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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 left Grassmann derivatives and define the supersymmetric covariant derivatives $D_{\pm }={\partial }_{{\theta }^{\pm }}-i{\theta }^{\pm }{\partial }_{\pm }$. Multiplication in the indicated order gives exactly the printed differential operator:Thus the equation is $i{D}_{-}{D}_{+}\Phi =W(\Phi )$ in two-dimensional N=(1,1) superspace.

Under a Lorentz boost, take $x^{\pm }\mapsto e^{\pm \omega }{x}^{\pm }$ and ${\theta }^{\pm }\mapsto e^{\pm \omega /2}{\theta }^{\pm }$. The derivatives ${\partial }_{\pm }$ have boost weights $\mp 1$, and $D_{\pm }$ have weights $\mp 1/2$. Consequently $D_{-}{D}_{+}$ has weight zero, and is a Lorentz scalar. Both $\Phi$ and the ordinary function $W(\Phi )$ are scalar superfields, so the equation is Lorentz invariant.

For supersymmetry, $\{{\partial }_{\theta },\theta \}=1$ and anticommutation of distinct odd coordinates giveMoving either supercharge through the two odd derivatives introduces two minus signs, hence $[Q_t,D_{-}{D}_{+}]=0$. If $\delta \Phi ={ϵ}^{+}{Q}_{+}\Phi +{ϵ}^{-}{Q}_{-}\Phi$, with odd constant parameters, the chain rule for an even scalar superfield givesTherefore a solution is mapped to a solution, proving supersymmetry invariance.

To compute the supersymmetric Liouville equation, order the odd coordinates as ${\theta }^{-}{\theta }^{+}$ and anticommute the fermionic component fields with both coordinates. In particular, ${\partial }_{{\theta }^{+}}({\theta }^{-}{\theta }^{+})=-{\theta }^{-}$. Direct application of the operator givesLet $N=\Phi -\varphi$. Only the two products of its degree-one terms survive in $N^2$; for example, $(i{\theta }^{-}{\psi }_{+})(i{\theta }^{+}{\psi }_{-})={\theta }^{-}{\theta }^{+}{\psi }_{+}{\psi }_{-}$. Hence $N^2=2{\theta }^{-}{\theta }^{+}{\psi }_{+}{\psi }_{-}$ and $N^3=0$. The exponential function expands exactly asEquating the constant terms fixes the auxiliary field, $\boxed{F=e^{\varphi }}$. The linear and quadratic terms then give the requested component field equations of motion:The signs here follow from the printed operator, the printed component expansion, and left Grassmann derivatives; no convention change has been made in eliminating the auxiliary field.

Here L-superfield means a left-chiral superfield: ${\bar{D}}_{\dot{\alpha }}\Phi =0$, and its complex conjugate is an antichiral superfield, $D_{\alpha }{\Phi }^{\ast }=0$. Interpret the two barred derivatives as the contracted scalar ${\bar{D}}^2={\bar{D}}_{\dot{\alpha }}{\bar{D}}^{\dot{\alpha }}$. The key point is chiral projection by squared supercovariant derivatives. Since the two components of $\bar{D}$ anticommute, any product of three barred derivatives repeats a component and is zero. Thusfor every superfield $X$, even when $X$ is antichiral. In particular, ${\bar{D}}^2{\Phi }^{\ast }$ is chiral.

The right side is chiral too. Because $\Phi$ is even, the graded derivation rule gives ${\bar{D}}_{\dot{\alpha }}({\Phi }^2)=({\bar{D}}_{\dot{\alpha }}\Phi )\Phi +\Phi ({\bar{D}}_{\dot{\alpha }}\Phi )=0$. Constant $m$ and $\lambda$ do not contribute derivatives. Consequentlyso applying the chiral constraint to the equation imposes no additional incompatible condition. This is the holomorphic closure of chiral superfields; a generic nonholomorphic right side would not satisfy it.

