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Use left Grassmann derivatives for the Grassmann variables. A chiral superfield satisfies ${\bar{\mathcal{D}}}_{\dot{\alpha }}\Phi =0$. In the derivative convention printed in the PDF, differentiating $\theta {\sigma }^{\mu }\bar{\theta }$ with respect to a barred coordinate brings a minus sign from moving the odd derivative through $\theta$. Consequently ${\bar{\mathcal{D}}}_{\dot{\alpha }}{y}^{\mu }=0$, so the general solution depends only on $y$ and $\theta$:Here $\varphi$ is a complex scalar field, $\psi$ a Weyl spinor and $F$ a complex auxiliary field. This is a finite Grassmann algebra expansion, not an assumption that the spacetime fields are constant.

For a completely convention-independent way to organize the ordinary-coordinate expansion, let $v^{\mu }=\theta {\sigma }^{\mu }\bar{\theta }$. Translation by $iv$ givesAll higher terms vanish because there are only two unbarred and two barred odd coordinates. For example take ${\sigma }^{\mu }=(I,\mathbf{\sigma })$, signature $(+---)$, ${\theta }^2={\theta }^{\alpha }{\theta }_{\alpha }$ and ${\bar{\theta }}^2={\bar{\theta }}_{\dot{\alpha }}{\bar{\theta }}^{\dot{\alpha }}$, with both lower epsilon tensors having ${ϵ}_{12}=1$. Then $v^{\mu }{v}^{\nu }=\frac{1}{2}{\theta }^2{\bar{\theta }}^2{\eta }^{\mu \nu }$ and the chiral-superfield component expansion readsThe frequently written positive quarter coefficient corresponds to the opposite convention for $\square$ or the spinor contractions; the preceding shift formula fixes the sign without ambiguity. The local TeX's barred derivative is damaged: the original PDF has ${\partial }_{\mu }$, not a barred-coordinate factor at its end.

Write $X=x+\sqrt{2}\theta G+{\theta }^2{F}_X$ in chiral coordinates. Multiplication gives the three component constraintsOn the regular branch where the ordinary commuting part of $F_X$ is nonzero, the nilpotent chiral superfield is thereforeThe goldstino $G$ and auxiliary field $F_X$ are independent; the scalar is a composite, not another independent degree of freedom. Its first two constraints follow because any product of three identical two-component odd spinor entries vanishes. It is essential not to divide by $F_X$ on a branch with zero commuting part: for example $X=0$ satisfies the nilpotency condition without breaking supersymmetry.

For a regular local action with no superspace derivatives in $K$ or $W$, nilpotency truncates the most general superpotential and Kähler potential toThe last inequality makes the kinetic Kähler metric healthy. In global four-dimensional N=1 supersymmetry, the constant and holomorphic pieces of $K$ integrate to zero, and $w_0$ does not affect the action. On the bosonic background $G=0$, the constraint sets $x=0$ and the auxiliary terms are $\kappa F_X{\bar{F}}_X+f{F}_X+\bar{f}{\bar{F}}_X$. Eliminating them givesA nonzero $f$ is necessary for this regular constrained branch to be an on-shell vacuum. It gives nonzero $F_X$ and supersymmetry breaking, with $G$ the goldstino. Nilpotency alone is not a proof that every possible branch breaks supersymmetry. With $f=0$ the above composite parametrization is singular on shell, rather than evidence for a regular supersymmetric vacuum containing an independent Goldstino.

If these data are instead embedded in supergravity, the constant and linear terms cannot simply be discarded independently: a Kähler transformation also transforms $W$. Before such a transformation, at $x=0$ the supergravity F-term potential isThus nonzero $F^X$ still breaks supersymmetry, but the vacuum energy need not be positive. This distinguishes the global answer from the local theory explicitly introduced in the next question.

