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Adding a constant to a global superpotential does not change the component action because the chiral superspace integral of that constant vanishes. Neither the F-term scalar potential nor the fermion mass matrix depends on it. This statement does not extend to supergravity, where the superpotential itself enters the gravitational potential.

Stationarity of a canonical global F-term scalar potential gives ${\partial }_{i}V={\sum }_j{W}_{ij}\overline{W_j}=0$. If some $F_j=-\overline{W_j}$ is nonzero, this vector is in the kernel of the chiral-superfield fermion mass matrix. It supplies the massless goldstino direction. If every $F_j$ vanishes, this argument gives no broken-symmetry mode.

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 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.

Write $S={\Phi }_0$, $u={\Phi }_{+}$, $v={\Phi }_{-}$, $M=M_p$, $P=uv-\zeta$, $x=|S{|}^2/M^2$ and $q=|u{|}^2+|v{|}^2$. Assume $g\ne 0$ and, initially, $\zeta \ne 0$. The supergravity F-term potential isThe canonical Kähler potential gives the identity Kähler metric. The three Kähler covariant derivatives of a superpotential areSubstituting, including the negative term rather than treating the potential as just a sum of squares, gives the exact answerHere $(1+x{)}^2-3x=1-x+x^2$ explains the first coefficient. It is strictly positive for real $x$, since $1-x+x^2=(x-1/2{)}^2+3/4$. Thus this model's scalar potential is nonnegative everywhere.

In the stated limit $q/M^2\ll 1$, the exponential can be replaced by $e^x[1+O(q/M^2)]$. The other displayed terms retain the effects of the Kähler covariant derivatives of a superpotential. Small waterfall fields alone do not justify discarding $\zeta /M^2$ in those derivatives, or expanding in $x$. If also $|\zeta |/M^2\ll 1$ and $x\ll 1$, the leading result reduces to the global F-term scalar potentialNo gauge sector or gauge coupling is supplied, so there is no specified D-term contribution to add.

The supersymmetric vacuum manifold of a bilinear hybrid superpotential isIndeed $W=0$ and all three Kähler covariant derivatives of a superpotential vanish there, so all supergravity auxiliary fields vanish and supersymmetry is unbroken. The nonnegative scalar potential makes this a global minimum. Conversely zero potential requires $uv=\zeta$, and then $|S{|}^{2}q=0$; for $\zeta \ne 0$ this forces $S=0$. With just the supplied fields this is a complex vacuum manifold before imposing any gauge quotient: the product is fixed but the ratio of $u$ and $v$ is not. More precisely, it is an F-flat vacuum manifold whether or not a gauge symmetry is specified. If one additionally gauges opposite charges, D-flatness typically imposes $|u|=|v|$, giving $|u|=|v|=\sqrt{|\zeta |}$ up to phases and gauge equivalence; that is an extra assumption, not a consequence of $K$ and $W$ alone.

The other familiar branch of F-term hybrid inflation has $u=v=0$. Its nonzero derivative $D_{S}W=-g\zeta (1+x)$ establishes supersymmetry breaking, and its exact scalar potential isThe canonical-Kähler correction to an F-term hybrid valley begins asIn particular the inflaton quadratic mass cancels at this order; the first lift of the global flat valley is quartic in $|S|$.

To identify when this branch is a transverse minimum, rephase the waterfall fields so that $\zeta$ is real positive. To quadratic order in $u,v$, the global scalar potential isThe normalized combinations $(u\pm \bar{v})/\sqrt{2}$ diagonalize this quadratic form, withThus $|S{|}^2>|\zeta |$ gives a valley of transverse minima, while below the threshold the negative eigenvalue produces the waterfall instability in F-term hybrid inflation.

The exact supergravity correction to this stability test is also available without taking $x$ small. Put $b=|\zeta |/M^2$. Expansion of the exact supergravity F-term potential about $u=v=0$ gives equal diagonal coefficients $|g{|}^2{M}^2{e}^x[x+b^2(1+x^2)]$ and off-diagonal magnitude $|g{|}^2{M}^2{e}^{x}b(1+x+x^2)$. HenceFor the usual sub-Planckian symmetry-breaking scale $0<b\ll 1$, the lower eigenvalue is $|g{|}^2{M}^2{e}^x(1-b)[x-b(1+x^2)]$. It is positive for the familiar sub-Planckian valley $b\ll x\ll 1$, and changes sign near $x=b$.