Both sides have zero Grassmann parity and are Lorentz scalars: the two dotted derivative indices are contracted with the invariant epsilon tensor, and $\Phi$ is a scalar superfield. Moreover, each covariant derivative anticommutes with every supercharge, so its squared contraction commutes with the supercharges. Hence the equation transforms covariantly under supersymmetry. It is the chiral field equation of motion of the Wess–Zumino model with superpotential $\mathcal{W}(\Phi )=m{\Phi }^2/2+\lambda {\Phi }^3/3$, for which ${\mathcal{W}}^{\prime }=m\Phi +\lambda {\Phi }^2$, using the normalization in the PDF. With mass dimensions $[\Phi ]=1$, $[\bar{D}]=1/2$, $[m]=1$ and $[\lambda ]=0$, each side has dimension two. No component field equations of motion are needed for this consistency argument.

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.

The chiral superfield condition is $\boxed{{\bar{D}}_{\dot{\alpha }}\Phi =0}$ for both dotted components. Use left Grassmann derivatives with the displayed supersymmetric covariant derivatives. The odd product rule gives $\partial (\theta {\sigma }^{\mu }\bar{\theta })/\partial {\bar{\theta }}^{\dot{\alpha }}=-(\theta {\sigma }^{\mu }{)}_{\dot{\alpha }}$, soThe chirality constraint therefore says that $\Phi$ is independent of $\bar{\theta }$ at fixed $y$. The Grassmann algebra has only two independent $\theta$ components, so every cubic product of them vanishes. Its general even scalar superfield consequently has a scalar coefficient, a term linear in $\theta$ with a Weyl spinor coefficient, and one quadratic term. Choosing the conventional component normalization gives the chiral-superfield component expansionHere $\varphi$ is a complex scalar field, $\psi$ an odd Weyl spinor and $F_{\mathrm{aux}}$ a complex nonpropagating auxiliary field. Conversely this expansion is annihilated by $\bar{D}$ because $y$ and $\theta$ are. At fixed ordinary $x$, the same result is the terminating translation $\Phi =e^{i\theta {\sigma }^{\mu }\bar{\theta }{\partial }_{\mu }}[\varphi (x)+\sqrt{2}\theta \psi (x)+(\theta \theta )F_{\mathrm{aux}}(x)]$; this retains the printed contraction convention without changing the order in the barred square.

For a local function of scalar superfields, initially allow $f(\Phi ,{\Phi }^{†})$. The chain rule givesFor arbitrary chiral component fields the last factor is not constrained to vanish, so chirality requires $f_{{\Phi }^{†}}=0$. Thus $\boxed{f\text{ must be holomorphic in the chiral fields}}$ on its domain, with no generic conjugate-field dependence. Its ordinary holomorphic derivatives must exist where its nilpotent expansion is evaluated; a pole or branch point there does not define a regular local superfield. Using $(\theta \psi {)}^2=-\frac{1}{2}(\theta \theta )(\psi \psi )$ gives the explicit holomorphic closure of chiral superfields:All three component coefficients are evaluated at $y$. For several chiral superfields, replace $f^{\prime }\psi$ by $f_i{\psi }_i$ and $f^{\prime \prime }\psi \psi$ by $f_{ij}{\psi }_i{\psi }_j$. This requirement concerns a function that is to be chiral for arbitrary field configurations; a special constant field configuration does not impose the same functional constraint.

The ungauged two-derivative supersymmetric action is obtained from a real Kähler potential and a holomorphic superpotential:The first integral is a D-term; the second is an F-term. The Kähler metric $K_{i\bar{j}}$ must be positive for positive scalar kinetic energy. The auxiliary part is $K_{i\bar{j}}{F}_i{\bar{F}}_{\bar{j}}+W_i{F}_i+{\overline{W}}_{\bar{j}}{\bar{F}}_{\bar{j}}$, so eliminating it gives the scalar potential $V=K^{i\bar{j}}{W}_i{\overline{W}}_{\bar{j}}$ in the bosonic sector.

