I understand the roughly understand the process of integrating out heavy degrees of freedom of a Lagrangian, namely, taking the action and performing the path integral over the high momentum modes.

However, in the case of the W boson this approach isn't really needed. You can just compare the four-Fermi interaction to the full theory result and you can find the coupling of the four-Fermi interaction. Is there always an "easy way" to accomplish this?

In my particular case, I am trying to understand this paper. We have two Higgs bosons and the interaction of the heavy triplet Higgs, $\xi = (\xi ^{++}, \xi^+ , \xi ^0) ^T$ with the leptons is \begin{equation} f _{ ij} \left[ \xi ^0 \nu _i \nu _j + \xi ^+ \left( \nu _i l _j + l _i \nu _j \right) / \sqrt{2} + \xi ^{ + + } l _i l _j \right] + h.c. \end{equation}

The heavy Higgs also interacts with the SM Higgs, $\Phi=(\phi^+,\phi^0)$ through: \begin{align} V_{\xi \phi} & = \lambda _3 (\Phi^\dagger \Phi )(\xi^\dagger \xi) +\mu \left( \xi ^0 \phi ^0 \phi ^0 + \sqrt{2} \xi ^- \phi ^+ \phi ^0 + \xi ^{ - - } \phi ^+ \phi ^+ \right) + h.c. \end{align}

After integrating out the heavy Higgs the Yukawa interactions take the form, \begin{equation} \frac{f_{ij} \mu }{M ^2} \left[ \phi^0\phi^0 \nu_i\nu_j - \phi^+\phi^0 (\nu_i l _j +l_i \nu_j ) + \phi^+ \phi^+ l _i l _j \right] +h.c. \end{equation}

Is the replacement $\xi \rightarrow \frac{\mu}{M^2} \phi \phi$ (with an obscure sign change) obvious?


This is basically second order perturbation theory. We are looking at diagrams where the external momenta are much smaller than the mass $M$. In Feynmann diagrams then the propogator for the heavy Higgs is basically $1/M^2$. This is a small parameter, so the leading contribution to the, say, $\ell \ell \rightarrow \phi \phi$ process is just a tree level exchange of an $\xi^{++}$. If I try to add more $\xi$ particles to the process this will cost me factor of $1/M^2$, and inserting other subdiagrams should just correspond just correspond to a renormalization of the parameters at low energies.

So at low energies the theory just has a quartic local interaction with magnitude $\mu f/M^2$. The sign comes from the factors of $i$ in the tree level diagram.

  • $\begingroup$ You mention that the sign comes from the factors of $i$, but doesn't each interaction correspond to a s-channel diagram and have the same $i$ factors? Why would the lepton-neutrino coupling get a relative minus sign? $\endgroup$ – JeffDror Feb 14 '14 at 14:38
  • $\begingroup$ @JeffDror: Perhaps I'm being stupid (and I'm not used to relativistic calculations), but don't you get a factor of $i$ from each of the vertices and another from the Green's function? So the whole thing is $-i f\mu/M^2$ $\endgroup$ – BebopButUnsteady Feb 14 '14 at 16:28
  • $\begingroup$ Yes that's right. But if I understand correctly you consider a $ \ell \ell \rightarrow \phi^+\phi^+$ to integrate out $ \xi^{++} $, then $ \ell\nu\rightarrow\phi^+\phi^0 $ to integrate out $\xi^{ + } $ and finally $ \nu \nu \rightarrow \phi ^0 \phi ^0 $ to integrate out $ \xi ^0 $. Which works very well to reproduce the correct results. However, the result in the paper has the integrated out Yukawas with a negative sign for the $ \phi ^+ \phi ^0 \nu _i l _j $ interaction. So I'm uncertain about how this can also be reproduced. Though maybe it can't using the ``easy way''? $\endgroup$ – JeffDror Feb 14 '14 at 16:45
  • $\begingroup$ con't: And each diagram seems to me to give the same factors of $i$. $\endgroup$ – JeffDror Feb 14 '14 at 16:46
  • $\begingroup$ My fault. I simply did not see that one term had a negative sign. I dont know why its there. $\endgroup$ – BebopButUnsteady Feb 14 '14 at 18:53

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