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Use $K_{abc}=e_a{}^{\mu }(\omega -\omega (e){)}_{\mu bc}$ and $T_{abc}=K_{bac}-K_{cab}$. For the symmetrized Dirac action $-e\bar{\chi }{\gamma }^{\mu }{\overleftrightarrow{\nabla }}_{\mu }\chi /2-em\bar{\chi }\chi$, varying the spin connection gives $-e\bar{\chi }{\gamma }^{\mu ab}\chi \delta {\omega }_{\mu ab}/4$. Only totally antisymmetric contorsion is sourced. Its action is $-e{K}^2/(2{\kappa }^2)-e{K}_{abc}\bar{\chi }{\gamma }^{abc}\chi /4$, hence the stated solution. Substitution generates $e{\kappa }^2(\bar{\chi }{\gamma }^{abc}\chi )(\bar{\chi }{\gamma }_{abc}\chi )/32$. The torsion is an auxiliary field, not an independent propagating degree of freedom.

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.

With $X^2=0$ and invertible $F_X$, the additional chiral superfield $Y=y+\sqrt{2}\theta \chi +{\theta }^2{F}_Y$ satisfying $XY=0$ has $y=(G\chi )/F_X-(GG)F_Y/(2F_X^2)$. Its Weyl spinor $\chi$ and auxiliary field $F_Y$ remain independent. This eliminates the independent scalar. It does not force $Y^2=0$.

A D-term is a full superspace integration of a real superfield. In four-dimensional N=1 supersymmetry, a Kähler potential contributes a D-term containing kinetic terms and quadratic auxiliary field terms.

For $K=-\log (S+\bar{S})-3\log (T+\bar{T})$ and $W=a{e}^{-\alpha S}+b$ with $\alpha >0$, the potential is $|b+[1+\alpha (S+\bar{S})]a{e}^{-\alpha S}{|}^2/[(S+\bar{S})(T+\bar{T}{)}^3]$. With nonzero $a,b$, a finite zero-energy vacuum exists exactly when $0<|b/a|\le 2e^{-1/2}$. To prove this, set $x=\alpha \mathrm{Re}S>0$; the magnitude equation is $|b/a|=(1+2x)e^{-x}$, whose derivative changes sign at $x=1/2$ and whose maximum is $2e^{-1/2}$. The phase fixes the imaginary part of $S$. At a zero-energy solution the $T$ auxiliary field remains nonzero, so supersymmetry is broken and both real components of $T$ are flat. If $a=b=0$, the family is instead supersymmetric.

A goldstino is the massless spin-one-half mode associated with spontaneously broken global supersymmetry. For F-term breaking, the fermion in a multiplet with nonzero auxiliary field transforms by an inhomogeneous term, $\delta G\supset \sqrt{2}ϵF$. In supergravity, the corresponding mode supplies the longitudinal degrees of freedom of a massive gravitino.

A hidden sector contains fields whose auxiliary fields acquire expectation values that break supersymmetry. Loop effects or suppressed interactions communicate the breaking to visible fields as effective soft supersymmetry breaking. This can avoid the canonical visible-sector tree-level supertrace mass sum rule.

A Higgs chiral doublet is a weak-doublet chiral superfield containing a Higgs complex scalar field, a higgsino and an auxiliary field. In the MSSM, $H_u$ has hypercharge $+1/2$ and $H_d$ has hypercharge $-1/2$. Their paired charges enable higgsino anomaly cancellation and the holomorphic need for two Higgs chiral doublets.

For $X=x+\sqrt{2}\theta G+{\theta }^{2}F$, the nilpotency condition gives $x^2=xG=0$ and $2xF=GG$. On the branch with invertible commuting part of $F$, $x=GG/(2F)$, while the goldstino $G$ and auxiliary field $F$ are independent. Products of three identical two-component odd spinor entries vanish, verifying the remaining constraints. The scalar is therefore composite. The branch hypothesis matters: $X=0$ also obeys the constraint and need not break supersymmetry.

