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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.

Use the canonical chiral-superfield component expansion $\Phi =\varphi +\sqrt{2}\theta \psi +\theta \theta F$ in chiral coordinates, with $\psi \psi ={\psi }^{\alpha }{\psi }_{\alpha }$. Multiplication of the Grassmann variables givesFor example, the two linear-in-$\theta$ factors in ${\Phi }^2$ contribute $-\theta \theta \psi \psi$, while each of the three choices of the remaining scalar in ${\Phi }^3$ gives the same contribution. Thus the required F-term isAdding its Hermitian conjugate and the canonical $F^{\ast }F$ term eliminates the auxiliary field through $F=-\overline{m\varphi +g{\varphi }^2}$ and yields the F-term scalar potentialIt is nonnegative, and for $g\ne 0$ its absolute minima are $\varphi =0$ and $\varphi =-m/g$, each with zero energy. They coincide when $m=0$. If $g=0,m\ne 0$, only $\varphi =0$ is a minimum; if $g=m=0$, every $\varphi$ is a minimum. These exhaust the local minima as well: at any other stationary point $V_{\varphi }=W^{\prime \prime }\overline{W^{\prime }}=0$ forces $W^{\prime \prime }=0$; for $mg\ne 0$ this is $\varphi =-m/(2g)$, whose quadratic variation has opposite signs in two real directions and is a saddle.

Expand around $\varphi =0$, writing $\varphi =h=(h_1+i{h}_2)/\sqrt{2}$. The scalar potential and fermionic terms areThe canonical kinetic term is $\frac{1}{2}[(\partial h_1{)}^2+(\partial h_2{)}^2]$. Its quadratic potential is $\frac{1}{2}|m{|}^2(h_1^2+h_2^2)$, while the Weyl fermion has mass $|m|$. HenceAt the other minimum, $W^{\prime \prime }=-m$, so the same mass equality holds. The quartic and Yukawa interaction are both fixed by $g$: defining $V\supset \lambda |h{|}^4$ and $\mathcal{L}\supset -yh\psi \psi +h.c.$ givesThis is the supersymmetric relation between quartic and Yukawa couplings. The printed equality of the two couplings refers to their common supersymmetric origin; literal equality of the conventional coefficients $\lambda$ and $y$ is not true for arbitrary $g$. If instead the Yukawa term is defined as $-\frac{1}{2}yh\psi \psi$, then $y=2g$ and $\lambda =|y{|}^2/4$.

To establish perturbative superpotential non-renormalization, use a Wilsonian effective action with a fixed positive infrared cutoff, so integrating out modes does not introduce massless infrared singularities. Promote $m,g$ to nondynamical chiral superfields, or spurions. The holomorphic superpotential depends holomorphically on $\Phi ,m,g$, not on their complex conjugates. Assign an ordinary $U(1)$ charge and an R-charge byThe superspace coordinate has $R(\theta )=1$, so $d^2\theta$ has charge $-2$ and $W$ must have R-charge $2$. Both original monomials satisfy the two selection rules. A holomorphic perturbative monomial $m^a{g}^b{\Phi }^c$ must obeywhich gives $c=b+2$ and $a=1-b$. Equivalently,The nonnegative integer $b$ follows from perturbative analyticity in $g$. Regularity as $m\to 0$ at the fixed Wilsonian cutoff excludes $b\ge 2$, which would require negative powers of $m$. The remaining quadratic coefficient is fixed to $c_0=1/2$ by the free theory at $g=0$. The coefficient $c_1$ is $1/3$: a three-point contribution at first order in $g$ has just one cubic vertex, uses all three legs externally, and cannot contain a loop. A loop correction would require additional interactions, whereas the selection rules and holomorphy have already excluded the corresponding monomials. ThereforeThis proof concerns the local Wilsonian superpotential. The Kähler potential can acquire a wave-function factor $Z$, and canonical normalization changes the couplings to $m_c=m/Z$ and $g_c=g/Z^{3/2}$. Such running, and nonlocal infrared terms in a one-particle-irreducible action, do not contradict the holomorphic result.

The barred notation denotes the antichiral supersymmetric covariant derivative. Write barred spinor indices as dotted indices, and use the stated plus-sign convention with left Grassmann derivatives:Their supercovariant derivative algebra with left derivatives gives $\{{\bar{D}}_{\dot{\alpha }},{\bar{D}}_{\dot{\beta }}\}=0$ and $\{D_{\alpha },{\bar{D}}_{\dot{\beta }}\}=2i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\partial }_{\mu }$. The constraint $\boxed{{\bar{D}}_{\dot{\alpha }}\Phi =0}$ defines a chiral superfield, also called an L-superfield.

To solve the constraint, put $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$. For a left Grassmann derivative, ${\partial }_{{\bar{\theta }}^{\dot{\alpha }}}(\theta {\sigma }^{\mu }\bar{\theta })=-{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }$. Therefore ${\bar{D}}_{\dot{\alpha }}{y}^{\mu }=0$ and ${\bar{D}}_{\dot{\alpha }}{\theta }^{\beta }=0$, so $\bar{D}$ becomes ordinary differentiation with respect to $\bar{\theta }$ at fixed $y$. Its general solution is the chiral-superfield component expansionHere $A$ is a complex scalar, $\chi$ a Weyl spinor, and $F$ a complex auxiliary field. Before imposing field equations of motion this is four real bosonic and four real fermionic components. Chirality means independence of $\bar{\theta }$ at fixed $y$, rather than at fixed $x$. The differential supercharges $Q_{\alpha }={\partial }_{{\theta }^{\alpha }}-i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\bar{\theta }}^{\dot{\beta }}{\partial }_{\mu }$ and ${\bar{Q}}_{\dot{\alpha }}={\partial }_{{\bar{\theta }}^{\dot{\alpha }}}-i{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }{\partial }_{\mu }$ anticommute with both covariant derivatives, so a supersymmetry transformation preserves the chiral constraint.

