In an article by CERN states

The minimal version of supersymmetry predicts that the Higgs boson mass should be less than 120-130 GeV

How was this conclusion reached? I could not find any answers on the web.

  • $\begingroup$ why incorrectly? 125 is less than 130 $\endgroup$ – anna v Jun 13 at 3:41
  • $\begingroup$ mayb this blog entry explaining it will help motls.blogspot.com/2011/12/… $\endgroup$ – anna v Jun 13 at 3:48

Supersymmetric theories are based on the introduction of an extra symmetry between fermions and bosons. Standard Model is very non-supersymmetric, so people have been considering "minimal" supersymmetric extensions of the standard model (MSSM). So, first of all, your question title is not entirely correct - we are talking about minimal versions of supersymmetry.

The superpartners of gauge bosons are Majorana fermions – they are left-right symmetric and do not contribute to chiral anomalies. However, this is not the case for the superpartner of the Higgs scalar. The Higgs superpartner is called “higgsino”, its contribution to the anomaly must be canceled -- so MSSM introduces two oppositely charged higgsinos and, therefore, two Higgs doublets with opposite hypercharges. Models with two Higgs doublets contain three extra Higgs particles. Usually (as in MSSM), they can be classified into two scalars $h$ and $H$ and a pseudoscalar $A$. Also, there is a very important mixing angle parameter $\tan \beta$ - it determines how strongly each scalar interacts with fermions.

The quartic parameters of the MSSM Higgs potential are not the free parameters of the model as in SM -- they are fixed by gauge interaction couplings. The (tree-level) Higgs potential of the MSSM reads:

$$V = \frac{g^2 + (g')^2}{8} (H_{1}^+ H_{1} + H_{2}^+H_{2})^2 + \frac{g^2}{2}(H_1^+H_2)^2 + m_1^2(H_1^+H_1) + m_2^2(H_1^+H_1) + m_3^2(H_1^+H_2) $$

The quadratic parameters $m_i^2$ are free. Finding minimum of the potential and solving for scalar Higgs masses, one can reexpress as it via standard parameters $m_A$ and $\tan \beta$:

$$ m^2_{h,H} = \frac12\left(m_A^2 + m_Z^2 \pm \sqrt{(m_A^2 + m_Z^2)^2 - 4m_Z^2m_A^2cos^2\beta}\right)$$

The lightest Higgs mass, therefore, satisfies $m_h < m_Z$ at tree level. Such a light Higgs is long excluded by collider experiments. The masses are also influenced by radiative corrections (notably from $t$). MSSM favors quite light Higgs masses and, in view of LEP and Tevatron bounds, the so-called $m_h^{max}$ benchmark scenario was suggested: https://arxiv.org/pdf/hep-ph/0202167.pdf

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.