Codex @codex  0

With left Grassmann derivatives, a chiral superfield is a finite expansion in chiral coordinates $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$:Here $\varphi$ is a complex scalar field, $\psi$ a Weyl spinor and $F$ an auxiliary field. The translation form of its ordinary-coordinate expansion isThe exponential terminates in the Grassmann algebra. To reduce its contractions, choose signature $(+---)$, ${\sigma }^{\mu }=(I,\mathbf{\sigma })$, left differentiation, and ${ϵ}_{12}=1$ for both lower spinor epsilon tensors. Define ${\theta }^2={\theta }^{\alpha }{\theta }_{\alpha }$ and ${\bar{\theta }}^2={\bar{\theta }}_{\dot{\alpha }}{\bar{\theta }}^{\dot{\alpha }}$. Then $v^{\mu }=\theta {\sigma }^{\mu }\bar{\theta }$ satisfies $v^{\mu }{v}^{\nu }=\frac{1}{2}{\theta }^2{\bar{\theta }}^2{\eta }^{\mu \nu }$. Expanding the translation givesThe last sign follows from $i^2/2=-1/2$ in the translation expansion. Defining the barred square in the reverse order, ${\bar{\theta }}^{\dot{\alpha }}{\bar{\theta }}_{\dot{\alpha }}$, negates it and writes the same term with a positive quarter coefficient. Spinor contractions and the Grassmann algebra convention must therefore accompany the reduced formula.

On the invertible-$F_X$ branch, use $y=(G\chi )/F_X-(GG)F_Y/(2F_X^2)$ and the two-component Grassmann algebra identity $(G\chi {)}^2=-\frac{1}{2}(GG)(\chi \chi )$. Then $y^2=-(GG)(\chi \chi )/(2F_X^2)$, $y^3=y^2\chi =0$ and $y^2{F}_Y-y\chi \chi =0$. These are the components of $Y^3=0$. Generally $y^2$ is nonzero, so cubic nilpotency must not be replaced by quadratic nilpotency. The allowed analytic holomorphic monomials are $1,X,Y,Y^2$.

The exterior algebra of a vector space $V$ combines scalars, vectors, and higher-dimensional oriented elements in one graded algebra. In a finite-dimensional vector space it iswith multiplication given by the exterior product. Degree zero is the scalar field, degree one is $V$, and degree two is the space of bivectors. Multiplication adds degrees: combining two vectors produces a bivector, rather than another ordinary vector.^[ref]

The essential rule is $v\wedge v=0$. Repeating the same direction cannot produce a nonzero oriented area. Bilinearity then gives $u\wedge v=-v\wedge u$: expand $(u+v)\wedge (u+v)=0$ and remove the two zero self-products. The alternating rule remains part of the definition in characteristic two, where sign reversal alone would not force a self-product to vanish.

If $V={\mathbb{R}}^2$ with standard basis $e_1,e_2$, every element of its exterior algebra has the formThere are four independent coefficients: one scalar, two vector components, and one area component. In ${\mathbb{R}}^3$, the respective degree dimensions are $1,3,3,1$, giving eight components in total. More generally, the exterior power basis gives total dimension ${\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)=2^n$.

This language is useful when direction and orientation matter together. Determinants measure the action on top-degree exterior powers; differential forms use exterior powers of a dual space to describe quantities integrated along paths, surfaces, and higher-dimensional regions. The Grassmann algebra viewpoint also makes the sign changes in anticommuting generators systematic. A single multiplication rule connects geometry and algebra.

A Grassmann derivative is an odd differentiation operation on a Grassmann algebra. A left Grassmann derivative obeys ${\partial }_{\theta }^L(ab)=({\partial }_{\theta }^{L}a)b+(-1{)}^{|a|}a({\partial }_{\theta }^{L}b)$ for homogeneous $a$. Right derivatives use the corresponding right-sided product rule. Their placement must be fixed when differentiating fermionic sources in a generating functional, since commuting ordinary derivatives would lose fermionic signs.

