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Take canonical kinetic terms, gauge coupling $g>0$, and a nonzero charge $q$. Define $e=gq$ and choose the convention in which the Fayet–Iliopoulos term contributes $\xi D$ to the Lagrangian. With one charged chiral superfield, gauge invariance permits no nonconstant polynomial superpotential; an irrelevant global constant gives $F=0$. The real auxiliary field part isIts equation of motion is $D=-(\xi +e|\varphi {|}^2)$. Eliminating it gives the D-term scalar potentialThe gaugino transformation has the form $\delta \lambda =iϵD+$ a field-strength term, with a convention-dependent overall phase. In a translationally invariant vacuum with zero field strength, a nonzero $⟨D⟩$ therefore gives an inhomogeneous shift of $\lambda$: no nonzero supersymmetry parameter leaves the vacuum invariant. The massless Goldstino is consequently the gaugino, not the matter fermion, since $F=0$.

Minimize the potential over $u=|\varphi {|}^2\ge 0$. If $q\xi <0$, the allowed value $u=-\xi /e$ sets $D=0$, so supersymmetry is unbroken while the internal gauge symmetry is Higgsed. If $\xi =0$, the origin also has $D=0$. For the broken branch, $q\xi >0$ and the formal zero of $D$ would require negative $u$. The actual minimum isThis is single charged-field D-term breaking. Nonzero FI parameter alone does not force breaking if a scalar can cancel it. The degenerate case $q=0$ is different: any $\xi \ne 0$ breaks supersymmetry in the vector sector while the neutral scalar remains flat; the sign test above assumes an actually charged field.

Expand around the broken vacuum:For $\varphi =({\varphi }_R+i{\varphi }_I)/\sqrt{2}$, both canonically normalized real scalars have squared mass $e\xi =gq\xi >0$. The matter Weyl spinor has no superpotential mass. The gauge boson remains massless because $⟨\varphi ⟩=0$. The gauge Yukawa coupling, proportional to ${\varphi }^{\ast }\lambda \psi$, gives no fermion mixing or mass at this vacuum, so the gaugino is also massless. The tree-level spectrum is thereforeand the chiral-multiplet splitting isThe vector and its gaugino happen to remain degenerate at zero mass even though supersymmetry is broken. If the sign convention for the FI contribution is reversed, the sign criterion changes accordingly; the invariant condition is that the scalar condensate cannot cancel the auxiliary source. This is the classical toy-model spectrum; a quantum completion of the single charged-fermion theory also has to cancel its gauge anomalies.

Four-dimensional N=1 supersymmetry is compatible with a chiral gauge spectrum. Each matter Weyl spinor belongs to a chiral superfield with a complex scalar in the same gauge representation, and each gauge boson belongs to a vector multiplet with an adjoint gaugino. Thus a supersymmetric extension of the Standard Model predicts squarks, sleptons and gauge-fermion partners, together with Higgs-sector partners. Exact unbroken supersymmetry would make partners mass-degenerate, so a phenomenologically useful model needs supersymmetry breaking.

The Minimal supersymmetric Standard Model uses two Higgs chiral doublets. Holomorphy of the superpotential prevents use of the conjugate of one doublet as an independent chiral field, so up-type and down-type Yukawa couplings require opposite-hypercharge doublets. Their higgsinos also cancel each other's gauge anomalies. The superpotential has the schematic formwith gauge contractions understood. Supersymmetry relates the scalar quartic interactions and Yukawa interactions to the same gauge and superpotential couplings, rather than allowing arbitrary independent choices.

A central motivation is the hierarchy problem. A nonsupersymmetric scalar mass can receive corrections proportional to a high ultraviolet cutoff squared. The paired bosonic and fermionic loops in exact supersymmetry cancel that quadratic sensitivity. Soft supersymmetry breaking introduces scalar squared masses, gaugino masses and appropriate scalar bilinear and trilinear terms without restoring the original quadratic divergence. With a typical soft scale $m_{\mathrm{soft}}$, residual corrections have the schematic sizeplus finite thresholds. This favors not excessively heavy partners if one wishes to avoid a new tuning, but does not derive their masses or guarantee that they are experimentally accessible. A hidden breaking sector can communicate soft terms through gauge or gravitational interactions.

