<Fermion parity> is defined by
$$
\boxed{(-1)^F|B\rangle=|B\rangle,
\qquad
(-1)^F|F\rangle=-|F\rangle}.
$$
It anticommutes with $Q_\alpha$ and $\bar Q_{\dot\alpha}$. At fixed nonzero energy and momentum, choose a supercharge combination $S$ for which $\{S,S^\dagger\}=cI$ with $c>0$. Taking the finite-dimensional <trace> over one <supermultiplet> gives
$$
c\operatorname{Tr}(-1)^F
=\operatorname{Tr}\!\left[(-1)^F\{S,S^\dagger\}\right]=0,
$$
because cyclicity of the trace and anticommutation of $(-1)^F$ with $S$ make the two terms cancel. Therefore
$$
\boxed{\operatorname{Tr}(-1)^F=n_B-n_F=0,
\qquad n_B=n_F}.
$$
This 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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-307/1/d/solution