# Why doesn't $h\rightarrow \gamma\gamma$ rule out two higgs doublet models?

In two higgs doublet models we introduce a down-type Higgs as well as an up type Higgs and typically (as I understand it) we take the down-type to be the 126 GeV resonance and the up type to be somewhat heavier, around the TeV scale.

At this point we've measured the 126 GeV resonance and its decays to $\gamma\gamma, \tau\tau,bb, WW , ...$. In order to predict the rate of $\gamma\gamma$ production correctly we need the top quarks to run in the loop, and hence a coupling between the top and the Higgs. Therefore, it seems the observation of this channel already eliminates the possibility that the observed Higgs is only of the down type since it couples to the up type quarks (I'm envisioning a SUSY scenario where only the down-type higgs only couples to the down-type quarks).

However, I've never heard anyone mention that this observation firmly rules out SUSY models, for example. Is it because $h\rightarrow \gamma\gamma$ can in principle be produced by some other particles in the loop with just the right masses to predict the rate given by the SM loop with tops or is there a better reason?

First of all, I'd like to make some remarks:

• The spectrum of a 2HDM (two-Higgs doublet model) is more complex than you think. There are 2 CP-even bosons ($h$, $H$), a charged scalar ($H^\pm$) and a CP-odd scalar ($A^0$). Usually, we identify the lightest CP-even boson as the scalar particle found at the LHC in 2012 and we assume that the other scalars are heavier than 126 GeV.

• When $h\to \gamma \gamma$ is computed in the Standard Model, there is also a contribution where $W^\pm$ bosons run in the loop.

• I think that the only channel with significant statistic is $\gamma\gamma$. In the strength signals listed in PDG, you'll see that $\tau\tau$, $bb$ and $ZZ$ were reported only at Tevatron. Maybe an experimental physicist could give us more details about these results!

Assuming that a model with two doublets provides a suitable description of nature, then there are two modifications that should be done to the computation of $h\to\gamma\gamma$ in the Standard Model:

1. The couplings of the SM higgs-like boson $h$ with the top and with $W^\pm$ bosons are modified by two new parameters: (i) the ratio of the two vev's $\tan \beta$ and (ii) the mixing angle of the two CP-even scalars $\alpha$ (cf table 2 of 1106.0034). For example $$g_{WWh}=g_{WWh}^{\text{SM}} \cos(\beta-\alpha)$$ and similar expressions can be found for the other couplings. In particular, we obtain SM-like couplings in the limit where $|\alpha-\beta|\to \pi/2$.

2. There is also a new contribution with a charged Higgs boson running in the loop, but this one decouples from the low-energy theory if $H^\pm$ is heavy enough.

So, as far as these new scalars are supposed to be heavy and the Higgs couplings with the top and $W^\pm$ are similar to those of the SM ($|\alpha-\beta|\to \pi/2$), then we can always be consistent with the measurement of $h\to\gamma\gamma$ without excluding this model. This is sometimes called the decoupling limit and there are several works in the literature which study the impact of LHC measurements in the parameters of a 2HDM (cf 1305.2424, for example).

• Good answer, see also these lectures by Haber, conference.ippp.dur.ac.uk/event/321/contribution/3/material/… The decoupling limit occurs for m_A >> m_W. – innisfree Nov 10 '14 at 12:35
• Thanks for your answer. I'm still a little confused about the decoupling limit. Sure if the new heavy fields are very heavy they should decouple, but the light Higgs in the MSSM still should only have couplings to the downtype fermions and the vector bosons. In this way it is very different from the non-SUSY case. If it doesn't couple to the top quarks then I'd expect that the MSSM prediction for the $h\rightarrow\gamma\gamma$ to be much smaller then expected due to largest contribution to the loop missing (even though it can still produce it using the $hW^+W^-$ coupling). – JeffDror Nov 10 '14 at 14:21
• The light Higgs in the MSSM (after EWSB, diagonalizing mass matrices etc) is an admixture of $H_u^0$ and $H_d^0$, by angle $\alpha$. From some trig identities, it works out that the couplings go like $\cos(\alpha-\beta)$ and $\sin(\alpha-\beta)$, and from decoupling that $\cos(\alpha-\beta)\to0$, etc., the basic stuff is in the above lecture – innisfree Nov 10 '14 at 16:28
• @innisfree: Oh, I see. I missed this point. Thanks that makes a lot more sense now. – JeffDror Nov 10 '14 at 17:14
• Thanks for the reference, @innisfree. Haber has good works about 2HDM! – Melquíades Nov 10 '14 at 18:14

Your guess goes into the correct direction. There are other particles contributing to $h \to \gamma \gamma$, such as $W^\pm$ bosons.

Moreover, these exact branching fractions depend on the parameters of the model. While a scenario with no coupling between the scalar at 125 GeV and the top looks very disfavored, the same model with different parameters might do just fine.

• This is too vague, the answer is that $h$ is an admixture of up- and down-type neutral Higgs, and that in the decoupling limit, the SM couplings are recovered. – innisfree Nov 10 '14 at 16:30