The altered particle spectrum also changes the renormalization group evolution of the gauge couplings. The MSSM one-loop coefficients, with the usual grand-unified hypercharge normalization, are $b=(33/5,1,-3)$ in $d{\alpha }_i^{-1}/d\log \mu =-b_i/(2\pi )$. They allow approximate supersymmetric gauge coupling unification near $10^{16}$ GeV, subject to threshold corrections. Top-Yukawa effects can drive the up-type Higgs squared mass negative and help produce radiative electroweak symmetry breaking. Neither phenomenon alone specifies a full breaking mechanism.

Without an additional symmetry, gauge invariance allows renormalizable baryon- or lepton-number violating superpotential terms. Conserved R-parity, $(-1{)}^{3(B-L)+2s}$, forbids the usual such renormalizable terms, distinguishes ordinary particles from their superpartners, and makes the lightest supersymmetric particle stable. A neutral stable lightest partner can be a dark matter candidate; its abundance and interactions require a separate calculation. Pair production followed by cascades to such a particle can give missing-momentum signals. R-parity is an extra model assumption, and does not by itself exclude all higher-dimensional proton-decay operators.

Possible low-energy consequences are a partner spectrum, controlled scalar-mass sensitivity, a constrained Higgs sector, modified coupling evolution and, with conserved R-parity, a stable lightest partner. Soft parameters also affect flavor-changing and CP-violating processes. Supersymmetry provides relations and mechanisms, not a unique numerical spectrum or an automatic solution to every flavor or cosmological problem.

Choose canonical vector normalization, $g>0$, and the convention that the Fayet–Iliopoulos term contributes $+\xi D$. In a constant Lorentz-invariant vacuum with vanishing gauge field strength, the gaugino transformation containsA nonzero $⟨D⟩$ therefore shifts a fermion by a constant under supersymmetry: no invariant vacuum can have this expectation. With only D-term breaking, the Goldstino is the gaugino $\lambda$. If F-term expectation values were present too, the Goldstino would instead be the corresponding linear combination of matter fermions and gauginos.

For a genuinely charged single chiral superfield, $q\ne 0$, gauge invariance forbids every nonconstant polynomial superpotential in that field. The renormalizable superspace action is consequentlywhere ${\mathcal{W}}_{\alpha }=-{\bar{D}}^2{D}_{\alpha }V/4$. A constant superpotential is dynamically irrelevant in global supersymmetry. Take the standard vector expansion with highest component ${\theta }^2{\bar{\theta }}^{2}D/2$. In the matter exponential, the term $2gqV$ then contributes $gq|\varphi {|}^{2}D$. The gauge kinetic chiral integral contributes $D^2/2$, and the FI term contributes $\xi D$. ThusThe auxiliary equations are $F=0$ and $D=-(gq|\varphi {|}^2+\xi )$. Completing the square givesThis is the single charged-field D-term breaking model. Its vacuum is found by minimizing over $\rho =|\varphi {|}^2\ge 0$, not by assuming that a nonzero FI constant must break supersymmetry.

For $q\xi >0$, the square cannot be canceled by any $\rho \ge 0$. It is minimized at the origin, whereSupersymmetry breaking occurs while the internal $U(1)$ gauge symmetry remains unbroken. Expanding the scalar potential about the origin givesThere is no superpotential fermion mass and no gaugino-matter mixing at $\varphi =0$. Therefore the chiral multiplet mass splitting isBoth real components of the complex scalar have the same positive squared mass; the gaugino is the massless Goldstino. With this sign convention and $\xi >0$, the breaking condition is $q>0$. Reversing the FI sign convention reverses this quoted charge inequality, while the condition that the scalar cannot cancel the auxiliary-field shift is unchanged.

For $q\xi <0$, the minimum instead satisfiesThe nonzero charged scalar expectation breaks the continuous internal gauge symmetry and preserves supersymmetry. To see the resulting supersymmetric Higgs mechanism, choose a real representative $v>0$ and expand $\varphi =v+(h+ia)/\sqrt{2}$. The scalar kinetic term supplies a vector mass, the D-term potential supplies the radial scalar mass, and the Yukawa interaction $-\sqrt{2}gq{\varphi }^{\ast }\lambda \psi +h.c.$ pairs the two Weyl fermions. They have the common squared massThe phase field $a$ is the absorbed Goldstone mode. There is no Goldstino in this supersymmetric Higgs vacuum; the fields assemble into a massive N=1 vector multiplet. This branch makes explicit that internal gauge breaking and supersymmetry breaking are different phenomena.