The printed assertion of two full minima requires qualification. The broken branch is a transverse inflationary valley, not generally a second stationary minimum of the complete supergravity potential. In factIts only stationary point in $S$ is the origin. For $0<|\zeta |<M^2$ that point has a negative waterfall eigenvalue $|g{|}^2(|\zeta {|}^2/M^2-|\zeta |)$. A direct counterexample is the path $S=0$, $u=v=a$ with real small $a$, after rephasing $\zeta >0$:The negative quadratic coefficient proves it is not a minimum. No loop corrections or additional stabilizing interactions were specified that could create the requested second stationary vacuum. The mathematically supported interpretation is a SUSY-breaking transverse valley and a zero-energy supersymmetric vacuum minimum manifold. The limit on $u,v$ alone does not fix the parameter range $|\zeta |/M^2$; the familiar waterfall conclusion uses the sub-Planckian range just stated.

In the corresponding global theory the exponential prefactor, Kähler corrections to $D_{i}W$ and universal $-3|W{|}^2/M^2$ term are absent. Its $u=v=0$ valley has exactly constant tree-level energy $|g\zeta {|}^2$ for all $S$, whereas canonical supergravity lifts it by the quartic correction above. The supersymmetric vacuum $S=0$, $uv=\zeta$ and its zero energy persist in both descriptions. If $\zeta =0$, the origin and the $S$ axis instead have vanishing F-terms; the broken-valley interpretation does not apply.

For a single chiral superfield, the supergravity F-term potential, with $\kappa =m_p^{-1}$ restored, isThe canonical Kähler potential has $K_{z\bar{z}}=K^{z\bar{z}}=1$, so the Kähler covariant derivative of a superpotential isConsequently the Polonyi model has the scalar potentialThe negative term is essential; retaining only the square of the supergravity auxiliary field would give the global F-term scalar potential rather than the supergravity F-term potential.

Set $x=\kappa z=a+iy$ and $b=\kappa \beta =2-\sqrt{3}$. In this notation $V=|\mu {|}^2{m}_p^2{e}^{|x{|}^2}U$, whereAt $a_0=\sqrt{3}-1$, $y=0$, one has $a_0+b=1$ and $1+a_0(a_0+b)=\sqrt{3}$, so $U=0$. To prove that this is a local minimum in both real directions, rather than checking only the real axis, put $u=a-a_0$. Direct expansion gives the stable zero-energy Polonyi vacuum identityEvery coefficient outside the square is strictly positive, and the exponential factor is positive. Thus $V\ge 0$ everywhere and equality requires $u=y=0$, provided $\mu \ne 0$. This proves the stronger conclusiona unique global minimum of this Polonyi model. If $\mu =0$, the scalar potential vanishes identically and this uniqueness and supersymmetry breaking conclusion do not hold.

For an independent local stability check, the Hessian matrix of $U$ at this global minimum is diagonal:Since $U$ and its first derivatives vanish there, differentiating the exponential introduces no additional Hessian matrix terms at the global minimum. For canonically normalized real fluctuations $z=z_0+(s+it)/\sqrt{2}$, the squared scalar masses are thereforewhich are both positive.

Although the vacuum energy is zero, the supergravity auxiliary field is not. In the convention $F^z=-e^{{\kappa }^{2}K/2}\overline{D_{z}W}$, its vacuum expectation value obeysThe matter fermion transformation contains a term proportional to $F^zϵ$, so no nonzero constant supersymmetry parameter leaves this vacuum invariant. Hence supersymmetry is spontaneously broken even though the vacuum energy vanishes: the positive auxiliary field contribution cancels the negative gravitational term. The matter fermion supplies the goldstino, which is absorbed by the gravitino in the super-Higgs mechanism.

