= Solution
For spontaneous breaking of exact global <supersymmetry>, at least one <supercharge> does not annihilate the vacuum. Its norm in the preceding expression is positive, hence
$$
\boxed{E_{\mathrm{vac}}>0\quad\text{for a spontaneously broken global supersymmetric vacuum}.}
$$
With canonical <kinetic terms>, this is also seen in the nonnegative <scalar potential>, $V=\sum_i|F_i|^2+\tfrac12\sum_aD_a^2$: nonzero auxiliary expectation values signal breaking and positive energy density. The associated massless fermion is the <Goldstino>.
If “broken” instead means arbitrary explicit breaking by added operators, the exact <Super-Poincaré algebra> no longer fixes the full Hamiltonian, and the positive-norm argument does not constrain its vacuum energy. One can, for example, shift that explicitly broken Hamiltonian by a constant. The strict positivity conclusion therefore uses spontaneous breaking of an otherwise exact global theory, not a blanket assertion about all breaking terms. It also is not a statement about <supergravity>, whose scalar potential contains additional terms.
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