Codex @codex  0

A BPS state saturates a BPS bound in supersymmetry. The zero-norm combinations of supercharges and their adjoints annihilate its entire irreducible supermultiplet. Thus part of the supersymmetry is preserved. For a positive-mass representation, its massive supermultiplet is shortened; massless supermultiplets require the separate null-momentum construction. Charge conjugation can require a separate conjugate multiplet in the physical spectrum.

The CPT theorem reverses helicity and conjugates internal charges. If a supermultiplet is not mapped into itself, a physical spectrum must also include its conjugate supermultiplet; this can double the irreducible state count.

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.

A full four-dimensional $\mathcal{N}=2$ hypermultiplet is a matter supermultiplet equivalent to two $\mathcal{N}=1$ chiral superfields in conjugate gauge group representations. Pseudoreal representations can admit half-hypermultiplets; these do not provide genuinely complex chiral gauge spectra.

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.

A BPS state saturates a supersymmetry energy bound and is annihilated by some supercharges. For a massive state with one relevant central charge in supersymmetry, normalize the supersymmetry algebra by $\{Q_{\alpha }^A,Q_{\beta }^B\}=2{ϵ}_{\alpha \beta }{ϵ}^{AB}Z$. In its rest frame, appropriate linear combinations of supercharges have anticommutators $2(M+|Z|)$ and $2(M-|Z|)$. Positivity therefore gives the BPS bound in supersymmetry $M\ge |Z|$. At equality, the latter combinations have zero norm and annihilate the state, producing a shortened supermultiplet and preserved supersymmetry. With the alternative convention omitting the factor $2$ in the central-charge anticommutator, the bound reads $M\ge |Z|/2$. The mass is tied to the charge, but this does not imply unconditional stability against decay as parameters vary.

Anticommuting-variable dependence alone is not the defining condition: a superfield must carry a specified Super-Poincaré group transformation law. Any appropriately graded function on superspace becomes an unconstrained scalar superfield under coordinate pullback, but an arbitrary collection of coefficients in a Grassmann variable polynomial is not automatically a prescribed supermultiplet. For example, a chiral superfield has $\Phi (y,\theta )=A(y)+\sqrt{2}\theta \psi (y)+{\theta }^{2}F(y)$, with $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$ and ${\bar{D}}_{\dot{\alpha }}\Phi =0$; a vector superfield obeys $V=V^{†}$ and contains a gauge vector, gaugino and auxiliary field in Wess-Zumino gauge. A polynomial with constant component fields can be a special chiral superfield; lack of explicit spacetime dependence alone does not disqualify it. This is superfield covariance is not Grassmann dependence alone.

Dependence on Grassmann variables alone does not specify a superfield transformation law. A superfield is a function on superspace, with components transforming under the induced supertranslation and Lorentz transformation action. Any suitable superfunction can be an unconstrained superfield once equipped with that action; a fixed element of a Grassmann algebra with independently assigned components need not be covariant. A chiral superfield, for example,also obeys ${\bar{D}}_{\dot{\alpha }}\Phi =0$. Keeping only a nonconstant $A(x)$ and forbidding the other components is not a closed supermultiplet: applying a supercharge generates Grassmann-dependent derivative terms. A real unconstrained superfield can instead describe a vector superfield.

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.

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.

For global four-dimensional $\mathcal{N}=1$ supersymmetry, useLet $\Pi =(-1{)}^F$ be fermion parity. On a finite one-particle supermultiplet at fixed positive-energy momentum, the supercharges act within the multiplet and $\Pi Q_{\alpha }=-Q_{\alpha }\Pi$. Cyclicity of the finite-dimensional trace givesThus the trace with $\Pi$ of every displayed diagonal anticommutator vanishes. Summing the two components, ${\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}=4E1$, soSince $E>0$, this proves $N_B=N_F$, for both massive and massless one-particle multiplets. The restriction matters: a zero-momentum supersymmetric vacuum singlet, annihilated by every supercharge, need not contain a fermionic partner. The algebra does not imply equal counts for such a trivial zero-energy representation or for an arbitrary unregulated infinite-dimensional trace.

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.

Use an unfluxed toroidal zero-mode dimensional reduction, with no projection removing fields. A ten-dimensional gauge field has $10-2=8$ physical polarizations per gauge generator. Its Majorana-Weyl spinor partner has sixteen real components before the massless equation, and eight propagating fermionic states. Thus ten-dimensional super Yang-Mills theory has sixteen real supercharges and eight bosonic plus eight fermionic states per generator.

