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An odd involution $A$ 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 $q$ with $q^2=0$ and $\{q,q^{†}\}=cI$, $c>0$, use $A=(q+q^{†})/\sqrt{c}$. 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 a massive four-dimensional N=1 supersymmetry representation, the two normalized raising supercharges transform as a spin-half doublet. Start with a Clifford vacuum of superspin $j=1/2$. At occupation levels zero, one and two, the ordinary spins areThe two-creation state has a spin-singlet oscillator factor because the creation operators anticommute. Thus the massive N=1 vector multiplet containsThere are four bosonic polarizations, $3+1$, and four fermionic polarizations, $2+2$, as required by boson-fermion degeneracy in a supermultiplet.

In the massless limit the helicity multiplicities become $+1,-1$, two copies each of $+1/2,-1/2$, and two zero-helicity states. They separate into the four irreducible helicity pairsWith the lower-helicity convention for superhelicity, a pair $(\kappa ,\kappa +1/2)$ has labels $\kappa =1/2,0,-1/2,-1$, respectively. A midpoint convention shifts each label by $1/4$; the explicit helicity pairs remove any naming ambiguity.

Pairing under the CPT theorem, the first and last pairs form a massless vector multiplet, and the middle two form a massless chiral multiplet. This is the massless limit of a massive N=1 vector multiplet. The longitudinal vector polarization supplies one of the chiral multiplet's two real scalar modes; the original spin-zero state supplies the other. Simply discarding the longitudinal polarization would lose a physical degree of freedom and give the wrong decomposition.

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.

There are two distinct parts of the hierarchy problem. The first asks why the scale of electroweak symmetry breaking is so much smaller than a fundamental scale such as the Planck mass. The second, the technical hierarchy problem, asks why that separation survives quantum corrections once it is chosen. With a fundamental Higgs boson, a mass term $m_H^2{H}^{†}H$ is allowed by the ordinary gauge symmetry, so heavy thresholds can contribute additive terms of order a heavy mass squared. This is different from the chiral protection of a fermion mass and the gauge protection of a vector mass.

For example, a cutoff description of the Standard Model gives the schematic one-loop sensitivityThe regulator-dependent quadratic term is not itself an observable, but a physical heavy particle of mass $M$ coupled to the Higgs likewise produces a matching correction of order coupling squared times $M^2/(16{\pi }^2)$. Maintaining $|m_H^2|\ll M^2$ then generically requires cancellation between large unrelated terms. This is the naturalness concern; changing the regulator does not eliminate the heavy-threshold issue.

Exact supersymmetry relates bosonic and fermionic couplings and masses. Their loop contributions have opposite signs, and the relations enforce supersymmetric cancellation of quadratic divergences. Boson-fermion degeneracy in a supermultiplet also transfers fermionic mass protection to its scalar partners. It is not sufficient merely to add arbitrary scalars: the supersymmetric coupling relations are what make the cancellation persist. Perturbative superpotential non-renormalization further protects holomorphic mass parameters, although wave-function renormalization still makes canonically normalized couplings run.

For a realistic separation between ordinary particles and their partners, supersymmetry must be broken. Soft supersymmetry breaking introduces masses and selected positive-mass-dimension interactions without reinstating the original ultraviolet quadratic divergence. A paired propagator difference behaves at high Euclidean momentum asso a residual scalar correction is of the orderThis soft scalar-mass sensitivity controls the technical problem if the relevant soft masses and thresholds are not too far above the weak scale. Very heavy partners can still require tuning through finite and logarithmic terms. Supersymmetric cancellations alone therefore do not guarantee a naturally small weak scale for an arbitrary broken spectrum.

The origin of the hierarchy needs further dynamics. In dynamical supersymmetry breaking, an asymptotically free hidden supersymmetry-breaking sector can generate an exponentially smaller scale by dimensional transmutation,If that sector actually has no supersymmetric vacuum, its breaking can be communicated to the visible sector to produce small soft terms. The exponential creates a hierarchy without specifying a tiny dimensionless coupling at the high scale, but the existence of breaking and its mediation are additional model-building requirements. In the MSSM, running Higgs soft masses can trigger electroweak symmetry breaking. The supersymmetric mu problem remains: the allowed superpotential mass $\mu H_u{H}_d$ must also be near the soft scale, and non-renormalization protects a chosen small $\mu$ without explaining its origin.

