Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 307 / 1 / d

Fermion parity is defined byIt anticommutes with $Q_{\alpha }$ and ${\bar{Q}}_{\dot{\alpha }}$. At fixed nonzero energy and momentum, choose a supercharge combination $S$ for which $\{S,S^{†}\}=cI$ with $c>0$. Taking the finite-dimensional trace over one supermultiplet givesbecause cyclicity of the trace and anticommutation of $(-1{)}^F$ with $S$ make the two terms cancel. ThereforeThis is boson-fermion degeneracy in a supermultiplet.

If explicit or soft supersymmetry breaking terms are added, the supercharge is no longer a conserved symmetry of the full Hamiltonian and states need not form representations of the supersymmetry algebra at equal energy. The positive anticommutator cannot be replaced by a constant on a purported multiplet, so the supertrace proof and mass degeneracy fail.