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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 $\mathcal{N}=1$ Super-Poincaré algebra, obtainingThe expectation in any state is thereforeThis proves energy positivity in global supersymmetry, with the energy origin fixed by the algebra.

If a ground state is invariant under supersymmetry, all its supercharge variations vanish and its energy is zero. Conversely, zero energy makes every nonnegative squared norm above zero, so every supercharge annihilates that vacuum. Thus strictly positive vacuum energy implies supersymmetry breaking: some supercharge fails to annihilate the vacuum. A finite-volume regulator can be used before passing to vacuum-energy density. This is a statement about the global supersymmetry algebra; it is not a positivity theorem for the general supergravity F-term potential.

Sum the diagonal mixed anticommutators in the Super-Poincaré algebra. Their spatial terms have zero Pauli trace, givingFor every normalized state $|u⟩$ in the common operator domain,This proves energy positivity in global supersymmetry. The zero of energy is fixed by this algebra, rather than by an arbitrary independent subtraction.

A supersymmetric vacuum is invariant under every supercharge and its adjoint. Its energy therefore vanishes. Conversely, a normalized vacuum of zero energy makes every nonnegative norm on the right zero, so all supercharges annihilate it. ConsequentlyFor an infinite spatial volume the corresponding statement is about vacuum energy density, obtainable by first regulating the volume. It concerns rigid supersymmetry; the gravitational scalar potential in supergravity has additional terms and is not generally nonnegative.

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 mostly-minus metric signature, $\eta =\mathrm{diag}(1,-1,-1,-1)$. The Weyl sigma matrices are ${\sigma }^{\mu }=(I,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, with the three Pauli matricesThe Pauli matrix multiplication law ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ immediately gives ${\sigma }^{\mu }{\bar{\sigma }}^{\nu }+{\sigma }^{\nu }{\bar{\sigma }}^{\mu }=2{\eta }^{\mu \nu }I$: the $00$ case is $2I$, the mixed cases cancel, and the spatial cases give $-2{\delta }_{ij}I$. The spinor indices of ${\sigma }_{\alpha \dot{\beta }}^{\mu }$ express the correspondence between a Lorentz four-vector and a Hermitian $2\times 2$ matrix. For $X=x^{0}I+x^i{\sigma }_i$, $\det X=(x^0{)}^2-|\mathbf{x}{|}^2$; $X\mapsto MX{M}^{†}$ preserves this determinant for $M\in \mathrm{SL}(2,\mathbb{C})$, yielding the Lorentz spinor double cover.

In four-dimensional N=1 supersymmetry, the odd supercharges $Q_{\alpha }$ and ${\bar{Q}}_{\dot{\alpha }}=Q_{\alpha }^{†}$ transform as conjugate Weyl spinors. Their Super-Poincaré algebra isThe ordinary translation and Lorentz generators satisfy the Poincare algebra, and the Lorentz generators act on the supercharges by the corresponding group representations on Weyl spinors. Tracing the mixed anticommutator gives $P_0=\frac{1}{4}{\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}$, proving energy positivity in global supersymmetry. For a massive rest-frame state, $P_{\mu }=(M,0,0,0)$, the rescaled charges $Q_{\alpha }/\sqrt{2M}$ obey the canonical anticommutation relations. Acting with their adjoints builds a supermultiplet containing states of alternating fermion parity.

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.