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A massive supermultiplet is a unitary irreducible representation of the Super-Poincaré algebra with positive timelike momentum. At rest, its nonzero supercharges become fermionic creation operators and fermionic annihilation operators. Without central charges in supersymmetry, four-dimensional extended supersymmetry has $2\mathcal{N}$ complex oscillators; a spin-$j$ Clifford vacuum gives $(2j+1)2^{2\mathcal{N}}$ states.

A massless supermultiplet is a positive-energy unitary irreducible representation of the Super-Poincaré algebra with null four-momentum. For finite helicity states in four dimensions, half the supercharges act trivially and the others form a fermionic Fock space.

For a normalized state in a unitary representation of the ordinary four-dimensional Super-Poincaré algebra, define $M_{ab}=⟨\{Q_a,Q_b\}⟩/2$ in a real Hermitian charge basis. Then $\Vert u\cdot Q\psi {\Vert }^2=u^{T}Mu$, so its annihilator is precisely the kernel of a linear map represented by the positive matrix $M$. In a suitable real basis its eigenvalues are the two eigenvalues of $⟨P^0⟩I+⟨\mathbf{P}⟩\cdot \mathbf{\sigma }$, each repeated twice. A nonzero kernel therefore has dimension two at nonzero null momentum expectation or four at zero expectation. Complex non-Hermitian lowering operators and algebras with domain-wall extensions are not covered by this argument.

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.

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.

Here a supersymmetry-preserving combination means a real combination of Hermitian Majorana spinor supercharges, in a positive-norm unitary representation of the ordinary Super-Poincaré algebra. This reality qualification is essential. Let $|\psi ⟩$ be normalized and in the charge domains, and define the real symmetric matrixFor real $u$, the Hermitian operator $q_u=u^a{Q}_a$ satisfiesTherefore $M$ is a positive semidefinite matrix, andThe reverse implication uses positivity: diagonalize $M$, so a zero quadratic form has zero component in every positive-eigenvalue direction. This converts the problem into nullity of real N=1 supercharges, not just a condition on the determinant of an operator-valued expression.

In the equivalent Weyl spinor form, $\{q_{\alpha },q_{\beta }^{†}\}=2{\sigma }_{\alpha \dot{\beta }}^m{P}_m$. After a fixed invertible change of real charge basis, $M$ is the realification of the Hermitian two-by-two matrixup to the harmless spatial-sign convention. A complex Hermitian matrix has the same eigenvalues in its realification, each repeated twice. The Pauli matrix multiplication law gives $(\mathbf{p}\cdot \mathbf{\sigma }{)}^2=|\mathbf{p}{|}^{2}I$, so these eigenvalues are $E+|\mathbf{p}|$ and $E-|\mathbf{p}|$, each twice. Positivity forces $E\ge |\mathbf{p}|$. If one nonzero real $u$ annihilates the state, $M$ has a kernel: either $E=|\mathbf{p}|>0$, with exactly two zero eigenvalues, or $E=|\mathbf{p}|=0$, with four. ThusFor a momentum eigenstate these are the null-momentum and zero-momentum cases; the expectation-matrix proof also covers states not assumed to have definite momentum. There are no massive states preserving a real charge in this unextended algebra.

If “linear combination” instead allows arbitrary complex coefficients, the unrestricted claim is false. In a massive fermionic Fock space with two lowering operators $a_1,a_2$ and a spin-half Clifford vacuum spectator, takeThen $a_1\psi =0$. The three vectors $a_2\psi$, $a_2^{†}\psi$ and $a_1^{†}\psi$ are linearly independent, by their occupation numbers and spectator spin labels. The complex annihilator is therefore only one-dimensional. Such a lowering operator is not a Hermitian generator of preserved real supersymmetry. The boxed conclusion applies to the physical real-charge interpretation; it also assumes no domain-wall charge in N=1 supersymmetry.

