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In the conventions $\{Q_{\alpha },Q_{\beta }^{†}\}=2{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$, the trace over the two spinor components of the Super-Poincaré algebra givesHence, on a normalized state in the domain of the supercharges,This proves energy positivity in global supersymmetry. The zero of energy is fixed by the algebra; adding an arbitrary constant to $H$ while keeping that algebra unchanged is not allowed.

For a finite-dimensional supermultiplet at fixed positive energy and four-momentum, choose a nonzero component $q$ with $\{q,q^{†}\}=cI$, $c>0$, and $q^2=0$. The Hermitian odd operatorIt is therefore an invertible map from the even fermion parity subspace to the odd subspace and conversely. This proves boson-fermion degeneracy in a supermultiplet, $n_B=n_F$, for every positive-energy physical particle supermultiplet. Equivalently, the supertrace of the supercharge anticommutator vanishes because odd operators interchange these two subspaces. A zero-energy vacuum is an exception to the literal claim about every possible representation: it can be a one-dimensional bosonic singlet with all supercharges acting trivially.

For a vacuum $|\Omega ⟩$, the sum-of-squares formula gives $E_0=0$ exactly when all supercharges and their adjoints annihilate it. If global supersymmetry is spontaneously broken and a vacuum exists, at least one such norm is nonzero, so the vacuum energy is positive. In infinite volume this is the statement about the vacuum energy density, understood with a finite-volume regulator. Explicit supersymmetry breaking need not preserve the algebra used in the proof. The conclusion also does not extend to the supergravity F-term potential, whose negative $-3e^K|W{|}^2$ term permits, for example, the zero-energy broken vacua of no-scale supergravity.

Use left Grassmann derivatives and place the odd constant transformation parameters on the left. A scalar superfield transformation can be writtenEquivalently, pull back the scalar function under the infinitesimal superspace translationThe term involving $\overline{ϵ}$ has the displayed sign because interchanging it with $\theta$ reverses the sign in the Grassmann algebra. Overall factors of $i$ in the definition of the supercharges can be changed together without changing this coordinate transformation.

For two even scalar superfields $S,T$, the odd generators obey the graded Leibniz rule, while the full parameter-weighted variation $\delta$ is even. ThusThis is precisely the transformation of a product of scalar superfields. The finite version is even clearer: if $S^{\prime }(z)=S(z^{\prime })$ and $T^{\prime }(z)=T(z^{\prime })$ for the same transformed superspace point, then $(ST{)}^{\prime }(z)=S(z^{\prime })T(z^{\prime })=(ST)(z^{\prime })$. No new representation law is required.

The expression $S(x,\theta ,\bar{\theta })=\varphi (x)$ needs a distinction between a configuration and a closed field content. As a function on superspace, it is a valid special configuration of an unconstrained superfield. But its variation isFor nonconstant $\varphi$, this generates nonzero components depending on Grassmann variables. Therefore the pure-scalar truncation is not a supersymmetric multiplet: an isolated spacetime scalar field with all its partners permanently set to zero is not closed under supersymmetry. This is the pure-scalar truncation of a superfield. A constant gauge-singlet $\varphi$ is the trivial invariant exception. Calling $\varphi (x)$ a special superfield configuration is consistent; claiming that its restricted field content remains a superfield representation is not.

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.

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.

The exponent in the original PDF is $e^{-\alpha S}$ with real $\alpha >0$, not the $e^{-aS}$ produced by the local TeX. Use that authoritative expression. PutThe domain conditions ensure a real Kähler potential and positive Kähler metric. There are no gauge multiplets specified, so the scalar potential is the supergravity F-term potentialThe Kähler covariant derivative of a superpotential and metric entries areState the inverse with its indices explicitly, since transposing the off-diagonal complex entries would change the answer:Here ${\sum }_j{K}_{i\bar{j}}{K}^{k\bar{j}}={\delta }_i^k$, so the displayed upper-index array is the transpose of the ordinary matrix inverse of the displayed lower-index array.

Substitution shows that the cross terms involving $W$ cancel and that the $T,C$ sector givesThe first term cancels the universal negative term. Thus the no-scale supergravity potential isThis positivity is an algebraic cancellation, not an assumption that the individual Kähler derivatives vanish.

Choose the supergravity auxiliary field convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$. Direct multiplication givesA conventional common phase or overall sign on the auxiliary fields changes none of the vanishing conditions. A supersymmetric configuration requires every $F^i$ to vanish. First $F^C=0$ implies $C=0$, then $F^T=0$ implies $w=0$, and $F^S=0$ implies $w_S=0$. Since $w_S=-\alpha a{e}^{-\alpha S}$, at a finite point of the physical domain this is possible only when $a=0$, followed by $b=0$. Consequently If $(a,b)\ne (0,0)$, every finite configuration has a nonzero auxiliary field, so any finite vacuum breaks supersymmetry. The printed request cannot hold for arbitrary parameters without this exception: $a=b=0$, $C=0$ is a zero-energy supersymmetric family with both $S$ and $T$ unfixed.