Now write $Y=y+\sqrt{2}\theta \chi +{\theta }^2{F}_Y$. The condition $XY=0$ becomesUsing the same nonzero-$F_X$ branch gives the chiral superfield constrained by a nilpotent superfieldThe spinor $\chi$ and the auxiliary $F_Y$ remain independent; the scalar is removed. Substituting this expression verifies the other two product constraints using two-component spinor identities. It does not impose $Y^2=0$. It does imply cubic nilpotency from a mixed chiral constraint: $Y^3=0$. Indeed $(G\chi {)}^2=-\frac{1}{2}(GG)(\chi \chi )$, soThese are exactly the three components needed for $Y^3=0$. The generic nonzero expression for $y^2$ shows why imposing quadratic nilpotency on $Y$ would lose allowed interactions.

The analytic chiral monomials consequently reduce to $1,X,Y,Y^2$. The most general superpotential isFor the Kähler potential, put $U=(X,Y,Y^2{)}^T$. Every allowed mixed monomial is included inThis includes $X\bar{X}$, $X\bar{Y}$, $X{\bar{Y}}^2$, $Y\bar{Y}$, $Y{\bar{Y}}^2$, $Y^2{\bar{Y}}^2$ and the necessary conjugates; products containing $XY$ or $\bar{X}\bar{Y}$ vanish. In a global theory the holomorphic $\ell$ terms again integrate to zero; in supergravity they can be changed by a Kähler transformation. Positivity requires the kinetic metric on the independent multiplets to be positive definite at the chosen background, not necessarily the entire coefficient matrix on the redundant composite list $U$. The most general claim here concerns regular analytic two-derivative actions; singular functions of the constrained fields or higher-derivative operators are outside it.

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

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

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

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

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

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

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

Use a positive-energy unitary representation with finite helicity states, excluding continuous-spin representations. In the specified frame the Pauli matrices giveFor every $A$ and every state $|\chi ⟩$,Positivity therefore gives $Q_2^A={\bar{Q}}_{\dot{2}A}=0$ on this massless supermultiplet. The equal-chirality anticommutator $\{Q_1^A,Q_2^B\}={ϵ}_{12}{Z}^{AB}$ then shows that every central charge in supersymmetry vanishes on this superalgebra representation.

Normalize the remaining supercharges as fermionic raising and lowering operators:They obey the canonical anticommutation relationsChoose a Clifford vacuum $|\lambda ⟩$ of helicity $\lambda \in \frac{1}{2}\mathbb{Z}$ annihilated by every $a_A$. Such a vector exists because the commuting fermion number operators $a_A^{†}{a}_A$ have eigenvalues $0$ and $1$, and lowering every occupied mode produces a nonzero empty state. In an irreducible representation this empty space has dimension one; multiple empty states produce a direct sum of copies. The states areThe canonical anticommutation relations make these states orthonormal for a normalized Clifford vacuum. In the helicity convention $[J_3,Q_1^A]=\frac{1}{2}{Q}_1^A$, its adjoint has $[J_3,a_A^{†}]=-\frac{1}{2}{a}_A^{†}$. Thus the Clifford vacuum is the highest-helicity state andReversing the rotation convention reverses the displayed ordering and leaves all counts unchanged. The level-$k$ states form the exterior power ${\Lambda }^k{\mathbb{C}}^{\mathcal{N}}$, which explains the binomial coefficient. Summing the binomial coefficients gives the helicity spectrum of a massless supermultiplet:Each application of a fermionic raising and lowering operator reverses fermion parity. The even and odd levels each contain $2^{\mathcal{N}-1}$ states, so a physical massless supermultiplet has $2^{\mathcal{N}-1}$ boson and $2^{\mathcal{N}-1}$ fermion states.

These formulas describe one irreducible representation of the massless Super-Poincaré algebra at fixed four-momentum. A reducible superalgebra representation can contain several such supermultiplets. Also, the CPT theorem sends helicity $h$ to $-h$ and conjugates internal charges. CPT completion of a supermultiplet may therefore require a second $2^{\mathcal{N}}$-state supermultiplet, with highest helicity $\mathcal{N}/2-\lambda$. A necessary condition for the original helicity spectrum of a massless supermultiplet to be self-conjugate is $\lambda =\mathcal{N}/4$; internal charges must also admit the conjugation in a realization compatible with the Spin-statistics theorem. Thus this helicity condition alone is not sufficient for CPT completion of a supermultiplet without doubling. In particular, the original irreducible count is not automatically the count after CPT completion of a supermultiplet.