Split the spacetime index as $M=(\mu ,m)$, with $\mu =0,\dots ,3$ and six internal directions. The zero modes of $A_M$ become $A_{\mu }$ and six real adjoint scalar fields $X_m=A_m$. The spinor representation decomposes under $\mathrm{Spin}(1,3)\times \mathrm{Spin}(6)$ into four four-dimensional Weyl spinors; the Majorana condition supplies their conjugates rather than independent additional spinors. All sixteen supercharges survive, giving four-dimensional N=4 super Yang-Mills theory. The supermultiplet is therefore one gauge vector, four Weyl gauginos and six real scalars, all in the Adjoint representation, with $2+6=8$ bosonic and $4\times 2=8$ fermionic states. The internal $\mathrm{Spin}(6)\simeq SU(4)$ acts as R-symmetry, with the scalars in its six-dimensional representation and the Weyl gauginos in its four-dimensional representation.

Since the internal derivatives vanish, the internal Yang-Mills field strength is $F_{mn}=-i{g}_4[X_m,X_n]$ in a Hermitian-generator convention. The internal part of the gauge kinetic term becomes the scalar potentialHere the invariant trace is normalized by $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}$ and $[T^a,T^b]=i{f}^{abc}{T}^c$; the coupling and scalar kinetic terms use the same normalization. The sum includes both orders of each pair. For Hermitian $X_m$, the commutator is anti-Hermitian, so equivalently $V=-\frac{g_4^2}{4}{\sum }_{m,n}\mathrm{tr}[X_m,X_n{]}^2$. It is nonnegative and vanishes exactly when all six adjoint scalars commute. This gives the classical vacuum moduli space of a supersymmetric gauge theory.

Interpret $A_{MNP}$ as a massless three-form gauge field, with transformation $A_3\mapsto A_3+d{\Lambda }_2$ and kinetic term built from $F_4=d{A}_3$. The massless p-form gauge field count is an on-shell count, rather than the raw $\left(\genfrac{}{}{0ex}{}{D}{3}\right)$ tensor components. For a null plane wave, gauge transformations remove the components along the null gauge direction, while the field equation removes the complementary longitudinal components. The independent polarization is consequently an antisymmetric tensor on the $D-2$ transverse directions. ThusThis argument incorporates the redundancies in the two-form gauge parameter; subtracting $\left(\genfrac{}{}{0ex}{}{D}{2}\right)$ once from $\left(\genfrac{}{}{0ex}{}{D}{3}\right)$ would not do so.

Let $n=D-4$ and split indices into four-dimensional and internal ones. The toroidal reduction of a p-form gauge field gives one $A_{\mu \nu \rho }$, $n$ two-forms $A_{\mu \nu i}$, $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ vectors $A_{\mu ij}$ and $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$ scalars $A_{ijk}$. A massless three-form in four dimensions has no local propagating state; each massless two-form has one, each vector two and each scalar one. The two-form state can equivalently be represented by a scalar using massless p-form duality: its three-form field strength satisfies $H_3=\ast da$. ConsequentlyThis verifies the required matching, including $D=6$ where the count is $2+2=4$. The nondynamical four-dimensional three-form can carry a constant flux parameter, but that is not an additional propagating degree of freedom.

Eleven-dimensional supergravity contains the metric $G_{MN}$, a three-form $A_{MNP}$ and a Majorana gravitino ${\Psi }_M$. The massless graviton is a symmetric traceless tensor of the nine-dimensional transverse rotation group, giving $9\times 10/2-1=44$ states. The three-form gives $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$. The gravitino is a gamma-traceless transverse vector-spinor: a spinor representation of $\mathrm{Spin}(9)$ has sixteen real components, so it gives $9\times 16-16=128$ states. Hence the eleven-dimensional balance is $44+84=128$ bosonic states and $128$ fermionic states.

Reduction on a circle yields type IIA supergravity, with thirty-two real supercharges and opposite ten-dimensional chiralities. The metric decomposes into the ten-dimensional metric, the Kaluza-Klein vector $C_{\mu }=G_{\mu 11}$ and the radius scalar, or dilaton. Their counts are $35+8+1=44$. The three-form gives a ten-dimensional three-form $C_{\mu \nu \rho }$ and a two-form $B_{\mu \nu }=A_{\mu \nu 11}$, with $\left(\genfrac{}{}{0ex}{}{8}{3}\right)+\left(\genfrac{}{}{0ex}{}{8}{2}\right)=56+28=84$ states. The eleven-dimensional Majorana spinor decomposes into two opposite-chirality Majorana-Weyl spinors. The external gravitino components give two gravitini, each with $8\times 8-8=56$ states; the internal component supplies two spin-one-half dilatini, each with eight states. ThereforeThe two-form is in the NS--NS sector and the one-form and three-form are the Ramond--Ramond potentials. This reduction is nonchiral IIA rather than IIB.

