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Split the free interaction picture field as $\varphi ={\varphi }^{(+)}+{\varphi }^{(-)}$, where the first part contains annihilation operators and the second creation operators. Time ordering places the operator at the later time on the left, while normal ordering places all creation operators on the left. For $x^0>y^0$, commuting the annihilation part of the first field through the creation part of the second givesThe commutator is the c-numberFor $y^0>x^0$, the same calculation interchanges $x,y$. Combining the two time orders proves the two-field Wick contraction identitywhereThe equality with the four-dimensional integral follows by closing the energy contour at the positive-energy pole for $x^0>y^0$ and the negative-energy pole for the reverse order. This fixes the scalar Feynman propagator pole prescription; in this convention $D_F$ itself includes the factor $i$.

Because these interactions have no derivatives, $H_{\mathrm{int}}=-\int {\mathcal{L}}_{\mathrm{int}}{d}^{3}x$. Expand the Dyson seriesApply the Wick theorem to each term together with the external insertions. Every paired field contributes a propagator, and the remaining normal-ordered vacuum expectation vanishes unless no fields remain. Representing the propagators by Fourier integrals and integrating each vertex position gives the four-momentum conservation delta function. The $3!$ and $4!$ contractions into labelled vertex slots cancel the factorials in the interaction density. The resulting scalar-field cubic and quartic Feynman vertices areEach vertex also has $(2\pi {)}^4{\delta }^{(4)}(\sum p)$, each independent internal momentum is integrated with $d^{4}p/(2\pi {)}^4$, and repeated equivalent contractions give the usual Feynman-diagram symmetry factor. For the unamputated correlation functions, retain the propagators joining external insertions to vertices. The displayed unnormalized numerator includes disconnected vacuum bubbles; dividing by $⟨0|S|0⟩$ cancels them. Amputated scattering amplitudes additionally use the LSZ reduction formula.

For stability, examine the classical scalar potentialThe tree-level boundedness criterion is $\lambda >0$, with arbitrary real cubic coupling. If $\lambda =0$, boundedness requires $\mu =0$ and $m^2\ge 0$; if $\lambda <0$, or $\lambda =0$ with $\mu \ne 0$, the potential is unbounded below. Thus a positive quartic coupling permits a stable ground-state vacuum, although it need not be the perturbative vacuum at $\varphi =0$.

For $\lambda >0$ and $m^2\ge 0$, the stability of a cubic-quartic scalar potential determines when zero is a global minimum:The square bracket is nonnegative for every field value exactly whenFor a nontrivial equality case there is a second degenerate minimum at $\varphi =2\mu /\lambda$. If $m^2>0$ and ${\mu }^2>3\lambda m^2$, zero remains a local minimum but is a false vacuum, with a lower minimum elsewhere. If $m^2=\mu =0$ and $\lambda >0$, the origin is the quartic minimum despite its vanishing quadratic curvature. These are classical or tree-level stability statements; renormalized couplings and the quantum effective potential refine the quantum analysis.

For a chiral superfield of mass dimension one, a renormalizable quantum field theory has a quadratic Kähler potential and a superpotential of degree at most three. Before canonical normalization one may writewith real $k_0$, andThe purely holomorphic terms in $K$, their conjugates and its constant have vanishing full superspace integration; they are Kähler transformations of the global theory. Rescaling the chiral superfield sets $Z=1$, so the physically general kinetic choice is $K={\Phi }^{†}\Phi$ up to these terms. The constant superpotential in global supersymmetry also has no component effect. The dimensions of $\ell ,m,g$ are respectively two, one and zero.

To eliminate the linear term, translate the complex chiral superfield as $\Phi ={\Phi }^{\prime }+s$. The shifted linear coefficient isIf $g\ne 0$, this quadratic has a root over the complex numbers; if $g=0$ but $m\ne 0$, take $s=-\ell /m$. The resulting linear term vanishes and the new quadratic coefficient is $m+2gs$. The canonical Kähler potential changes only by holomorphic, antiholomorphic and constant terms, which do not change the global kinetic action. This is the superpotential critical-point shift.