For the conventional bounds on extended supersymmetry, suppose all massless states satisfy $|h|\le s_{\max }$. The helicity spectrum of a massless supermultiplet occupies an interval of width $\mathcal{N}/2$, whereas the available interval has width $2s_{\max }$. HenceAn ordinary four-dimensional renormalizable quantum field theory without gravity uses interacting massless fields of spin angular momentum at most one. Massless spin angular momentum $3/2$ requires supergravity, whose gravitational coupling has negative mass dimension and is a nonrenormalizable interaction by power counting in quantum field theory. Setting $s_{\max }=1$ gives $\mathcal{N}\le 4$ for conventional renormalizable theories. The bound is attained by four-dimensional N=4 super Yang-Mills theory: choosing $\lambda =1$ givesfor its supersymmetric vector multiplet.

Allowing gravity permits $s_{\max }=2$, so $\mathcal{N}\le 8$ for conventional theories including gravity. The four-dimensional N=8 supergravity supergravity multiplet attains the bound with $\lambda =2$:It has $256$ states, $128$ boson and $128$ fermion states. The conventional claim about a maximum “in general” assumes local interacting four-dimensional theories in Minkowski spacetime with a finite spectrum of fields and no massless spin angular momentum above two. It is not a theorem excluding arbitrary free higher-spin angular momentum constructions or all theories in other spacetime settings.

Finally, a chiral gauge spectrum means that the independent left-handed Weyl spinors occur in complex gauge group representations without obligatorily paired conjugate left-handed Weyl spinors. In four-dimensional N=1 supersymmetry, an irreducible matter supermultiplet with helicities $(0,-\frac{1}{2})$ can carry a complex gauge group representation $R$. Its CPT completion of a supermultiplet supplies the right-handed antiparticle in $\bar{R}$, as every left-handed Weyl spinor requires; it does not supply a second independent left-handed Weyl spinor in $\bar{R}$. Consequently a chiral gauge spectrum is possible.

For $\mathcal{N}=2$, the matter hypermultiplet has helicities $(\frac{1}{2},0,0,-\frac{1}{2})$ before internal charge conjugation. If it carries a complex gauge group representation $R$, its CPT completion of a supermultiplet adds the conjugate hypermultiplet. Equivalently, the resulting full hypermultiplet contains two $\mathcal{N}=1$ chiral superfields, one in $R$ and one in $\bar{R}$. These give a vectorlike gauge spectrum. Half-hypermultiplets can occur in pseudoreal representations, which also do not produce genuinely complex chiral gauge spectra. A supersymmetric vector multiplet has Adjoint representation, hence real gauge quantum numbers. Since every $\mathcal{N}\ge 2$ algebra contains an $\mathcal{N}=2$ subalgebra, the same obstruction applies to greater extended supersymmetry.

Therefore a conventional chiral gauge spectrum is possible only for $\mathcal{N}=0$ or $\mathcal{N}=1$. Here $\mathcal{N}=0$ means no supersymmetry; among theories with actual supersymmetry, only $\mathcal{N}=1$ qualifies. These are necessary structural conditions, not a guarantee that a chosen chiral gauge spectrum satisfies anomaly cancellation.

Use the canonical Kähler potential and normalize superspace integration by $\int d^2\theta {\theta }^2=1$. The superspace Lagrangian density isThe full superspace integration produces a D-term, while the chiral superfield integral produces an F-term. Work in chiral coordinates for the chiral-superfield component expansion, so the omitted spacetime-derivative terms do not enter the extraction of $F$.

