 Solution

A BPS state saturates a supersymmetry energy bound and is annihilated by some supercharges. For a massive state with one relevant central charge in supersymmetry, normalize the supersymmetry algebra by $\{Q_{\alpha }^A,Q_{\beta }^B\}=2{ϵ}_{\alpha \beta }{ϵ}^{AB}Z$. In its rest frame, appropriate linear combinations of supercharges have anticommutators $2(M+|Z|)$ and $2(M-|Z|)$. Positivity therefore gives the BPS bound in supersymmetry $M\ge |Z|$. At equality, the latter combinations have zero norm and annihilate the state, producing a shortened supermultiplet and preserved supersymmetry. With the alternative convention omitting the factor $2$ in the central-charge anticommutator, the bound reads $M\ge |Z|/2$. The mass is tied to the charge, but this does not imply unconditional stability against decay as parameters vary.