Is the QFT amplitude always proportional to the tree-level amplitude?

A physics professor told me that: "The QFT amplitude is always proportional to the tree-level amplitude, even for non-renormalizable theories". i.e.

$$$$\mathcal{M}^{n-loop}=\mathcal{M}^{tree}\times f(\lbrace p\rbrace)$$$$

I could not find this claim in any of the QFT textbooks however. Is this true? If so, is there any easy way to see that this must be the case? I have heard that in some cases the tree level amplitude vanishes, so if the above statement is true, should it actually be stated that the amplitude is always proportional to the first non-vanishing loop order?

• Well, if you take your equation as the definition of the function $f({p})$, then it is true - but this statement has no physical meaning. I don't think the statement of your professor is true in any sense, or at least useful in any way. Nov 9 '20 at 15:20
• I agree that if we take it as the definition of f(p) then the statement is a tautology. What would be interesting is if after calculating the n-loop amplitude one finds that the co-efficient of say some log(s), log(t) and log (u) terms have a common factor of the tree amplitude, or lowest non-vanishing loop level amplitude. This probably does not happen, but I would just like confirmation that it does not.
– Luke
Nov 10 '20 at 20:03

No, $$\mathcal{M}^{\text{n-loop}} = \mathcal{M}^{\text{tree}}\times f(p,\dots)$$ is not true because there are real processes that have non-zero amplitude but no tree-level diagrams.

The most famous, perhaps, is a Higgs boson decaying to two photons. This ended up being one of the most sensitive signatures in the search for the Higgs. There is no tree-level diagram for this because the photon is massless, but there are one-loop diagrams.

Another example is photon-photon scattering. There are no tree-level, $$\mathcal{O}(\alpha)$$ diagrams for this process but there are one-loop, $$\mathcal{O}(\alpha^2)$$ diagrams.

It is also not the case that $$\mathcal{M}^{n\text{-loop}} = \mathcal{M}^{\text{min-loop}}\times f(p,\dots)$$. For example, consider electron-positron scattering in QED. The lowest-level interaction is the electron and position interacting through single photons, but the full process includes a channel where the electron and positron combine briefly into a positronium resonance before decaying again. Since positronium has a mass its propagator cannot be proportional to a photon propagator. That is $$\mathcal{M}^{\text{min}} = \frac{f_1}{s}+\frac{f_2}{t}$$ but $$\mathcal{M}^{\text{full}} = \frac{f_1}{s}+\frac{f_2}{t}+\frac{f_3}{s-m_{Ps}^2+im_{Ps}\Gamma_{Ps}}+\dots$$ where $$m_{Ps}$$ and $$\Gamma_{Ps}$$ are the mass and decay rate of positronium. The full amplitude has different poles than the lowest-level amplitude, so it can't be proportional.

• Great answer. The property $\cal M^\text{n-loop}\propto \cal M^\text{min-loop}$ is interesting; if there are no resonances, would we conclude that it is satisfied?
– user21299
Feb 19 at 3:36

I would also like to add to Luke's answer, the example of amplitude in Yang-Mills with all $$n$$-gluons of positive helicity $$\mathcal{A}_n(++ \ldots+)$$ - where the tree diagram vanish, and only the 1- and multi-loop diagrams are non-zero.

• Thanks I would still like to confirm that the $n$-loop level amplitude is not proportional to the lowest non-vanishing loop level amplitude.
– Luke
Nov 10 '20 at 20:00