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?    
 A: 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.  
A: 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. 
