I am trying to calculate the amplitude for a decay $\phi \to e^+e^-$ under a Yukawa interaction $\mathcal{L}_I = -g\phi \bar{\psi}\psi$ to one-loop order (with massless fermions for simplicity).

If I'm not wrong, there are 4 diagrams that contribute to 1 loop, three diagrams involving self-energy corrections (i.e. inserting a loop into the external lines) and an extra diagram with vertex correction (a $\phi$ field exchanged by $e^+$ and $e^-$).

I have no problem calculating the integrals, but I'm not sure if the condition I use for renormalization is correct. Following the example of QED, to apply on-shell renormalization I used the following conditions;

The scalar propagator in the limit $p^2 \to M^2$ should be $\frac{i}{p^2-M^2}$

The fermion propagator in the limit $\not{\!p} \to 0$ should be $\frac{i}{\not{p}}$

The vertex function in the limit $p^2 \to M^2$ should be $-ig$. ($p$ is the momentum of the scalar particle.)

Now, because the self-energy diagrams are all in external legs, the first two corrections mean that those diagrams vanish. But the third condition tells that the vertex correction must also vanish when the scalar particle is on-shell (as in my diagram). Therefore all the diagrams here vanish trivially due to renormalization conditions.

Is this analysis correct? Or did I make some mistake in the renormalization part?

EDIT: I think that the fact I'm working with massless fermions is irrelevant to the discussion. Also, I'm considering a general Yukawa interaction, not related to the Higgs, so even for massless fermions there is still a non-zero interaction.


3 Answers 3


Since you want to use on-shell renormalization I will think of your scalar as some physical field, i.e. its quanta represent proper mass-eigenstates. Furthermore, I will assume a minimal model in the sense that there are kinetic terms for the fermion and the scalar and that there is a simple mass term for the scalar. I assume that you are using the real on-shell scheme, which is sufficient for one-loop calculations anyway. Since the fermions are massless there is no on-shell condition needed to fix the mass renormalization and we are hence left with the condition $$ \lim_{p^2\to 0}\left\{\frac{\not{p}}{p^2} \widetilde{\text{Re}} \left(\Gamma^{\bar f f}_{R,ii}(-p,p)\right)u_i(p)\right\}=u_i(p),$$ which assures that close to the pole of the renormalized propagator the propagator is given by its lowest-order expression. For your scalar particle we need two on-shell conditions. One to renormalize the field and one for the mass-renormalization. These are given by $$ \lim_{p^2\to M^2}\left\{\frac{1}{p^2-M^2} \widetilde{\text{Re}} \left(\Gamma^{\phi\phi}_{R}(-p,p)\right)\right\}=1,$$ $$ \lim_{p^2\to M^2} \widetilde{\text{Re}} \left(\Gamma^{\phi\phi}_{R}(-p,p)\right)=0.$$ The last renormalization conditions needed is the one for your coupling $g$. For example in QED there would be the condition that the electric charge is given by the Thompson limit. (That the Thompson limit in fact matches the QED charge with the classical charge is assured by Thirrings theorem.) In your case you therefore need to tell me what a suiting renormalization limit for $g$ would be. In the OP this condition is given by $$ \lim_{p^2\to 0} \left(\Gamma^{\phi\bar f f}_{R}(M,-p,p)\right)=g,\qquad (1)$$ which in fact assures that the renormalized vertex corrections should vanish for an on-shell scalar. And since OP did not specify the theory OP is working in further, there is no relation between the renormalization of the coupling and the renormalization of the masses and fields.

So far so good. But there is one thing which really should be stressed whenever talking about renormalization.

Renormalzation is the procedure to make the connection between a theory parameter, which we use in a perturbative manner, and a measurement.

The renormalization conditions you chose assure that the decay $\phi\to\bar f f$ rate exactly fixes the coupling $g$. Hence you are not able to calculate any corrections to that decay! You chose it as an input-parameter.