The claim needs an exception: if $m=g=0$ but $\ell \ne 0$, the superpotential is purely linear and no invertible translation can remove it. More generally an invertible regular field redefinition cannot turn a nowhere-zero derivative into a critical point. The theory with $W=\ell \Phi$ has $F=-\bar{\ell }$, a constant positive scalar potential $|\ell {|}^2$, and flat F-term breaking with a linear superpotential. Below, $m$ and $g$ denote the quadratic/cubic coefficients after the valid shift, as intended by the question.

Expand $\Phi =\varphi +\sqrt{2}\theta \psi +{\theta }^{2}F$ in chiral coordinates. The highest component of $W(\Phi )$ is $W^{\prime }(\varphi )F-\frac{1}{2}{W}^{\prime \prime }(\varphi )\psi \psi$, while the full-superspace kinetic term contains $F^{\ast }F$. Thus the auxiliary field part of the component Lagrangian isVarying $F$ and $F^{\ast }$ independently givesSubstitution gives ${\mathcal{L}}_F=-|m\varphi +g{\varphi }^2{|}^2$, and hence the F-term scalar potentialIt is nonnegative and vanishes at $\varphi =0$, and also at $\varphi =-m/g$ when $g\ne 0$. These are supersymmetric vacua, since the auxiliary field vanishes. Thus the quadratic/cubic model does not spontaneously break supersymmetry. If $m=0$ but $g\ne 0$, zero is the sole critical point; if both vanish, every $\varphi$ is a zero-energy vacuum. Nonzero $m,g$ give the usual two vacua, coinciding when $m=0$.

For the loop argument, promote the coefficients to background chiral spurions. One convenient pair of independent formal $U(1)$ charge assignments isBoth symmetries act nontrivially on all three of $\Phi ,m,g$; the second is an R-symmetry. The superspace integration measure $d^2\theta$ has R-charge $-2$, so both tree monomials give invariant actions. These are spurionic selection rules, not necessarily actual symmetries after fixing numerical couplings. The R-row is an ordinary-charge redefinition of the usual Wess–Zumino spurion charge assignment.

A local Wilsonian superpotential is a holomorphic function of the chiral fields and chiral spurions. The ratio $z=g\Phi /m$ is neutral under both charge assignments and has dimension zero, whereas $m{\Phi }^2$ has exactly the charges and dimension of $W$. The general symmetry-allowed form, on a patch with $m\ne 0$ and up to a physically irrelevant global constant, is thereforewith $f$ holomorphic where that description is regular. Symmetry by itself does not determine $f$.

For a regular perturbative local Wilsonian effective action retaining the same elementary field and an infrared cutoff, expand $f(z)={\sum }_n{c}_n{z}^n$. Its term of order $n$ is $c_n{g}^n{m}^{1-n}{\Phi }^{n+2}$. Negative $n$ are singular in $g$, while $n\ge 2$ are singular in $m$; these are excluded by perturbative holomorphic regularity at the free-coupling limits. Thus only the quadratic and cubic terms survive. Their regular holomorphic coefficients cannot acquire coupling-dependent invariant corrections. At $g=0$ the quadratic theory is Gaussian, fixing $c_0=1/2$ exactly. A contribution linear in $g$ to the cubic vertex has only one cubic interaction: three external lines exhaust that vertex, leaving no contraction to form a loop. Thus $c_1=1/3$ as well in the present normalization. This supplies the holomorphy argument for superpotential non-renormalization rather than concluding non-renormalization from symmetry alone.

The Kähler potential can still undergo wave-function renormalization. If its quadratic coefficient becomes $Z$, canonical normalization gives $m_c=m/Z$ and $g_c=g/Z^{3/2}$, so physical masses and interactions can run despite superpotential non-renormalization. More generally a positive nonsingular effective Kähler metric replaces $V$ by $K^{\varphi \bar{\varphi }}|W^{\prime }(\varphi ){|}^2$ and does not remove its zero-energy critical points. Infrared-singular nonlocal one-particle-irreducible terms and eliminating the entire massive field are outside this Wilsonian statement.