A four-dimensional Majorana spinor gravitino has $4\times 4=16$ real components, minus four local supersymmetry gauge functions, leaving twelve before using any Euler-Lagrange field equation. A vierbein has sixteen real components, minus six local Lorentz transformation and four diffeomorphism gauge functions, leaving six. Equivalently, a symmetric metric tensor has $10-4=6$. The propagating graviton and massless gravitino each have two states on shell. Six bosonic auxiliary field components complete an off-shell supergravity multiplet.

Old-minimal four-dimensional supergravity adds a complex scalar $M$ and a real vector $b_{\mu }$ to the vierbein and gravitino. These six real bosonic auxiliary field components have no propagating kinetic terms. They balance the off-shell component count of minimal supergravity and permit the local supersymmetry algebra to close without imposing the propagating Euler-Lagrange field equations.

Anticommuting-variable dependence alone is not the defining condition: a superfield must carry a specified Super-Poincaré group transformation law. Any appropriately graded function on superspace becomes an unconstrained scalar superfield under coordinate pullback, but an arbitrary collection of coefficients in a Grassmann variable polynomial is not automatically a prescribed supermultiplet. For example, a chiral superfield has $\Phi (y,\theta )=A(y)+\sqrt{2}\theta \psi (y)+{\theta }^{2}F(y)$, with $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$ and ${\bar{D}}_{\dot{\alpha }}\Phi =0$; a vector superfield obeys $V=V^{†}$ and contains a gauge vector, gaugino and auxiliary field in Wess-Zumino gauge. A polynomial with constant component fields can be a special chiral superfield; lack of explicit spacetime dependence alone does not disqualify it. This is superfield covariance is not Grassmann dependence alone.

For a massless supermultiplet, put the momentum along the third axis. The matrix $2\sigma \cdot p$ in the supersymmetry algebra has rank one, so only one fermionic oscillator acts nontrivially. It relates states whose helicities differ by $\frac{1}{2}$. Including the antiparticles required by CPT completion of a supermultiplet gives the following complete spectra.

A chiral multiplet contains a complex scalar $A$ and a Weyl spinor $\psi$. Its scalar particle and antiparticle have helicity $0$; its fermion particle and antiparticle have helicities $-\frac{1}{2}$ and $+\frac{1}{2}$. Therefore $n_B=2=n_F$, with maximum absolute helicity $\frac{1}{2}$.

A supersymmetric vector multiplet contains a massless gauge vector $A_{\mu }$, with helicities $+1,-1$, and a Majorana spinor gaugino $\lambda$, with helicities $+\frac{1}{2},-\frac{1}{2}$. Again $n_B=2=n_F$, now with maximum absolute helicity $1$. The chiral $F$ and vector $D$ auxiliary fields contribute no on-shell states.

The Higgs mechanism is compatible with unbroken supersymmetry. A supersymmetric vacuum may have nonzero scalar vacuum expectation values that break the internal gauge symmetry while obeying both F-flatness and D-flatness. For example, oppositely charged chiral superfields with $W=0$, no Fayet–Iliopoulos term and equal nonzero scalar magnitudes have $F=D=0$ but break the Abelian gauge group. In the supersymmetric Higgs mechanism, a massless vector multiplet combines with the chiral multiplet along the broken gauge direction. One real scalar becomes the longitudinal vector polarization; the other remains a physical real scalar, and the gaugino combines with the Higgsino into a Dirac spinor. The resulting massive vector multiplet has $3+1=4$ bosonic and $4$ fermionic states with a common mass, as in part (i). Any other uneaten chiral multiplets remain separately in the spectrum.

The irreducible massive N=1 vector multiplet consists of one massive vector, one real scalar, and two massive spin-$\frac{1}{2}$ fermions. Thus its on-shell boson and fermion counts areThe two fermions can be written as two Majorana spinors or one Dirac spinor. All these states have the same mass when supersymmetry is unbroken; auxiliary fields are not extra propagating states.