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.

The general holomorphic form. Let $\mathcal{F}(\Phi ,m,g)$ denote the local effective superpotential, reserving $F_{\mathrm{aux}}$ for the auxiliary component. The holomorphy argument for superpotential non-renormalization permits dependence on the chiral spurions, not on their conjugates. The dimensionless combination $z=g\Phi /m$ is neutral under both formal symmetries, whereas $m{\Phi }^2$ has the required dimension and charges. Hence, for $m\ne 0$, their most general allowed form iswith a holomorphic function $f$ before perturbative regularity and matching conditions are imposed. Equivalently, a monomial $m^a{g}^b{\Phi }^c$ must obeySolving gives $a=1-b$, $c=b+2$. Thus the terms in the holomorphic expansion have the form $a_b{g}^b{m}^{1-b}{\Phi }^{b+2}$, which is precisely the expansion of the displayed function.

What is and is not renormalized. Apply the non-renormalization theorem to a local Wilsonian effective action retaining the elementary field and a nonzero infrared cutoff. Perturbative coefficients are regular as $g\to 0$ and $m\to 0$: no massless infrared modes have been integrated all the way to zero momentum. Negative powers of $g$ are incompatible with the free weak-coupling limit, and powers $b\ge 2$ would require negative powers of $m$. Only the quadratic and cubic structures survive this regularity test. Their coefficients cannot acquire a loop correction here: a quadratic term with no $g$ is the free-theory mass term, while a cubic term only linear in $g$ is already the tree interaction. A loop renormalizing that cubic term requires additional interaction insertions. Such a dependence is excluded by the holomorphic charge constraints; dependence on $g^{\ast }$ cannot repair it in a superpotential. Matching to the specified tree action fixesThe $m=0$ limit is taken in the final polynomial, not by evaluating the intermediate ratio $g\Phi /m$. The holomorphic Wilsonian superpotential and its parameters $m,g$ receive no independent perturbative vertex renormalization. Symmetries alone would only give the arbitrary function $f$; regularity and the free/tree matching are necessary to reach the stronger conclusion.

The Kähler potential is not protected by that theorem. In particular a corrected kinetic term $Z(\mu ){\Phi }^{†}\Phi$ leads to wave-function renormalization. Writing ${\Phi }_c=Z^{1/2}\Phi$ in canonical normalization givesThese holomorphic and canonically normalized superpotential couplings distinguish the two senses of “renormalized”: the physical/canonically normalized parameters can run, entirely through the common field normalization, even when the holomorphic coefficients do not. The scalar effective potential can consequently receive quantum corrections through the Kähler potential. Nor does the local perturbative statement automatically apply to infrared-singular one-particle-irreducible actions or to integrating out whole massive fields. Non-renormalization protects the local holomorphic F-term, not the complete quantum action or every physically normalized mass and coupling.

With left Grassmann derivatives, differentiating $\theta {\sigma }^{\mu }\bar{\theta }$ with respect to $\bar{\theta }$ introduces a minus sign. ConsequentlyAt fixed $y$, the chirality condition becomes independence of $\bar{\theta }$. There are only two independent Grassmann variables ${\theta }^{\alpha }$, so the chiral-superfield component expansion terminates:Here $\varphi$ is a complex scalar field, $\psi$ a Weyl spinor, and $F$ a complex auxiliary field. Its four real off-shell bosonic components, two in $\varphi$ and two in $F$, match the four real off-shell fermionic components. Equivalently, in ordinary coordinates the complete expansion is fixed without ambiguous contraction signs byThe exponential terminates because of the Grassmann algebra.

All derivatives below are left Grassmann derivatives. Decompose the supersymmetric covariant derivatives into odd piecesWhen these operators act on an arbitrary superfield, the graded Leibniz rule givesThe anticommutator of the two pure Grassmann derivatives vanishes. Also $\{A_{\alpha },B_{\dot{\beta }}\}=0$: the spacetime derivatives commute, while the $\theta$ and $\bar{\theta }$ coefficients anticommute. Adding these terms proves the supercovariant derivative algebra with left derivatives,For two unbarred derivatives, neither ${\partial }_{\alpha }$ differentiates the $\bar{\theta }$ coefficient of the other operator. Their coefficient products anticommute, so $\boxed{\{D_{\alpha },D_{\beta }\}=0}$. The same argument with barred and unbarred coordinates exchanged gives $\boxed{\{{\bar{D}}_{\dot{\alpha }},{\bar{D}}_{\dot{\beta }}\}=0}$. These signs follow the derivatives printed in this paper; a convention replacing $\bar{D}$ by $-\bar{D}$ reverses the mixed sign as well.

Set $y^{\mu }=x^{\mu }+i{\theta }^{\alpha }{\sigma }_{\alpha \dot{\beta }}^{\mu }{\bar{\theta }}^{\dot{\beta }}$. Left differentiation givesThe chain rule therefore makes ${\bar{D}}_{\dot{\alpha }}$ simply ${\bar{\partial }}_{\dot{\alpha }}$ at fixed $y,\theta$. The chiral superfield constraint ${\bar{D}}_{\dot{\alpha }}\Phi =0$ means that $\Phi$ is independent of $\bar{\theta }$ in these coordinates. Since there are only two independent entries of $\theta$, its Grassmann algebra expansion terminates at degree two. Naming the coefficients gives the chiral-superfield component expansionHere $\varphi$ is a complex scalar field, $\psi$ a Weyl spinor, and $F$ a nonpropagating auxiliary field; the $\sqrt{2}$ is their conventional normalization.

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.

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.