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In the Standard Model, the Higgs mass doesn't really have any theoretical constraints. It could have basically any value and nothing 'breaks'.

However, in MSSM models, we often see the tree level constraint $$m_h < m_z \cos(2\beta)$$ which is modified by 1-loop stop corrections.

What is happening differently in the SUSY EWSB to give this constraint? Is it because of SUSY or would you also get this by having two Higgs doublets with different VEVs?

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up vote 7 down vote accepted

First, it's not really true that nothing breaks for any mass in the Standard Model. Because of the "renormalization group flows", i.e. the dependence of the couplings on the characteristic energy scales, most of the values of the Higgs mass are inconsistent and imply that the theory breaks down almost immediately at higher energies.

In particular, even the mass 126 GeV seen at the LHC implies that the potential becomes unbounded from below at energy scales beneath the Planck scale, see e.g.

http://motls.blogspot.com/2012/10/higgs-living-near-cliff-of-instability.html?m=1

126 GeV is just too light for the Standard Model – too low quartic coupling, vulnerable to the risk of going negative. Even if there were no instability, the Standard Model would naturally predict $m_h\sim \Lambda$ where the Greek letter indicates the highest energy scale where some new physics exists, probably the Planck scale. The fact that the Higgs boson is much lighter is a sign of fine-tuning and it's been generally believed that an explanation – new physics near the Higgs mass – should exist to "explain" the fine-tuning – the LHC doesn't seem to confirm this belief, at least so far.

Now the MSSM issue

The Higgs sector in the MSSM is even more constrained because one isn't inserting and one can't be inserting the quartic coupling for the Higgs, $\lambda h^4/4$, by hand, so one can't freely adjust the value of the coefficient $\lambda$ – and it is this coefficient that determines the Higgs mass (an increasing function of $\lambda$) given a fixed Higgs vev. That's because the Higgs field is a charged chiral superfield and one can't write down any terms to the superpotential etc. that would do so. For example, the superpotential $W$ that would produce a quartic $V$ would have to be cubic in $h$ but cubic expressions in a field with electroweak charges clearly can't be gauge-invariant.

Because of the symmetry-induced restrictions imposed by supersymmetry on the chiral superfields, the quartic couplings come from one specific place, namely the $D^2$ terms in the potential – from the "squared" supersymmetric terms of the type $h^\dagger e^V h$ which have to be there with a fixed coefficient because they are needed to make the $h$ kinetic terms gauge invariant. Consequently, the coefficient of $h^4$ terms (various types involving either $h_u$ or $h_d$) isn't a new $\lambda$ but it is some combination of the gauge couplings $g^2$ and $g^{\prime 2}$. We know what they roughly are, from the W-boson and Z-boson masses, and they're just a bit smaller than "of order one". Consequently, the Higgs quartic couplings are also "just a bit smaller" than of order one and the Higgs mass must be close to the Higgs vevs, too.

For the precise calculation, one has to diagonalize the Higgs mass matrices, and so on, but the tree-level inequalities one may derive include $m_{h_0}\leq m_A$, $m_{h_0}\leq m_Z$, $m_{H^\pm}\geq m_W$, and others. Those would predict a very light lighter Higgs, but pretty much unlimited heavy heavier Higgs. However, one-loop corrections already allow $m_{h_0}$ to be raised significantly above the Z-boson mass.

See also a new text I wrote about the construction of supersymmetric Lagrangians:

http://motls.blogspot.com/2012/11/supersymmetric-lagrangians.html?m=1

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Nice answer :-). I first planned to boldly write somthing too here, because the higgs sector of the MSSM is nicely explained in the phenomenology chapter of my SUSY demystified book. But now it is no longr needed that I lean myself out of the window :-P –  Dilaton Nov 2 '12 at 12:09
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