# Is there some explanation for $y_t=1$

The Yukawa coupling of the top quark is Dirac-natural in a too excellent way, it is within one sigma experimentally, and within 99.5% in absolute value, of being equal to one. Without some symmetry, it seems too much for a quantity that is supposed to come down from GUT/Planck scale via the renormalization group. Is there some explanation for this?

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I don't know of any explanation for this but I'd be interested to hear if someone else comes up with something. – David Z Sep 19 '11 at 17:17
Search on "large top yukawa" returns some theories... – Mitchell Porter Sep 20 '11 at 2:28
@MitchellPorter yep, a "order one" yukawa was expected, even predicted, in some setups. But one thing is a range, say, 0.2.. 20, and a very different thing is 0.995 pm 0.005 – arivero Sep 20 '11 at 9:33
Sounds like a question suitable for the theoreticalphysics.stackexchange.com , imo – anna v Dec 12 '11 at 6:45
It is OK to me, if the moderators forward them to the other SE. – arivero Dec 12 '11 at 8:47

In a new paper, Rodejohann and Zhang write (pages 13 to 14) that in the standard model (with massless neutrinos), the top Yukawa can never RG-evolve to exactly 1, but that this becomes possible once you have massive neutrinos. Then it will grow beyond 1 as you continue to still higher energies. But they also write that attaining the exact value 1 could indicate "the restoration of certain kinds of Yukawa unifications or flavor symmetries". So if you can find a form of symmetry breaking which sometimes occurs when a coupling is exactly unity, and then use it appropriately in a GUT or other model of new physics... then you will have an explanation.

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I think that some predictions of the mass of top quark were based on similar arguments, but not sure. Then, there was also the topic of an "infrared fixed point" of the RG; and also the point of getting the sign of the higgs field crossing zero just when the yukawa of the top was near one, or so. – arivero Mar 21 '12 at 20:58

This is a very naive answer or, in fact, it is not an answer. Among all numbers of order one, is not $y_t=1$ the most likely value, i.e., the statically expected value? Why do we need an explanation for $y_t=0.995$ and not for, say, $y_t=0.629$?

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Perhaps the criticism them should be on the hep-ph modellers who are not courageous enough to quote a variance for their "order 1" predictions. – arivero Nov 19 '12 at 10:14
Or, given that we are in the 99.5 % level, it could be more appropiate to ask if 1000 is the most likely value between the numbers of "order 1000" – arivero Nov 23 '12 at 17:25
@arivero Thanks, you have been so generous with the bounty. – Diego Mazón Nov 26 '12 at 17:10
Well, I say it was a bounty for new answers, if they were giving some new food for thought. Even if naive, (likeliness becomes a subtle concept in probability), it is a decent starting point to argue about. – arivero Nov 27 '12 at 19:00

Some lecture notes, e.g. http://arxiv.org/pdf/0906.1271v4.pdf p. 149, are more concrete about the "order one" condition by asking that the $J=0$ channel of the diagram $t\bar t \to ZZ$ that emits two $Z$ must be canceled with the diagram that actually aniquilates $t \bar t$ into $H$ and then decays to two $Z$. The first diagram goes as $\alpha^2 m_t^2 / m_Z^4$, and the second one goes as $Y_t^2 / m_Z^4$. This is, it seems, the unitarity argument of Llewellyn Smith and Christopher Hubert (CERN-TH-1665)

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