 Energy positivity in global supersymmetry

For $\{Q_{\alpha },Q_{\beta }^{†}\}=2{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$, summing diagonal components gives $4H$. The expectation of each anticommutator is $\Vert Q_{\alpha }v{\Vert }^2+\Vert Q_{\alpha }^{†}v{\Vert }^2$, so $H$ is a positive semidefinite operator. A vacuum has zero energy exactly when every supercharge annihilates it. Spontaneously broken global supersymmetry with an existing vacuum therefore gives positive vacuum energy density. This statement fixes the energy zero through the algebra and does not apply to the general supergravity F-term potential.