Extra-dimensional proposals use geometry to explain why observable gravitational and weak scales differ. In large extra dimensions, assume gravity propagates in $4+n$ dimensions with fundamental scale $M_{\ast }$, while ordinary fields are localized on a brane. Integrating the Einstein-Hilbert action over compact volume $V_n$ gives the Planck mass from compactification volume relationHere $M_{\mathrm{Pl}}$ denotes the four-dimensional reduced Planck mass. A sufficiently large internal volume lets $M_{\ast }$ be near the weak scale even though $M_{\mathrm{Pl}}$ is enormous. The Higgs need then be stable only up to the lower fundamental cutoff. This explains the weakness of long-distance gravity geometrically but trades the scale question for the origin and stabilization of a large compact volume. Ordinary matter is localized to avoid an unwanted low-mass tower of its own Kaluza-Klein modes; gravitational modes, departures from four-dimensional gravity and bulk energy loss impose phenomenological requirements.

The Randall–Sundrum model instead uses a warped five-dimensional interval, conventionally the doubled $S^1/{\mathbb{Z}}_2$ orbifold with fixed branes at $0$ and $L$:For a scalar confined to the distant brane, the kinetic and mass terms carry factors $e^{-2kL}$ and $e^{-4kL}$. Canonical rescaling therefore producesA moderately large $kL$ yields an exponentially small observable mass without a comparably large compact-volume factor. The inter-brane size is the radion, so one still needs a mechanism fixing its expectation value. Consistent curvature, brane tensions and ultraviolet physics are also part of the model. Warping is a redshift mechanism; it does not provide the same cancellation of arbitrary scalar loops as supersymmetry.

Finally, protecting a Higgs mass is not the same as solving the vacuum-energy hierarchy. Exact global supersymmetry has zero-energy supersymmetric vacua, but broken supersymmetry generically leaves vacuum energy set by its breaking dynamics. In supergravity, positive auxiliary-field contributions coexist with a negative superpotential term, and obtaining a very small cosmological constant requires further structure or cancellation. Neither soft supersymmetry nor the two geometric mechanisms automatically resolves that problem. The mechanisms can also coexist in one higher-dimensional supersymmetric theory.

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.

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.

Fix a positive-mass unitary irreducible representation at rest, $P^{\mu }=(M,0,0,0)$ with $M>0$. Use the central charge in supersymmetry conventionThe conjugate relation is $\{{\bar{Q}}_{\dot{\alpha }A},{\bar{Q}}_{\dot{\beta }B}\}={ϵ}_{\dot{\alpha }\dot{\beta }}\overline{Z^{AB}}$, with the dotted epsilon tensor the conjugate of the undotted one. The supercharges commute with $P_{\mu }$ and the central charges. Their Lorentz transformation is that of a Weyl spinor; together with the ordinary Poincare algebra, these relations specify the Super-Poincaré algebra. Take ${ϵ}_{12}=1$ and write $Z^{12}=2z$. This makes the normalization of the mass bound explicit: a convention that writes $2{ϵ}_{\alpha \beta }{Z}^{AB}$ instead has $z=Z^{12}$.

When $Z=0$, define four fermionic annihilation operators and their fermionic creation operators byIn the rest frame they obey the canonical anticommutation relations $\{a_i,a_j^{†}\}={\delta }_{ij}$ and $\{a_i,a_j\}=0$. Choose a spin-zero Clifford vacuum annihilated by all four $a_i$. Acting with each distinct creator at most once constructs the fermionic Fock spaceThe five levels have the binomial coefficient multiplicities $1,4,6,4,1$, so the massive supermultiplet contains $2^4=16$ states. The four creators are two copies of the spin-one-half SU(2) representation. Their second exterior power isHere the first factor carries physical spin and the second counts the two supersymmetries. Thus level two contains one spin-one triplet and three spin-zero singlets; levels one and three each contain two spin-one-half doublets. The final spin content isEven levels give eight boson states and odd levels eight fermion states, exhibiting boson-fermion degeneracy in a supermultiplet.