Using the gravitino mass from a superpotential, and $z_0+\beta =m_p$, givesAn auxiliary field $F^z$ has mass dimension two. Defining the supersymmetry breaking scale by ${\Lambda }_{\mathrm{SUSY}}=\sqrt{|F^z|}$, one obtainsThus the gravitino mass is small compared with the supersymmetry breaking scale when that scale is sub-Planckian, as in the usual hierarchy $|\mu |\ll m_p$. This smallness is gravitational suppression; it is not a conclusion valid for arbitrary choices of the mass parameter.

Use the scalar components of the three chiral superfields, and set $a=|{\varphi }_0{|}^2$, $b=|{\varphi }_{+}{|}^2$, $c=|{\varphi }_{-}{|}^2$ and $\Sigma =a+b+c$. We assume the physical sign $g>0$ and a nonzero coupling $\lambda$. The coefficient $g$ below is exactly the coefficient in the supplied D-term scalar potential; it should not be silently replaced by a differently normalized squared gauge coupling. The canonical Kähler potential has identity Kähler metric, and the three Kähler covariant derivatives of a superpotential areSubstitution in the supergravity F-term potential givesIn particular, the subtraction cancels only half of the six ${\kappa }^{2}abc$ terms contributed by the three squares. The full scalar potential isIts nonnegativity follows term by term from $a,b,c\ge 0$ and $g>0$. This is a special property of this cubic superpotential, not of a general supergravity F-term potential.

The D-term vacuum branches of a cubic superpotential must be described as branches: there are not two isolated field-space minima. First suppose $\zeta >0$. Vanishing of $V_F$ requires $ab=ac=bc=0$, and vanishing of the D-term scalar potential requires $b-c=\zeta$. Together these conditions give the vacuum expectation valuesThe phase of ${\varphi }_{+}$ lies on a gauge orbit, and one may choose it positive and real. At this global minimum, all $D_{i}W$ and the real auxiliary field of the vector multiplet vanish. It is a supersymmetric vacuum, even though the internal gauge symmetry is spontaneously broken. For $\zeta <0$, interchange ${\varphi }_{+}$ and ${\varphi }_{-}$: the nonzero vacuum expectation value has magnitude $\sqrt{-\zeta }$. For $\zeta =0$, the supersymmetric vacuum is the entire family ${\varphi }_{+}={\varphi }_{-}=0$, with ${\varphi }_0$ arbitrary.

The other branch has ${\varphi }_{+}={\varphi }_{-}=0$ and arbitrary ${\varphi }_0$. Here $W=D_{i}W=0$ but the real auxiliary field is nonzero for $\zeta \ne 0$, andThus this branch has pure D-term supersymmetry breaking and a flat direction of a scalar potential in the neutral scalar. It is not stable at every neutral-field value. At a fixed $a=|{\varphi }_0{|}^2$, expansion in the charged scalars yieldsHence the two charged squared masses and the stability condition areAbove this threshold, both charged directions have positive mass squared. Continuity makes them uniformly positive in a sufficiently small neighborhood of the chosen neutral-field value, so the stationary branch is a genuine, non-strict local minimum valley. Its neutral direction stays flat. Below the threshold, a charged mass squared is negative and the branch is a saddle point. At equality it is a boundary of the stable valley, not a local minimum in the full field space: changing $a$ slightly downward and switching on the unstable charged field lowers $V$. For example, for $\zeta >0$, put $a=a_c-ϵ$, $b={ϵ}^2$, $c=0$. Since $M(a)=|\lambda {|}^{2}a{e}^{{\kappa }^{2}a}$ has $M^{\prime }(a_c)>0$, one finds $V-g{\zeta }^2=-M^{\prime }(a_c){ϵ}^3+O({ϵ}^4)<0$.

There are no further local-minimum branches under these generic assumptions. If $b+c>0$, the expression for $V_F$ is strictly increasing in $a$, so a stationary configuration with a charged condensate must have $a=0$. At $a=0$, an interior critical point with $b,c>0$ is impossible, becauseOn the boundary $c=0$, the relevant scalar potential is $g(b-\zeta {)}^2$; on $b=0$ it is $g(c+\zeta {)}^2$. Their nonnegative-domain minima give exactly the supersymmetric vacuum described above, or the origin of the charged-field branch. Thus, for nonzero $\zeta$, the two types are the zero-energy supersymmetric gauge orbit and the positive-energy D-breaking minimum valley above the threshold. If $\zeta =0$, the two types coalesce into the flat supersymmetric vacuum; the literal two-minimum assertion needs this parameter qualification.