Set $\delta \Phi =\sqrt{2}\theta \psi +{\theta }^{2}F$. A Taylor expansion of the superpotential givesHigher terms vanish because there are only two Grassmann variables ${\theta }^{\alpha }$. With $\theta \psi ={\theta }^{\alpha }{\psi }_{\alpha }$ and $\psi \psi ={\psi }^{\alpha }{\psi }_{\alpha }$, their anticommutation gives $(\theta \psi {)}^2=-\frac{1}{2}{\theta }^2\psi \psi$. ThusThe canonical Kähler potential contributes $F^{\ast }F$. Consequently the part of the Lagrangian density containing the auxiliary field isThe fermion terms coming from the superpotential areThe Weyl spinor kinetic term and the complex scalar field kinetic term are canonical because $K={\Phi }^{†}\Phi$.

Treat $F$ and $F^{\ast }$ as independent variables in their algebraic Euler-Lagrange equations. They giveEquivalently,After eliminating the auxiliary field, the Lagrangian density contains $-V$, so the F-term scalar potential isIt is nonnegative for arbitrary complex $m,g,\phi$, since it is a modulus squared. A supersymmetric vacuum satisfies F-flatness, so its zeros are:If $m=0$ and $g\ne 0$ these coincide; if $g=0$ and $m\ne 0$ only the origin remains; if $m=g=0$ every constant $\phi$ is a supersymmetric vacuum.

To compare masses, expand around a supersymmetric vacuum $v$ satisfying $W^{\prime }(v)=0$, writing $\phi =v+\eta$. With $M=W^{\prime \prime }(v)=m+2gv$,Writing $\eta =(A+iB)/\sqrt{2}$ gives the scalar field mass term $-\frac{1}{2}|M{|}^2(A^2+B^2)$; the Weyl spinor has physical mass $|M|$. Therefore the supersymmetric mass degeneracy isIn particular, at $v=0$ the boson and fermion masses are both $|m|$. At the second supersymmetric vacuum $v=-m/g$ the sign of $M$ reverses, but the physical mass is again $|m|$. Equality here concerns fluctuations about a supersymmetric vacuum, not arbitrary backgrounds with $W^{\prime }(v)\ne 0$.

In the convention $V\supset \lambda |\phi {|}^4$ and ${\mathcal{L}}_{\mathrm{Yukawa}}=-y\phi \psi \psi +h.c.$, the same superpotential coefficient gives $\lambda =|g{|}^2$ and $y=g$. The supersymmetric relation between quartic and Yukawa couplings is thereforeIf instead the Yukawa interaction is written $-\frac{1}{2}{y}_3\phi \psi \psi +h.c.$, then $y_3=2g$ and the same physical relation is $\lambda =|y_3{|}^2/4$. The numerical factor depends on the definition of the Yukawa coupling.

The perturbative non-renormalization theorem applies to the local holomorphic superpotential in the Wilsonian effective action, using regularization in quantum field theory that preserves supersymmetry. There are two complementary ways to see why it applies here.

First, in supergraph calculations in perturbative quantum field theory, the algebra of supersymmetric covariant derivatives puts any candidate local correction to the superpotential into a full $d^4\theta$ superspace integration. It therefore corrects a D-term, such as the Kähler potential, rather than a local chiral $d^2\theta$ F-term. Turning such a contribution into an apparent chiral integral would require nonlocal inverse spacetime derivatives, schematically $1/\square$. These are excluded from the local Wilsonian effective action with a fixed nonzero momentum cutoff. This gives the diagrammatic non-renormalization theorem.