Spontaneous supersymmetry breaking preserves the action and conserved supercharges, but the vacuum fails to be invariant. This differs from explicit breaking, which changes the symmetry of the action itself. In global four-dimensional N=1 supersymmetry, summing diagonal entries of the Super-Poincaré algebra writes the Hamiltonian as a sum of positive charge anticommutators. Its expectation is a sum of squared norms. Consequently a zero-energy vacuum is supersymmetric, whereas an existing vacuum with broken global supersymmetry has positive energy density. The energy zero is fixed by the algebra. Neither this positivity statement nor the particle-multiplet mass relations should be transferred without qualification to supergravity.

For supersymmetric quantum mechanics, let $H=\{q,q^{†}\}/2$, $q^2=0$, and let fermion parity anticommute with both charges. On a positive-energy eigenspace, $(q+q^{†})/\sqrt{2E}$ is an odd involution: its square is one, it preserves energy, and it reverses parity. Thus every positive-energy bosonic state has a fermionic partner and conversely. Their contributions cancel in the Witten indexFor a discrete spectrum with a well-defined trace, this proves independence of $\beta$. Continuous changes of parameters also preserve the index if the trace stays well defined and no state escapes to infinity or crosses an unregulated continuum threshold. A nonzero index guarantees an unbroken supersymmetric ground state; a zero index is inconclusive. For example, de Rham supersymmetric quantum mechanics on a circle has a bosonic constant zero-form and a fermionic constant one-form, both normalizable zero modes, so it is unbroken with index zero.

A more explicit one-dimensional family illustrates how normalizability determines the answer. DefinePositivity of these Hamiltonian operators shows that a zero mode must solve $A{\psi }_{-}=0$ or $A^{†}{\psi }_{+}=0$. Thus the only candidates areFor $w(x)=x$, the Gaussian minus-sector wavefunction is normalizable and its reciprocal is not, giving an unbroken model with $I=1$ when the minus sector is assigned even parity. For $w(x)=x^2+a$, $a>0$, the exponent $x^3/3+ax$ tends to opposite infinities at the two ends, so neither candidate is normalizable. Both partner potentials $[w^2\mp w^{\prime }]/2$ grow quartically; their spectra are discrete and their lowest energies cannot be zero. The lowest states are therefore paired at positive energy, $I=0$, and supersymmetry is spontaneously broken. This is supersymmetric factorization and zero-mode normalizability. In quantum mechanics the paired ground states illustrate the charge algebra; a relativistic massless-particle conclusion requires field-theoretic hypotheses.

In relativistic global field theory, Goldstone's theorem has a fermionic version. If a conserved supercurrent charge does not annihilate the vacuum, its Ward identity has a nonzero vacuum contact term. Fourier transforming, the divergence of the current correlator cannot tend to zero smoothly at zero momentum. Its spectral representation therefore contains a massless fermionic pole: the Goldstino, with spin one-half. This statement assumes a Poincaré-invariant vacuum and the usual local unitary field-theory conditions. It does not require a bosonic Goldstone particle for the fermionic symmetry. In supergravity the Goldstino can instead supply the longitudinal polarization of a massive gravitino.

For canonical chiral superfields ${\Phi }_i$, elimination of their auxiliary fields givesA supersymmetric vacuum solves every $W_i=0$. At a stationary broken vacuum,The same matrix $W_{ij}$ is the chiral-fermion mass matrix, so the nonzero vector $\overline{W_j}$ is a zero mode. This derives the Goldstino zero mode from vacuum stationarity. The transformation $\delta {\psi }_i=\sqrt{2}ϵF_i+\cdots$ independently identifies that direction through its inhomogeneous shift.

For one canonical chiral field, $W=fZ$, $f\ne 0$, has $U=|f{|}^2$ everywhere, a massless fermion and a classically flat scalar direction. It is flat F-term breaking with a linear superpotential, rather than an isolated vacuum. In contrast, a nonconstant polynomial $W^{\prime }$ of positive degree has a complex root and hence a supersymmetric vacuum; in particular a generic quadratic or cubic single-field superpotential cannot remove all such roots. This argument assumes canonical nonsingular kinetic terms and does not claim that every runaway or nonpolynomial model has a vacuum.

