Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 50 / 3

For global four-dimensional $\mathcal{N}=1$ supersymmetry, useLet $\Pi =(-1{)}^F$ be fermion parity. On a finite one-particle supermultiplet at fixed positive-energy momentum, the supercharges act within the multiplet and $\Pi Q_{\alpha }=-Q_{\alpha }\Pi$. Cyclicity of the finite-dimensional trace givesThus the trace with $\Pi$ of every displayed diagonal anticommutator vanishes. Summing the two components, ${\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}=4E1$, soSince $E>0$, this proves $N_B=N_F$, for both massive and massless one-particle multiplets. The restriction matters: a zero-momentum supersymmetric vacuum singlet, annihilated by every supercharge, need not contain a fermionic partner. The algebra does not imply equal counts for such a trivial zero-energy representation or for an arbitrary unregulated infinite-dimensional trace.

The same Super-Poincaré algebra gives the operator identityFor a normalized physical state $|v⟩$ in a positive-norm Hilbert space,An unbroken supersymmetric vacuum must be annihilated by every supercharge and its adjoint, hence has zero energy. Conversely, zero expectation of this positive sum forces each norm separately to vanish. Therefore a vacuum with strictly positive energy necessarily breaks supersymmetry. For an infinite-volume field theory the corresponding statement is about vacuum energy density, obtained with an appropriate finite-volume regulator. The zero of energy is fixed by the algebra; it cannot be arbitrarily shifted while preserving this identity. This is the global-supersymmetry result and does not assert positivity of the general local-supergravity scalar potential.

For the unheaded extended-algebra request, fix the central-charge normalization explicitly:The central charges are scalar numbers on an irreducible massive representation. A unitary skew-diagonalization of an antisymmetric matrix puts $Z$ into two-by-two blocks $\left(\begin{array}{cc}0& z_r\\ -z_r& 0\end{array}\right)$ and, for odd $\mathcal{N}$, a leftover zero entry. Since this change of supercharge basis is unitary, it preserves the mixed anticommutator. Go to the rest frame $P^{\mu }=(M,0,0,0)$.

For one block set $q_{\alpha }=Q_{\alpha }^1$ and $r_{\alpha }={ϵ}_{\alpha \gamma }(Q_{\gamma }^2{)}^{†}$, with ${ϵ}_{12}=1$. Direct use of the algebra givesWrite $z=|z|e^{i\phi }$ and form the combinations suggested by the hint:ThenEach diagonal anticommutator has nonnegative expectation because it is the sum of squared norms of an operator and its adjoint. Therefore $M-|z_r|\ge 0$ for every block, provingIf the algebra is instead written with $\{Q,Q\}=ϵZ$ without the factor two, the same proof gives $M\ge \frac{1}{2}{\max }_r|Z_r|$. This is the normalization used by some definitions of the BPS bound in supersymmetry; the physical bound is unchanged by the naming of the central charge.

A massive BPS state saturates at least one block bound. At saturation $\{B_{\alpha },B_{\alpha }^{†}\}=0$, so both $B_{\alpha }$ and $B_{\alpha }^{†}$ annihilate every state in its multiplet. The two complex inactive oscillators are four preserved real supersymmetries per saturated block. If $r$ blocks saturate, the preserved fraction is $r/\mathcal{N}$ and only $2\mathcal{N}-2r$ complex fermionic oscillators remain active. For a spin-$j$ Clifford vacuum,compared with $(2j+1)2^{2\mathcal{N}}$ for a long massive multiplet. Thus BPS states have an exact mass-charge relation, preserve part of the supersymmetry, and belong to shortened representations. A physical CPT-invariant spectrum may additionally require the conjugate-charge multiplet. The short representation protects the saturated mass relation while it exists; it does not prohibit a decay at a threshold where central-charge phases align and the sum of daughter masses equals the bound.