Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 53 / 4 / b / Solution

Let $\Gamma =(-1{)}^F$ be fermion parity. It acts as $+1$ on bosons and $-1$ on fermions, and anticommutes with every odd supercharge. In a finite physical supermultiplet at fixed positive energy $E$, trace cyclicity givesIndeed, cycling $Q_{\alpha }$ in the second term and using $Q_{\alpha }\Gamma =-\Gamma Q_{\alpha }$ cancels the first term. Summing diagonal spinor components of the Super-Poincaré algebra gives ${\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}=4H$, since the Pauli matrices are traceless. Hence supertrace pairing at positive energy yieldsThis is boson-fermion degeneracy in a supermultiplet. Equivalently, a nonzero charge combination with positive anticommutator gives an invertible parity-changing map between the two sectors. The positive-energy qualification is important: a zero-energy supersymmetric vacuum may be a one-state bosonic representation and need not have a fermionic vacuum partner.

For any normalizable state $|v⟩$, the same algebra gives energy positivity in global supersymmetry:It vanishes exactly when every supercharge and its adjoint annihilates the state. In particular, for an existing translation-invariant vacuum, this provesIn infinite volume, interpret the statement as one about vacuum energy density, using finite-volume normalization first. The energy zero is fixed by the supersymmetry algebra; adding an arbitrary constant to the Hamiltonian while leaving that algebra unchanged is not allowed. For a canonical matter and gauge action, the same positivity is seen from $V={\sum }_i|F_i{|}^2+\frac{1}{2}{\sum }_a{D}_a^2$, so a nonzero auxiliary-field expectation breaks the symmetry. This global Minkowski statement does not apply unchanged to the supergravity F-term potential, which can have negative-energy supersymmetric vacua.