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For ungauged chiral superfields, the two-derivative supersymmetric action is a real full-superspace integral plus a holomorphic half-superspace integral:The real Kähler potential $K$ determines the Kähler metric and kinetic terms; the superpotential $W$ determines Yukawa interactions and the F-term scalar potential. For canonical normalization, $K={\sum }_i{\Phi }_i^{†}{\Phi }_i$. Berezin integration is normalized to extract the highest component, $\int d^2\theta {\theta }^2=1$, and the conjugate measure is chosen so the canonical term is $F_i^{\ast }{F}_i$. The D-term and F-term highest components vary by spacetime total derivatives, making the action invariant under supersymmetry. Gauging would replace the canonical bilinear by its gauge-covariant version and add a gauge kinetic F-term; no gauge multiplet is needed for the chiral theory here.

In four spacetime dimensions, canonical mass dimensions areSince a Lagrangian has mass dimension four, a power-counting renormalizable quantum field theory requires $[K]=2$ and $[W]=3$, with couplings of nonnegative mass dimension. Up to a positive constant kinetic matrix, field redefinitions and Kähler transformations, the renormalizable ungauged kinetic term is canonical. Holomorphic additions to $K$ integrate to a boundary term. The general renormalizable superpotential isThe coefficients have dimensions $3,2,1,0$, respectively; $m_{ij}$ and $y_{ijk}$ may be taken symmetric. The constant $W_0$ does not affect a global chiral theory. Higher polynomial degree in $W$, nontrivial higher-degree terms in $K$, or higher-derivative interactions are nonrenormalizable interactions and can instead be retained in an effective field theory.

For canonical $K$, extract the ${\theta }^2{\bar{\theta }}^2$ component of ${\Phi }_i^{†}{\Phi }_i$. The part depending on the auxiliary fields is $F_i^{\ast }{F}_i$. To extract the F-term of the superpotential, expand around the complex scalar fields:All higher terms contain at least three identical-type Grassmann variables and vanish. In the convention $(\theta {\psi }_i)(\theta {\psi }_j)=-\frac{1}{2}{\theta }^2{\psi }_i{\psi }_j$, the highest component is $W_i{F}_i-\frac{1}{2}{W}_{ij}{\psi }_i{\psi }_j$. ThereforeHere $W_i=\partial W/\partial {\varphi }_i$ and $W_{ij}={\partial }_i{\partial }_{j}W$. The Yukawa interaction is displayed to distinguish it from the terms actually containing $F$. With signature $(+---)$, the other canonical terms are ${\partial }_{\mu }{\varphi }_i^{\ast }{\partial }^{\mu }{\varphi }_i+i{\bar{\psi }}_i{\bar{\sigma }}^{\mu }{\partial }_{\mu }{\psi }_i$, up to a total derivative.

Take a canonical Kähler potential $K={\Phi }^{†}\Phi$ and the Wess–Zumino model superpotentialThe F-term component givesEliminating the auxiliary field by $F=-\overline{W^{\prime }(\varphi )}$ gives the F-term scalar potential $V=|W^{\prime }(\varphi ){|}^2$. ConsequentlyThis is the supersymmetric relation between quartic and Yukawa couplings: there is one holomorphic parameter $g$, with the quartic coefficient $|g{|}^2$ and the Yukawa coupling $g$ in the displayed normalization. For real $g$ the quartic coefficient is the $g^2$ of the question. They cannot be varied independently while retaining this canonical supersymmetry.

The relation produces the supersymmetric cancellation of quadratic divergences. Bosonic and fermionic Feynman diagrams correcting a scalar mass have equal ultraviolet quadratic coefficients and opposite signs, so the mass is not driven directly to the ultraviolet cutoff scale. In a theory with soft supersymmetry breaking, residual corrections instead have the schematic sizewith coefficients and logarithms depending on the spectrum. Thus superpartner masses near the electroweak scale can stabilize the Higgs boson mass against a much larger cutoff, addressing the radiative part of the hierarchy problem. Heavy sparticles leave a corresponding residual sensitivity; the cancellation does not by itself explain every mass scale.