Use natural units and the unreduced Planck mass $M_P=G^{-1/2}$; it is $\sqrt{8\pi }$ times the reduced Planck mass. A homogeneous canonical inflaton has ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Vacuum domination by a scalar field requires ${\dot{\varphi }}_i^2/2\ll V({\varphi }_i)$, with other matter and the curvature term negligible in the Friedmann equation. The equation-of-state parameter is then near $-1$, and the Friedmann acceleration equation gives accelerating expansion.

For sustained slow-roll inflation, Hubble friction must dominate the inflaton acceleration: $|\ddot{\varphi }|\ll 3H|\dot{\varphi }|$, with the scalar potential changing only slightly in one Hubble time. In terms of the potential slow-roll parameter and second potential slow-roll parameter, this requiresFor the quartic scalar potential,Thus $|{\varphi }_i|\gg M_P$ supplies a broad slow-roll range when $\lambda \ll 1$. At an initial density of order the Planck density, potential domination givesso both potential slow-roll parameters are small for small $\lambda$. Small initial kinetic energy is a separate condition; the height of the scalar potential alone does not enforce it. These classical equations describe the post-Planck evolution under the model assumptions, rather than resolving quantum gravity at the initial endpoint.

Take ${\varphi }_i>0$; the negative branch follows by the symmetry $\varphi \mapsto -\varphi$. Dropping kinetic energy from the Friedmann equation and acceleration from the inflaton equation of motion givesWriting $b=M_P\sqrt{\lambda /(6\pi )}$, the quartic-potential slow-roll solution isIndeed, $\dot{\varphi }\simeq -b\varphi$, and integrating $H=\dot{a}/a$ gives $\log (a/a_i)=\pi ({\varphi }_i^2-{\varphi }^2)/M_P^2$. These expressions apply during slow-roll inflation; their formal late-time saturation must not be used beyond the end of that approximation.

The exact end of accelerating expansion is ${ϵ}_H=-\dot{H}/H^2=1$, equivalently ${\dot{\varphi }}^2=V$ for the scalar-dominated flat model. The quartic-inflation endpoint estimate uses the leading potential condition ${ϵ}_V\simeq 1$, givingThe numerical coefficient is an endpoint estimate: the slow-roll approximation breaks down there. Imposing $|{\eta }_V|=1$ estimates when slow roll loses accuracy and gives a different order-one coefficient, not a different exact acceleration condition.

The slow-roll e-fold count isThe quartic-inflation expansion from Planck density combines this initial estimate with the conventional endpoint to giveFor small $\lambda$ the dominant term is enormous; the order-one endpoint uncertainty does not affect that conclusion.

To estimate the reheating temperature, assume prompt, efficient reheating converts the end-of-inflation energy into a thermal radiation bath. With the same endpoint convention, $V_e\simeq \lambda M_P^4/(4{\pi }^2)$, and equating this to the radiation energy density givesThe quartic-inflation prompt-reheating estimate fixes the scaling; the endpoint kinetic energy changes the coefficient by order one. Without an inflaton coupling and its thermalization history, $V$ alone cannot specify the actual reheating temperature.

Finally, during cosmic inflation the comoving Hubble radius decreases:An initially causally connected patch is enlarged enough that the present observable region can lie within it. Its portions may subsequently be far outside one another's Hubble radii while retaining their common earlier thermal history. This solves the horizon problem if the number of e-folds is sufficiently large, conventionally of order sixty for the relevant thermal history; $N\sim 2\pi /\sqrt{\lambda }$ can readily exceed that. The necessary count depends on reheating and later expansion. The causal mechanism and the conversion back to a hot universe are discussed in David Tong's account of inflation.

