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Feynman diagrams with one or more loops in an interacting QFT are diagrammatic representation of corrections to the Green's functions and amplitudes beyond the lowest order in perturbation theory (called tree level). If I wish I can painfully work with perturbation series and never use Feynman diagrams. So for me the full set up is quantum and loops are just higher order effects instead of quantum corrections to something classical.

Is it not clear why only the loop effects be quantum corrections when the full theory is quantum?

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Yes, in some sense you are correct, but it depends on what you are calculating. In an interacting theory, one can already have quantum interactions at the tree-level. Consider for instance Compton scattering. Here one starts with a photon and an electron and ends with a photon and an electron. In between, one would have tree level diagrams that involve interactions. Each vertex is represented by an h-bar, which is usually associated with a quantum process. At higher orders, one also gets loops, which are then referred to as "quantum corrections." But it does not mean that only the corrections are "quantum" in this case.

On the other hand, if you are computing the two-point function (Green function), then, at tree-level, there are no interactions. Consider for instance, the case where you have one photon at the beginning and one photon at the end. This is equivalent to the classical theory and therefore does not represent the quantum theory at tree-level. When you go to higher orders in this case, you get the quantum corrections, because these self-energy diagrams, would include the vertices associated with the quantum interactions.

Hope this clarifies things a bit.

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I partly disagree with the accepted answer. The loopwise expansion is the same as the $\hbar$ expansion and the lowest order of it, i.e. the tree level, corresponds to $\hbar^0$, which means classical.

The thing is that propagators have factors of $\hbar$ too, and if one counts them correctly the result is what I claimed. Here's a proof: the number of loops in a diagram is $$ L = I - V + 1\,, $$ calling $I$ the internal edges. The partition function in the path integral formulation reads $$ Z[J] = \int \mathcal{D}\phi\,e^{\frac{i}\hbar (S + \int J\phi)}\,. $$ Assuming $S = \tfrac12\int \phi \,G^{-1} \,\phi + \lambda S_I(\phi)$, where $G$ is the propagator, one has $$ Z[J] = \exp\left(\frac{\lambda i}{\hbar} \,S_I\!\left(-i\hbar\frac{\delta}{\delta J}\right)\right)\exp\left(-\frac{i}{2\hbar} \int J(x) G(x,y)J(y) \right)\,. $$ Each term in the expansion of the first exponential corresponds to a certain number of vertices (which come with a $\hbar^{-1}$) and each derivative with respect to $J$ is a propagator (which comes with $\hbar$). So one has $$ Z[J] = \frac{1}{\hbar}\sum_{\mathrm{Diagrams}}(\mbox{Diagram with $I$ edges and $V$ vertices}) \,\hbar^{I-V + 1}\,, $$ that is $$ Z[J] = \frac{1}{\hbar}\sum_{\mathrm{Diagrams}}(\mbox{Diagram with $L$ loops}) \,\hbar^{L}\,. $$ The $1/\hbar$ is overall and has to be factored out when comparing with the classical answer.

Nothe that there is also an interpretation in terms of the WKB (or semiclassical) approximation. That is a systematic expansion in powers of $1/\hbar$ and at the lowest order it corresponds to taking the saddle value of the quantum action. Famously the saddle value is the solution to the classical equations of motion.

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  • $\begingroup$ Do you happen to know in what sense a tree-level correlation function relates to a quantity in classical field theory? For example, one might have $\langle \phi(0) \phi(r) \rangle \sim e^{-mr}$. What is the corresponding classical field theory statement? $\endgroup$ – knzhou Jun 27 at 15:47
  • $\begingroup$ Would you have a (non-trivial) $\hbar^0$ term in the case of Compton scattering (one where the energies of the initial and final particles are not the same)? $\endgroup$ – flippiefanus Jun 27 at 15:53
  • $\begingroup$ I think that both these processes are a solution to the classical equations of motion that stem from the action, am I wrong? $\endgroup$ – MannyC Jun 27 at 16:01
  • $\begingroup$ @MannyC In natural units, tree level Compton scattering amplitude is proportional to $e^4$ which has mass dimension $[M]^0$. Therefore, tree level amplitude is proportional to $\hbar^2 c^2$. $\endgroup$ – mithusengupta123 Jun 27 at 17:02
  • $\begingroup$ Yes but the observable is the cross section. $\sigma = e^4/E^2$. In cgs the units work out ($\mathrm{cm}^2$). In terms of $\alpha$ it would be $\alpha^2 \hbar^2 c^2 / E^2$ but it makes sense to take the classical limit as $\hbar\to0$ and $e$ fixed, not $\alpha$ fixed. Because the classical force between two charges depends on $e$. $\endgroup$ – MannyC Jun 27 at 17:33

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