I read that in the MSSM with mSUGRA boundary conditions, the mass spectrum of the model is determined by five parameters at the GUT scale: $m_0$ (universal scalar mass), $m_{1/2}$ (universal gaugino mass), $A_0$ (universal trilinear coupling), $\tan \beta$ (ratio of Higgs vevs), and sign[$\mu$] (Higgs/Higgsino mass parameter). I have two questions: First, why is it the sign of $\mu$ rather than $\mu$? Second, why $\mu$ can take the two opposite signs?


2 Answers 2


I guess I know the answer now. In the MSSM, once we minimise the scalar potential of the Higgs, we obtain: $$M_z^2 = 2[-|\mu|^2 + \frac{1} {\tan^2\beta - 1 } (m_{H_d}^2 - \tan^2\beta m_{H_u}^2)]$$ For simplicity, in the large $\tan\beta$ limit, this can be written as, $$M_z^2 = -2 (|\mu|^2 + m_{H_u}^2)$$ Where $M_z$ is known from experiments, and the soft-breaking parameter $m_{H_u}$ is determined by the GUT-scale input parameters (universal scalar $m_0$, universal gauigino $m_{1/2}$, and a universal trilinear $A_0$) So there is a constraint on $\mu$ (i.e. for a choice of input parameters, we want $\mu$ to take the value that gives the correct $M_z$). Solving for $\mu$: $$|\mu| = \pm \sqrt{-0.5 M_z^2 - m_{H_u}^2}$$ So, the sign of $\mu$ is free but not the value of $\mu$, and the solution could be positive or negative. The absolute value appearing here is because, in the universality condition where everything is real, $\mu$ must be real.

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    $\begingroup$ you don't honor your nick, you are the kind of member we want on this site, the kind that likes to get his hands dirty with the problems $\endgroup$
    – lurscher
    Commented May 21, 2012 at 16:26
  • $\begingroup$ @stupidity Can you paste the reference from where you were learning this? $\endgroup$
    – Student
    Commented Sep 19, 2012 at 19:13
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    $\begingroup$ @Anirbit I believe I was reading Martin's Susy primer: arxiv.org/abs/hep-ph/9709356. $\endgroup$
    – stupidity
    Commented Sep 19, 2012 at 20:50
  • $\begingroup$ The second equation doesn't make sense, you have a negative Z mass? $\endgroup$
    – user788171
    Commented Sep 19, 2012 at 23:23
  • $\begingroup$ @user788171 $m_{H_u}^2$ is negative at the weak scale, allowing electroweak symmetry breaking radiatively. I suggest that you read Martin's reference that I mentioned earlier. $\endgroup$
    – stupidity
    Commented Sep 20, 2012 at 0:06

That's correct - EWSB imposes $\partial V / \partial H_u = 0$ and $\partial V / \partial H_d = 0$, leading to (see Martin p.91)

$m_{H_u}^2 + |\mu|^2 - b \cot \beta - m_Z^2/2 \cos 2\beta =0$

$m_{H_d}^2 + |\mu|^2 - b \tan \beta + m_Z^2/2 \cos 2\beta =0$

If we input the soft-breaking masses $m_{H_u}$ and $m_{H_d}$, $\tan \beta$, and the $Z$-boson mass $m_Z$, we can solve for $b$, soft-breaking bi-linear, and $|\mu|$, but the sign of $\mu$ remains undetermined.

The CMSSM/mSUGRA uses this convenient parametrization which trades $b$ and $|\mu|$ as inputs for $m_Z$, sgn $\mu$ and $\tan \beta$ as inputs.


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