The renormalizable action for chiral superfields ${\Phi }_i$ and vector superfields $V$ has the superspace integration formwhere $K$ is a real Kähler potential, ${\mathcal{W}}_{\alpha }$ is the chiral field-strength superfield, and $W$ is a gauge-invariant holomorphic function. For the renormalizable theory choose canonical positive kinetic terms, a polynomial $W$ of degree at most three, and field-independent gauge kinetic coefficients $f_{ab}$. We prove the non-renormalization theorem for the local Wilsonian effective action, with a nonzero infrared cutoff and a regulator preserving supersymmetry. The elementary chiral superfields are retained; this is not a claim that eliminating whole massive fields leaves the same polynomial.

Write $W={\sum }_A{\lambda }_A{\mathcal{O}}_A(\Phi )$, treating masses and interaction coefficients alike as background chiral spurions. A local F-term is chiral even when these backgrounds vary, so its coefficient depends holomorphically on $\Phi ,{\lambda }_A$ and the holomorphic gauge couplings. Dependence on ${\bar{\lambda }}_A$ would not be chiral. Such dependence can enter D-terms or terms with supercovariant derivatives instead. With the infrared cutoff kept fixed, the perturbative local coefficients are regular power series around vanishing masses and interactions. This removes the inverse-mass and infrared singular terms that would spoil the following spurion selection rule for perturbative non-renormalization.

Assign every elementary ${\Phi }_i$ R-charge zero, every ${\lambda }_A$ R-charge two, and $\theta$ R-charge one. The superpotential must have R-charge two because $d^2\theta$ has R-charge minus two. Holomorphy excludes the conjugate spurions, so regularity and this formal R-symmetry require exactly one factor of a superpotential spurion in any perturbative candidate. Thus $W_{\mathrm{eff}}$ is linear in the ${\lambda }_A$, with the ordinary flavor spurion charges fixing the permitted field monomials. An anomalous formal R-symmetry is implemented by also transforming the holomorphic gauge coupling; its anomaly must not simply be ignored.

Gauge interactions cannot provide a perturbative coefficient modifying this conclusion. With $\tau =\vartheta /(2\pi )+4\pi i/g^2$, perturbative diagrams are independent of the topological angle $\vartheta$. A holomorphic coefficient invariant under continuous real shifts of $\tau$ is constant, by the Cauchy-Riemann equations. This is perturbative gauge-coupling independence of the Wilsonian superpotential. Contributions involving $e^{2\pi i\tau }$ are nonperturbative and are outside this argument. For an Abelian factor the same angle-independent holomorphy argument applies. Hence a putative perturbative coefficient linear in ${\lambda }_A$ is independent of the gauge coupling and can be evaluated as that coupling tends to zero.

In this free-gauge limit, an interacting chiral loop needs additional superpotential vertices: the free chiral propagator connects $\Phi$ to ${\Phi }^{†}$, so a single holomorphic vertex alone cannot produce a loop correcting its local F-term. Extra vertices introduce extra spurions or their conjugates, both excluded by the preceding charge and holomorphy argument. The surviving coefficient is its tree value. Therefore

For an explicit charge check, take $W=m{\Phi }^2/2+g{\Phi }^3/3$ in the Wess–Zumino model. Give $(\Phi ,m,g)$ ordinary charges $(1,-2,-3)$ and R-charges $(1,0,-1)$. A holomorphic monomial $m^a{g}^b{\Phi }^n$ of the required charges satisfies $n-2a-3b=0$ and $n-b=2$, giving $a=1-b$, $n=b+2$. Perturbative regularity requires $a,b\ge 0$, hence only $(a,b,n)=(1,0,2)$ and $(0,1,3)$ survive. The free limit fixes their coefficients to $1/2$ and $1/3$; there is no room for a coupling-dependent loop correction. This illustrates the holomorphy argument for superpotential non-renormalization explicitly.

The Kähler potential can acquire wave-function renormalization. Restoring canonical field normalization changes physical masses and Yukawa couplings, so their running does not contradict the boxed statement about holomorphic coordinates. Gauge kinetic F-terms have their own renormalization and are not the superpotential. Infrared effects in the one-particle-irreducible quantum effective action, tree-level elimination of retained fields, and genuine nonperturbative effects must likewise be distinguished from this perturbative Wilsonian effective action theorem.

