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It is often said that quantum effects only become manifest in loops, and all tree-level calculations are classical. I am trying to figure out to what extent this claim is true. I know the claim arises from the expansion of the quantum effective action as a power series in $\hbar$, with the leading term being the classical action. Another way to see this is by counting the powers of $\hbar$ in a Feynman graph and noting that there is one power per loop. This is all fine.

However, when it comes to calculating an actual, tangible cross section at the classical level, I believe the claim breaks down. Take, for instance, pair production ($\gamma\gamma\rightarrow e^+e^-$), a process that has a tree-level contribution in the quantum theory. As far as I can tell, this process cannot occur at all in the classical theory. Consider the QED equation of motion for the fermions, $$(i\gamma^\mu\partial_\mu-m)\psi=e\gamma^\mu A_\mu\psi.$$

Classical scattering can be thought of as the (deterministic) interaction between two wavepackets. If the process is $\gamma\gamma\rightarrow e^+e^-$, then initially we must have $\psi=0$ and the field $A_\mu$ will be the sum of two localized wavepackets which will move toward each other and eventually overlap. However, we see from the above equation that if $\psi=0$ initially, then it must remain zero at later times, since the source term is proportional to $\psi$ itself. In other words, there is no way to excite the $\psi$ field.

Is the above argument correct? Is pair production impossible in classical field theory? Is the claim that quantum effects are in loop corrections just sloppy, handwavy nonsense?

Note: I know there are related questions here, here, here and here, but the answers do not go beyond what I mention in the first paragraph.

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  • $\begingroup$ I took a look at some of the links you mentioned and I don't understand how this post fails to answer your question. Could you provide a bit more information? $\endgroup$ Commented Jan 28, 2022 at 2:35
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    $\begingroup$ @NíckolasAlves The post does not mention pair production at all, but Compton scattering. The post claims that in this case, "the fully relativistic classical cross section should be exactly equal to the tree level quantum cross section." This might be true for Compton, but my point is that pair production cannot even occur in the classical theory, and thus identifying the tree-level result for pair production with some classical process is impossible. I would like to know whether this reasoning is correct. $\endgroup$
    – LordDirac
    Commented Jan 28, 2022 at 9:05
  • $\begingroup$ You should consider the condition $\psi=0$ at the initial space-time as a boundary condition about time ( i.e. you have to solve the Dirac equation under the specific boundary condition $\psi(x)\rightarrow0\ (x\rightarrow-\infty$)). Also, one can solve such a problem by using the Green functoion (e.g., see this link.) $\endgroup$
    – Siam
    Commented Jan 28, 2022 at 9:05
  • $\begingroup$ @Siam Does the $x$ in your reply denote the spatial coordinates? At which time is this boundary condition evaluated? It'd be great if you could provide some details in an answer. Thanks! $\endgroup$
    – LordDirac
    Commented Jan 28, 2022 at 9:16
  • $\begingroup$ In addition, the classical limit means simply taking the limit of $\hbar\rightarrow 0$, but the Dirac equation has the effect of a quantum theory in some meaning because the Klein-Gordon eq. is constructed from the Schrodinger eq. and Einstein’s formula. Reflecting these conditions, the “classical” action in the sense of a path integral have somehow strange properties (e.g., the mass of particles is proportional to $\hbar$, and the interaction term is quantum mechanical). But actions of classical field theory is classical because our dynamical variables ($\phi$, $\psi$ …) are not quantized. $\endgroup$
    – Siam
    Commented Jan 28, 2022 at 9:17

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