For the single-field Wess–Zumino model, take the canonical Kähler potential $K={\Phi }^{†}\Phi$ and spacetime signature $(+---)$. Component superspace integration givesSince $W^{\prime }=m\varphi +g{\varphi }^2$ and $W^{\prime \prime }=m+2g\varphi$, the auxiliary equation is $F_{\mathrm{aux}}=-(m\varphi +g{\varphi }^2{)}^{\ast }$. Completing its square leaves $-|W^{\prime }{|}^2$ in the Lagrangian. Hence the tree-level effective potential isFor nonzero $m,g$, its two zero-energy supersymmetric vacua are $\varphi =0$ and $\varphi =-m/g$. The potential is nonnegative because it is an absolute square.

The Trilinear scalar vertices in the Wess–Zumino model come fromWith all field legs incoming, expanding $e^{iS}$ and differentiating each cubic monomial givesThe factor $2!$ counts the two identical legs; there is no $1/2!$ in ${\mathcal{L}}_3$ to cancel it. A factor enforcing four-momentum conservation is implicit. These labels specify fields in an all-incoming Feynman rule; physical outgoing particles are assigned by crossing. There are no ${\varphi }^3$ or $({\varphi }^{\ast }{)}^3$ scalar vertices from this on-shell potential.

Equivalently, if $m,g$ are real and $\varphi =(A+iB)/\sqrt{2}$ with canonically normalized real scalars $A,B$, then ${\mathcal{L}}_3=-mgA(A^2+B^2)/\sqrt{2}$. Differentiation gives $\boxed{V_{AAA}=-3\sqrt{2}img,V_{ABB}=-\sqrt{2}img}$, with vanishing $AAB$ and $BBB$ vertices. This is the same pair of complex-field interactions in a real basis.

For the non-renormalization argument, promote $m,g$ to nondynamical chiral spurions. Define one ordinary $U(1)$ and one R-symmetry by the Wess–Zumino spurion charge assignment:

Each superpotential term is neutral under the ordinary symmetry. Under the R-symmetry, both terms have charge two, which is canceled by the charge $-2$ of $d^2\theta$. In particular, $m$ is R-neutral as required. The scalar, spinor and auxiliary components of $\Phi$ have R-charges $1,0,-1$. These are formal symmetries when the couplings transform as spurions; replacing them by fixed nonzero numbers need not leave an actual continuous field symmetry.

A local loop-corrected Wilsonian effective action superpotential $F_{\mathrm{eff}}(\Phi ,m,g)$ must be holomorphic in all three chiral variables and obey the same charges. The ratio $z=g\Phi /m$ is neutral under both symmetries, while $m{\Phi }^2$ has the required charges and mass dimension three. Therefore, on a patch with $m\ne 0$, the most general symmetry-allowed form iswith $h$ holomorphic on its domain. This form follows directly for a monomial ${\Phi }^a{m}^b{g}^c$: ordinary charge gives $a-2b-3c=0$ and R-charge gives $a-c=2$, hence $a=c+2$, $b=1-c$. Each allowed monomial is $g^c{m}^{1-c}{\Phi }^{c+2}$ and already has mass dimension three. Charge symmetry alone leaves $h$ undetermined.

To constrain perturbative loops, retain the original elementary chiral superfield and a nonzero infrared cutoff, rather than integrating out the entire field. A regular perturbative superpotential can be expanded about $g=0$, so $h(z)={\sum }_{n\ge 0}{c}_n{z}^n$ andTerms with $n\ge 2$ have poles at $m=0$. They are excluded by regularity of the local Wilsonian coefficients in the massless limit with the infrared cutoff fixed. At $g=0$ the theory is Gaussian, so the quadratic coefficient remains $c_0=1/2$. The term linear in $g$ is the single original cubic vertex: a diagram with three external fields and only that one interaction vertex has no internal loop. Holomorphy disallows coefficients involving $g^{\ast }$ or $m^{\ast }$, so it cannot conceal the usual nonholomorphic wave-function corrections in $c_1$. Matching the one-vertex term gives $c_1=1/3$. Thus the holomorphy argument for superpotential non-renormalization givesThe infrared qualification matters: a singular one-particle-irreducible expression or the action after eliminating an entire massive multiplet is not the same local Wilsonian effective action for the retained field.