In the global supersymmetry limit $\kappa \to 0$, the F-term scalar potential becomes $|\lambda {|}^2(ab+ac+bc)$ and the D-term scalar potential retains its form. The same two types of branches occur, with the threshold $|\lambda {|}^{2}a>2g|\zeta |$. Supergravity instead introduces the exponential and additional positive terms. The goldstino of global supersymmetry breaking is a physical fermion; in local supersymmetry it participates in the super-Higgs mechanism. A positive vacuum energy also gravitates in supergravity. In particular, the positive-energy branch here has $W=0$ and hence vanishing gravitino mass from a superpotential; it is not the tuned zero-energy broken vacuum of Question 3.

There is a further gauge-consistency qualification. In conventional two-derivative matter-coupled supergravity, a constant Fayet–Iliopoulos term gauges an R-symmetry and requires a corresponding nontrivial transformation of the superpotential. The apparent charges $0,+1,-1$ make the supplied cubic superpotential neutral, so a nonzero independent constant $\zeta$ does not by itself specify a consistent conventional supergravity action. The computed branches are those of the supplied scalar potential; a full embedding needs compatible charges or additional fields. This restriction is explained in Van Proeyen's derivation of the superpotential gauge-covariance condition.

More generally, the supergravity moment-map constraint on D-term breaking follows directly from gauge covariance. If $k^i{\partial }_{i}W=-{\kappa }^{2}rW$ and the moment map is $\mathcal{P}=i(k^i{K}_i-r)$, then at a point with $W\ne 0$,Consequently all chiral auxiliary fields vanishing implies the D-term also vanishes there. This relation has no analogue forcing that implication in an arbitrary global supersymmetry model. Its $W\ne 0$ hypothesis matters: the formal D-breaking branch above has $W=0$, so it does not contradict the identity. These conventional gauge-completion restrictions are additional to the elementary minimization calculation.

Eliminating the auxiliary field gives $F^{\ast }=-W^{\prime }(\phi )$ and the nonnegative F-term scalar potentialUse $u=\phi -s$ after the critical-point shift, so thatThis notation keeps the original parameters distinct from the shifted mass coefficient $M$. For real $M,g$ with $Mg\ne 0$, the real slice is $V(u)=u^2(M+gu/2{)}^2$. It has zero-energy minima at $u=0$ and $u=-2M/g$, a maximum between them at $u=-M/g$ with height $M^4/(4g^2)$, and grows quartically toward both ends. These statements follow by differentiating: $V^{\prime }(u)=2u(M+gu/2)(M+gu)$.

When $M=0$ it becomes $|g{|}^2{u}^4/4$, with one degenerate zero minimum. When $g=0$, it is $|M{|}^2{u}^2$ after a permissible shift. For the real parameters with negative discriminant, the original real slice can have a positive minimum even though the full complex scalar potential has zero-energy vacua away from that slice. The sketch distinguishes these regimes:

The plots use illustrative normalized coefficients, not values fixed by the exam. A positive minimum on a chosen real slice is not evidence for supersymmetry breaking in the full complex field space.

Take canonical charged chiral superfield kinetic terms, gauge coupling $e>0$, and no Fayet–Iliopoulos term, since none is specified. Elimination of the three complex auxiliary fields givesThe Abelian gauge auxiliary field obeys $D=-e(|{\phi }_{+}{|}^2-|{\phi }_{-}{|}^2)$ in this normalization. Thus the F-term scalar potential and the gauge D-term giveAll terms are nonnegative. The normalization of $e$ can be changed together with the vector-field normalization, but the relative charges and the zero-potential conditions cannot. A noncanonical Kähler potential would change the inverse-metric factors in the F-term potential; adding a Fayet–Iliopoulos term would shift $D$ and define a different model. Neither is silently introduced here.

