0
$\begingroup$

In $e^{+}e^{-} \rightarrow \mu^{+}\mu^{-}$ when we take into account $\gamma$ and $Z_0$ contributions interference term appears.

We usually just add amplitudes of contributions from different Feynman diagrams for one process. Interference term in $\gamma/Z$ contribution appears from the fact that we also than square or amplitude. But if we take into account $H$ contribution there is no interference term between H and $\gamma/Z$. My first guess was that it is due to the fact that Higgs doesn't couple to photon, but it couples to Z, so it can't be an answer.

So, what is the reason that H doesn't have interference term with photon and Z?

Improve this question
$\endgroup$
5
  • $\begingroup$ Do you meant that we can add only interactions through interaction particles in total cross sections? And why Higgs can not produce the same final state? I know, that Higgs can decay into fermions. As I understand if Higgs can decay into fermions than fermion annihilation (unlikely) can produce Higgs. And we can have contribution from this process. Why shouldn't we take it into account? It is not important for cross-section how what was a particular process. Cross section is just a probability of going from initial to final state. $\endgroup$
    – watmann
    Commented Oct 21, 2017 at 12:36
  • $\begingroup$ Can you specify your question further? The Higgs mediation amplitude goes as $Y_eY_\mu\sim m_e m_\mu /v^2 \sim 10^{-9}$ so, in practice, it is comically uncompetitive. Are you probing a theoretical high energy limit of thousands of TeV's? In any case, can you write down the amps, at least in the ultra-high energy limit, and support your impression they do not interfere? Have you monitored the equivalence theorem in the Z-exchange? $\endgroup$
    – Cosmas Zachos
    Commented Oct 21, 2017 at 14:57
  • $\begingroup$ I have removed my comments as they are not helpful for future readers. $\endgroup$
    – anna v
    Commented Oct 21, 2017 at 16:30
  • $\begingroup$ @CosmasZachos Unfortunately, I am not that advanced yet. The question was motivated by one of the exercises we were discussing in particle physics class. It was mentioned that higgs contribution doesn't interfere. I tried to understand why at home, but I couldn't find anything that can forbid interference to appear (assuming that we don't neglect it). $\endgroup$
    – watmann
    Commented Oct 22, 2017 at 10:33
  • $\begingroup$ It is an issue of numbers, not principle. As you learn about the longitudinal component of the Z, which is essentially the light brother of the Higgs, the equivalence theorem will explain its role in the exchanges. You should, however, take this teaching moment to remind yourself why the Yukawa couplings have the size indicated and replicate the part per billion number above. $\endgroup$
    – Cosmas Zachos
    Commented Oct 22, 2017 at 13:16

1 Answer 1

Reset to default
1
$\begingroup$

Is the following what you had in mind?

enter image description here

Then the last amplitude with the Higgs is suppressed at least by a factor of $m_em_\mu/m_W^2$ (just considering the vertices) compared to the second one with the $Z$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ In your 3 diagrams, the first one represents electromagnetic interaction mediated by the photon and the second represents weak interactions mediated by the Z-bozon. What type of interactions does the third diagram represent? The virtual Higgs there appears to be mediating some interactions representing a force, but of which kind? $\endgroup$
    – safesphere
    Commented Oct 21, 2017 at 17:50
  • $\begingroup$ I don't know that it has an accepted "official" name but you could call it the Higgs force. This question of yours has been asked on this Q&A in the past but answers have been wishy-washy or downright wrong (i.e. denying the 3rd diagram) iirc. The thing is that it is so weak that it is irrelevant in practice, except for the top quark (because of the mass ratio I reminded in my answer). $\endgroup$
    – user154997
    Commented Oct 22, 2017 at 1:18
  • $\begingroup$ Yes, that's exactly what I had in mind. So, there is nothing that forbids interference and the term to exist, but it is just a negligible effect, so no need to treat it. $\endgroup$
    – watmann
    Commented Oct 22, 2017 at 10:27
  • $\begingroup$ In summary, yes. $\endgroup$
    – user154997
    Commented Oct 22, 2017 at 10:33

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.