Second, the holomorphy argument for superpotential non-renormalization treats $m$ and $g$ as background chiral spurions, so the Wilsonian effective action superpotential is a holomorphic function of $\Phi ,m,g$, with no dependence on $m^{\ast }$ or $g^{\ast }$. Assign an ordinary $U(1)$ spurion charge and an R-charge as follows:The superpotential has ordinary charge zero and R-charge two, since $d^2\theta$ has R-charge minus two. A prospective holomorphic monomial $m^a{g}^b{\Phi }^c$ must therefore satisfyFor perturbative corrections, regularity at $m=0$ and $g=0$ requires nonnegative powers of the couplings. Regularity in $m$ is justified by keeping a nonzero Wilsonian momentum cutoff and retaining $\Phi$ as a light field. Hence only $c=2$ and $c=3$ survive: the existing $m{\Phi }^2$ and $g{\Phi }^3$ structures. The quadratic coefficient is fixed by the free theory at $g=0$, where there are no interaction loops. The cubic coefficient is fixed at first order in $g$, where a connected three-field interaction has only its tree-level Feynman diagram; loop corrections would require further powers of the interaction couplings, which the holomorphy argument for superpotential non-renormalization charge constraints forbid. ThusA field-independent constant is physically irrelevant in this rigid model; the spurion charges would in any case require $m^3/g^2$, which is not regular at $g=0$.

The Kähler potential can still receive wave-function renormalization. For example, $K=Z{\Phi }^{†}\Phi +\cdots$ gives the canonical field normalization ${\Phi }_c=Z^{1/2}\Phi$, and consequentlyThese canonically normalized parameters can run with the renormalization scale; this does not contradict the holomorphic non-renormalization theorem. When massless modes are included in the full quantum effective action, nonlocal terms associated with infrared divergences can also imitate an F-term. Distinguishing that action from the local Wilsonian effective action is part of the theorem's convention.

Write $m=|m|e^{i\theta }$. Parity reverses the spatial coordinates and must exchange the two chiral Weyl spinors. One convenient choice isThe identities ${\sigma }^0={\bar{\sigma }}^0$ and ${\sigma }^i=-{\bar{\sigma }}^i$ exchange the two kinetic terms after $\mathbf{x}\mapsto -\mathbf{x}$. The mass bilinear $m{\bar{\psi }}_R{\psi }_L$ maps to its complex conjugate $m^{\ast }{\bar{\psi }}_L{\psi }_R$, so the two mass terms are exchanged and the action is invariant. More generally, independent unit phases $a,b$ may multiply the two transformations provided $b^{\ast }a=m^{\ast }/m$.

In four spacetime dimensions a Weyl spinor has mass dimension $3/2$, so each chiral current $J_L^{\mu }={\bar{\psi }}_L{\bar{\sigma }}^{\mu }{\psi }_L$ or $J_R^{\mu }={\bar{\psi }}_R{\sigma }^{\mu }{\psi }_R$ has dimension three. Since the interaction is $(g{J}_L+g^{\prime }{J}_R{)}^2$, dimensional homogeneity of the Lagrangian givesEquivalently, the expanded four-fermion interaction has coefficients $g^2,g{g}^{\prime },g^{\prime 2}$ of mass dimension minus two.

Parity exchanges $J_L$ and $J_R$. The square is invariant precisely when the two same-chirality coefficients agree, $g^2=g^{\prime 2}$, while the mixed coefficient is already symmetric. Therefore

A chiral superfield obeysExamples in the Minimal supersymmetric Standard Model include the quark, lepton and Higgs chiral superfields. Such a multiplet contains a complex scalar $\varphi$, a two-component Weyl spinor $\psi$, and a complex auxiliary field $F$; only the scalar and fermion propagate on shell.

The chiral coordinate $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$ is annihilated in the required combination by ${\bar{D}}_{\dot{\alpha }}$, so the solution has the simple formTaylor-expand each component about $x$. Nilpotence truncates the series, and the Grassmann identitygives the chiral-superfield component expansionTherefore

With ${ϵ}_{0123}=+1$, define the Pauli-Lubanski pseudovector byFor a translation $a^{\mu }$ and a Lorentz transformation $\Lambda =\exp (\omega )$, a left-handed Weyl spinor transforms asEquivalently, the active field at a fixed point uses ${\psi }^{\prime }(x)=S(\Lambda )\psi ({\Lambda }^{-1}(x-a))$.