With the stipulated vanishing vacuum expectation value of $C$, define $A(S)=b+(1+\alpha s)a{e}^{-\alpha S}$. The potential reduces to $|A{|}^2/(s{t}^3)$. Its derivative along $t$ is $-3V/t$, so a finite stationary point must have $A=0$. Such a point is a global minimum of the full nonnegative potential, since both squares vanish at $C=0$. The finite zero-energy vacuum for a single exponential superpotential therefore satisfiesFor nonzero $a,b$, write $S=\sigma +i\chi$ and $x=\alpha \sigma >0$. The modulus and phase conditions areThe function $(1+2x)e^{-x}$ increases up to $x=1/2$ and then decreases to zero; its maximum is $2e^{-1/2}$. Hence a finite minimum exists exactly whenThere is one positive solution for $0<\rho \le 1$, two for $1<\rho <2e^{-1/2}$, and one coalesced solution at the upper bound. The formal $x=0$ solution at $\rho =1$ is outside $s>0$. At each allowed $x$, the phase fixes $\chi$ modulo $2\pi /\alpha$. Equivalently $x=-\frac{1}{2}-W_k(-\rho /(2\sqrt{e}))$, using the real branches of the Lambert W function and retaining only $x>0$.

At any of these nontrivial minima, $w=-\alpha sa{e}^{-\alpha S}\ne 0$. The auxiliary fields becomeThis is a supersymmetry breaking Minkowski minimum in which the nonzero auxiliary field belongs to $T$. Both real components of $T$ are exact flat directions of a scalar potential at the minimum. They change the metric and auxiliary-field magnitudes but not the zero potential. Generically the two real components of $S$ are fixed. At the coalesced solution $x=1/2$, the radial quadratic restoring term vanishes, but the leading restoring term is quartic; this is not an additional exact flat direction of a scalar potential. The matter field $C$ likewise has a positive leading quartic potential, not an exact flat direction of a scalar potential, despite its zero quadratic mass here.

The exceptional parameter cases must also be stated. If $a=0,b\ne 0$ or $a\ne 0,b=0$, no finite zero-energy minimum with $C=0$ exists. The same holds when $|b/a|>2e^{-1/2}$. Because the $t$ derivative is nonzero at any positive-energy point on $C=0$, none is a finite minimum there; the energy approaches zero along the runaway $t\to \infty$. If $a=b=0$, all physical $S,T$ at $C=0$ give supersymmetric zero-energy minima. Thus an unconditional finite minimum or unconditional breaking would be a false claim for the printed arbitrary parameters.

For the homogeneous model, assume $\Gamma >0$ is twice differentiable, of degree one in the real moduli ${\tau }_i$, and that its Hessian matrix for $K$ is invertible on the sector considered. Differentiate $K=-3\log \Gamma$:The two Euler theorem for homogeneous functions identities in the question implyMultiplying by the inverse gives $K_{ij}^{-1}{\Gamma }_j/\Gamma ={\tau }_i/3$, so contracting once more yieldsThis is the no-scale identity from degree-one homogeneity. It does not assert that every degree-one function gives a positive or invertible metric: for example $\Gamma ={\tau }_1+{\tau }_2$ has a rank-one Kähler Hessian, and its inverse is undefined. The inverse hypothesis is necessary.

For complex moduli with ${\tau }_i=T_i+{\bar{T}}_i$, the same derivatives are the mixed Kähler metric; choosing real parts with a factor of two only introduces factors that cancel in the contraction. If $W$ is independent of these moduli, $D_{i}W=K_{i}W$, and their contribution is $3e^K|W{|}^2$. It cancels the universal $-3e^K|W{|}^2$, leaving no scalar potential from this isolated no-scale sector. Other chiral sectors can still contribute positive terms. For this conclusion in a larger theory, the displayed metric must be the appropriate decoupled no-scale block, or the full inverse metric must itself satisfy the corresponding identity; arbitrary mixed additions to $K$ do not inherit the cancellation automatically.

Use a positive-energy unitary representation with finite helicity states, excluding continuous-spin representations. In the specified frame the Pauli matrices giveFor every $A$ and every state $|\chi ⟩$,Positivity therefore gives $Q_2^A={\bar{Q}}_{\dot{2}A}=0$ on this massless supermultiplet. The equal-chirality anticommutator $\{Q_1^A,Q_2^B\}={ϵ}_{12}{Z}^{AB}$ then shows that every central charge in supersymmetry vanishes on this superalgebra representation.