To obtain rather than assume this spectrum, use the rest-frame supersymmetry algebra $\{Q_{\alpha },Q_{\beta }^{†}\}=2M{\delta }_{\alpha \beta }$. The two normalized fermionic creation operators form a spin-$\frac{1}{2}$ doublet. Start with a spin-$\frac{1}{2}$ fermionic Clifford vacuum annihilated by the two lowering operators. The zero- and two-creation-operator levels each have spin $\frac{1}{2}$, whereas the one-creation-operator level hasIt is bosonic, giving exactly the scalar and vector above. A higher-spin Clifford vacuum would introduce spin above $1$, while a spin-zero Clifford vacuum gives maximum spin only $\frac{1}{2}$.

Take canonical kinetic terms, gauge coupling $g>0$, and define the Fayet–Iliopoulos term by $+\xi D$. If $\varphi$ is the scalar component of the charge-$q$ chiral superfield, the relevant auxiliary field Lagrangian isThe algebraic equation eliminates $D$ and gives the D-term scalar potentialSince the superpotential vanishes, $F=0$. In a homogeneous vacuum, the supersymmetry transformation of the gaugino includesThus $D_0\ne 0$ gives an inhomogeneous shift: no nonzero constant supersymmetry parameter leaves this vacuum invariant. The gaugino $\lambda$ is the goldstino, because it is the field shifting under the broken supercharge; there is no $F$ contribution here.

For $q\ne 0$, minimizing over $x=|\varphi {|}^2\ge 0$ separates three cases. If $q\xi <0$, the allowed value $x=-\xi /(gq)$ gives $D_0=V_0=0$: supersymmetry is unbroken while the internal gauge symmetry is Higgsed. If $q\xi >0$, the unconstrained zero of $D$ would have negative $x$, so the minimum is $\varphi =0$ with $D_0=-\xi$ and $V_0={\xi }^2/2$. If $\xi =0$, the minimum $\varphi =0$ is supersymmetric. Consequently, with this explicitly stated sign convention,Reversing the sign used to define the Fayet–Iliopoulos term reverses the printed sign criterion, not the physics. If $q=0$, the scalar is decoupled and flat: any $\xi \ne 0$ still breaks supersymmetry through $D_0=-\xi$, but the charged-field criterion does not apply.

On the broken branch, expand about $\varphi =0$:Writing $\varphi =({\varphi }_1+i{\varphi }_2)/\sqrt{2}$ gives both real scalar modes the squared mass $gq\xi$. There is no superpotential fermion mass and no vacuum expectation value to induce gaugino–matter-fermion mixing. The Abelian gauge symmetry is unbroken, so its vector is also massless. The complete mass spectrum of single charged-field D-term breaking isThe original chiral multiplet therefore has the squared-mass splittingThe original vector multiplet has zero mass splitting, although the nonzero $D_0$ means it is not an unbroken-supersymmetry representation. As a check, the tree-level supertrace mass sum rule gives $\mathrm{Str}{M}^2=2gq\xi$ when the two real scalar degrees of freedom are counted. On the $q=0$ broken branch all the propagating fields are massless.

There is a necessary scope qualification: one nonzero-charge Weyl fermion has an Abelian cubic gauge anomaly proportional to $q^3$ and a mixed gravitational gauge anomaly proportional to $q$. These are the classical vacuum and spectrum calculations requested by the model; a consistent quantum gauge theory needs an anomaly-cancelling completion, whose additional matter must also be included in its vacuum analysis.

Use Planck units and restrict to the physical Kähler metric domainPut $A=a{e}^{-\alpha S}$ and $w=A+b$, so that $W=C^3+w$. There are no gauged interactions specified, hence no additional D-term scalar potential. The supergravity F-term potential is $V=e^K(K^{i\bar{j}}{D}_{i}W\overline{D_{j}W}-3|W{|}^2)$, with $e^K=1/(x{y}^3)$. Its Kähler covariant derivatives of a superpotential areThe nonzero metric entries areThe determinant of this $T,C$ block is $9/y^3>0$, so it is positive definite. With the inverse-index convention $K_{i\bar{j}}{K}^{k\bar{j}}={\delta }_i^k$, the raised components areKeeping these complex-conjugate off-diagonal entries in their correct index positions is essential.