If $\xi =0$ and $q\ne 0$, the minimum of $g^2{q}^2|\varphi {|}^4/2$ is $\varphi =0$. Both supersymmetry and the continuous gauge symmetry are then unbroken, and the fields have zero tree-level masses at that vacuum. For the degenerate uncharged case $q=0$ with no superpotential interactions, a nonzero $\xi$ still breaks supersymmetry in the vector sector, but the free chiral multiplet stays unsplit and cannot Higgs the gauge field. An uncharged field would allow additional superpotential terms, outside the genuinely charged model above.

The calculation is classical. A single charged Weyl fermion has nonzero Abelian gauge and mixed gravitational anomaly coefficients proportional to $q^3$ and $q$; a complete quantum gauge theory needs an anomaly-canceling completion. The tree-level branches above refer to the specified field content; extra fields in such a completion can change the vacuum structure.

Assume global four-dimensional N=1 supersymmetry with canonical kinetic terms and no explicit gaugino mass. Write $W_i={\partial }_{i}W$ and $W_{ij}={\partial }_i{\partial }_{j}W$. To match the displayed stationarity equation, use $F^i=\overline{W_i}$; this differs by a minus sign from the eliminated auxiliary field in the component convention of Solution 2. The real D-terms are $D^a=-g^a({\bar{\varphi }}_i(T^a{)}^i{}_j{\varphi }^j+{\xi }^a)$, with Hermitian gauge generators. A constant Fayet–Iliopoulos term is permitted only for an appropriate Abelian factor. The scalar potential isDifferentiate with respect to ${\varphi }^i$, holding $\bar{\varphi }$ independent:Thus stationary point conditions gives the first requested identity. Gauge invariance of the superpotential gives $W_i(T^a{)}^i{}_j{\varphi }^j=0$. Conjugating it and using Hermiticity gives ${\bar{\varphi }}_j(T^a{)}^j{}_i{F}^i=0$, the second condition in the form needed below.

At the vacuum, define $H_{ij}=⟨W_{ij}⟩$ and $G_{ia}=g^a⟨{\bar{\varphi }}_j⟩(T^a{)}^j{}_i$. The two conditions combine asHere the upper equation is stationarity and the lower equation is conjugated gauge invariance. The chiral-superfield fermion mass matrix contributes $-\frac{1}{2}{H}_{ij}{\psi }^i{\psi }^j$, and the gauge Yukawa interaction contributes the mixed chiral-fermion/gaugino bilinear. With canonical Weyl fermion and gaugino kinetic terms, and choosing the gaugino phase to put the mixed bilinear in real-sign convention, these terms areThe factor $\sqrt{2}$ is important: the matrix directly acting on the printed $(F,D)$ is the same mass bilinear with rescaled gauginos, rather than literally the canonical matrix. In detail, for $S=\mathrm{diag}(I,\sqrt{2}I)$,Since supersymmetry breaking makes this vector nonzero, there is at least one zero mass eigenvalue. A normalized massless fermion isWith the consistent gaugino phase, the inhomogeneous vacuum supersymmetry transformation of $(\psi ,\lambda )$ is proportional to the same vector $(F,D/\sqrt{2})$. Therefore $\eta$ shifts by a nonzero constant times the transformation parameter: it is the Goldstino, the Goldstone fermion of broken global supersymmetry. The argument guarantees at least one Goldstino zero mode, not the absence of other massless fermions.

For the second calculation, first take only canonical chiral superfields and F-term breaking. At a stationary vacuum $v$, let $h=\varphi -v$, $m_{ij}=W_{ij}(v)$ andThese follow by differentiating $V={\sum }_k|W_k{|}^2$ twice. The quadratic scalar potential isThe doubled scalar mass matrix has blocksTransforming to the $2N$ canonically normalized real components of $h$ preserves its trace, so the sum of real scalar mass squares is $\mathrm{tr}{M}_s^2=2\mathrm{tr}A$. The Weyl fermion mass squares are the eigenvalues of $m^{†}m=A$ and their sum is $\mathrm{tr}A$. Each Weyl mass eigenstate has two physical spin states. The printed supertrace has scalar sign $-1$ and fermion weight $+2$, henceThe frequently used boson-minus-fermion definition has the opposite overall sign and gives the same zero. The mixing block $B$ can split scalar masses but cannot change this trace. This proves the tree-level supertrace mass sum rule; it uses canonical kinetic terms and a renormalizable global theory, rather than a general supergravity or noncanonical effective theory.