A renormalizable three-field O'Raifeartaigh model gives a less degenerate illustration:The auxiliary equations require $W_2=m{\varphi }_1=0$, then $W_1=hX{\varphi }_1+m{\varphi }_2=0$, leaving $W_X=f\ne 0$. They cannot vanish simultaneously. For real positive $f,h,m$ with $m^2>fh$, the potential obeysTherefore its global tree-level minimum has ${\varphi }_1={\varphi }_2=0$, arbitrary $X$, and energy $f^2$. The massless ${\psi }_X$ is the Goldstino, while $X$ is a pseudomodulus. At $X=0$, the scalar mass-squared spectrum is $0,0,m^2,m^2,m^2+fh,m^2-fh$, and the fermion masses are $0,m,m$. This exhibits both the stability condition and the loss of ordinary supermultiplet mass degeneracy in a broken vacuum. Quantum corrections can lift the pseudomodulus and must be analysed rather than assumed away; the noncompact classical valley also prevents an unqualified finite-volume Witten index trace argument. The index, the vacuum equations and the Goldstino theorem consequently answer complementary questions.

The same Super-Poincaré algebra gives the operator identityFor a normalized physical state $|v⟩$ in a positive-norm Hilbert space,An unbroken supersymmetric vacuum must be annihilated by every supercharge and its adjoint, hence has zero energy. Conversely, zero expectation of this positive sum forces each norm separately to vanish. Therefore a vacuum with strictly positive energy necessarily breaks supersymmetry. For an infinite-volume field theory the corresponding statement is about vacuum energy density, obtained with an appropriate finite-volume regulator. The zero of energy is fixed by the algebra; it cannot be arbitrarily shifted while preserving this identity. This is the global-supersymmetry result and does not assert positivity of the general local-supergravity scalar potential.

Work with metric $\eta =\mathrm{diag}(1,-1,-1,-1)$, ${\sigma }^{\mu }=(I,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, where $\mathbf{\sigma }$ are the Pauli matrices. The supercharges of four-dimensional N=1 supersymmetry form one Weyl spinor $Q_{\alpha }$ and its conjugate ${\bar{Q}}_{\dot{\alpha }}$. Dotted and undotted indices cannot be interchanged; the conjugate charge in the printed mixed relation has a dotted index.

A direct derivation starts from superspace coordinates $x^{\mu },{\theta }^{\alpha },{\bar{\theta }}^{\dot{\alpha }}$, with the four odd coordinates generating a Grassmann algebra. Use left Grassmann derivatives, soOdd derivatives anticommute with one another, and multiplication by distinct odd coordinates also anticommutes. The operators specified by the question can be writtenTheir coefficients do not depend on $x$, hence $[{\mathcal{Q}}_{\alpha },{\mathcal{P}}_{\mu }]=[{\bar{\mathcal{Q}}}_{\dot{\alpha }},{\mathcal{P}}_{\mu }]=0$. Within $\{{\mathcal{Q}}_{\alpha },{\mathcal{Q}}_{\beta }\}$, the two odd derivative terms anticommute; the derivatives of the barred-coordinate coefficients vanish; and the two barred-coordinate multiplication terms cancel because the coordinates anticommute while the spacetime derivatives commute. ThusThis realizes the minimal algebra with no extra tensorial charges.

For the mixed anticommutator, split each operator into its odd derivative and odd-coordinate parts. The derivative/derivative anticommutator is zero. The coordinate/coordinate terms also cancel, since $\theta \bar{\theta }=-\bar{\theta }\theta$. The two remaining pieces areAdding gives the superspace differential realization of N=1 supercharges:In particular the differentiated-coordinate contributions have the same sign, not opposite signs. The graded product rule is the essential step; treating these as ordinary commuting-coordinate derivatives would give an incorrect algebra.

