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Explicit breaking changes the action so that the former supercharges are not conserved symmetries of the full Hamiltonian. Spontaneous breaking preserves the action and algebra but has a vacuum that some supercharge does not annihilate. Energy positivity in global supersymmetry then implies positive vacuum energy density, and a Goldstino accompanies the broken symmetry. That exact-algebra energy argument does not apply to an arbitrarily explicitly broken Hamiltonian. The physical particle spectrum in a broken vacuum need not assemble into the finite equal-mass supermultiplets of an invariant vacuum.

Sum the diagonal spinor entries of the supercharge anticommutator. Since $\mathrm{tr}{\sigma }^i=0$ and $\mathrm{tr}{\sigma }^0=2$, the Hamiltonian isFor a normalized vacuum, energy positivity in global supersymmetry follows fromIn the unbroken case, all supercharges annihilate the vacuum, giving $E_{\mathrm{vac}}=0$. Conversely, zero energy forces each nonnegative norm to vanish, so the vacuum is supersymmetric. The algebra fixes the additive zero of energy here. For an infinite homogeneous vacuum, use a finite-volume regulator and interpret the result as its vacuum energy density.

Every global minimum found for nonzero $\lambda$ has $F_{+}=F_0=F_{-}=D=0$. By energy positivity in global supersymmetry, its zero energy means all supercharges annihilate it. ThereforeThe arbitrary neutral expectation $v$ is a supersymmetric modulus; the nonzero derivatives that would break supersymmetry vanish even though $v$ itself need not vanish. In the exceptional $\lambda =0$ case, every D-flatness minimum likewise has zero F- and D-auxiliaries, so supersymmetry remains unbroken even on the gauge-Higgsed branch. The supersymmetric Higgs mechanism permits internal gauge breaking without supersymmetry breaking. No extra constant or linear superpotential term is needed to find a zero-energy vacuum in this model.

In the conventions $\{Q_{\alpha },Q_{\beta }^{†}\}=2{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$, the trace over the two spinor components of the Super-Poincaré algebra givesHence, on a normalized state in the domain of the supercharges,This proves energy positivity in global supersymmetry. The zero of energy is fixed by the algebra; adding an arbitrary constant to $H$ while keeping that algebra unchanged is not allowed.

For a finite-dimensional supermultiplet at fixed positive energy and four-momentum, choose a nonzero component $q$ with $\{q,q^{†}\}=cI$, $c>0$, and $q^2=0$. The Hermitian odd operatorIt is therefore an invertible map from the even fermion parity subspace to the odd subspace and conversely. This proves boson-fermion degeneracy in a supermultiplet, $n_B=n_F$, for every positive-energy physical particle supermultiplet. Equivalently, the supertrace of the supercharge anticommutator vanishes because odd operators interchange these two subspaces. A zero-energy vacuum is an exception to the literal claim about every possible representation: it can be a one-dimensional bosonic singlet with all supercharges acting trivially.

For a vacuum $|\Omega ⟩$, the sum-of-squares formula gives $E_0=0$ exactly when all supercharges and their adjoints annihilate it. If global supersymmetry is spontaneously broken and a vacuum exists, at least one such norm is nonzero, so the vacuum energy is positive. In infinite volume this is the statement about the vacuum energy density, understood with a finite-volume regulator. Explicit supersymmetry breaking need not preserve the algebra used in the proof. The conclusion also does not extend to the supergravity F-term potential, whose negative $-3e^K|W{|}^2$ term permits, for example, the zero-energy broken vacua of no-scale supergravity.

Supersymmetry extends spacetime symmetry by odd supercharges that interchange boson and fermion states. Their Super-Poincaré algebra pairs positive-energy states into supermultiplets and gives energy positivity in global supersymmetry. Superspace and superfields package the component fields into representations of this symmetry.