A classical supersymmetric mechanical model takes its coordinates in a Grassmann algebra: the position is even and the fermionic coordinates are odd. Time differentiation is even, while variation with respect to an odd coordinate uses Grassmann derivatives. For example, moving an odd variation to the left and integrating by parts gives $\delta \int \psi \dot{\psi }dt=2\int \delta \psi \dot{\psi }dt$.

For the real variant $L={\dot{x}}^2/2+U^2/2-{\psi }_1{\dot{\psi }}_1/2+{\psi }_2{\dot{\psi }}_2/2+U^{\prime }{\psi }_1{\psi }_2$, the Euler-Lagrange equations areThe conserved even energy is $E={\dot{x}}^2/2-U^2/2-U^{\prime }{\psi }_1{\psi }_2$ and the odd supercharges are $Q_1=\dot{x}{\psi }_1-U{\psi }_2$, $Q_2=\dot{x}{\psi }_2-U{\psi }_1$. For example, ${\dot{Q}}_i=U^{\prime \prime }{\psi }_1{\psi }_2{\psi }_i=0$, since each odd element squares to zero. Also $d({\psi }_1{\psi }_2)/dt=U^{\prime }({\psi }_2^2+{\psi }_1^2)=0$, which proves $\dot{E}=0$.

A Grassmann variable is an odd generator $\theta$ of a Grassmann algebra. In characteristic zero it obeys ${\theta }^2=0$, so functions of finitely many Grassmann variables have finite Taylor expansions.

Left differentiation with respect to an odd Grassmann variable is the odd linear operation ${\partial }_{\theta }^L\theta =1$, with zero derivative on every other generator. For a homogeneous element $a$ of parity $|a|$, it obeys the graded product ruleFor independent odd generators, ${\partial }_{\bar{\theta }}^L(\theta \bar{\theta })=-\theta$. This sign comes from moving the derivative through the first odd factor in the Grassmann algebra. It fixes chirality in the chiral-superfield component expansion; an ordinary commuting-variable product rule would give the wrong sign.

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.

Work with metric $\eta =\mathrm{diag}(1,-1,-1,-1)$, ${\sigma }^{\mu }=(I,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, where $\mathbf{\sigma }$ are the Pauli matrices. The supercharges of four-dimensional N=1 supersymmetry form one Weyl spinor $Q_{\alpha }$ and its conjugate ${\bar{Q}}_{\dot{\alpha }}$. Dotted and undotted indices cannot be interchanged; the conjugate charge in the printed mixed relation has a dotted index.

A direct derivation starts from superspace coordinates $x^{\mu },{\theta }^{\alpha },{\bar{\theta }}^{\dot{\alpha }}$, with the four odd coordinates generating a Grassmann algebra. Use left Grassmann derivatives, soOdd derivatives anticommute with one another, and multiplication by distinct odd coordinates also anticommutes. The operators specified by the question can be writtenTheir coefficients do not depend on $x$, hence $[{\mathcal{Q}}_{\alpha },{\mathcal{P}}_{\mu }]=[{\bar{\mathcal{Q}}}_{\dot{\alpha }},{\mathcal{P}}_{\mu }]=0$. Within $\{{\mathcal{Q}}_{\alpha },{\mathcal{Q}}_{\beta }\}$, the two odd derivative terms anticommute; the derivatives of the barred-coordinate coefficients vanish; and the two barred-coordinate multiplication terms cancel because the coordinates anticommute while the spacetime derivatives commute. ThusThis realizes the minimal algebra with no extra tensorial charges.

For the mixed anticommutator, split each operator into its odd derivative and odd-coordinate parts. The derivative/derivative anticommutator is zero. The coordinate/coordinate terms also cancel, since $\theta \bar{\theta }=-\bar{\theta }\theta$. The two remaining pieces areAdding gives the superspace differential realization of N=1 supercharges:In particular the differentiated-coordinate contributions have the same sign, not opposite signs. The graded product rule is the essential step; treating these as ordinary commuting-coordinate derivatives would give an incorrect algebra.

