# mSUGRA boundary conditions and the MSSM

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?

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.

• 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 May 21, 2012 at 16:26
• @stupidity Can you paste the reference from where you were learning this? Sep 19, 2012 at 19:13
• @Anirbit I believe I was reading Martin's Susy primer: arxiv.org/abs/hep-ph/9709356. Sep 19, 2012 at 20:50
• The second equation doesn't make sense, you have a negative Z mass? Sep 19, 2012 at 23:23
• @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. 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.