To exhibit the no-scale supergravity cancellation, write the $T,C$ norm asHere $D_{C}W+\bar{C}{D}_{T}W=3C^2$. The $3|W{|}^2$ term cancels the negative term in the supergravity F-term potential. The result is the nonnegative scalar potentialThis is the matter-logarithm no-scale potential, specialized to the exponential superpotential.

Choose the supergravity auxiliary field convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$. Substitution into the full inverse Kähler metric givesAn overall auxiliary-field phase is conventional. The $C$ dependence of $F^T$ cancels between the two metric terms. If all three auxiliary fields vanished at a finite physical point, $F^C=0$ would force $C=0$, $F^T=0$ would force $A+b=0$, and $F^S=0$ would then force $\alpha xA=0$. Thus $a=b=0$ is necessary. For any nontrivial choice of $a,b$, supersymmetry is broken at every finite physical point. The allowed exception $a=b=0$, $C=0$ is supersymmetric; the statement cannot hold for all arbitrary parameters without this qualification.

The no-scale volume runaway criterion also determines whether the minimum is attained. For fixed $S,C$, differentiation with respect to $y$ givesIt is negative unless both numerators vanish. Therefore a finite stationary vacuum must satisfyEvery such point is a global minimum with $V_{\min }=0$. For $a,b\ne 0$, setting $r=|b/a|$ and $z=\alpha x>0$ reduces existence toThe derivative of the magnitude function is $(1-z)e^{-z/2}/2$. It rises from $1$ at the excluded boundary $z=0$ to $2e^{-1/2}$ at $z=1$, then falls to zero. Consequently a finite zero-energy minimum exists exactly whenor in the separate trivial family $a=b=0$. There is one positive root for $0<r\le 1$, two for $1<r<2e^{-1/2}$, and one double root at equality. If only one of $a,b$ vanishes, or $r$ exceeds this maximum, there is no finite minimum: the infimum is zero as $\mathrm{Re}T\to \infty$. Calling this runaway a finite zero-energy vacuum would be incorrect.

At a nontrivial zero-energy minimum, $W=A+b=-\alpha xA\ne 0$, and the auxiliary expectation values reduce toThis explicitly verifies supersymmetry breaking with zero vacuum energy in no-scale supergravity. Both real components of $T$ are exact flat directions of a scalar potential on the minimum locus. Away from that locus, $\mathrm{Im}T$ is still absent from the potential, but $\mathrm{Re}T$ is not flat. At a minimum, $C$ is lifted at quartic order: at fixed $S,T$, $V=3|C{|}^4/[x(T+\bar{T}{)}^2]+O(|C{|}^6)$. A zero quadratic mass does not make $C$ an exact flat direction. The quartic stabilization at a double no-scale root likewise gives a vanishing quadratic mass, but no exact flat direction, for $\mathrm{Re}S$ at $z=1$. For $a=b=0$, the whole $C=0$ family has both $S$ and $T$ flat and supersymmetry unbroken.

Finally, the gravitino mass from a superpotential isAt a nontrivial zero-energy minimum this becomesIt depends on the unfixed volume modulus, so the model does not predict one numerical mass. For $a=b=0$, $C=0$, the gravitino is massless.

In the ordinary local two-derivative matter class, the general ungauged supersymmetric action for a chiral superfield isThe Kähler potential $K$ is real, and the superpotential $W$ is a holomorphic function. Adding $f(Z)+\bar{f}(\bar{Z})$ to $K$ changes the action only by a boundary term: full superspace integration of a chiral or antichiral function vanishes modulo that term. This is a Kähler transformation. A positive Kähler metric $G=K_{z\bar{z}}$ is required for a nonsingular positive-norm kinetic term, not just for derivative counting.

To see the differential order, the bosonic component Lagrangian in mostly-plus signature containsThe auxiliary field equation is algebraic: $F=-G^{-1}\overline{W_z}$ on a purely bosonic configuration. It gives the scalar potential $U=G^{-1}|W_z{|}^2$. The scalar equations have at most two spacetime derivatives, the fermion equations from the sigma-model kinetic term have one, and the remaining interaction terms do not raise this order. The fermionic corrections to the auxiliary equation are still algebraic.

