Every positive-energy supermultiplet contains equally many bosonic and fermionic states. A supertrace of a positive supercharge anticommutator proves the equality, provided supersymmetry is exact and the representation is finite-dimensional at fixed momentum.
For a finite physical supermultiplet at fixed positive energy, fermion parity anticommutes with each odd supercharge. Trace cyclicity makes . Summing the Super-Poincaré algebra diagonal entries replaces the anticommutator by , giving equal bosonic and fermionic state counts. A zero-energy vacuum is an exception to the division by , so it need not have a partner. The argument requires an invariant fixed-momentum representation, which is the step lost under explicit supersymmetry breaking.
An odd involution reverses fermion parity and is its own inverse, so it identifies the bosonic and fermionic subspaces. They have equal dimension in a finite-dimensional representation. For a supercharge with and , , use . This gives a direct proof of boson-fermion degeneracy in a supermultiplet at positive energy. A zero-energy supersymmetric vacuum need not have an accompanying fermionic vacuum.
In an unbroken supersymmetric vacuum, physical boson and fermion partners have equal masses because supercharges commute with four-momentum. For one canonical chiral superfield, fluctuations about have mass for both the complex scalar field and Weyl spinor.
For one canonical chiral field at , the quadratic scalar potential is . Writing gives two canonically normalized real scalars with squared mass . The chiral-superfield fermion mass matrix has the same physical mass. A complex bilinear coefficient or its algebraic square is not itself a signed physical squared mass.

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