= Supertrace pairing at positive energy
{title2=$4E(n_B-n_F)=0$}
For a finite physical <supermultiplet> at fixed positive energy, <fermion parity> anticommutes with each odd <supercharge>. Trace cyclicity makes $\operatorname{Tr}[(-1)^F\{Q,Q^\dagger\}]=0$. Summing the <Super-Poincaré algebra> diagonal entries replaces the anticommutator by $4E$, giving equal bosonic and fermionic state counts. A zero-energy vacuum is an exception to the division by $E$, so it need not have a partner. The argument requires an invariant fixed-momentum representation, which is the step lost under explicit supersymmetry breaking.
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