For four-dimensional renormalizability, assign engineering dimensions $[Z]=1$, $[d^4\theta ]=2$, and $[d^2\theta ]=1$. Hence, up to a Kähler transformation and a constant field normalization, the restrictions areA mixed higher-degree term in $K$ or a term of degree above three in $W$ introduces a negative-mass-dimension coupling. The constant $w_0$ has no local effect in global supersymmetry, as in the constant superpotential in global supersymmetry.

The boxed general-action statement uses the conventional two-derivative matter interpretation. It is not a classification of all derivative-dependent effective superfield actions by the order of their bosonic equations alone. For example, a scalar Lagrangian quartic in first derivatives still yields a second-order Euler–Lagrange equation, and supersymmetric derivative-dependent interactions can contain such terms. Their auxiliary and fermionic equations require separate checking. This distinction is explicit in the original construction of supersymmetric derivative-dependent scalar actions. Such actions are outside the ordinary $K+W$ class used for the renormalizable answer here.

Spontaneous supersymmetry breaking preserves the action and conserved supercharges, but the vacuum fails to be invariant. This differs from explicit breaking, which changes the symmetry of the action itself. In global four-dimensional N=1 supersymmetry, summing diagonal entries of the Super-Poincaré algebra writes the Hamiltonian as a sum of positive charge anticommutators. Its expectation is a sum of squared norms. Consequently a zero-energy vacuum is supersymmetric, whereas an existing vacuum with broken global supersymmetry has positive energy density. The energy zero is fixed by the algebra. Neither this positivity statement nor the particle-multiplet mass relations should be transferred without qualification to supergravity.

For supersymmetric quantum mechanics, let $H=\{q,q^{†}\}/2$, $q^2=0$, and let fermion parity anticommute with both charges. On a positive-energy eigenspace, $(q+q^{†})/\sqrt{2E}$ is an odd involution: its square is one, it preserves energy, and it reverses parity. Thus every positive-energy bosonic state has a fermionic partner and conversely. Their contributions cancel in the Witten indexFor a discrete spectrum with a well-defined trace, this proves independence of $\beta$. Continuous changes of parameters also preserve the index if the trace stays well defined and no state escapes to infinity or crosses an unregulated continuum threshold. A nonzero index guarantees an unbroken supersymmetric ground state; a zero index is inconclusive. For example, de Rham supersymmetric quantum mechanics on a circle has a bosonic constant zero-form and a fermionic constant one-form, both normalizable zero modes, so it is unbroken with index zero.

A more explicit one-dimensional family illustrates how normalizability determines the answer. DefinePositivity of these Hamiltonian operators shows that a zero mode must solve $A{\psi }_{-}=0$ or $A^{†}{\psi }_{+}=0$. Thus the only candidates areFor $w(x)=x$, the Gaussian minus-sector wavefunction is normalizable and its reciprocal is not, giving an unbroken model with $I=1$ when the minus sector is assigned even parity. For $w(x)=x^2+a$, $a>0$, the exponent $x^3/3+ax$ tends to opposite infinities at the two ends, so neither candidate is normalizable. Both partner potentials $[w^2\mp w^{\prime }]/2$ grow quartically; their spectra are discrete and their lowest energies cannot be zero. The lowest states are therefore paired at positive energy, $I=0$, and supersymmetry is spontaneously broken. This is supersymmetric factorization and zero-mode normalizability. In quantum mechanics the paired ground states illustrate the charge algebra; a relativistic massless-particle conclusion requires field-theoretic hypotheses.

In relativistic global field theory, Goldstone's theorem has a fermionic version. If a conserved supercurrent charge does not annihilate the vacuum, its Ward identity has a nonzero vacuum contact term. Fourier transforming, the divergence of the current correlator cannot tend to zero smoothly at zero momentum. Its spectral representation therefore contains a massless fermionic pole: the Goldstino, with spin one-half. This statement assumes a Poincaré-invariant vacuum and the usual local unitary field-theory conditions. It does not require a bosonic Goldstone particle for the fermionic symmetry. In supergravity the Goldstino can instead supply the longitudinal polarization of a massive gravitino.

