= Solution
Sum the diagonal spinor entries of the <supercharge> anticommutator. Since $\operatorname{tr}\sigma^i=0$ and $\operatorname{tr}\sigma^0=2$, the Hamiltonian is
$$
H=\frac14\sum_{\alpha=1}^2\{Q_\alpha,Q_\alpha^\dagger\}.
$$
For a normalized vacuum, <energy positivity in global supersymmetry> follows from
$$
\boxed{E_{\mathrm{vac}}=\frac14\sum_\alpha
\left(\|Q_\alpha|\mathrm{vac}\rangle\|^2+
\|Q_\alpha^\dagger|\mathrm{vac}\rangle\|^2\right)\geq0.}
$$
In the unbroken case, all <supercharges> annihilate the vacuum, giving \b[$E_{\mathrm{vac}}=0$]. Conversely, zero energy forces each nonnegative norm to vanish, so the vacuum is supersymmetric. The algebra fixes the additive zero of energy here. For an infinite homogeneous vacuum, use a finite-volume regulator and interpret the result as its <vacuum energy> density.
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