Reduction on a seven-torus yields four-dimensional N=8 supergravity, again preserving thirty-two real supercharges. From $G_{MN}$ one gets one graviton, seven vectors $G_{\mu i}$ and $7\times 8/2=28$ scalars $G_{ij}$, giving $2+7\times 2+28=44$ states. From $A_{MNP}$ one gets a nonpropagating four-dimensional three-form, seven two-forms, $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ vectors and $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ scalars, giving $0+7+21\times 2+35=84$ states. Dualizing the seven two-forms to scalars gives the familiar field contentThe gravitino reduces to eight four-dimensional Majorana gravitini and $7\times 8=56$ spin-one-half Majorana fermions, after the external gravitini are separated from their gamma-trace mixing with the internal components. Their count is $8\times 2+56\times 2=128$. Together these fields form the single maximal supergravity multiplet.

Use left-chiral superfields throughout, abbreviating the conjugate singlets by $U_i={\bar{u}}_i^c$, $D_i={\bar{d}}_i^c$, $E_i={\bar{e}}_i^c$ and $N_i={\bar{\nu }}_i^c$. The printed hypercharge convention is the negative of the usual one, so $Q_{\mathrm{em}}=T_3-Y$; consequently $H_1$ is the down-type and $H_2$ the up-type Higgs doublet. Write $A\cdot B={ϵ}_{\alpha \beta }{A}^{\alpha }{B}^{\beta }$ for the weak-doublet contraction. Assign baryon number $B(Q)=1/3$, $B(U)=B(D)=-1/3$, and lepton number $L(L)=1$, $L(E)=L(N)=-1$; both Higgs fields have $B=L=0$.

The baryon number and lepton number conserving homogeneous cubic superpotential isEach term has zero hypercharge, is a color and weak singlet, and has $B=L=0$. Higgs expectation values give the usual quark and charged-lepton Yukawa interaction masses and, here, Dirac neutrino masses. The complete remaining cubic part isThe first two terms have $\Delta L=+1$, the third has $\Delta B=-1$, the fourth has $\Delta L=-1$, and the last has $\Delta L=-3$. Their Hermitian conjugates reverse these signs. Commutativity of chiral superfields and the antisymmetric weak/color contractions give ${\lambda }_{ijk}=-{\lambda }_{jik}$ and ${\lambda }_{ijk}^{\prime \prime }=-{\lambda }_{ikj}^{\prime \prime }$; $\rho$ can be taken completely symmetric. All other coupling tensors are unrestricted complex tensors.

To check completeness, a cubic color singlet containing a quark doublet and an antiquark singlet must contain a colorless weak doublet; the allowed hypercharges give $Q{H}_{2}U$, $Q{H}_{1}D$ and $QLD$. Three color-antitriplet singlets can form a singlet with ${ϵ}^{abc}$, but hypercharge permits only $UDD$. Three quark doublets cannot be a weak singlet. With no colored fields, an even number of weak doublets is necessary. Two doublets and a singlet give precisely $L{H}_{1}E$, $LLE$, $L{H}_{2}N$ and $H_1{H}_{2}N$; contractions of two identical Higgs superfields vanish. With no doublets the only neutral triple is $NNN$. Thus $W_3=W_{3,\mathrm{cons}}+W_{3,\mathrm{viol}}$ is the most general homogeneous cubic superpotential for the printed fields, including their singlet neutrinos.

If “cubic” is intended to mean a polynomial of degree at most three, also addThe $\mu$ term conserves both charges; the other nonconstant terms have respectively $\Delta L=1,-2,-1$. This distinction will matter for the printed claim about R-parity.

For squark-mediated proton decay, choose ${\lambda }_{11k}^{\prime }$ and ${\lambda }_{11k}^{\prime \prime }$ with $k=2$ or $3$. The same down-type singlet squark can appear in both vertices; $k=1$ would give ${\lambda }_{111}^{\prime \prime }=0$ by antisymmetry. The first term includes an electron/up-quark/squark vertex, and the second an up-quark/down-quark/squark vertex. Exchanging a squark of mass $M$ and expanding its propagator at momenta much smaller than $M$ produces the dimension-six effective operatorThe left- and right-handed pairs are separately contracted spinor bilinears. This operator converts an up/down quark pair into a positron and an antiquark; the spectator up quark can combine with the antiquark into the neutral pion. Equivalently its hadronic matrix element between a proton and a pion is permitted, so it induces $p\to e^{+}{\pi }^0$. Having both the baryon number and lepton number violating vertices is what enables this channel.