For nonzero $z=|z|e^{i\phi }$, pair the two supersymmetries before normalizing. DefineThe rest-frame algebra gives $\{q_{\alpha },r_{\beta }^{†}\}=2z{\delta }_{\alpha \beta }$ and consequentlyFor $M>|z|$, divide $b^{\pm }$ by $\sqrt{2(M\pm |z|)}$ to recover four normalized oscillators. Positivity of the Hilbert space norm proves the BPS bound in supersymmetry, sinceThereforeFor a BPS state saturating this bound, both $b_{\alpha }^{-}$ and their adjoints annihilate every state of the representation. Four of the eight real supercharges act trivially. Only the two $b^{+}$ oscillators remain, so a scalar Clifford vacuum gives a shortened massive supermultiplet with two spin-zero states and one spin-one-half doublet:This is the irreducible count at fixed central charge. A CPT completion of a supermultiplet can require the charge-conjugate multiplet as well, doubling it to eight states; the full hypermultiplet count must not be confused with this four-state count. Saturation is a representation-theoretic relation between mass and conserved charge: shortening persists while the state remains BPS, although its existence can change across parameter space.

For even $\mathcal{N}=2n$, unitary skew-diagonalization of an antisymmetric matrix brings the central-charge matrix to blocks $2z_r\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, $r=1,\dots ,n$. Repeat the oscillator construction separately for each block. Every block supplies two oscillators with norm $2(M+|z_r|)$ and two with norm $2(M-|z_r|)$, givingIf $k$ blocks saturate, $2k$ complex oscillators disappear, leaving $2\mathcal{N}-2k$. Starting with spin $j_0$, the irreducible state count and preserved fraction areThe largest massive shortening, $k=n$, preserves one-half. The ordinary massive representation has $k=0$. This rest-frame discussion excludes $M=0$, where the massless supermultiplet construction uses a null momentum instead.

Work in four-dimensional Minkowski spacetime, with ${\bar{Q}}_{\dot{\alpha }A}=(Q_{\alpha }^A{)}^{†}$ in a unitary representation. The usual Super-Poincaré algebra has the Poincare algebra and possible internal Lorentz scalars as its even generators, and only the stated Weyl spinors as its odd generators. This closure assumption is essential: Lorentz covariance alone would also permit additional tensor-valued generators. It is the setting selected by the Haag–Łopuszański–Sohnius theorem for ordinary interacting relativistic theories.

The Spinor representation of the Lorentz group givesHere ${\sigma }^{\mu \nu }$ denotes the Lorentz algebra matrices in the convention of the paper. Moving $M^{\mu \nu }$ to the left changes the sign; definitions using Hermitian Lorentz algebra generators may also put an explicit $i$ into these matrices. The conjugate relation acts on the dotted Weyl spinor indices.

The tensor product of group representationsis a Lorentz four-vector. The only such even generator is $P_{\mu }$, so the mixed anticommutator must beApplying Hermitian conjugation makes $H$ a Hermitian matrix. Positivity of $\{S,S^{†}\}$ for every linear combination $S$ of supercharges makes $H$ a positive semidefinite matrix on a positive-energy physical superalgebra representation. Remove any null supercharges, which act trivially, and use a unitary diagonalization of a normal matrix followed by rescaling to obtain $H^A{}_B=2{\delta }^A{}_B$. Thus $\mathcal{N}$ counts the independent nontrivial supercharges.

For equal chirality, the tensor product of group representations isConsequently the tentative equal-chirality anticommutator can contain ${ϵ}_{\alpha \beta }{Z}^{AB}$ and a symmetric spinor tensor $T_{\alpha \beta }^{\mu \nu }{M}_{\mu \nu }{Y}^{AB}$, where $T$ is the intertwining operator for Lorentz covariance onto $(1,0)$. Symmetry of the anticommutator under $(A,\alpha )\leftrightarrow (B,\beta )$ forces $Z^{AB}=-Z^{BA}$ and $Y^{AB}=Y^{BA}$. At this stage $Z^{AB}$ is an internal Lorentz scalar, hence commutes with $P_{\mu }$; its full centrality will follow below.

