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At fixed nonzero four-momentum, take the finite physical supermultiplet and write $P_F=(-1{)}^F$ for fermion parity. Each supercharge is odd, so $P_F{Q}_{\alpha }=-Q_{\alpha }{P}_F$ and the same holds for its adjoint. By the cyclic property of the trace,Consequently $\mathrm{Tr}[P_F\{Q_{\alpha },Q_{\alpha }^{†}\}]=0$. Sum the two spinor components of the $\mathcal{N}=1$ Super-Poincaré algebra. Since ${\mathrm{Tr}}_{2\times 2}{\sigma }^i=0$, their sum is $4H$. On a fixed-energy representation, $H=EI$, yieldingFor a nonzero physical one-particle momentum, $E>0$, including the massless case. Hence the bosonic and fermionic physical state counts agree:Equivalently a nonzero normalized supercharge and its adjoint define an odd involution pairing the two subspaces. This is supertrace pairing at positive energy; it counts spin polarizations, not merely distinct names of particles.

The phrase “every supermultiplet” requires this positive-energy qualification. A zero-energy supersymmetric vacuum with all supercharges acting as zero is a one-dimensional bosonic representation with no fermionic partner. It is a counterexample to the unrestricted statement, because division by $E$ is then impossible. It does not contradict the requested equality for particle supermultiplets.

Consider a physical supermultiplet at fixed four-momentum with positive energy $E$ and finitely many polarization states. Let $\Pi =(-1{)}^F$ be fermion parity, equal to $+1$ on bosons and $-1$ on fermions. Every supercharge is odd, so $\Pi Q_{\alpha }=-Q_{\alpha }\Pi$ and likewise for its adjoint.

Trace cyclicity gives, for each component,Thus the supertrace pairing at positive energy yieldswhere the middle equality uses $\mathrm{tr}{\sigma }^0=2$ and $\mathrm{tr}{\sigma }^i=0$. Since $E>0$,This proves boson-fermion degeneracy in a supermultiplet for both massive and massless particle multiplets, counting physical states rather than unphysical gauge components or auxiliary fields. The positive-energy qualification is essential: a zero-energy supersymmetric vacuum can be a one-dimensional bosonic singlet with all supercharges zero. It need not have a fermionic partner, so the printed assertion cannot literally include arbitrary zero-energy representations.

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.

Use the usual normalization of the Super-Poincaré algebra, with ${\sigma }^{\mu }=(1,\mathbf{\sigma })$ and ${\bar{Q}}_{\dot{\alpha }}=Q_{\alpha }^{†}$ under the corresponding index convention:The supercharges are odd operators: they turn bosonic states into fermionic states and vice versa. If $\Pi =(-1{)}^F$ is fermion parity, this statement is $\Pi Q_{\alpha }{\Pi }^{-1}=-Q_{\alpha }$, soThe same relation holds for the conjugate supercharges.

For a finite-dimensional physical supermultiplet at fixed four-momentum with energy $E>0$, let $n_B,n_F$ count physical bosonic and fermionic states. Cyclicity of the ordinary trace and the parity anticommutation implySumming over the two spinor indices givesThis supertrace pairing at positive energy proves boson-fermion degeneracy in a supermultiplet for massive as well as massless positive-energy representations. It counts on-shell polarizations, not merely the names of fields. The requirement $E>0$ matters: a zero-energy supersymmetric vacuum can be a bosonic singlet without a paired fermionic vacuum.

If supersymmetry-breaking operators are explicitly added to the Lagrangian, the original supercharges generally no longer commute with the full Hamiltonian. They are not conserved symmetries generating finite fixed-energy physical supermultiplets; their original anticommutator does not equal the full translation generator with the breaking terms included. Thus the step replacing the parity-weighted anticommutator by $4E\Pi$ on a closed physical representation fails. The odd parity relation alone does not force energy degeneracy or an equal number of physical states at each mass.

This is explicit versus spontaneous supersymmetry breaking. In spontaneous breaking the action still has conserved supercharges, but the vacuum is not annihilated by them. Acting on particle excitations about that vacuum involves the broken-vacuum/Goldstino sector, so an ordinary finite particle multiplet above an invariant vacuum is no longer the correct pairing argument. The vacuum-energy statements below refer to an exact globally supersymmetric Hamiltonian, including the spontaneously broken case; they are not positivity claims for an arbitrary explicitly broken Hamiltonian.