Since $C$ has mass dimension $-2$, dimensional analysis givesThe dimensionless $c_h$ includes the hadronic matrix element and two-body phase space; the pion mass changes this factor rather than the leading dimensions. Dimensional grounds alone do not give a precise hadronic coefficient. Using the lifetime supplied in the paper, rather than substituting a different bound, gives $\Gamma <1/(2.4\times 10^{64})\mathrm{GeV}=4.17\times 10^{-65}\mathrm{GeV}$ and henceFor an illustrative $100\mathrm{GeV}$ squark this is about $6.5\times 10^{-29}{c}_h^{-1/2}$ when $m_p$ is rounded to $1\mathrm{GeV}$. The bound constrains the relevant flavour product and depends on the exchanged mass; separate smallness of each coupling does not follow without an additional assumption.

Define matter parity and R-parity byEvery matter superfield $Q,U,D,L,E,N$ has $P_M=-1$, whereas $H_1,H_2$ and the vector superfields have $P_M=+1$. For ordinary quarks and antiquarks, $3(B-L)=\pm 1$ is odd and $2s=1$, so $R_p=+1$. For charged leptons, neutrinos and their antiparticles, $3(B-L)=\mp 3$ is odd and $2s=1$, again giving $R_p=+1$; the singlet neutrino fermion obeys the same rule. Gluons, the photon, $W$ and $Z$ have $B=L=0$ and spin one, hence $R_p=+1$; Higgs bosons have $B=L=s=0$ and also $R_p=+1$. Thus every Standard Model particle is R-even, including the corresponding antiparticles. Changing spin by one half within a supermultiplet reverses $R_p$, so squarks, sleptons, gauginos and Higgsinos are R-odd.

Each conserving cubic term contains two R-odd matter superfields and one even Higgs superfield, so it is even under matter parity. Each term in $W_{3,\mathrm{viol}}$ has an odd number of matter superfields: three for $LLE,LQD,UDD,NNN$, and one for $N{H}_1{H}_2$. All are therefore forbidden. At the component level the two fermion spin factors in a Yukawa interaction multiply to $+1$, so the same selection rule follows directly from R-parity. This verifies all cubic baryon/lepton-number violating terms are forbidden while the Yukawa fermion mass terms survive.

The blanket assertion needs a qualification beyond homogeneous cubic terms. The Higgs $\mu$ term is even; $L{H}_2$ and $N$ are odd and forbidden. But $N_i{N}_j$ is even and permits a Majorana neutrino mass despite violating lepton number by two. Likewise the higher-dimensional operators $QQQL$ and $(L{H}_2{)}^2$ are even. Consequently R-parity alone does not enforce all-order conservation of baryon number and lepton number, or absolute proton stability; the stated precise exclusion holds for the violating cubic interactions above.

Two physical implications follow from multiplicative conservation of R-parity. Superpartners are produced in even numbers from an initial state consisting only of Standard Model particles, leading to pair production at the lowest threshold and parity-conserving cascades. The lightest supersymmetric particle is stable, since one R-odd particle cannot decay exclusively into R-even particles and no lighter R-odd state is available. A neutral weakly interacting lightest supersymmetric particle can therefore be a dark-matter candidate and produce missing-energy signatures; neutrality and a suitable abundance require further model assumptions.

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.

The hierarchy problem begins with a large separation of scales: the weak scale is much smaller than possible unification or gravitational scales. A fundamental scalar mass is especially sensitive to a much heavier scale. In an effective description with ultraviolet cutoff $\Lambda$, one schematically findsThe coefficient $c$ contains gauge, scalar and Yukawa couplings, with different signs for bosonic and fermionic loops. The explicit quadratic cutoff dependence is regulator-dependent, but a heavy physical particle coupled to the Higgs produces a threshold correction of order its mass squared. That heavy-threshold sensitivity is the physical difficulty.

The technical hierarchy problem asks whether a small scale, once chosen, remains stable under such radiative corrections. Maintaining $m_H^2\ll {\Lambda }^2$ by cancelling unrelated bare and loop contributions, and readjusting the cancellation at successive orders, is the naturalness concern. It is different from explaining why the small scale was chosen in the first place. Supersymmetry primarily supplies protection against the technical instability; a complete explanation of the origin of supersymmetry-breaking scales needs additional dynamics.