To determine the commutator with translations, Lorentz covariance and the absence of additional odd generators leave onlyand the relation obtained by Hermitian conjugation. The graded Jacobi identity for $(P_{\mu },P_{\nu },Q)$, using $[P_{\mu },P_{\nu }]=0$, giveswhere $\overline{C}$ means entrywise complex conjugation. This equation alone would not imply $C=0$. Contract the graded Jacobi identitywith ${ϵ}^{\alpha \beta }$. The $M$ term drops out because its spinor tensor is symmetric, and $[P_{\mu },Z^{AB}]=0$ makes the remaining left side vanish. Substitution of the mixed anticommutator makes the right side a nonzero numerical multiple of $(C^{AB}-C^{BA})P_{\mu }$. Thus $C=C^T$. It follows that $\overline{C}=C^{†}$ and $C{C}^{†}=0$, whence $C=0$. Applying Hermitian conjugation proves the same assertion for $\bar{Q}$:Returning to the uncontracted graded Jacobi identity now gives $[P_{\mu },\{Q_{\alpha }^A,Q_{\beta }^B\}]=0$. Since the Lorentz algebra generators do not commute with translations, the tentative $M$ term must have $Y^{AB}=0$. HenceIn particular $Z^{11}=0$ for $\mathcal{N}=1$. The dotted-dotted anticommutator follows by Hermitian conjugation applied to the second relation.

The graded Jacobi identity for $(Q^A,Q^B,{\bar{Q}}_C)$, together with $[P,Q]=0$, gives $[Z^{AB},{\bar{Q}}_C]=0$. As a Lorentz scalar, $Z^{AB}$ commutes with $M_{\mu \nu }$, and it already commutes with $P_{\mu }$. Closure and Lorentz covariance make $[Z^{AB},Q_{\alpha }^C]=D^{AB,C}{}_D{Q}_{\alpha }^D$ for some coefficients. Apply the graded Jacobi identity once more:Thus $D=0$. Since every $Z$ and $Z^{†}$ is itself an anticommutator of supercharges, the $Z$'s also commute with each other and their conjugates. Therefore $Z^{AB}$ are central charges in supersymmetry of the displayed algebra. Additional R-symmetry automorphisms are not among the generators specified here.

For boson-fermion degeneracy in a supermultiplet, fix a physical four-momentum $p$ with $p_0=E>0$, and take the finite internal space of states at that four-momentum. Let $\Gamma =(-1{)}^F$ be fermion parity. It anticommutes with each supercharge and its adjoint. For any fixed $A$, summing the diagonal spinor indices givesbecause ${\sigma }^0=1$ and the Pauli matrices are traceless. Cyclicity of the operator trace and $\Gamma Q=-Q\Gamma$ implyThe supertrace of the preceding summed anticommutator therefore vanishes:This counts physical boson and fermion degrees of freedom, including polarizations. The phrase “any representation” needs a qualification: a one-dimensional even supersymmetric vacuum with $P=Q=Z=0$ has one boson and no fermion. Thus the equality applies to positive-energy physical supermultiplets with finite state counts at fixed four-momentum, not to arbitrary abstract superalgebra representations or unregulated infinite-dimensional operator traces.

Fermion parity is defined byIt anticommutes with $Q_{\alpha }$ and ${\bar{Q}}_{\dot{\alpha }}$. At fixed nonzero energy and momentum, choose a supercharge combination $S$ for which $\{S,S^{†}\}=cI$ with $c>0$. Taking the finite-dimensional trace over one supermultiplet givesbecause cyclicity of the trace and anticommutation of $(-1{)}^F$ with $S$ make the two terms cancel. ThereforeThis is boson-fermion degeneracy in a supermultiplet.

If explicit or soft supersymmetry breaking terms are added, the supercharge is no longer a conserved symmetry of the full Hamiltonian and states need not form representations of the supersymmetry algebra at equal energy. The positive anticommutator cannot be replaced by a constant on a purported multiplet, so the supertrace proof and mass degeneracy fail.