After reading a few textbooks on Quantum Field Theory there's something that's always struck me as bizarre. Take a scattering process in QED like $\gamma$,e$^-$ $\rightarrow$ $\gamma$,e$^-$. The leading order contribution to this process starts at tree-level. If we assume the incoming photon is randomly polarized, that the incoming electron has a random spin, and that we are insensitive to the photon's final polarization/the electron's final spin, then the differential cross section for $\gamma$,e$^-$ $\rightarrow$ $\gamma$,e$^-$ in the laboratory frame where the electron is initially at rest is given by the Klein–Nishina formula.

The thing is, I constantly read in various textbooks that the tree-level contribution to a scattering process corresponds to the contribution of the 'classical' field theory to said process, and that truly 'quantum' effects begin at next to leading order (almost always involving loop diagrams). But with a process like $\gamma$,e$^-$ $\rightarrow$ $\gamma$,e$^-$, the tree-level contribution contains effects that are simply not predicted by classical electromagnetism. The differential cross section predicted by classical electromagnetism is equal to $r_e^2(1+\frac{\cos(\theta)^2}{2})$, whereas the differential cross section predicted by the Klein–Nishina formula has a dependence on the energy of the incoming photon, which goes reduces to the classical differential cross section as Planck's constant goes to zero. So clearly there's something predicted in the tree level cross section that is missed by classical electromagnetism. So the idea that 'quantum' predictions begin at next to leading order seems erroneous at best.

Am I completely off here? Is the idea that 'quantum' effects start at next to leading order only meant to be taken as a heuristic way of thinking about things? Or am I just overthinking things?

  • $\begingroup$ +1, great question. As more food for thought, the process $\gamma \gamma \to e^+ e^-$ has a tree-level contribution in QED, but does it occur at all in classical electrodynamics? $\endgroup$
    – gj255
    Jul 27, 2017 at 22:30
  • $\begingroup$ It most certainly does not. I was thinking about that actually; it seems that many processes involving annihilation/pair production have no classical counterparts. I've also been thinking a lot about the limit of quantum field theory as planks constant goes to zero. I've often heard that in this limit you recover a classical field theory; but clearly that's not true since the theory can still have things like pair production and annihilation that the classical field theory cannot predict. $\endgroup$ Jul 28, 2017 at 2:47
  • $\begingroup$ This is probably related (actually, I think it answers your question, because people usually mean quantum corrections to the effective action when they say that quantum effects manifest themselves at 1-loop): physics.stackexchange.com/q/337898 I agree that the terminology is kinda misleading. You are absolutely right, even the tree-level diagrams describe the dynamics of the quantum theory. $\endgroup$ Jul 28, 2017 at 3:50
  • $\begingroup$ I suspect the reason for this has something to do with the fact that the momenta of virtual particles is unconstrained in QFT. I think this will have to be taken into account somehow in order to obtain the real "classical" limit. But I agree that the "tree-level is classical" argument has always bothered me. $\endgroup$
    – Philip
    Jul 28, 2017 at 3:56
  • 1

1 Answer 1


Quantum effects vanish when $\hbar \rightarrow 0$. The analysis of the $\hbar$ powers in the vertices and propagators results in a simple rule asserting that the contribution of a diagram containing $N-$ loops to the amplitudes is proportional to $\hbar^{N}$. Thus, we should expect that the classical amplitudes to be given exactly by the tree level diagrams.

However, there is one exception in the correspondence rule between the tree level diagrams and classical amplitudes. This exception is explained in the following work by: Holstein and Donoghue. Please see also, previous works of the same authors cited in the article, where more cases were analyzed.

The exception of the correspondence rule occurs when the loop diagram contains two or more massless propagators. In this case, it was observed by Holstein and Donoghue that contributions to the classical amplitudes occur at the one loop level due a certain non-analytical term in the momentum space. This tem can be recognized to contain $\hbar$ to the zeroth power when the loop diagram is expressed in terms of the momenta rather than the wave numbers as usually implicitly done when loop diagrams are solved. Holstein and Donoghue show that this term does not exist in the case of massive propagators where there is no contribution of the loop diagrams to the classical amplitudes.

