# 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?

• 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

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, $y_t=0.629$?
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)