For canonical chiral superfields ${\Phi }_i$, elimination of their auxiliary fields givesA supersymmetric vacuum solves every $W_i=0$. At a stationary broken vacuum,The same matrix $W_{ij}$ is the chiral-fermion mass matrix, so the nonzero vector $\overline{W_j}$ is a zero mode. This derives the Goldstino zero mode from vacuum stationarity. The transformation $\delta {\psi }_i=\sqrt{2}ϵF_i+\cdots$ independently identifies that direction through its inhomogeneous shift.

For one canonical chiral field, $W=fZ$, $f\ne 0$, has $U=|f{|}^2$ everywhere, a massless fermion and a classically flat scalar direction. It is flat F-term breaking with a linear superpotential, rather than an isolated vacuum. In contrast, a nonconstant polynomial $W^{\prime }$ of positive degree has a complex root and hence a supersymmetric vacuum; in particular a generic quadratic or cubic single-field superpotential cannot remove all such roots. This argument assumes canonical nonsingular kinetic terms and does not claim that every runaway or nonpolynomial model has a vacuum.

A renormalizable three-field O'Raifeartaigh model gives a less degenerate illustration:The auxiliary equations require $W_2=m{\varphi }_1=0$, then $W_1=hX{\varphi }_1+m{\varphi }_2=0$, leaving $W_X=f\ne 0$. They cannot vanish simultaneously. For real positive $f,h,m$ with $m^2>fh$, the potential obeysTherefore its global tree-level minimum has ${\varphi }_1={\varphi }_2=0$, arbitrary $X$, and energy $f^2$. The massless ${\psi }_X$ is the Goldstino, while $X$ is a pseudomodulus. At $X=0$, the scalar mass-squared spectrum is $0,0,m^2,m^2,m^2+fh,m^2-fh$, and the fermion masses are $0,m,m$. This exhibits both the stability condition and the loss of ordinary supermultiplet mass degeneracy in a broken vacuum. Quantum corrections can lift the pseudomodulus and must be analysed rather than assumed away; the noncompact classical valley also prevents an unqualified finite-volume Witten index trace argument. The index, the vacuum equations and the Goldstino theorem consequently answer complementary questions.

Ordinary electromagnetism uses a one-form gauge potential, its gauge transformation and its curvature. On superspace the corresponding object is a superspace gauge connection, with both vector and spinor components. Write ${\nabla }_A=D_A+i{\mathcal{A}}_A$. Its graded commutator contains the flat-frame torsion plus supercurvature. The conventional constraints set the two undotted-spinor curvatures, the two dotted-spinor curvatures and the conventional mixed spinor curvature to zero; the mixed derivative algebra then determines the vector connection. The remaining independent gauge curvature is packaged into a chiral field-strength superfield. A real vector superfield $V=V^{†}$ is a prepotential solving these constraints, rather than four unrelated ordinary gauge potentials.

Use the left-derivative conventions of Question 5, with $D^2=D^{\alpha }{D}_{\alpha }$ and ${\bar{D}}^2={\bar{D}}_{\dot{\alpha }}{\bar{D}}^{\dot{\alpha }}$. An Abelian supergauge transformation has a chiral parameter $\Lambda$, and one consistent normalization isThe parameter is chiral so that the transformed matter is still a chiral superfield; its complex components remove more of the prepotential than an ordinary real gauge parameter. In Wess-Zumino gauge, one convenient expansion isIts components are the gauge potential $A_m$, the gaugino $\lambda$ and the real auxiliary field $D$. Wess-Zumino gauge removes the extra scalar and spinor components of a general real superfield. Supersymmetry transformations usually leave that gauge slice and require a compensating supergauge transformation. For a residual chiral parameter with real lowest component $a(x)$, the expansion gives $A_m^{\prime }=A_m+2{\partial }_{m}a$ and ${\varphi }^{\prime }=e^{-2iqa}\varphi$; thus $\alpha =2a$ is the ordinary Maxwell gauge parameter.