The example given in the question electron-photon scattering does not suffer from the above problem. The expression given in the question is valid in the particular case of an electron in nonrelativistic motion as emphasized in Jackson's book "Classical electrodynamics" section 14.7. The fully relativistic classical cross section should be exactly equal to the tree level quantum (Klein-Nishina ) cross section.


Answer to the first follow up question

In order to perform correctly the $\hbar$ expansion, the fields should be scaled by appropriate powers $\hbar$ in order to make their commutation relations proportional to $\hbar$, so they will commute in the $\hbar \rightarrow 0$ limit. The coupling constants should be scaled accordingly. For example the Dirac Lagrangian: $$\mathcal{L}_D = c \bar{\psi}\bigg(\gamma^{\mu} (i \hbar \partial_{\mu} – e A_{\mu}) – m c \bigg)\psi$$ We need to absorb the $\hbar$ coming from the momentum operator into the field, thus by redefining $$\mathcal{L}_D = c \bar{\tilde{\psi}}\bigg(\gamma^{\mu} (i \partial_{\mu} – \frac{e }{\hbar}A_{\mu}) – \frac{ m}{\hbar} c \bigg)\tilde{\psi}$$ Thus we need to take $$ \tilde{e} = \frac{e}{\hbar}, \quad \tilde{m} = \frac{m}{\hbar} $$ To be fixed as $\hbar \rightarrow 0$.

Using the scaled electric charge, the fine structure constant takes the form: $$\alpha = \frac{e^2}{4 \pi \epsilon_0 \hbar c} = \frac{\tilde{e} \hbar}{4 \pi \epsilon_0 c}$$ Now, $\alpha$ is linear in $\hbar$, therefore vanishes in the classical limit. Please see the following work by Brodsky and Hoyer, for the scaling of the various fields and coupling constant.

Answer to the second follow up question:

In their original paper from 1929 Klein and Nishina computed the Compton cross section using Dirac's equation in the background of a classical radiation field. They did not use a quantized electromagnetic field. Therefore, they did not use QED. Please see the derivation also in Yazaki's article(There is a pdf version inside).

In my opinion, the only reason that they had to use the Dirac equation is to take into account the spin. I am quite sure that using more modern classical models for spin such as the Bargmann-Michel-Telegdi theory can be used to provide a fully classical derivation of the Klein-Nishina result. I couldn't find a reference that anyone has performed this work. I am sure it could be a nice project to do.

  • $\begingroup$ Does the vertex correction to the electron's magnetic moment go to zero as planks constant goes to zero? Also; what is the fully relativistic classical cross section in this case? Where is it derived (I've never seen anything like it derived from a classical theory). $\endgroup$ Jul 30, 2017 at 18:45
  • $\begingroup$ @chuxley I have added answers to your follow up questions $\endgroup$ Jul 31, 2017 at 14:31
  • 1
    $\begingroup$ Thank you very much. I wasn't aware that the Klein Nishina formula could be derived without quantum field theory. Your comment on spin is sort of interesting as I've read a project on compton scattering in Scalar QED. At low energies there's actually no evidence of spin in the total cross sections (The ratio between the klein nishina total cross section and the scalar QED total cross section is 1 + O(h^2*w^2/(m^2*c^4)), where w is the incoming photon's frequency, m is mass of the scalar particle/electron, c is the speed of light and h planks reduced constant. $\endgroup$ Aug 4, 2017 at 1:45
  • $\begingroup$ I don't see how Planck's constant could enter into a classical version of Klein-Nishina result, especially given chuckstables note that at lowest order in Planck's constant, scalar QED gives the same cross-section. Then BMT wouldn't be applicable and we wouldn't have a place for Planck's constant through e.g. the magnetic moment. Also, classical scattering of a plane wave off of a point charge has been found exactly in the relativistic limit, in papers like "Classical theory of the scattering of intense laser radiation by free electrons" or Zangwill's Electrodynamics without Planck's constant. $\endgroup$
    – user196574
    Feb 3, 2020 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.