Consequently the holomorphic superpotential and its parameters m and g receive no perturbative vertex renormalization. The Kähler potential does undergo wave-function renormalization. If its quadratic term is $Z{\Phi }^{†}\Phi$, then canonical normalization gives ${\Phi }_c=Z^{1/2}\Phi$ andThese canonically normalized running couplings can therefore be renormalized despite the non-renormalization theorem for the holomorphic superpotential. Loop corrections can also change the effective scalar potential through the corrected Kähler metric. The distinction between holomorphic coefficients and canonically normalized couplings is essential to answering the renormalization question.

Chirality and the component expansion. A chiral superfield is constrained byUse left Grassmann derivatives. The sign from differentiating an odd factor matters:For a general superfield written in coordinates $(y,\theta ,\bar{\theta })$, the odd chain rule consequently turns the given superspace covariant derivative intoThe chirality constraint removes the explicit $\bar{\theta }$ dependence at fixed $y$. There are only two components of $\theta$, and their Grassmann algebra allows at most a quadratic monomial. Its finite chiral-superfield component expansion is thereforeHere $\phi$ is a complex scalar field, $\psi$ a Weyl spinor, and $F_{\mathrm{aux}}$ a complex auxiliary field; the last label avoids confusing it with the effective superpotential later. The factor $\sqrt{2}$ is the standard canonical component normalization. In ordinary coordinates the same statement isThis translation exponential terminates because its shift is nilpotent; it displays the full component dependence without an unstated convention for the barred spinor square.

A superspace covariant derivative obeys the graded product rule. Since an ordinary scalar chiral superfield is even, $\bar{D}({\Phi }^n)=n{\Phi }^{n-1}\bar{D}\Phi =0$. Linear combinations prove the polynomial claim. More generally, a nonsingular holomorphic function of chiral fields is chiral; inserting conjugate fields generally spoils this holomorphic closure of chiral superfields.

The superspace action. For a real Kähler potential and a holomorphic superpotential, the global chiral-field action has Lagrangian densityFull superspace integration gives a D-term, and chiral superspace integration an F-term. The action is real, and its supersymmetry variations are spacetime total derivatives. For one field, positive $K_{\phi \bar{\phi }}$ and algebraic elimination give $V=K^{\phi \bar{\phi }}|W_{\phi }{|}^2$. The Wess–Zumino model uses the canonical choice $K={\Phi }^{†}\Phi$ at tree level.

Scalar potential and the vertex. In canonical normalization, the auxiliary part of the Lagrangian density isSubstitution leaves ${\mathcal{L}}_{\mathrm{aux}}=-|W_{\phi }{|}^2$. Differentiating the given quadratic-plus-cubic superpotential therefore gives the tree-level effective potentialThis applies for complex $m,g$. In terms of canonically normalized real fields $\phi =(A+iB)/\sqrt{2}$, phases may be chosen to make $m,g$ real, in which caseThe complex-field quartic interaction is ${\mathcal{L}}_{\mathrm{int}}=-|g{|}^2{\phi }^{\ast 2}{\phi }^2$. There are two identical external legs of each field type. Differentiating with respect to those four fields, or counting the Wick attachments to the vertex, produces the factor $2!2!$:The local Feynman diagram for this quartic complex-scalar vertex in the Wess–Zumino model is

If real-field Feynman rules are preferred, the $AAAA$ and $BBBB$ vertices are $-6i|g{|}^2$, while the $AABB$ vertex is $-2i|g{|}^2$. These are the same interaction in a different component basis. They should not be confused with a convention that absorbs $2!2!$ into the coefficient of the complex quartic term.

Spurion symmetries. Treat $m,g$ as chiral spurions. An ordinary $U(1)$ acts on $\Phi ,m,g$ with charges $(1,-2,-3)$ and leaves $\theta$ neutral. An R-symmetry gives $\theta$ charge one and $\Phi ,m,g$ charges $(1,0,-1)$. Thus

Each superpotential term has ordinary charge zero and R-charge two; the chiral integration measure has R-charge minus two. In particular $m$ is neutral under the specified R-symmetry. These are formal transformations of fields and parameters together, not two exact symmetries of a theory with arbitrary fixed nontransforming numerical couplings. This Wess–Zumino spurion charge assignment is useful because it constrains possible quantum terms.