Normalize the remaining supercharges as fermionic raising and lowering operators:They obey the canonical anticommutation relationsChoose a Clifford vacuum $|\lambda ⟩$ of helicity $\lambda \in \frac{1}{2}\mathbb{Z}$ annihilated by every $a_A$. Such a vector exists because the commuting fermion number operators $a_A^{†}{a}_A$ have eigenvalues $0$ and $1$, and lowering every occupied mode produces a nonzero empty state. In an irreducible representation this empty space has dimension one; multiple empty states produce a direct sum of copies. The states areThe canonical anticommutation relations make these states orthonormal for a normalized Clifford vacuum. In the helicity convention $[J_3,Q_1^A]=\frac{1}{2}{Q}_1^A$, its adjoint has $[J_3,a_A^{†}]=-\frac{1}{2}{a}_A^{†}$. Thus the Clifford vacuum is the highest-helicity state andReversing the rotation convention reverses the displayed ordering and leaves all counts unchanged. The level-$k$ states form the exterior power ${\Lambda }^k{\mathbb{C}}^{\mathcal{N}}$, which explains the binomial coefficient. Summing the binomial coefficients gives the helicity spectrum of a massless supermultiplet:Each application of a fermionic raising and lowering operator reverses fermion parity. The even and odd levels each contain $2^{\mathcal{N}-1}$ states, so a physical massless supermultiplet has $2^{\mathcal{N}-1}$ boson and $2^{\mathcal{N}-1}$ fermion states.

These formulas describe one irreducible representation of the massless Super-Poincaré algebra at fixed four-momentum. A reducible superalgebra representation can contain several such supermultiplets. Also, the CPT theorem sends helicity $h$ to $-h$ and conjugates internal charges. CPT completion of a supermultiplet may therefore require a second $2^{\mathcal{N}}$-state supermultiplet, with highest helicity $\mathcal{N}/2-\lambda$. A necessary condition for the original helicity spectrum of a massless supermultiplet to be self-conjugate is $\lambda =\mathcal{N}/4$; internal charges must also admit the conjugation in a realization compatible with the Spin-statistics theorem. Thus this helicity condition alone is not sufficient for CPT completion of a supermultiplet without doubling. In particular, the original irreducible count is not automatically the count after CPT completion of a supermultiplet.

For the conventional bounds on extended supersymmetry, suppose all massless states satisfy $|h|\le s_{\max }$. The helicity spectrum of a massless supermultiplet occupies an interval of width $\mathcal{N}/2$, whereas the available interval has width $2s_{\max }$. HenceAn ordinary four-dimensional renormalizable quantum field theory without gravity uses interacting massless fields of spin angular momentum at most one. Massless spin angular momentum $3/2$ requires supergravity, whose gravitational coupling has negative mass dimension and is a nonrenormalizable interaction by power counting in quantum field theory. Setting $s_{\max }=1$ gives $\mathcal{N}\le 4$ for conventional renormalizable theories. The bound is attained by four-dimensional N=4 super Yang-Mills theory: choosing $\lambda =1$ givesfor its supersymmetric vector multiplet.

Allowing gravity permits $s_{\max }=2$, so $\mathcal{N}\le 8$ for conventional theories including gravity. The four-dimensional N=8 supergravity supergravity multiplet attains the bound with $\lambda =2$:It has $256$ states, $128$ boson and $128$ fermion states. The conventional claim about a maximum “in general” assumes local interacting four-dimensional theories in Minkowski spacetime with a finite spectrum of fields and no massless spin angular momentum above two. It is not a theorem excluding arbitrary free higher-spin angular momentum constructions or all theories in other spacetime settings.

Finally, a chiral gauge spectrum means that the independent left-handed Weyl spinors occur in complex gauge group representations without obligatorily paired conjugate left-handed Weyl spinors. In four-dimensional N=1 supersymmetry, an irreducible matter supermultiplet with helicities $(0,-\frac{1}{2})$ can carry a complex gauge group representation $R$. Its CPT completion of a supermultiplet supplies the right-handed antiparticle in $\bar{R}$, as every left-handed Weyl spinor requires; it does not supply a second independent left-handed Weyl spinor in $\bar{R}$. Consequently a chiral gauge spectrum is possible.