To obtain the Lorentz part, let $x$ transform as a four-vector and $\theta ,\bar{\theta }$ in conjugate spinor representations. DefineThe Pauli matrix multiplication law gives the intertwining identityIt shows that the two terms in ${\mathcal{Q}}_{\alpha }$ transform together as a covariant Weyl spinor: the coordinate derivative transforms in the dual spinor representation, while the barred-coordinate/spacetime-derivative term transforms in that same representation through this identity. Infinitesimally, with $U=\exp [-\frac{i}{2}{\omega }_{\mu \nu }{M}^{\mu \nu }]$, its transformation is $U{Q}_{\alpha }{U}^{-1}=Q_{\alpha }-\frac{1}{2}{\omega }_{\mu \nu }({\Sigma }^{\mu \nu }{)}_{\alpha }{}^{\beta }{Q}_{\beta }$. Comparing with the first-order expansion of $U$ givesIf the paper's spin matrix is denoted ${\sigma }^{\mu \nu }=i{\Sigma }^{\mu \nu }$, this is exactly its first displayed relation. Defining that spin matrix without the factor $i$ instead requires an explicit $i$ in the commutator; this is a generator convention, not a different algebra. The conjugate relation follows by Hermitian conjugation.

Combining the Lorentz transformation with the differential computations gives the Super-Poincaré algebra in the question's notation:The even generators retain the ordinary Poincare algebra. As a closure check, the even superspace variation $\delta =i(ϵ\mathcal{Q}-\overline{ϵ}\bar{\mathcal{Q}})$ gives $\delta \theta =ϵ$, $\delta \bar{\theta }=\overline{ϵ}$ and $\delta x^{\mu }=i\theta {\sigma }^{\mu }\overline{ϵ}-iϵ{\sigma }^{\mu }\bar{\theta }$. Two such variations commute to a translation by $2i({ϵ}_1{\sigma }^{\mu }{\overline{ϵ}}_2-{ϵ}_2{\sigma }^{\mu }{\overline{ϵ}}_1)$, with zero commutator on the odd coordinates, reproducing the same normalization and showing why the anticommutator produces momentum.

In four dimensions, extended supersymmetry has $\mathcal{N}>1$ independent Weyl spinor supercharges, or $4\mathcal{N}$ real generators. With the ordinary particle Super-Poincaré algebra,The central charges in supersymmetry commute with spacetime generators and the supercharges; an R-symmetry can act on their internal indices. Antisymmetry makes a scalar central charge impossible for $\mathcal{N}=1$.

At positive timelike momentum, the rest-frame supercharges act as fermionic oscillators. Without central charges, a spin-$j$ Clifford vacuum generates $(2j+1)2^{2\mathcal{N}}$ states. For a massless supermultiplet, positivity makes half the supercharges act trivially. The remaining $\mathcal{N}$ creation operators generate $2^{\mathcal{N}}$ states with helicity levels $\lambda -k/2$ and multiplicities $\left(\genfrac{}{}{0ex}{}{\mathcal{N}}{k}\right)$. A separate CPT completion of a supermultiplet may be necessary. Keeping all helicities between $-1$ and $1$ gives $\mathcal{N}\le 4$ for a theory without gravity; allowing helicity two gives $\mathcal{N}\le 8$.

A nonzero central charge in supersymmetry changes the oscillator norms. For an $\mathcal{N}=2$ central-charge block, write $Z^{12}=2z$. Suitable combinations of a charge and the adjoint of the other have anticommutators $2(M+|z|)$ and $2(M-|z|)$. Their positivity givesA BPS state saturates this BPS bound in supersymmetry. The combinations with zero anticommutator annihilate it, preserving part of supersymmetry and giving a shortened BPS supermultiplet. In this example a scalar-seeded massive multiplet has four states instead of sixteen, before any required CPT completion. More central-charge blocks allow different fractions of preserved supersymmetry. The mass-charge relation is protected as long as the short state exists; this does not guarantee stability against decay everywhere in parameter space.