In the MSSM, trying to make all squarks and sleptons heavy through direct canonical tree-level F-term breaking encounters this constraint. In an unbroken electric-charge and color sector, the scalar mass-squared trace is twice the trace for its ordinary fermionic partners. More explicitly, take a unit vector $u$ minimizing $u^{†}Au$. For $h=e^{i\alpha }u$, choose $\alpha$ so that $\mathrm{Re}(e^{2i\alpha }{u}^{T}Bu)\le 0$. The curvature along this real scalar direction is then at most $u^{†}Au$, so at least one scalar mass square is no larger than the smallest fermion mass square in that sector. For one isolated pair the scalar masses are $m_f^2\pm |B|$. Thus light quarks can force an unacceptably light squark, or a tachyon, under these assumptions. A hidden supersymmetry-breaking sector with mediation produces effective soft supersymmetry breaking; radiative effects, noncanonical kinetic terms, or supergravity can change the assumptions and allow a viable spectrum. The trace over an arbitrary enlarged theory alone would not locate the light state in the visible sector; the MSSM conclusion uses the conserved-charge block structure.

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.

Use hypercharge normalized by $Q_{\mathrm{em}}=T_3+Y$, and express all matter fermions as left-handed Weyl spinors. Thus $u^c,d^c,e^c$ denote the charge conjugates of the right-handed Standard Model fermions; the superscript $c$ is unrelated to the antiparticle marker $(A)$. Each chiral superfield contains a complex scalar field and a Weyl spinor, while each vector superfield contains a gauge boson and a gaugino. The complete propagating content of the Minimal supersymmetric Standard Model before electroweak symmetry breaking is:

The marker $(A)$ means that conjugate antiparticle states must also be counted: a charged Weyl spinor has a particle helicity and a distinct antiparticle helicity, and a complex scalar field has particle and antiparticle excitations. The real adjoint gauge bosons and adjoint Majorana fermions already contain their conjugate states, so no extra marker is attached to them. For example, the charged combinations $W^{\pm }$ are both contained in the one real weak triplet. There are no right-handed-neutrino chiral superfields, no gravitino, and no propagating auxiliary fields in this field table. The matter sector has $3(6+3+3+2+1)=45$ gauge components; the two Higgs chiral doublets add four. Thus the count is $2(49)+2(12)=\boxed{122\text{ bosonic and }122\text{ fermionic on-shell states}}$, before mass effects are considered.

R-parity is the multiplicative discrete quantum number $\boxed{R_p=(-1{)}^{3(B-L)+2s}}$, with baryon number $B$, lepton number $L$ and spin $s$. Ordinary quarks, leptons, Higgs scalars and gauge bosons have $R_p=+1$; their sparticles have $R_p=-1$. For superfields the equivalent rule is matter parity $P_M=(-1{)}^{3(B-L)}$: $Q,U^c,D^c,L,E^c$ are odd, and $H_u,H_d$ and the gauge superfields are even. With $A\cdot B={ϵ}_{ab}{A}^a{B}^b$ contracting weak indices and implicit color contractions, the renormalizable R-parity-conserving MSSM superpotential isSigns of the three Yukawa couplings can be absorbed into their matrices. Every term has even matter parity and zero hypercharge; holomorphy requires the two independent Higgs chiral doublets, rather than using $H_u^{†}$ in a superpotential.

Conserved R-parity forces an even number of odd particles in a process whose incoming particles are all ordinary particles, so colliders produce sparticles in pairs or another even multiplicity. Their decay chains end in the stable lightest supersymmetric particle. A neutral, weakly interacting lightest supersymmetric particle, commonly a neutralino after electroweak symmetry breaking, escapes the detector and gives missing transverse momentum; it is also a possible dark matter constituent. Stability alone does not establish its relic abundance, and a charged or colored stable lightest supersymmetric particle is not the usual viable dark matter candidate. Conserved R-parity forbids the renormalizable interactions responsible for the decay below. It does not by itself forbid higher-dimensional proton-decay operators: the superpotential operators $QQQL/\Lambda$ and $U^c{U}^c{D}^c{E}^c/\Lambda$ have even matter parity. Thus the renormalizable R-parity-conserving MSSM protects the proton, while ultraviolet physics can still induce its decay.

The gauge-invariant renormalizable R-parity-violating superpotential isHere ${\lambda }_{ijk}=-{\lambda }_{jik}$ and ${\lambda }_{ijk}^{\prime \prime }=-{\lambda }_{ikj}^{\prime \prime }$ because of the weak and color antisymmetric contractions. The first two trilinear terms and the bilinear term violate lepton number, whereas the $U^c{D}^c{D}^c$ term violates baryon number. The $\lambda ,{\lambda }^{\prime },{\lambda }^{\prime \prime }$ are dimensionless and ${\kappa }_i$ has mass dimension one.

