Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-48/1/solution
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 48 1 Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-06
Use the minimal four-dimensional Super-Poincaré algebra, with , and commuting translations. Lorentz covariance and closure on the existing supercharges permit onlyThe graded Jacobi identity for therefore givesFor example, with , and , the coefficient is a nonzero multiple of . Independence of the translation generators forces . Its adjoint gives the barred result. Thus every supercharge commutes with four-momentum:Consequently . If has a particular mass-shell condition, the nonzero state has the same one. This gives supersymmetric mass degeneracy within an unbroken physical supermultiplet. The qualification matters: after spontaneous supersymmetry breaking, the chosen vacuum is not annihilated by the supercharges, and its one-particle excitations need not constitute degenerate supermultiplets.
For the O'Raifeartaigh model, write , , . Field phases allow to be positive. The canonical Kähler potential and elimination of the auxiliary fields give the F-term scalar potentialA supersymmetric vacuum would require , hence , while could not vanish. Thus F-flatness is impossible. More precisely, with ,The stated strict inequality makes every nonconstant term nonnegative. The entire classical vacuum family isHere , while . The Weyl spinor is the massless goldstino, and the complex scalar field is a pseudomodulus with two zero tree-level squared masses. The hierarchy separates the massive fields from the breaking scale, but it does not itself select along this flat direction.
At the representative vacuum , the chiral-superfield fermion mass matrix isIts physical masses are its singular values: . The two massive Weyl spinors can be combined into one massive Dirac spinor. For , , the real scalar squared masses areIn particular, there is no tachyon under the stated inequality. The splitting of the masses displays supersymmetry breaking directly.
For completeness, the mass spectrum can be given throughout the classical vacuum family, rather than silently fixing the pseudomodulus. A phase rotation makes real and nonnegative without changing physical masses. The massive fermion squared masses areThe real and imaginary scalar sectors have two-by-two scalar squared-mass matriceswhose eigenvalues areBoth determinants are and their traces are positive, so stability holds at every finite . Quantum corrections can lift a pseudomodulus; these formulas describe the classical, tree-level spectrum requested here.
Define the physical mass supertrace byEach real scalar contributes once and each Weyl spinor twice with a minus sign. At this gives . At general , . Hence the tree-level mass supertrace vanishes despite broken supersymmetry:This is the tree-level supertrace mass sum rule, not a claim that the individual masses are equal.
For direct breaking in the Minimal supersymmetric Standard Model, the difficulty is both field content and the tree-level spectrum. A linear gauge-invariant superpotential term needs a gauge-singlet chiral superfield, which the minimal model does not contain. Moreover, with canonical kinetic terms and purely neutral F-term breaking, without D-term mass shifts, the MSSM tree-level sfermion mass constraint applies separately to conserved charge sectors. For one electron pair it gives : both scalar partners cannot be heavy. This is an illustrative charge-sector consequence under those assumptions, not an unrestricted inference from the total supertrace alone. A separate hidden supersymmetry-breaking sector communicating effective soft supersymmetry breaking through loops or suppressed operators avoids the direct canonical tree-level obstruction.
New to topics? Read the docs here!