Recently in the CERN courier www.cerncourier.com it was reported on the decay of the new boson found at the LHC (shortly called Higgs-boson $H$ in the following) into a $b\bar{b}$-pair. I wonder if also the decay of the Higgs-boson into a $t\bar{t}$-pair already was found.

So my first question is: Was $H \rightarrow t\bar{t}$ already found ? If yes, that would have profound implications I guess.

My second question is: If this first question is true, would it mean that supersymmetry would be ruled out then, since according to supersymmetry quarks of the same generation should couple to 2 different Higgs-bosons (which would be no longer true if the answer of my 1.question would be yes)?

In fact, this is a re-edit of the question "Check on Supersymmetry by measuring the Higgs-coupling to heavy fermions."

  • $\begingroup$ Hi Frederic, Physics.SE is not intended to be a forum but a question-answer site, so it would really help if you would ask a clear question, like "How could we tell the difference between these two different things experimentally?" or "Is this reasoning correct?" or what have you. $\endgroup$ – CR Drost Jun 18 '17 at 16:28
  • $\begingroup$ The mass of the top , is 175 GeV. The mass of the Higgs is 125GeV. tt_bar needs 350 GeV . There is no way to find a decay mode of the measured Higgs into tt_bar. $\endgroup$ – anna v Aug 21 '17 at 14:07
  • $\begingroup$ Okay, I got that. But if it were possible kinematically (for instance at CLIC) to measure/find $H\rightarrow t\bar{t}$, would my conclusion be correct (actually, this is the part of my question I am most curious about) ? $\endgroup$ – Frederic Thomas Aug 21 '17 at 14:32

As for your first question, as @anna pointed out, $H\to t\bar{t}$ is not possible with the Standard Model Higgs: $m_H < 2m_{t\bar{t}}$. A Higgs decaying into $t\bar{t}$ must be neutral. It turns out that the Minimal Supersymmetric Model (MSSM) does provide two neutral Higgs, a pseudo-scalar Higgs and two charged Higgs. There is therefore a good incentive to investigate such models, which are part of a larger class called Two-Higgs Doublet Models (2HDM). So far, LHC has failed to discover any of those.

As for your second question, you are making a subtle mistake. The neutral components of the two Higgs field developing non-zero vacuum expectation values, let's call them by $H_1^0$ and $H_2^0$ do indeed couple only to down-type and up-type quarks, respectively. However, those are not the physical field $H$ and $H'$ we can observe. They are related by a rotation:

$$\begin{pmatrix}H\\H'\end{pmatrix}=\begin{pmatrix}\cos\varphi&\sin\varphi\\ -\sin\varphi&\cos\varphi\end{pmatrix}\begin{pmatrix}H_1^0\\H_2^0\end{pmatrix}$$

As a result, the both of the neutral Higgs can couple to $q\bar{q}$ for any type of quark. So can the pseudo-scalar Higgs. And of course, the two charged Higgs couple to $u\bar{d}$, $s\bar{c}$, or $b\bar{t}$ and the charge conjugates of those for the other Higgs charge.

I did learn most of what I have just explained in [1], which I remember finding rather didactic.

[1] Abdelhak Djouadi, The Anatomy of Electro-Weak Symmetry Breaking. II: The Higgs bosons in the Minimal Supersymmetric Model. https://arxiv.org/abs/hep-ph/0503173

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  • $\begingroup$ Thank you for the answer! So there is a mixing angle. Is this angle a fundamental parameter or is it not possible to derive it from a (more) fundamental theory ? BTW in the very wide-spread SUSY-variant mSUGRA this angle does not seem to appear. Can I learn more from this observation ? $\endgroup$ – Frederic Thomas Aug 21 '17 at 17:22
  • $\begingroup$ iirc, this mixing angle is related to the mass of the two neutral Higgs. This is all in the paper I cited, which I can't check at the moment. As for mSUGRA, I am not a specialist but your claim surprises me, as mSUGRA is just one flavour of the kind of Higgs model I have just wrote about, one with special constraints on the soft breaking terms. $\endgroup$ – user154997 Aug 21 '17 at 17:32
  • $\begingroup$ so trying to understand what you mean: you guess that mSUGRA kind of predicts this mixing angle ? $\endgroup$ – Frederic Thomas Aug 21 '17 at 17:41
  • $\begingroup$ Yes. I need to read again the reference I gave you though, as it has been a long time, but I am pretty sure. $\endgroup$ – user154997 Aug 21 '17 at 17:58
  • $\begingroup$ Thank you again. I already learnt a lot today. Don't feel obliged to read the article. $\endgroup$ – Frederic Thomas Aug 21 '17 at 18:18

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