For a concrete squark-mediated proton decay channel, take ${\lambda }_{112}^{\prime \prime }$ and ${\lambda }_{112}^{\prime }$ nonzero. Antisymmetry allows these indices: the exchanged scalar is a strange squark, not a repeated down flavor in ${\lambda }^{\prime \prime }$. Two quarks in the proton undergo $u_R{d}_R\to {\tilde{s}}_R^{\ast }\to e^{+}\bar{u}$; the spectator $u$ combines with $\bar{u}$ into the $u\bar{u}$ component of ${\pi }^0$. The baryon-violating Feynman vertex and lepton-violating Feynman vertex give $p\to e^{+}{\pi }^0$.

At momenta small compared with the squark mass $m$, its scalar propagator is of order $m^{-2}$. Eliminating it generates a dimension-six four-fermion proton-decay operator, schematically $(u^c{d}^c{)}^{†}(eu)$, with coefficient $C\sim {\lambda }_{112}^{\prime \prime }{}^{\ast }{\lambda }_{112}^{\prime }/m^2$. Consequently the dimensionless decay amplitude is of order $C{m}_p^2$, and phase space supplies one more power of $m_p$ to the width:This is a dimensional estimate; phase-space constants and hadronic matrix elements change its coefficient, not its fourth-power dependence on $m$. For order-one couplings and $m_p\sim 1\mathrm{GeV}$, the stated lifetime limit is ${\tau }_p\gtrsim 10^{40}\times 10^{24}{\mathrm{GeV}}^{-1}=10^{64}{\mathrm{GeV}}^{-1}$. Therefore $m^4\gtrsim 10^{64}{\mathrm{GeV}}^4$, giving $\boxed{m\gtrsim 10^{16}\mathrm{GeV}}$.

Such a mass is incompatible with the intended weak-scale MSSM explanation of the hierarchy problem: squarks that cancel the Higgs-sector sensitivity to high scales must not introduce a new enormous splitting from their light quark partners. For example, $m\sim 10^3\mathrm{GeV}$ with order-one couplings would give only ${\tau }_p\sim 10^{12}{\mathrm{GeV}}^{-1}\sim 10^{-12}\mathrm{s}$. The field content alone does not mathematically prohibit extremely heavy superpartners, but that choice sacrifices the weak-scale naturalness motivation. The usual resolution is conserved R-parity, or sufficiently suppressed simultaneous baryon number and lepton number violation, rather than making the exchanged squark this heavy.

Use the mostly-plus Clifford algebra and Riemann curvature tensor conventions of the preceding solutions, with $e=\det (e_{\mu }{}^a)$, ${\kappa }^2=8\pi G$, and a Majorana gravitino. An action for minimal four-dimensional supergravity in the exact transformation normalization given isThe kinetic coefficient is tied to the normalization of the supersymmetry transformations: if ${\Psi }_{\mu }=2{\psi }_{\mu }$ and $\eta =2ϵ$, this becomes the canonical coefficient $-1/2$, with $\delta e_{\mu }{}^a=(\kappa /2)\bar{\eta }{\gamma }^a{\Psi }_{\mu }$ and $\delta {\Psi }_{\mu }={\nabla }_{\mu }\eta /\kappa$. Mixing those two normalizations would spoil the cancellation.

To the retained order, use the torsion-free connection. More generally the 1.5-order formalism allows its variation to be omitted after imposing its own algebraic equation; the resulting contorsion is quadratic in fermions and first affects the displayed action at quartic order. The metric tensor variation induced by the vierbein is $\delta g_{\mu \nu }=4\kappa \overline{ϵ}{\gamma }_{(\mu }{\psi }_{\nu )}$, so the Einstein-Hilbert action varies asFor the fermionic action, varying both Majorana factors gives the same contribution after integration by parts and the Grassmann bilinear interchange rule. ThusThe vector-index Riemann curvature tensor contribution vanishes by the first Bianchi identity. The spin-index contribution is determined byFor the last equality, expand the antisymmetric three-gamma/two-gamma product: the totally antisymmetric Riemann curvature tensor terms vanish by Bianchi, and the surviving terms are the Ricci contraction minus half its trace. Hencewhich cancels the gravitational variation. Variations of vierbeins, gamma matrices and the connection in the fermionic kinetic term carry three dynamical fermions. They lie beyond this linear-fermion cancellation and are cancelled by the omitted four-fermion completion. Thus the action is invariant up to boundary terms at the order specified, rather than claiming that a truncated action is exactly invariant at every fermion order.