First complexify the Clifford algebra in $D=2n$ spacetime dimensions. Multiplying appropriate generators by $i$ converts the signature to the Euclidean form $\{{\Gamma }_a,{\Gamma }_b\}=2{\delta }_{ab}$ without changing the complex module dimension. Pair the generators intoThe Clifford algebra relations giveThe commuting occupation operators $N_j=b_j^{†}{b}_j$ have eigenvalues zero or one. Choose a nonzero state killed by every $b_j$; repeated annihilation reaches such a state in any finite irreducible module. Acting with distinct creation operators produces the $2^n$ occupation statesTheir distinct occupation patterns make them linearly independent, and their span is invariant under every Clifford generator. Irreducibility therefore makes it the entire Dirac module. This derives the even-dimensional Clifford spinor dimension:The chirality matrix is a phase-adjusted product of all $2n$ generators. It squares to one, anticommutes with each generator and acts as occupation parity up to an overall sign. Its two eigenspaces each have $2^{n-1}$ states. The even Clifford algebra, and hence the connected spin group, preserves them, giving irreducible Weyl spinors of complex dimension $2^{D/2-1}$. A Majorana spinor imposes a reality condition only in signatures where it exists; simultaneous Majorana and Weyl conditions are not available in every even dimension.

In Lorentzian ten dimensions, the complex Dirac module has dimension $32$, and a Weyl module dimension $16$. A Majorana-Weyl spinor is allowed and has $16$ real components before its equation of motion. The massless Dirac equation leaves eight physical gaugino polarizations. A ten-dimensional massless gauge field also has $10-2=8$ physical polarizations. Thus ten-dimensional super Yang-Mills theory has eight bosonic and eight fermionic states per gauge generator, with sixteen real supercharges.

For dimensional reduction, take a flat six-torus with periodic fermions and retain fields independent of its coordinates. Split $A_M$ into $A_{\mu }$ and six internal components $A_m$, $m=4,\dots ,9$. The latter are four-dimensional real adjoint scalar fields. Under the spin groups of the two factors, a ten-dimensional chiral spinor decomposes asfor $\mathrm{Spin}(1,3)\times \mathrm{Spin}(6)$, with $\mathrm{Spin}(6)\simeq SU(4)$. The Majorana condition relates these conjugate pieces, leaving four independent four-dimensional Weyl spinors. The supersymmetry parameters reduce in exactly the same fashion: sixteen real charges are four four-dimensional supersymmetries. The resulting field content isThe on-shell counts are $2+6=8$ bosonic and $4\times 2=8$ fermionic states, as required. The internal spin group supplies the $SU(4)$ R-symmetry of the reduced action.

To derive its scalar potential, use Hermitian gauge matrices and a positive trace inner product, withIf the internal volume is $V_6$, integration gives $g_4^2=g_{10}^2/V_6$. Since internal derivatives vanish,With mostly-minus spacetime signature, $F_{\mu m}{F}^{\mu m}$ has an extra minus sign and the double sum over mixed indices gives a factor two. Rescale $A_{\mu }=g_4{a}_{\mu }$ and $A_m=g_4{X}_m$ to canonical fields. The resulting bosonic Lagrangian iswhere $D_{\mu }{X}_m={\partial }_{\mu }{X}_m-i{g}_4[a_{\mu },X_m]$. A commutator of Hermitian matrices is anti-Hermitian, so the last term is minus a nonnegative scalar potential:The sum is over ordered internal pairs; a sum over $m<n$ would instead have coefficient $g_4^2/2$. Trace normalization merely changes the matched gauge-coupling convention.

