 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, henceWith canonical kinetic terms, this is also seen in the nonnegative scalar potential, $V={\sum }_i|F_i{|}^2+\frac{1}{2}{\sum }_a{D}_a^2$: 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.