The main obstruction to unbroken extended supersymmetry at low energies is the chirality constraint on extended supersymmetry. An $\mathcal{N}=2$ vector multiplet contains an additional adjoint complex scalar. A full hypermultiplet contains two $\mathcal{N}=1$ matter chiral superfields in conjugate gauge representations $R$ and $\overline{R}$. Its charged fermion spectrum is therefore vectorlike. Half-hypermultiplets in pseudoreal representations do not supply genuinely complex chirality. The $\mathcal{N}=4$ gauge multiplet is still more restrictive: it has four adjoint Weyl spinors and six adjoint real scalars. These spectra and their constrained interactions cannot directly reproduce the chiral electroweak matter and arbitrary Yukawa couplings of the Standard Model.

At most unbroken $\mathcal{N}=1$ supersymmetry can directly describe the observed chiral four-dimensional gauge spectrum. Extended supersymmetry can be useful in a higher-energy theory, in an internal sector or before compactification, but it must be reduced or broken to obtain that low-energy spectrum. This is a statement about unbroken symmetry of the relevant sector, not a prohibition on extended supersymmetric ultraviolet descriptions.

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.

Use ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, ${\sigma }^{\mu }=(I,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, where $\mathbf{\sigma }$ are the Pauli matrices. The translation and Lorentz generators obey the Poincare algebraOne can derive the additional relations by realizing the supercharges as differential operators on superspace. With left Grassmann derivatives and $P_{\mu }=i{\partial }_{\mu }$, takeUsing $\{{\partial }_{{\theta }^{\alpha }},{\theta }^{\beta }\}={\delta }_{\alpha }{}^{\beta }$, the two mixed derivative-coordinate terms each give $i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\partial }_{\mu }$. The remaining terms cancel by Grassmann anticommutation. Equal-chirality terms cancel in the same way. HenceAll coefficients are independent of $x$, so direct differentiation gives $[P_{\mu },Q_{\alpha }]=[P_{\mu },{\bar{Q}}_{\dot{\alpha }}]=0$. This also shows that supercharges commute with translations and preserve the four-momentum of a state.

For the Lorentz generators, define the left-handed spinor matricesThe identity ${\sigma }^{\mu }{\bar{\sigma }}^{\nu }+{\sigma }^{\nu }{\bar{\sigma }}^{\mu }=2{\eta }^{\mu \nu }I$ gives their Lorentz commutation relations. Acting on the spinor superspace coordinates and their dual derivatives then gives, in a column-component convention,Here dotted index values are identified with the conjugate undotted values, and ${\bar{Q}}_{\dot{\alpha }}=Q_{\alpha }^{†}$. The second formula also follows directly by taking the adjoint of the first, using Hermitian $M_{\mu \nu }$. For example, $S_{12}={\sigma }^3/2$, so $[J_3,Q_1]=-Q_1/2$ and $[J_3,Q_2]=Q_2/2$ for $J_3=M_{12}$. This explicit component convention avoids reversing signs by mixing state matrices with operator-component matrices. Covariance of the mixed anticommutator follows by applying these matrices to ${\sigma }^{\rho }{P}_{\rho }$.

These are the minimal Super-Poincaré algebra relations. In the ordinary particle algebra a possible scalar term in $\{Q_{\alpha },Q_{\beta }\}$ would be proportional to ${ϵ}_{\alpha \beta }$, but an anticommutator is symmetric in its two labels; with only one supersymmetry such a central term vanishes. Extended algebras can instead use an antisymmetric internal index tensor. Tensor charges associated with extended objects are not part of this minimal particle algebra.