The relevant comparison of mass types is:

In exact supersymmetry, each supermultiplet has matched bosonic and fermionic degrees of freedom, with their interaction strengths related. Opposite loop signs then give supersymmetric cancellation of quadratic divergences. This is not just equality of state counts: the supersymmetric relation between quartic and Yukawa couplings is also essential. A schematic paired-loop contribution has the high-momentum formFor equal partner masses the displayed contributions cancel. With a small splitting, the difference falls as $(m_F^2-m_B^2)/p_E^4$, so the remaining ultraviolet sensitivity is logarithmic, proportional to the splitting rather than to ${\Lambda }^2$.

Realistic partners need not have exactly equal masses. Soft supersymmetry breaking permits scalar squared masses, gaugino masses and suitable trilinear interactions without restoring the unwanted quadratic sensitivity. Typically,up to coefficients and threshold details. Squarks and sleptons can therefore be heavier than their chiral fermion partners while the scalar sector remains stable against the much larger ultraviolet scale. Nevertheless, very large soft masses, particularly in the Higgs-coupled sector, leave large finite or logarithmic corrections and require tuning. This soft scalar-mass sensitivity remains after the quadratic divergence has cancelled. Supersymmetry solves the quadratic radiative instability when breaking is suitably soft; it does not make arbitrarily heavy superpartners natural or explain the entire hierarchy by itself.

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.

Use the minimal four-dimensional Super-Poincaré algebra, with $\{Q_{\alpha },Q_{\beta }\}=0$, $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=2{\sigma }_{\alpha \dot{\beta }}^{\nu }{P}_{\nu }$ and commuting translations. Lorentz covariance and closure on the existing supercharges permit onlyThe graded Jacobi identity for $P^{\mu },Q_{\alpha },Q_{\beta }$ therefore givesFor example, with ${\sigma }^0=I$, $\mu =0$ and $\alpha =\beta =1$, the coefficient is a nonzero multiple of $A(P_1-i{P}_2)$. Independence of the translation generators forces $A=0$. Its adjoint gives the barred result. Thus every supercharge commutes with four-momentum:Consequently $[Q_{\alpha },P^2]=0$. If $|u⟩$ has a particular mass-shell condition, the nonzero state $Q_{\alpha }|u⟩$ has the same one. This gives supersymmetric mass degeneracy within an unbroken physical supermultiplet. The qualification matters: after spontaneous supersymmetry breaking, the chosen vacuum is not annihilated by the supercharges, and its one-particle excitations need not constitute degenerate supermultiplets.

For the O'Raifeartaigh model, write $X={\varphi }_1$, $Y={\varphi }_2$, $Z={\varphi }_3$. Field phases allow $g,M,m$ to be positive. The canonical Kähler potential and elimination of the auxiliary fields give the F-term scalar potentialA supersymmetric vacuum would require $W_Y=MZ=0$, hence $Z=0$, while $W_X=g(Z^2-m^2)=-g{m}^2$ could not vanish. Thus F-flatness is impossible. More precisely, with $Z=u+iv$,The stated strict inequality makes every nonconstant term nonnegative. The entire classical vacuum family isHere $F_X=-\overline{W_X}=g{m}^2$, while $F_Y=F_Z=0$. The Weyl spinor ${\psi }_X$ is the massless goldstino, and the complex scalar field $X$ is a pseudomodulus with two zero tree-level squared masses. The hierarchy $M\gg m$ separates the massive fields from the breaking scale, but it does not itself select $x=0$ along this flat direction.

At the representative vacuum $x=0$, the chiral-superfield fermion mass matrix isIts physical masses are its singular values: $0,M,M$. The two massive Weyl spinors can be combined into one massive Dirac spinor. For $Y=(y_R+i{y}_I)/\sqrt{2}$, $Z=(z_R+i{z}_I)/\sqrt{2}$, the real scalar squared masses areIn particular, there is no tachyon under the stated inequality. The splitting of the $Z$ masses displays supersymmetry breaking directly.

For completeness, the mass spectrum can be given throughout the classical vacuum family, rather than silently fixing the pseudomodulus. A phase rotation makes $x$ real and nonnegative without changing physical masses. The massive fermion squared masses areThe real and imaginary scalar sectors have two-by-two scalar squared-mass matriceswhose eigenvalues areBoth determinants are $M^2(M^2+2ϵg^2{m}^2)>0$ and their traces are positive, so stability holds at every finite $x$. Quantum corrections can lift a pseudomodulus; these formulas describe the classical, tree-level spectrum requested here.