To obtain the Lorentz part, let $x$ transform as a four-vector and $\theta ,\bar{\theta }$ in conjugate spinor representations. DefineThe Pauli matrix multiplication law gives the intertwining identityIt shows that the two terms in ${\mathcal{Q}}_{\alpha }$ transform together as a covariant Weyl spinor: the coordinate derivative transforms in the dual spinor representation, while the barred-coordinate/spacetime-derivative term transforms in that same representation through this identity. Infinitesimally, with $U=\exp [-\frac{i}{2}{\omega }_{\mu \nu }{M}^{\mu \nu }]$, its transformation is $U{Q}_{\alpha }{U}^{-1}=Q_{\alpha }-\frac{1}{2}{\omega }_{\mu \nu }({\Sigma }^{\mu \nu }{)}_{\alpha }{}^{\beta }{Q}_{\beta }$. Comparing with the first-order expansion of $U$ givesIf the paper's spin matrix is denoted ${\sigma }^{\mu \nu }=i{\Sigma }^{\mu \nu }$, this is exactly its first displayed relation. Defining that spin matrix without the factor $i$ instead requires an explicit $i$ in the commutator; this is a generator convention, not a different algebra. The conjugate relation follows by Hermitian conjugation.

Combining the Lorentz transformation with the differential computations gives the Super-Poincaré algebra in the question's notation:The even generators retain the ordinary Poincare algebra. As a closure check, the even superspace variation $\delta =i(ϵ\mathcal{Q}-\overline{ϵ}\bar{\mathcal{Q}})$ gives $\delta \theta =ϵ$, $\delta \bar{\theta }=\overline{ϵ}$ and $\delta x^{\mu }=i\theta {\sigma }^{\mu }\overline{ϵ}-iϵ{\sigma }^{\mu }\bar{\theta }$. Two such variations commute to a translation by $2i({ϵ}_1{\sigma }^{\mu }{\overline{ϵ}}_2-{ϵ}_2{\sigma }^{\mu }{\overline{ϵ}}_1)$, with zero commutator on the odd coordinates, reproducing the same normalization and showing why the anticommutator produces momentum.

The gauge-invariant Yang-Mills action integrates over physically equivalent configurations along each gauge orbit. At the quadratic level its gauge-field kernel has longitudinal zero directions: in momentum space $K_{\mu \nu }(k)=-k^2{\eta }_{\mu \nu }+k_{\mu }{k}_{\nu }$ satisfies $K_{\mu \nu }{k}^{\nu }=0$. It therefore has no inverse from which to construct a gauge-boson propagator. Gauge fixing both removes the redundant orbit integration and supplies an invertible quadratic operator away from the ordinary physical propagator poles.

For the covariant condition ${\partial }^{\mu }{A}_{\mu }=0$, the Faddeev-Popov determinant is that of ${\partial }^{\mu }{D}_{\mu }$. Its representation by odd Faddeev-Popov ghost fields and antighost fields produces the displayed ghost term. In a nonabelian theory $D_{\mu }$ depends on the gauge field, so the determinant is not an irrelevant constant: ghost interactions must be retained. The even Nakanishi-Lautrup field $b$ writes gauge fixing in a form with off-shell BRS symmetry. For $\xi \ne 0$, its algebraic equation is $b=-\partial \cdot A/\xi$, giving $-(\partial \cdot A{)}^2/(2\xi )$ after elimination. At $\xi =0$ it imposes the gauge condition as a Lagrange multiplier.