The Clebsch-Gordan decomposition isIndeed,The symmetric spinor is the $(1,0)$ representation, while its antisymmetric part is the Lorentz scalar $(0,0)$.

In the conventions used below, the nonzero brackets involving the supercharges areand $\{Q_{\alpha },Q_{\beta }\}=\{{\bar{Q}}_{\dot{\alpha }},{\bar{Q}}_{\dot{\beta }}\}=0$. These equations, together with the Poincare algebra, are the four-dimensional $\mathcal{N}=1$ Super-Poincaré algebra without central charges.

A coordinate transformation $x\mapsto Px$ conjugates an infinitesimal generator, hence $M^{\mu \nu }\mapsto P{M}^{\mu \nu }P$. It fixes the rotation generators $J_i$ and negates the boosts $K_i$. Therefore it exchanges $A_i$ and $B_i$, and the parity action on a Lorentz representation isAn irreducible representation is parity invariant precisely when $j_L=j_R$. If $j_L\ne j_R$, parity invariance requires the reducible sum $(j_L,j_R)\oplus (j_R,j_L)$. Thus the two Weyl-spinor representations $(1/2,0)$ and $(0,1/2)$ are exchanged, while their direct sum is the parity-invariant Dirac spinor.

In the chiral coordinate $y^{\mu }=x^{\mu }+i\bar{\theta }{\bar{\sigma }}^{\mu }\theta$, a chiral superfield isHere $\varphi$ is a complex scalar, ${\psi }_{\alpha }$ is a two-component Weyl spinor, and $F$ is a complex auxiliary field. Off shell, $\varphi$ and $F$ contribute two real bosonic components each, while $\psi$ has two complex Grassmann components. Thus there are four real bosonic and four real fermionic degrees of freedom.

A quark field is a Dirac field or a chiral Weyl spinor carrying the colour fundamental representation of $SU(3)$. Its three colour components transform as ${\psi }^i\mapsto U^i{}_j{\psi }^j$ under local gauge transformations. The colour index is independent of the Lorentz spinor index, and the weak representations of its left and right chiral components differ in the Standard Model.

An irreducible complex Clifford algebra module in even spacetime dimension $2k$ has dimension $2^k$ and splits under the connected Spin group into two Weyl spinor modules of dimension $2^{k-1}$. In odd dimension $2k+1$, an irreducible spinor still has dimension $2^k$ but has no Weyl splitting. Reality conditions depend on signature and are additional to these complex counts.

With left Grassmann derivatives, define ${\mathcal{P}}_{\mu }=-i{\partial }_{\mu }$, ${\mathcal{Q}}_{\alpha }=-i{\partial }_{{\theta }^{\alpha }}-{\sigma }_{\alpha \dot{\gamma }}^{\mu }{\bar{\theta }}^{\dot{\gamma }}{\partial }_{\mu }$ and ${\bar{\mathcal{Q}}}_{\dot{\beta }}=i{\partial }_{{\bar{\theta }}^{\dot{\beta }}}+{\theta }^{\gamma }{\sigma }_{\gamma \dot{\beta }}^{\mu }{\partial }_{\mu }$. The anticommutators of the derivative parts and of the coordinate parts vanish. Each mixed derivative/coordinate pair contributes $-i{\sigma }^{\mu }{\partial }_{\mu }$ by the graded product rule, giving the displayed mixed anticommutator. Equal-chirality anticommutators vanish, and all these operators commute with translations. Under Lorentz transformations their spinor indices transform in conjugate Weyl spinor representations, completing the minimal Super-Poincaré algebra. A convention for left derivatives is essential to the signs.

In an unbroken supersymmetric vacuum, physical boson and fermion partners have equal masses because supercharges commute with four-momentum. For one canonical chiral superfield, fluctuations about $W^{\prime }(v)=0$ have mass $|W^{\prime \prime }(v)|$ for both the complex scalar field and Weyl spinor.

A supersymmetry generator is an odd Weyl spinor generator of the Super-Poincaré algebra. Acting on a physical state reverses fermion parity while preserving its four-momentum.