The Abelian field strength is the Abelian field-strength chiral projectionThree barred derivatives vanish because there are only two anticommuting dotted components, so ${\bar{D}}_{\dot{\beta }}{W}_{\alpha }=0$. Gauge invariance follows directly: $D_{\alpha }\bar{\Lambda }=0$ by antichirality, while the mixed derivative algebra gives ${\bar{D}}^2{D}_{\alpha }\Lambda =0$ for chiral $\Lambda$. In chiral coordinates its component structure iswhere ${\sigma }^{mn}=({\sigma }^m{\bar{\sigma }}^n-{\sigma }^n{\bar{\sigma }}^m)/4$ fixes the tensor normalization. Overall gaugino phase choices do not change the gauge-invariant field strength or its Bianchi constraint.

Reality of $V$ and the derivative algebra imply the superspace Yang-Mills Bianchi identityIndeed, the two sides are the same differential operator on $V$, because $D^{\alpha }{\bar{D}}^2{D}_{\alpha }={\bar{D}}_{\dot{\alpha }}{D}^2{\bar{D}}^{\dot{\alpha }}$. This is the superspace counterpart of $dF=0$, and its components contain ${\partial }_{[m}{F}_{np]}=0$. It is a defining curvature constraint, not the Maxwell equations. Varying the pure gauge action additionally sets the common real superfield $D^{\alpha }{W}_{\alpha }$ to zero; matter supplies a current instead.

The Supersymmetric Yang-Mills action and minimal matter coupling areThe chiral phase factors cancel in $\bar{\Phi }{e}^{2qV}\Phi$. The component matter derivative is $({\partial }_m+iq{A}_m)\varphi$, which transforms covariantly under the residual gauge transformation just derived. The same superspace term generates the associated matter–gaugino Yukawa coupling and $qD|\varphi {|}^2$ auxiliary coupling, so minimal coupling is not only replacing an ordinary derivative in the scalar action. The gauge action contains $-F_{mn}{F}^{mn}/(4g^2)$, the gaugino kinetic term and $D^2/(2g^2)$. Rescaling the vector multiplet to canonical kinetic normalization gives the physical coupling $qg$. The holomorphic superpotential must be gauge invariant; a single field of nonzero continuous Abelian charge admits no nonconstant polynomial superpotential without additional fields. For several fields, invariance requires ${\sum }_i{q}_i{\Phi }_i{\mathcal{W}}_i=0$. A Fayet–Iliopoulos term is additionally possible for an Abelian factor, shifting the algebraic $D$ equation.

For a non-Abelian gauge group, $V$ and $\Lambda$ are Lie-algebra-valued, and multiplication cannot be reordered. For matter in a specified representation, setThe same matter quadratic form remains invariant. Its curvature isAt first order in an Abelian $V$, this reduces to $-{\bar{D}}^2{D}_{\alpha }V/4$, fixing the factor $1/8$. In the chiral gauge frame, $\bar{\nabla }=\bar{D}$ and ${\nabla }_{\alpha }=e^{-2V}{D}_{\alpha }{e}^{2V}$ as an operator. The gauge-covariant Bianchi identity is ${\nabla }^{\alpha }{W}_{\alpha }={\bar{\nabla }}_{\dot{\alpha }}{\tilde{\bar{W}}}^{\dot{\alpha }}$, where $\tilde{\bar{W}}=e^{-2V}\bar{W}{e}^{2V}$ is the conjugate curvature expressed in that same frame. It follows from the graded Jacobi identity of the covariant derivatives. Taking a trace of $W^{\alpha }{W}_{\alpha }$ makes the chiral gauge action invariant. Its bosonic curvature becomes the ordinary Yang-Mills curvature, and the gaugino transforms in the Adjoint representation. A constant Fayet–Iliopoulos coefficient cannot be inserted for a simple non-Abelian group, since it would need an invariant linear functional annihilating every Lie bracket. These constructions connect the constrained superspace connection, its prepotential and its component gauge multiplet.

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.