For the nilpotence proof, write the adjoint bracket as $[X,Y{]}^a=f^{abc}{X}^b{Y}^c$, with coefficients in a Grassmann algebra. If $|X|,|Y|$ are their parities, this is a graded Lie bracket:Thus two odd fields have a symmetric bracket, in particular $[c,c]$ need not vanish. The transformations become $s{A}_{\mu }=D_{\mu }c$, $sc=-[c,c]/2$, $s\bar{c}=-b$, $sb=0$, and $s$ obeys the left graded Leibniz rule.

Because $D_{\mu }$ is even, $D_{\mu }[c,c]=[D_{\mu }c,c]+[c,D_{\mu }c]=2[D_{\mu }c,c]$. Varying both the field and the connection in $D_{\mu }c$ now givesFor the ghost,The last equality is the graded Jacobi identity applied to three copies of the odd field $c$. In components the remaining expression is $\frac{1}{2}{f}^{abc}{f}^{bde}{c}^d{c}^e{c}^c$; its coefficient is antisymmetrized over $c,d,e$, and the ordinary Jacobi identity for the structure constants makes that antisymmetrization zero. Finally $s^2\bar{c}=-sb=0$ and $s^{2}b=0$. HenceThis is off-shell nilpotence of the Yang-Mills BRST quartet.

To verify invariance of the complete Lagrangian, $s{F}_{\mu \nu }=[F_{\mu \nu },c]$, obtained from $[D_{\mu },D_{\nu }]c=[F_{\mu \nu },c]$. Thus $s(F_{\mu \nu }^a{F}^{\mu \nu a})=0$ because the product of the two curvature components is symmetric in their color labels while $f^{abc}$ is antisymmetric. The remaining terms vary asThe last term vanishes by the already proved $s^{2}A=0$, and the first and third terms cancel pointwise. ThereforeEquivalently, the covariant gauge-fixing density as a BRST variation is $s\Psi$ with $\Psi =-{\bar{c}}^a(\partial \cdot A^a+\xi b^a/2)$. The graded sign from moving $s$ past $\bar{c}$ produces the positive ghost term, and $s^2\Psi =0$ proves its invariance. No integration-by-parts qualification is needed for the density as written.

Let $Q=Q_{\mathrm{BRS}}$ be the BRS charge. The physical state space is its BRST cohomology, usually at ghost number zero:Nilpotence ensures $\mathrm{im}Q\subseteq \ker Q$. A state $Q|\chi ⟩$ is orthogonal to every closed state because $⟨\psi |Q|\chi ⟩=⟨Q\psi |\chi ⟩=0$, and its norm is $⟨\chi |Q^2|\chi ⟩=0$. Thus adding an exact state changes neither physical inner products nor amplitudes. Self-adjointness here is relative to the stated indefinite Hermitian form; a nonzero self-adjoint nilpotent operator could not exist on a positive-definite state space. Positivity must be proved on the quotient, not assumed on the gauge/ghost space.

Fix a color label and a nonzero momentum $k$, and take the usual nontrivial asymptotic charge, $\alpha \ne 0$. Nilpotence on a vector state gives $Q|c⟩=0$, because at least one component of $k$ is nonzero. Nilpotence on the antighost state then givesThe two coefficients are complex conjugates as required by self-adjointness: using the given ghost pairing,Thus the antighost-to-vector coefficient is ${\alpha }^{\ast }$, as the printed bar notation indicates. These conditions are the massless shell condition and the consistency of the charge with the inner products.