First assume global four-dimensional N=1 supersymmetry, canonical positive Kähler potential, and tree level. Write $W_i={\partial }_{i}W$ and $m_{ij}=W_{ij}$ at the stationary vacuum. The chiral-superfield fermion mass matrix is the symmetric matrix $m$, and the sum of squared Weyl spinor masses is $\mathrm{tr}(m^{†}m)$. From the F-term scalar potential $V={\sum }_k|W_k{|}^2$, its mixed and holomorphic Hessian matrix blocks areFor canonically normalized real and imaginary parts of the scalar fields, the real scalar mass matrix has trace twice the mixed trace. Its holomorphic blocks can split the two real scalar masses but do not change their sum. HenceA Weyl spinor has two spin states, so this proves the tree-level supertrace mass sum rule,The paper defines its supertrace with $(-1{)}^{2j+1}$, the negative of the conventional boson-minus-fermion supertrace used here. The asserted zero is the same in either convention. The result requires the stated canonical tree-level hypotheses; a noncanonical Kähler metric, supergravity, or radiative corrections can change the sum rule.

If vector multiplets are also present, the gauge contributions to the F-term supertrace cancel rather than being omitted. At $D_a=0$, define $C={\sum }_a{g}_a^2\Vert T_{a}v{\Vert }^2$ for the scalar expectation vector $v$. Differentiating $\frac{1}{2}{\sum }_a{D}_a^2$ adds $2C$ to the real-scalar trace. The symmetric gaugino-matter mass matrix has off-diagonal entries $\sqrt{2}{g}_a{T}_{a}v$, giving an extra $4C$ in its fermion squared-mass trace. The covariant scalar kinetic term gives vector squared-mass trace $2C$. Thus their contribution is $2C-2(4C)+3(2C)=0$. The gauge-theory result uses all spin states, including the vector weight three; it is not obtained by dropping massive gauge partners.

The MSSM tree-level sfermion mass constraint explains the phenomenological difficulty with direct visible-sector breaking. If F-term breaking is neutral under an unbroken electric and color gauge symmetry, and no D-term shifts are present, the same trace argument applies within each conserved-charge fermion block. Its corresponding scalar partners cannot all have squared masses above the mean of the light fermion squared masses. Thus canonical tree-level visible-sector supersymmetry breaking alone cannot make every squark and other scalar partner heavy while leaving the observed fermions light. The usual effective soft supersymmetry breaking terms arise after communicating breaking from a separate sector; integrating out that sector, noncanonical interactions and radiative effects evade the hypotheses. The global trace identity alone would not identify a particular light squark without the conserved-charge block argument.

For the specified O'Raifeartaigh model, phases may be chosen so that $\lambda ,\mu ,M$ are real and positive. Its superpotential derivatives areand its F-term scalar potential isFor fixed $x,z$, choose $y=-2\lambda xz/M$ to minimize the final square. Since $|z^2-{\mu }^2{|}^2\ge (|z{|}^2-{\mu }^2{)}^2$,Thus $M^2>2{\lambda }^2{\mu }^2$ makes $z=y=0$ a global minimum with arbitrary $x$. The hierarchy $M\gg \mu$ ensures this for fixed perturbative $\lambda$, rather than for arbitrarily large $\lambda$. The origin is one member of this classically flat family, with $V_0={\lambda }^2{\mu }^4>0$ and $F_X=\lambda {\mu }^2$. The complex field $x$ is a pseudomodulus.

At the origin the chiral-superfield fermion mass matrix, in the $({\psi }_X,{\psi }_Y,{\psi }_Z)$ basis, isIts physical fermion masses are $0,M,M$; the two massive Weyl spinors form one massive four-component fermion. The massless ${\psi }_X$ is the goldstino. More generally, stationarity gives $W_{ij}\overline{W_j}=0$, so the nonzero auxiliary field direction is a null vector of $m$, proving the goldstino zero mode from vacuum stationarity.

Writing $z=(z_R+i{z}_I)/\sqrt{2}$ and similarly for $x,y$, the quadratic scalar potential isThe mass spectrum of the quadratic-cubic O'Raifeartaigh model is thereforeThe scalar squared-mass sum is $4M^2$, while twice the fermion squared-mass sum is also $4M^2$, so the supertrace is zero, as required. The two massless real $x$ modes are tree-level pseudomoduli, which can be lifted by a quantum effective potential; their tree-level masslessness is not the exact symmetry protection enjoyed by the goldstino.

