The generators for the odd part of the
superalgebra have relations: : \begin{align} \{Q_\alpha^A , \bar{Q}_{\dot{\beta} B} \} & = 2 \sigma_{\alpha \dot{\beta}}^m P_m \delta^A_B\\ \{Q_\alpha^A , Q_\beta^B \} & = 2 \epsilon_{\alpha \beta} \epsilon^{A B} \bar{Z}\\ \{ \bar{Q}_{\dot{\alpha} A} , \bar{Q}_{\dot{\beta} B} \} & = -2 \epsilon_{\dot{\alpha} \dot{\beta}} \epsilon_{AB} Z\\ \end{align} where: \alpha \dot{\beta} are the Lorentz group indices, A and B are
R-symmetry indices. Take linear combinations of the above generators as follows: : \begin{align} R_\alpha^A & = \xi^{-1} Q_\alpha^A + \xi \sigma_{\alpha \dot{\beta}}^0 \bar{Q}^{\dot{\beta} B}\\ T_\alpha^A & = \xi^{-1} Q_\alpha^A - \xi \sigma_{\alpha \dot{\beta}}^0 \bar{Q}^{\dot{\beta} B}\\ \end{align} Consider a state ψ which has 4 momentum (M,0,0,0). Applying the following operator to this state gives: : \begin{align} (R_1^1 + (R_1^1)^\dagger )^2 \psi & = 4 ( M + Re(Z\xi^{2}) ) \psi\\ \end{align} But because this is the square of a Hermitian operator, the right hand side coefficient must be positive for all \xi. In particular the strongest result from this is : \begin{align} M \geq |Z|\\ \end{align} ==Example applications==