A general one-particle vector representative is $|\epsilon ⟩={\epsilon }^{\mu }|A_{\mu }⟩$. It is closed precisely when $k_{\mu }{\epsilon }^{\mu }=0$. The vector state $k^{\mu }|A_{\mu }⟩=Q|\bar{c}⟩/\bar{\alpha }$ is exact, so representatives obeyThe ghost state is also exact: choose $v^{\mu }$ with $\alpha v\cdot k=1$, then $Q(v^{\mu }|A_{\mu }⟩)=|c⟩$. The antighost state is not closed because its image is a nonzero longitudinal vector. More explicitly, closure of a general sum of vector, ghost and antighost states forces the antighost coefficient to vanish and the vector polarization to be orthogonal to $k$; its ghost component disappears in the quotient. Thus the full one-particle cohomology already reduces toWith the mostly-plus Minkowski metric, choose a Lorentz frame with $k^{\mu }=(\omega ,0,0,\omega )$, $\omega >0$. Closure gives ${\epsilon }^3={\epsilon }^0$, and the norm isSubtracting a multiple of $k$ sets ${\epsilon }^0={\epsilon }^3=0$. Consequently positive one-particle BRST cohomology consists of two transverse positive-norm polarizations per color, represented by $|A_1⟩,|A_2⟩$. A nonzero class has strictly positive norm; the zero-norm longitudinal direction was precisely the exact direction removed.

The qualifications $k\ne 0$ and $\alpha \ne 0$ are essential to this one-particle argument. The zero-momentum sector requires separate treatment, and the trivial-charge case cannot yield the asserted removal of negative-norm vector states from the stated relations alone. Abstract nilpotence and self-adjointness by themselves do not guarantee positive cohomology.

Odd elements of a Grassmann algebra anticommute, and in particular ${\psi }_i^2=0$. For variations with fixed endpoints, move each odd variation to the left before applying integration by parts. For example,The interaction varies as ${\delta }_{{\psi }_1}({\psi }_1{\psi }_2)=\delta {\psi }_1{\psi }_2$ and ${\delta }_{{\psi }_2}({\psi }_1{\psi }_2)=-\delta {\psi }_2{\psi }_1$. Thus variation of the action gives the Euler-Lagrange equationsAll products of $U$ and its derivatives commute with the odd variables. These signs are essential for this variant of Grassmann-valued classical supersymmetric mechanics.

The fermion bilinear is constant along a solution:For the energy, direct differentiation now givesSimilarly the given supercharge, denoted $Q_1$, satisfiesA second conserved supercharge issince ${\dot{Q}}_2=(\ddot{x}-U{U}^{\prime }){\psi }_2=U^{\prime \prime }{\psi }_1{\psi }_2{\psi }_2=0$. The two charges are independent functions on the unrestricted phase space: their coefficient matrix on $({\psi }_1,{\psi }_2)$ has determinant ${\dot{x}}^2-U^2$, and is invertible wherever its even body is nonzero. They can become dependent on a restricted family of trajectories.

Under $\dot{x}=U$ and ${\psi }_2=-{\psi }_1$, the bilinear ${\psi }_1{\psi }_2$ vanishes. Differentiating the first-order equation gives $\ddot{x}=U^{\prime }\dot{x}=U{U}^{\prime }$, which is the bosonic Euler-Lagrange equation. The first fermionic equation becomes ${\dot{\psi }}_1=-U^{\prime }{\psi }_1$; differentiating ${\psi }_2=-{\psi }_1$ gives ${\dot{\psi }}_2=U^{\prime }{\psi }_1$, so the second fermionic equation holds too. This establishes the first-order zero-energy solution in Grassmann supersymmetric mechanics without imposing an additional equation.

Let $x_0=x(0)$ and ${\psi }_{10}={\psi }_1(0)$. For $U=1+cx$ and $c\ne 0$, solve $d(x+1/c)/dt=c(x+1/c)$ and ${\dot{\psi }}_1=-c{\psi }_1$ to obtainFor $c=0$, the nonsingular limiting answer is $\boxed{x(t)=x_0+t,{\psi }_1(t)={\psi }_{10},{\psi }_2(t)=-{\psi }_{10}}$. These formulas also hold for even initial positions in the Grassmann algebra by ordinary polynomial evaluation. Finally, $U(x(t))=(1+c{x}_0)e^{ct}$, so the conserved quantities areThe individual odd supercharges generally do not vanish, although their sum vanishes on this restricted solution.

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