The origin attains zero potential, so the global minima are exactlyFor a compact connected gauge group, a commuting tuple is simultaneously conjugate into a Cartan subalgebra. Write $X_m={\sum }_{a=1}^r{x}_m^a{H}_a$, where $r$ is the rank. Every choice of the $6r$ real numbers $x_m^a$ has zero potential. After the residual Weyl group identifications, the classical commuting-scalar vacua of four-dimensional N=4 Yang-Mills theory form $({\mathbb{R}}^6{)}^r/\mathcal{W}$ in the four-dimensional theory. For $SU(N)$ one may describe them as six simultaneous diagonal matrices, with tracelessness and permutations of their common eigenvalue labels. Varying these eigenvalues changes gauge-invariant configurations while leaving $V=0$, proving the existence of genuine flat directions of a scalar potential, not just gauge-equivalent directions. Generic values break the gauge group to its maximal torus; coincident eigenvalue vectors produce enhanced unbroken gauge symmetry. If one keeps a finite compact torus rather than the four-dimensional limit, large gauge transformations add periodic identifications of the Wilson-line coordinates.

Choose canonical vector normalization, $g>0$, and the convention that the Fayet–Iliopoulos term contributes $+\xi D$. In a constant Lorentz-invariant vacuum with vanishing gauge field strength, the gaugino transformation containsA nonzero $⟨D⟩$ therefore shifts a fermion by a constant under supersymmetry: no invariant vacuum can have this expectation. With only D-term breaking, the Goldstino is the gaugino $\lambda$. If F-term expectation values were present too, the Goldstino would instead be the corresponding linear combination of matter fermions and gauginos.

For a genuinely charged single chiral superfield, $q\ne 0$, gauge invariance forbids every nonconstant polynomial superpotential in that field. The renormalizable superspace action is consequentlywhere ${\mathcal{W}}_{\alpha }=-{\bar{D}}^2{D}_{\alpha }V/4$. A constant superpotential is dynamically irrelevant in global supersymmetry. Take the standard vector expansion with highest component ${\theta }^2{\bar{\theta }}^{2}D/2$. In the matter exponential, the term $2gqV$ then contributes $gq|\varphi {|}^{2}D$. The gauge kinetic chiral integral contributes $D^2/2$, and the FI term contributes $\xi D$. ThusThe auxiliary equations are $F=0$ and $D=-(gq|\varphi {|}^2+\xi )$. Completing the square givesThis is the single charged-field D-term breaking model. Its vacuum is found by minimizing over $\rho =|\varphi {|}^2\ge 0$, not by assuming that a nonzero FI constant must break supersymmetry.

For $q\xi >0$, the square cannot be canceled by any $\rho \ge 0$. It is minimized at the origin, whereSupersymmetry breaking occurs while the internal $U(1)$ gauge symmetry remains unbroken. Expanding the scalar potential about the origin givesThere is no superpotential fermion mass and no gaugino-matter mixing at $\varphi =0$. Therefore the chiral multiplet mass splitting isBoth real components of the complex scalar have the same positive squared mass; the gaugino is the massless Goldstino. With this sign convention and $\xi >0$, the breaking condition is $q>0$. Reversing the FI sign convention reverses this quoted charge inequality, while the condition that the scalar cannot cancel the auxiliary-field shift is unchanged.

For $q\xi <0$, the minimum instead satisfiesThe nonzero charged scalar expectation breaks the continuous internal gauge symmetry and preserves supersymmetry. To see the resulting supersymmetric Higgs mechanism, choose a real representative $v>0$ and expand $\varphi =v+(h+ia)/\sqrt{2}$. The scalar kinetic term supplies a vector mass, the D-term potential supplies the radial scalar mass, and the Yukawa interaction $-\sqrt{2}gq{\varphi }^{\ast }\lambda \psi +h.c.$ pairs the two Weyl fermions. They have the common squared massThe phase field $a$ is the absorbed Goldstone mode. There is no Goldstino in this supersymmetric Higgs vacuum; the fields assemble into a massive N=1 vector multiplet. This branch makes explicit that internal gauge breaking and supersymmetry breaking are different phenomena.

If $\xi =0$ and $q\ne 0$, the minimum of $g^2{q}^2|\varphi {|}^4/2$ is $\varphi =0$. Both supersymmetry and the continuous gauge symmetry are then unbroken, and the fields have zero tree-level masses at that vacuum. For the degenerate uncharged case $q=0$ with no superpotential interactions, a nonzero $\xi$ still breaks supersymmetry in the vector sector, but the free chiral multiplet stays unsplit and cannot Higgs the gauge field. An uncharged field would allow additional superpotential terms, outside the genuinely charged model above.

