I am very curious as to how the classical field theory emerges from quantum field theory. In free quantum field theories, coherent states have classical properties, namely that the observables' uncertainties could be made relatively small: $\Delta \phi << \langle \phi \rangle$. So this is one notion in which you could associate classical behavior to a QFT state. But if a QFT is interacting then a state that starts out coherent loses coherence in the future. But we clearly observe classical electrodynamics which is an interacting theory. So, how does classical behavior emerge from interacting QFTs?

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    $\begingroup$ This is a classic question, answered in many many textbooks, usually called decoherence, or quantum to classical transition. So please be precise about what you misunderstood in the quantum to classical transition. A coherent state is not really quantum. It's at the border between quantum and classical : it's the most quantum of the classical state (it cannot be described by a classical field), and the most classical of the quantum state (it cannot be destroyed, being the eigenstate of the annihilation operator). $\endgroup$
    – FraSchelle
    Commented Oct 17, 2017 at 8:14
  • $\begingroup$ $$\hbar \rightarrow 0 $$ of coherent state path integral leads to classical field equations. $\endgroup$
    – Sunyam
    Commented Oct 17, 2017 at 18:47
  • $\begingroup$ @confused, thanks...could you provide a reference where this calculation is explicitly done? $\endgroup$
    – arvinshm
    Commented Oct 17, 2017 at 19:12

1 Answer 1


Suppose free quantum system. It can be well described in terms of creation-annihilation operators $\hat{c}_{\mathbf k}^{\dagger},\hat{c}_{\mathbf k}$. The classical description is good approximation of this quantum system if the number of particles $N$ in this system is macroscopical, i.e. for the system treated and the state $|N\rangle$ we have $$ \hat{c}_{\mathbf k}^{\dagger}|N\rangle = \sqrt{N+1}|N+1\rangle \approx \sqrt{N}|N+1\rangle + \mathcal{O}(N^{-\frac{1}{2}}), $$ $$ \hat{c}_{\mathbf k}|N\rangle = \sqrt{N}|N-1\rangle \approx \sqrt{N}|N+1\rangle \Rightarrow $$ $$ [\hat{c}_{\mathbf k},\hat{c}_{\mathbf k}^{\dagger}]|N\rangle = \mathcal{O}(N^{-\frac{1}{2}}) \to 0 \ \ \text{ as } \ \ N\to \infty $$ Then the creation-annihilation operators can be treated as ordinary c-numbers. The coherent state is just a state on which the equality $[\hat{c}_{\mathbf k},\hat{c}_{\mathbf k}^{\dagger}] = 0$ is exact. Parametrizing it by the number of particles $N$, $|N\rangle_{\text{coh}}$, we have $$ \tag 1 |N\rangle_{\text{coh}} = e^{-\frac{N}{2}+\sqrt{N}\hat{c}_{\mathbf k}^{\dagger}}|0\rangle $$ In this case $$ \hat{c}_{\mathbf{k}}|N\rangle_{\text{coh}} = \hat{c}^{\dagger}_{\mathbf k}|N\rangle_{\text{coh}} = \sqrt{N}|N\rangle_{\text{coh}}, $$ the quantum system can be equivalently described in terms of classical variables after considering the VEV on the coherent state of all the quantum operators. Say, for the field operator $$ \hat{\phi}(x) = \sum_{\mathbf k}\frac{1}{\sqrt{2\omega_{\mathbf k}V}}(e^{ikx}\hat{c}_{\mathbf k}+e^{-ikx}\hat{c}^{\dagger}_{\mathbf k}), $$ where $V$ is the volume of the system and $\omega_{\mathbf k} = \sqrt{k^{2}+m^{2}}$, the VEV on the coherent state $(1)$ for, say, $\mathbf k = 0$ is $$ \langle N|_{\text{coh}}\hat{\phi}(x)|N\rangle_{\text{coh}} = \sqrt{\frac{N}{mV}}\cos(mt) $$ Let's now turn on quantum interactions, assuming for simplicity that they're sufficiently small, i.e. parametrized by dimensionless couplings $\alpha << 1$. Then the interactions can be viewed as a perturbation around the coherent vacuum. Their destruction of the coherent state (and therefore the classical description) can therefore be described as decays and rescatterings of the particles from the coherent state, because of which the particles leave the condensate. Deriving the corresponding reaction rate $\Gamma$ (using Fermi golden rule for this), one can then estimate the time $t \simeq N\Gamma^{-1}$ for which the quantum interaction will lead to decay of the condensate. If this time is sufficiently large (often this is the case) in compare to time scales relevant for description of classical processes, then approximately the classical description holds.

  • $\begingroup$ I thought OP's question was about recovering classical electrodynamics (i.e. Maxwell's equations) from QED, not recovering the classical particle interpretation. $\endgroup$ Commented Oct 17, 2017 at 10:58
  • $\begingroup$ @SolenodonParadoxus : at least the large part of the question was about the decoherentization, which is the topic of my answer. Moreover, as for the Maxwell equations, many phenomena, for example, quantum theory of dispersion of light, within which the EM field is treated as classical field because of macroscopically large number of quanta, are also related to the coherent states. $\endgroup$
    – Name YYY
    Commented Oct 17, 2017 at 13:29
  • $\begingroup$ Well I'm not saying that what you wrote isn't useful, just that OP (and me) was hoping to get an answer for a slightly different question, which is – how does classical field theory (equations of motion) arise as a $\hbar \rightarrow 0$ limit of the interacting QFT. Maybe you can extend your answer to also answer this? $\endgroup$ Commented Oct 17, 2017 at 17:52
  • $\begingroup$ @NameYYY, thanks for your answer. I have 3 questions: 1. what is the justification for approximating the state of a condensate with a large particle number by \ket{N}? 2. In what sense are \ket{N} and \ket{N+1} approximately the same for large N, because in the Hilbert space, they are orthogonal states. Moreover, does your answer imply that we only see the classical behavior of "weakly" interacting QFTs because otherwise the decoherence time will be very quick for the classical description to remain good throughout a typical classical process? $\endgroup$
    – arvinshm
    Commented Oct 17, 2017 at 18:13
  • $\begingroup$ @arvinshm : 1,2. The justification is just the requirement that the quantum field $\hat{\phi}(x)$ has non-zero condensate in the given quantum state (see, for example, superconductivity BCS theory). If, instead, we would use the "usual" $N$-particle state $|N\rangle$, then due to orthogonality of $|N\rangle$, $|N'\rangle$ we'll obtain zero condensate. 3. This depends on the details of the theory. In particular, as long as interactions are not weak, you can't start from the free theory expansion of $\phi$. $\endgroup$
    – Name YYY
    Commented Oct 18, 2017 at 7:47

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