For a positive-energy massless supermultiplet, choose $p^{\mu }=(E,0,0,E)$ with $E>0$, so $p_{\mu }=(E,0,0,-E)$ andPositivity of the Hilbert-space norm implies that both $Q_1$ and $Q_1^{†}$ annihilate every state: their squared norms sum to zero. SetFor momentum along the positive third axis, helicity is $h=J_3$. The Lorentz commutators give $[h,a]=a/2$ and $[h,a^{†}]=-a^{†}/2$. Choose a highest-helicity Clifford vacuum $|\lambda ⟩$ with $a|\lambda ⟩=0$. Its partner $a^{†}|\lambda ⟩$ has unit norm if the first state does and has helicity $\lambda -1/2$. No further state is generated because $(a^{†}{)}^2=0$. ThereforeThe two states have opposite fermion parity. A CPT completion of a supermultiplet adds helicities $(-\lambda ,-\lambda +1/2)$ and conjugate internal charges when necessary. Taking $\lambda =1/2$ produces a complex scalar and a Weyl spinor after CPT completion; taking $\lambda =1$ produces the massless vector multiplet. This construction assumes ordinary finite-helicity particle representations, rather than continuous-spin representations.

Use four-dimensional Minkowski signature $(+---)$, ${\sigma }^{\mu }=(I,{\sigma }^i)$ and ${\bar{Q}}_{\dot{\alpha }}=(Q_{\alpha }{)}^{†}$. The odd supercharges transform as a Weyl spinor and its conjugate. In the ordinary particle Super-Poincaré algebra, Lorentz covariance makes the mixed anticommutator a vector, since $(1/2,0)\otimes (0,1/2)=(1/2,1/2)$. The even vector generator is the translation generator $P_{\mu }$, so the anticommutator is a real normalization constant times ${\sigma }^{\mu }{P}_{\mu }$. Unitarity makes that constant positive, and rescaling the charges sets it to two.

The same-chirality anticommutator is symmetric in its two spinor indices. A Lorentz-scalar term would require ${ϵ}_{\alpha \beta }$ and hence an antisymmetric internal charge label, unavailable for $\mathcal{N}=1$. A term in the Lorentz generators would violate the translation Jacobi identity. Excluding additional tensorial brane charges, as in the ordinary particle algebra, the same-chirality anticommutators therefore vanish. Lorentz covariance could initially allow $[P_{\mu },Q_{\alpha }]=a{\sigma }_{\mu \alpha \dot{\gamma }}{\bar{Q}}^{\dot{\gamma }}$. The graded Jacobi identity with $P_{\mu },Q_{\alpha },Q_{\beta }$, using their zero same-chirality anticommutator, then forces $a=0$. Thus the part of the Super-Poincaré algebra involving the odd generators isThe last identity means that a supercharge preserves momentum and mass.

The remaining odd-generator commutators express their spinor transformation under rotations and boosts. One consistent lower-index component convention isThe barred formulas follow by Hermitian conjugation. Component coefficient matrices act in the dual representation; the signs therefore must be kept consistent, as in the Jacobi sign test for supercharge components.

One can also derive and check the odd brackets directly in superspace. With left Grassmann derivatives, takeThe two cross terms in the mixed anticommutator each give $-i{\sigma }^{\mu }{\partial }_{\mu }$, totaling $2{\sigma }^{\mu }{P}_{\mu }$. Equal-chirality terms vanish because the Grassmann derivatives anticommute and their coordinate dependence is of opposite chirality. All coefficients are independent of $x$, so the charges commute with translations. This realizes the displayed algebra rather than merely postulating its brackets.

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.