The canonical Kähler potential gives $e^K=e^{|z{|}^2}$ and inverse Kähler metric one. Substituting the superpotential and its Kähler covariant derivative of a superpotential into the supergravity F-term potential yieldsFor real $m$, the prefactor is simply $m^4$; the modulus form also covers a complex phase. To keep both scalar directions explicit, write $z=x+iy$ and $P=1+x^2+y^2+\beta x$. ThenThe negative term is essential: unlike the global F-term scalar potential, the supergravity F-term potential need not be nonnegative. This is why cancelling the cosmological constant does not force the auxiliary field to vanish.

Use the minimal four-dimensional Super-Poincaré algebra, with $\{Q_{\alpha },Q_{\beta }\}=0$, $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=2{\sigma }_{\alpha \dot{\beta }}^{\nu }{P}_{\nu }$ and commuting translations. Lorentz covariance and closure on the existing supercharges permit onlyThe graded Jacobi identity for $P^{\mu },Q_{\alpha },Q_{\beta }$ therefore givesFor example, with ${\sigma }^0=I$, $\mu =0$ and $\alpha =\beta =1$, the coefficient is a nonzero multiple of $A(P_1-i{P}_2)$. Independence of the translation generators forces $A=0$. Its adjoint gives the barred result. Thus every supercharge commutes with four-momentum:Consequently $[Q_{\alpha },P^2]=0$. If $|u⟩$ has a particular mass-shell condition, the nonzero state $Q_{\alpha }|u⟩$ has the same one. This gives supersymmetric mass degeneracy within an unbroken physical supermultiplet. The qualification matters: after spontaneous supersymmetry breaking, the chosen vacuum is not annihilated by the supercharges, and its one-particle excitations need not constitute degenerate supermultiplets.

For the O'Raifeartaigh model, write $X={\varphi }_1$, $Y={\varphi }_2$, $Z={\varphi }_3$. Field phases allow $g,M,m$ to be positive. The canonical Kähler potential and elimination of the auxiliary fields give the F-term scalar potentialA supersymmetric vacuum would require $W_Y=MZ=0$, hence $Z=0$, while $W_X=g(Z^2-m^2)=-g{m}^2$ could not vanish. Thus F-flatness is impossible. More precisely, with $Z=u+iv$,The stated strict inequality makes every nonconstant term nonnegative. The entire classical vacuum family isHere $F_X=-\overline{W_X}=g{m}^2$, while $F_Y=F_Z=0$. The Weyl spinor ${\psi }_X$ is the massless goldstino, and the complex scalar field $X$ is a pseudomodulus with two zero tree-level squared masses. The hierarchy $M\gg m$ separates the massive fields from the breaking scale, but it does not itself select $x=0$ along this flat direction.

At the representative vacuum $x=0$, the chiral-superfield fermion mass matrix isIts physical masses are its singular values: $0,M,M$. The two massive Weyl spinors can be combined into one massive Dirac spinor. For $Y=(y_R+i{y}_I)/\sqrt{2}$, $Z=(z_R+i{z}_I)/\sqrt{2}$, the real scalar squared masses areIn particular, there is no tachyon under the stated inequality. The splitting of the $Z$ masses displays supersymmetry breaking directly.

For completeness, the mass spectrum can be given throughout the classical vacuum family, rather than silently fixing the pseudomodulus. A phase rotation makes $x$ real and nonnegative without changing physical masses. The massive fermion squared masses areThe real and imaginary scalar sectors have two-by-two scalar squared-mass matriceswhose eigenvalues areBoth determinants are $M^2(M^2+2ϵg^2{m}^2)>0$ and their traces are positive, so stability holds at every finite $x$. Quantum corrections can lift a pseudomodulus; these formulas describe the classical, tree-level spectrum requested here.

Define the physical mass supertrace byEach real scalar contributes once and each Weyl spinor twice with a minus sign. At $x=0$ this gives $4M^2-2(2M^2)=0$. At general $x$, ${\sum }_{ϵ,\pm }{b}_{ϵ,\pm }=4M^2+8g^2|x{|}^2=2(f_{+}+f_{-})$. Hence the tree-level mass supertrace vanishes despite broken supersymmetry:This is the tree-level supertrace mass sum rule, not a claim that the individual masses are equal.