The calculation is classical. A single charged Weyl fermion has nonzero Abelian gauge and mixed gravitational anomaly coefficients proportional to $q^3$ and $q$; a complete quantum gauge theory needs an anomaly-canceling completion. The tree-level branches above refer to the specified field content; extra fields in such a completion can change the vacuum structure.

There is a sign inconsistency in the printed model. For the ordinary cross product let $J(a)z=a\times z$. The vector triple-product identity gives $[J(a),J(b)]=J(a\times b)$. Consequently the printed $D_{\mu }={\partial }_{\mu }+eJ(A_{\mu })$ has gauge curvatureIts gauge curvature cannot instead contain the printed negative quadratic term. For a noncommuting pure gauge potential for this gauge covariant derivative, $F^{+}=0$ whereas $F^{-}=-2e{A}_{\mu }\times A_{\nu }\ne 0$. Thus the printed gauge kinetic term assigns a nonzero value to a connection obtained by applying the gauge-field transformation law to the zero connection. Literally the printed action has global rotations but not the requested local triplet gauge symmetry. The cross-product convention for an adjoint covariant derivative makes the minimal repair explicit: change the gauge field strength quadratic term to a plus sign, retaining the printed gauge covariant derivative. Changing the sign in the gauge covariant derivative instead would be another consistent convention. The remaining calculation uses the first repair.

The corrected theory has local SO(3) gauge symmetry, equivalently the adjoint-field realization of the SU(2) Lie algebra; its adjoint scalar fields and gauge fields are insensitive to the center of a group of SU(2). Explicitly, $\Phi \mapsto R(x)\Phi$ andBoth $D_{\mu }\Phi$ and $F_{\mu \nu }$ transform as triplets, so their dot products and the scalar potential are invariant. Assume $\lambda >0$, $e\ne 0$ and take $v\ge 0$ without loss of generality. For $v>0$ the minima have ${\Phi }^2=v^2$. Choose the scalar-field vacuum ${\Phi }_0=v(0,0,1)$, whose stabilizer consists of rotations around the third axis: the residual gauge symmetry is SO(2), locally U(1).

In unitary gauge write $\Phi =(0,0,v+h)$, $A_{\mu }=A_{\mu }^3$ and $W_{\mu }^{\pm }=(A_{\mu }^1\mp i{A}_{\mu }^2)/\sqrt{2}$. These are the neutral massless gauge field, two charged massive gauge fields and a real radial scalar field. DefineSubstitution gives $F_{\mu \nu }^3=f_{\mu \nu }-ie{C}_{\mu \nu }$ and $(F_{\mu \nu }^1\mp i{F}_{\mu \nu }^2)/\sqrt{2}=G_{\mu \nu }^{\pm }$. The entire physical-field Lagrangian density is thereforeIn particular $D_{\mu }\Phi =(e(v+h)A_{\mu }^2,-e(v+h)A_{\mu }^1,{\partial }_{\mu }h)$; this directly verifies the scalar field kinetic and vector mass terms. Expanding the scalar potential gives $\lambda v^2{h}^2/2+\lambda v{h}^3/2+\lambda h^4/8$. Comparing the quadratic terms with standard real scalar field and charged-vector mass terms yields the adjoint triplet Higgs spectrum,This also rewrites the purely vector interactions, which are already contained in the displayed gauge field strengths rather than omitted from the theory.

The original symmetry remains unbroken at $v=0$. The unique minimum is then $\Phi =0$, with three massless gauge fields and three massless real scalar field components; the quartic scalar field interaction remains. The physical degrees of freedom number $3\times 2+3=9$. At $v>0$ the two broken generators supply two Goldstone bosons, eaten to give longitudinal gauge-boson polarizations of the charged vectors. The physical count is $2\times 3+2+1=9$: two massive vectors, one massless vector and one radial scalar field. The scalar components eaten in unitary gauge are not additional physical particles.