Choose a positive-energy massless momentum $P^{\mu }=(E,0,0,E)$. Then ${\sigma }^{\mu }{P}_{\mu }=E(I-{\sigma }^3)$, so the Super-Poincaré algebra givesUnitarity forces $Q_1$ and its adjoint to vanish on this representation: the sum of their squared state norms is zero. Set $a=Q_2/\sqrt{4E}$. The surviving operators obeyThe helicity operator is $h=J_3$ for this momentum. In the component convention above, $[h,a]=a/2$ and $[h,a^{†}]=-a^{†}/2$. A highest-helicity state $|\lambda ⟩$ with $a|\lambda ⟩=0$ therefore generates exactly two states,one bosonic and one fermionic. This is the $\mathcal{N}=1$ helicity spectrum of a massless supermultiplet. The CPT theorem reverses helicity and conjugates internal charges, so CPT completion of a supermultiplet adds $(-\lambda ,-\lambda +1/2)$ when that pair is not already present.

For the usual integer and half-integer helicities up to spin two, the CPT-complete contents areThe chiral superfield describes one complex scalar and one Weyl fermion; the vector multiplet has a massless gauge vector and its gaugino. The gravitino multiplet contains a spin-one vector and a spin-three-halves field, while the supergravity multiplet contains the graviton and gravitino. All have two physical bosonic and two physical fermionic helicity states after CPT completion. The general two-state construction also describes higher-helicity formal multiplets; their existence as representations alone does not establish a consistent interacting higher-spin theory.

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.

A BPS state is a state whose mass saturates a lower bound set by conserved charges in the Super-Poincaré algebra. Saturation leaves some linear combinations of supercharges unbroken and produces a shortened massive supermultiplet. Both statements follow from the positivity of the physical Hilbert-space norm, rather than from an assumed classical field configuration.

For four-dimensional extended supersymmetry, choose the normalizationThe conjugate algebra is obtained by Hermitian conjugation. The antisymmetry of the central charge in supersymmetry matrix follows because the anticommutator is symmetric in the combined labels $(\alpha ,A)$ while ${ϵ}_{\alpha \beta }$ is antisymmetric. These central charges commute with momentum and supercharges, so their eigenvalues are fixed in an irreducible massive representation. In the rest frame, $P_{\mu }=(M,0,0,0)$ and $\{Q_{\alpha }^A,(Q_{\beta }^B{)}^{†}\}=2M{\delta }^{AB}{\delta }_{\alpha \beta }$.

For any operator $S$ and physical state $|v⟩$,Apply this to $S_{\alpha }^A=Q_{\alpha }^A-{ϵ}_{\alpha \beta }{U}^{AB}(Q_{\beta }^B{)}^{†}$, where $U$ is unitary. This is the rest-frame version of the positive operator constructed from $Q-\Gamma$. Summing over spinor and internal labels givesThe two diagonal terms each contribute $4\mathcal{N}M$, because $U^{†}U=I$; each mixed term contains ${\sum }_{\alpha \beta }{ϵ}_{\alpha \beta }^2=2$. For $\mathcal{N}=2$, write $Z^{12}=|Z|e^{i\phi }$ and choose $U=e^{i\phi }\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Then $\mathcal{H}=16M-8|Z|$, so $M\ge |Z|/2$. The factor of one-half reflects the stated convention with $Z$, rather than $2Z$, in the equal-chirality anticommutator.

To obtain the sharp general bound and exhibit the shortening, use unitary skew-diagonalization of an antisymmetric matrix to put $Z$ into two-by-two blocks $Z_r\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, with a possible zero row and column for odd $\mathcal{N}$. In one block, relabel its two internal indices $1,2$ and write $Z_r=|Z_r|e^{i{\phi }_r}$. DefineDirect substitution, using ${ϵ}_{12}=1$, givesand all annihilator-annihilator anticommutators vanish. The minus operators must have nonnegative norm in every block, hence the BPS bound in supersymmetry isEquivalently, defining $z_r=Z_r/2$ gives $M\ge {\max }_r|z_r|$. Applying the positive $Q-\Gamma$ construction to each block gives the same bound; applying only its full trace need not isolate the largest block. For $M>0$ and no central charge saturation, each operator can be normalized by $\sqrt{2M\pm |Z_r|}$ to an ordinary fermionic annihilation operator. The unmatched internal index, if any, contributes two more such operators.