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.

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.

For the stated two-dimensional N=(2,2) supersymmetry conventions, define the supersymmetric covariant derivativesThe terms in which a Grassmann derivative hits the explicit Grassmann coordinate cancel the spacetime-derivative terms, while derivatives involving different signs act on independent coordinates. Therefore

The twisted chiral superfield constraints $D_{-}U={\bar{D}}_{+}U=0$ are solved by the twisted chiral coordinatesThe superfield depends only on $({\tilde{y}}^{\pm };{\theta }^{+},{\bar{\theta }}^{-})$ and has the finite Grassmann variable expansionwhere every component on the right is evaluated at $\tilde{y}$; numerical $\sqrt{2}$ factors may be absorbed into the component definitions.

For one chiral superfield $\Phi$ and one twisted chiral superfield $U$, the most general local two-derivative supersymmetric action isHere $K$ is real, $W$ is a holomorphic superpotential, and $\tilde{W}$ is a holomorphic twisted superpotential. A full superspace integral, a chiral F-term, and a twisted F-term each vary by a spacetime or Berezin total derivative, so all three are supersymmetric.

Choose the Vector R-symmetry and Axial R-symmetry conventionswith conjugate coordinates transforming oppositely, and assign compatible charges to $\Phi$ and $U$. The D-term is invariant when $K$ is neutral, up to a generalized Kähler transformation. The chiral measure has vector R-charge $-2$ and axial charge zero, whereas the twisted chiral measure has axial R-charge $-2$ and vector charge zero. Hence classical invariance requirestogether with neutrality of $K$. Equivalently, $W$ and $\tilde{W}$ must be quasi-homogeneous with these charges; absent suitable charge assignments, the corresponding superpotential breaks that R-symmetry.

The Kähler potential $K(\Phi ,{\Phi }^{†})$ is a real function integrated over all four fermionic coordinates, $\int d^4\theta K$. Its complex Hessian gives the scalar Kähler metric and therefore the kinetic terms. The superpotential $W(\Phi )$ is holomorphic and is integrated over chiral superspace, $\int d^2\theta W+h.c.$; its derivatives determine Yukawa couplings and the F-term scalar potential.

For the canonical Kähler potential $K={\Phi }^{†}\Phi$, extracting the ${\theta }^2{\bar{\theta }}^2$ component of the supplied chiral-superfield component expansion and integrating by parts gives the bosonic actionin the mostly-plus convention. Extracting the ${\theta }^2$ component of a holomorphic function givesso its bosonic term is $F{W}^{\prime }(\varphi )$, with the Hermitian conjugate understood in a real action. For several fields, eliminating each algebraic auxiliary field by $F_i^{\ast }=-W_i$ produces $V_F={\sum }_i|W_i{|}^2$.

There is a conflict in the printed question: with dynamical $X$ and the displayed superpotential, $F_X=\tilde{\varphi }\varphi =M$ forces $M=0$, so the $N=1$ theory has no vacuum branch parametrized by nonzero $M$. The natural intended calculation is the Kähler quotient of the $N=1$ D-flat SQED matter fields before imposing that extra F-term.

On this quotient, $|\varphi |=|\tilde{\varphi }|$ and the gauge-invariant coordinate is $M=\tilde{\varphi }\varphi$. Hence $|\varphi {|}^2=|\tilde{\varphi }{|}^2=|M|$, and the canonical Kähler potential descends toThe associated Kähler metric isIt is smooth for $M\ne 0$ and has a conical singularity at $M=0$, locally $\mathbb{C}/{\mathbb{Z}}_2$. Physically, the charged fields become massless and the U(1) gauge symmetry is restored there, so integrating out the vector multiplet to obtain a sigma-model metric ceases to be valid.

Write the scalar components as ${\varphi }_i,{\tilde{\varphi }}_i,x$. The superpotential givesChoosing the sign of the Fayet–Iliopoulos term so that positive $\zeta$ favors the positively charged fields, the full Supersymmetric quantum electrodynamics potential isChanging the FI sign convention exchanges $\varphi$ with $\tilde{\varphi }$ in the descriptions below.