= Solution
Consider a physical <supermultiplet> at fixed four-momentum with positive energy $E$ and finitely many polarization states. Let $\Pi=(-1)^F$ be <fermion parity>, equal to $+1$ on bosons and $-1$ on fermions. Every <supercharge> is odd, so $\Pi Q_\alpha=-Q_\alpha\Pi$ and likewise for its adjoint.
Trace cyclicity gives, for each component,
$$
\operatorname{Tr}(\Pi Q_\alpha Q_\alpha^\dagger)
=\operatorname{Tr}(Q_\alpha^\dagger\Pi Q_\alpha)
=-\operatorname{Tr}(\Pi Q_\alpha^\dagger Q_\alpha).
$$
Thus the <supertrace pairing at positive energy> yields
$$
0=\sum_{\alpha=1}^2\operatorname{Tr}\bigl(\Pi\{Q_\alpha,Q_\alpha^\dagger\}\bigr)
=\operatorname{Tr}(4E\Pi)=4E(n_B-n_F),
$$
where the middle equality uses $\operatorname{tr}\sigma^0=2$ and $\operatorname{tr}\sigma^i=0$. Since $E>0$,
$$
\boxed{n_B=n_F.}
$$
This proves <boson-fermion degeneracy in a supermultiplet> for both massive and massless particle multiplets, counting physical states rather than unphysical gauge components or auxiliary fields. The positive-energy qualification is essential: a zero-energy <supersymmetric vacuum> can be a one-dimensional bosonic singlet with all supercharges zero. It need not have a fermionic partner, so the printed assertion cannot literally include arbitrary zero-energy representations.
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