4), 867. http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.62.867

A second kind of fermionic coherent states is obtained by treating the coefficients in your expression as Grassmann variables. This gives a generalized classical limit in which one has classical anticommuting variables - the same kind as one has in fermionic path integrals.

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  • $\begingroup$ Can you explain how anticommuting variables can appear classically? I thought classical field theory was exclusively built out of real numbers. How would you measure anything Grassmann-valued? $\endgroup$ – knzhou Apr 3 '16 at 18:31
  • $\begingroup$ @knzhou: Anticommuting variables are ''generalized classical'' - the algebra of exterior forms has this property. This is also the basis of so-called supermanifolds, where differential geometry is generalized to the anticommuting case. It is all a bit abstract, but everything generalizes neatly. $\endgroup$ – Arnold Neumaier Apr 4 '16 at 7:37
  • $\begingroup$ One can measure only the even variables, which commute. These form a subalgebra of the full algebra. Note that in expressions for fieldtheoretic physical observables, Fermionic operators always occur in pairs, hence the observables are even. $\endgroup$ – Arnold Neumaier Apr 4 '16 at 7:39

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