Your renormalization condition defines $-ig$ as the value of the three-point diagram to all orders at $p^2 \to M^2$ so of course the loop diagrams disappear. Alternatively you could have set the renormalization point at some off-shell momenta $p^2 \to \bar{M}^2$, which would render your loop corrections to the physical decay rate non-zero. This would shift the definition of the coupling constant $g$ from being based on the decay rate to some other (not neccessarily physical) diagram.

To understand how this works without the interference of divergent integrals you could try computing the three-point diagram in $d=3$ where everything except the two-point diagrams converge. Up to 1-loop the three-point diagram will have no divergences and you can obtain a finite quantity in terms of the bare coupling constant and the renormalized masses. The choice of a renormalization scheme in this case then is just a choice on how you want to define $g$. Not renormalizing at all is fine too and you can figure out exactly what $g$ is supposed to mean in that context.

  1. As a general strategy, write down the most general renormalizable Lagrangian (i.e. including all terms up to operator dimension $4$) in terms of your favourite fields, respecting all space-time symmetries and possibly also some additional symmetries (like parity, discrete symmetries, etc.).

  2. In your case, the Lagrangian (with a real scalar field $\phi$ and a Dirac field $\psi$) is given by $$ \mathcal{L} = \frac{1}{2}(\partial_\mu \phi \partial^\mu \phi -M^2 \phi^2)+ \bar{\psi}(i \partial \! \! \!/-m)\psi -g \bar{\psi}\psi \phi -\frac{\kappa}{3!} \phi^3 -\frac{\lambda}{4!}\phi^4+ \mathcal{L}_{\rm counter},$$ with $$\mathcal{L}_{\rm counter}= \delta_0 \phi + \delta_1 \partial_\mu\phi \partial^\mu\phi +\delta_2 \phi^2 +\delta_3 \bar{\psi}\partial \! \! \! /\psi + \delta_5 \bar{\psi} \psi + \delta_6 \bar{\psi} \psi \phi + \delta_7 \phi^3 + \delta_8 \phi^4, $$ where the parameters $M, m, g, \kappa, \lambda$ are renormalized (finite) quantities and the coefficients $\delta_i$ of the conterterms have to be tuned (order by order in the loop expansion) in such a way that your favourite renormalization conditions are met.

  3. At one loop order, the Yukawa interaction receives the lowest order contribution $-i g$, a one-loop contribution ($\mathcal{O}(\hbar)$) of the form $i(\alpha /\epsilon + \rm finite)$ (in dimensional regularization with $d=4-2 \epsilon$) and the counterterm contribution $i \delta_6$ of $\mathcal{O}(\hbar)$. Imposing your renormalization condition, you obtain $\delta_6 = - \alpha /\epsilon + \ldots$, where the dots stand for finite contributions from the loop integral.

  4. As (by your renormalization condition) you have chosen the coupling $g$ such that the invariant matrix element of the decay $\phi \to e^+ e^-$ is $g \, \bar{u}(p_1) v(p_2)$ to all orders of the loop expansion, the contribution from the counterterm and the loop contribution (at $p^2 =M^2$) must, of course, cancel. This is not strange, this was just your definition of $g$ (in terms of the observable decay rate). On the other hand, if you consider e.g. the contribution of virtual $\phi$ exchange ($p^2 \ne M^2$) to fermion-fermion scattering, the $p^2$ dependence of the fermion-fermion-$\phi$ vertex will appear at higher orders ($ g \to g F(p^2)$, effectively).

  • $\begingroup$ thank you for your answer. My question is regarding the correction to the $\phi \to e^- e^+$ decay. I know how to regularize and renormalize the infinities since this it's done in almost all textbooks. As I say in my question, using the renormalization scheme defined in my post, I get that the finite contribution is zero, after cancelling infinities. This seems odd to me and makes me think I might be doing something wrong. $\endgroup$
    – Gaussian97
    Commented Feb 14, 2023 at 8:54
  • $\begingroup$ @Gaussian97 Thank you for your reply. I have added a further explanation of this point in my answer. $\endgroup$
    – Hyperon
    Commented Feb 14, 2023 at 9:21

Your Answer

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

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