For $\mathcal{N}=2$, the matter hypermultiplet has helicities $(\frac{1}{2},0,0,-\frac{1}{2})$ before internal charge conjugation. If it carries a complex gauge group representation $R$, its CPT completion of a supermultiplet adds the conjugate hypermultiplet. Equivalently, the resulting full hypermultiplet contains two $\mathcal{N}=1$ chiral superfields, one in $R$ and one in $\bar{R}$. These give a vectorlike gauge spectrum. Half-hypermultiplets can occur in pseudoreal representations, which also do not produce genuinely complex chiral gauge spectra. A supersymmetric vector multiplet has Adjoint representation, hence real gauge quantum numbers. Since every $\mathcal{N}\ge 2$ algebra contains an $\mathcal{N}=2$ subalgebra, the same obstruction applies to greater extended supersymmetry.

Therefore a conventional chiral gauge spectrum is possible only for $\mathcal{N}=0$ or $\mathcal{N}=1$. Here $\mathcal{N}=0$ means no supersymmetry; among theories with actual supersymmetry, only $\mathcal{N}=1$ qualifies. These are necessary structural conditions, not a guarantee that a chosen chiral gauge spectrum satisfies anomaly cancellation.

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 Polonyi model with canonical Kähler potential and $W=m^2(z+\beta )$, $\beta >0$, a constant supersymmetric vacuum with $m\ne 0$ requires real $z=(-\beta \pm \sqrt{{\beta }^2-4})/2$ and exists exactly for $\beta \ge 2$. Its supergravity F-term potential is negative because $W\ne 0$. For $0<\beta <2$, the supergravity auxiliary field cannot vanish. If $m=0$, the model is flat with unbroken supersymmetry. This parameter-dependent statement is sharper than asserting that every linear superpotential breaks supersymmetry in supergravity.

A function $S(x,\theta ,\bar{\theta })=\varphi (x)$ is an allowed special configuration of an unconstrained superfield. The subspace where all other components remain zero is not invariant: $\delta S=(i\theta {\sigma }^{\mu }\overline{ϵ}-iϵ{\sigma }^{\mu }\bar{\theta }){\partial }_{\mu }\varphi$ produces components depending on Grassmann variables whenever $\varphi$ is nonconstant. Thus an isolated spacetime scalar field is not closed under supersymmetry. A constant gauge-singlet value is a trivial invariant exception.

For one charged chiral superfield, canonical kinetic terms, no nonconstant gauge-invariant superpotential, and $g>0$, the sign of $q\xi$ selects the vacuum. If $q\xi >0$, the minimum has $\varphi =0$, positive vacuum energy, unbroken internal gauge symmetry and a gaugino Goldstino; the scalar squared mass is $gq\xi$ while its chiral fermion is massless. If $q\xi <0$, a scalar condensate cancels the auxiliary field, preserving supersymmetry while Higgsing the gauge group. These are classical statements; a consistent quantum charged spectrum must cancel its gauge anomalies.

In the critical RNS string, a spacetime supersymmetry generator is $Q_A=\oint e^{-{\varphi }_{\mathrm{gh}}/2}{S}_A$. The RNS spin field has weight $5/8$ and the superghost exponential has weight $3/8$, making a weight-one current. The GSO projection ensures the locality needed to act on retained vertex operators. Its operator products generate translations and relate NS sector and R sector states.

In four-dimensional supersymmetry, a finite-helicity massless supermultiplet has helicity interval of width $\mathcal{N}/2$. If every state obeys $|h|\le s$, that interval fits inside one of width $2s$, giving $\mathcal{N}\le 4s$. The bounds $4$ for $s=1$ and $8$ for $s=2$ are attained by the four-dimensional N=4 super Yang-Mills theory vector multiplet and the four-dimensional N=8 supergravity supergravity multiplet. Spin bounds alone do not prove that an arbitrary action is renormalizable.

The local gauge choice $h_{\mu \nu }=e^{2\omega }{\eta }_{\mu \nu }$ and ${\chi }_{\mu }=0$ for an RNS string. Diffeomorphisms and Weyl symmetry fix the metric, while local supersymmetry and super-Weyl symmetry fix the gravitino. Stress-tensor and supercurrent constraints survive the gauge choice and eliminate longitudinal degrees of freedom. Higher-genus surfaces retain metric and spin-structure moduli.

The supercurrent is the spinor-valued Noether current of rigid supersymmetry. Its spatial integral is a supercharge. Localizing the supersymmetry parameter produces ${\partial }_{\mu }\overline{ϵ}{S}^{\mu }$ in the variation of the action; a gravitino coupling cancels this term and starts the Noether gauging procedure.

When local supersymmetry is spontaneously broken, the gravitino absorbs the goldstino and becomes massive. The two goldstino states provide its longitudinal spin-$1/2$ polarizations. With only global supersymmetry breaking, the goldstino remains a physical massless particle. An ordinary supersymmetric Higgs mechanism instead makes an internal spin-one gauge boson massive and can leave supersymmetry unbroken.