For the broken phase all interactions containing the physical scalar field areMultiplying interaction coefficients by $i$ and differentiating with respect to the external fields includes the identical-field factorials. The scalar vertices of an adjoint triplet Higgs model areFor example, the $h^3$ rule contains $3!$, whereas the $hh{W}^{+}{W}^{-}$ rule contains only $2!$ from its two identical scalar fields. There are no $hAA$ or $hhAA$ vertices: the radial scalar field is neutral under the residual U(1). There is no derivative scalar-vector vertex in this physical spectrum in unitary gauge.

For completeness, if the unbroken phase is chosen, its physical scalar fields are the three components ${\varphi }_a$, not a single radial $h$. Expanding its gauge-covariant kinetic term givesWith all momenta incoming and Fourier transform convention $e^{-ipx}$, the three unbroken-phase rules areThe first rule follows by differentiating once on each possible scalar: for $a=3,b=1,c=2$, the interaction is $e{A}_{\mu }^3({\varphi }_1{\partial }^{\mu }{\varphi }_2-{\varphi }_2{\partial }^{\mu }{\varphi }_1)$, which fixes the momentum sign. The remaining rules follow from the quadratic gauge interaction and $-\lambda ({\varphi }_a{\varphi }_a{)}^2/8$. There are no unbroken-phase cubic scalar field Feynman vertices.

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.

Assume global four-dimensional N=1 supersymmetry with canonical kinetic terms and no explicit gaugino mass. Write $W_i={\partial }_{i}W$ and $W_{ij}={\partial }_i{\partial }_{j}W$. To match the displayed stationarity equation, use $F^i=\overline{W_i}$; this differs by a minus sign from the eliminated auxiliary field in the component convention of Solution 2. The real D-terms are $D^a=-g^a({\bar{\varphi }}_i(T^a{)}^i{}_j{\varphi }^j+{\xi }^a)$, with Hermitian gauge generators. A constant Fayet–Iliopoulos term is permitted only for an appropriate Abelian factor. The scalar potential isDifferentiate with respect to ${\varphi }^i$, holding $\bar{\varphi }$ independent:Thus stationary point conditions gives the first requested identity. Gauge invariance of the superpotential gives $W_i(T^a{)}^i{}_j{\varphi }^j=0$. Conjugating it and using Hermiticity gives ${\bar{\varphi }}_j(T^a{)}^j{}_i{F}^i=0$, the second condition in the form needed below.

At the vacuum, define $H_{ij}=⟨W_{ij}⟩$ and $G_{ia}=g^a⟨{\bar{\varphi }}_j⟩(T^a{)}^j{}_i$. The two conditions combine asHere the upper equation is stationarity and the lower equation is conjugated gauge invariance. The chiral-superfield fermion mass matrix contributes $-\frac{1}{2}{H}_{ij}{\psi }^i{\psi }^j$, and the gauge Yukawa interaction contributes the mixed chiral-fermion/gaugino bilinear. With canonical Weyl fermion and gaugino kinetic terms, and choosing the gaugino phase to put the mixed bilinear in real-sign convention, these terms areThe factor $\sqrt{2}$ is important: the matrix directly acting on the printed $(F,D)$ is the same mass bilinear with rescaled gauginos, rather than literally the canonical matrix. In detail, for $S=\mathrm{diag}(I,\sqrt{2}I)$,Since supersymmetry breaking makes this vector nonzero, there is at least one zero mass eigenvalue. A normalized massless fermion isWith the consistent gaugino phase, the inhomogeneous vacuum supersymmetry transformation of $(\psi ,\lambda )$ is proportional to the same vector $(F,D/\sqrt{2})$. Therefore $\eta$ shifts by a nonzero constant times the transformation parameter: it is the Goldstino, the Goldstone fermion of broken global supersymmetry. The argument guarantees at least one Goldstino zero mode, not the absence of other massless fermions.