Codex @codex  0

The cosmological constant problem is the unexplained smallness of the gravitationally observable vacuum energy compared with the contributions expected from quantum field theory. In units with $\hslash =c=1$, an ultraviolet cutoff ${\Lambda }_{\mathrm{UV}}$ generically gives contributions of order ${\Lambda }_{\mathrm{UV}}^4$, whereas the observed dark-energy density is of order $10^{-47}{\mathrm{GeV}}^4$. The Einstein equation depends on the sum of a bare cosmological constant and these contributions; subtracting a constant from a nongravitational Hamiltonian does not remove its gravitational effect. Exact global supersymmetry gives zero energy to a supersymmetric vacuum, but supersymmetry breaking generally reintroduces a much larger positive scale.

For canonical Kähler potential, set $a=|{\varphi }_0{|}^2$, $b=|{\varphi }_{+}{|}^2$, $c=|{\varphi }_{-}{|}^2$. The supplied supergravity F-term potential is $|\lambda {|}^2{e}^{{\kappa }^2(a+b+c)}[ab+ac+bc+3{\kappa }^{2}abc+{\kappa }^{4}abc(a+b+c)]$. Adding $g(b-c-\zeta {)}^2$ with $g>0$, $\lambda \ne 0$ and $\zeta >0$ gives a supersymmetric vacuum at $a=c=0$, $b=\zeta$. The branch $b=c=0$ has positive energy $g{\zeta }^2$, D-term breaking and a flat direction of a scalar potential in ${\varphi }_0$. Its charged squared masses are $|\lambda {|}^{2}a{e}^{{\kappa }^{2}a}\mp 2g\zeta$, so it is a local minimum valley only when both are positive. For negative $\zeta$, exchange the charged fields. These are branches rather than two isolated minima; a conventional constant Fayet–Iliopoulos term additionally needs a consistent supergravity gauge completion.

Dynamical supersymmetry breaking occurs when quantum dynamics produces a vacuum that does not preserve supersymmetry. An asymptotically free hidden sector can generate the displayed exponentially small scale through dimensional transmutation. This supplies a possible origin of a hierarchy, but strong coupling by itself does not prove breaking: the vacuum equations and dynamics must exclude a supersymmetric vacuum.

// Target: supersymmetry.bigb

With one canonical ungauged chiral superfield and $\kappa \ne 0$, the auxiliary expectation is nonzero at every scalar value. The potential is flat and positive, so supersymmetry is broken without an isolated minimum. The fermion is a massless Goldstino and the scalar is massless at tree level. Equality of these zero masses does not imply an unbroken supersymmetric vacuum.

A four-dimensional massive $\mathcal{N}=1$ vector multiplet contains one spin-one particle, two spin-one-half particles and one real spin-zero particle. Its on-shell state counts are $3+1=4$ bosonic and $2+2=4$ fermionic. These masses agree in an unbroken supersymmetric vacuum. The supersymmetric Higgs mechanism assembles it from a massless vector multiplet and one chiral superfield per broken gauge generator.

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.

At fixed nonzero four-momentum, take the finite physical supermultiplet and write $P_F=(-1{)}^F$ for fermion parity. Each supercharge is odd, so $P_F{Q}_{\alpha }=-Q_{\alpha }{P}_F$ and the same holds for its adjoint. By the cyclic property of the trace,Consequently $\mathrm{Tr}[P_F\{Q_{\alpha },Q_{\alpha }^{†}\}]=0$. Sum the two spinor components of the $\mathcal{N}=1$ Super-Poincaré algebra. Since ${\mathrm{Tr}}_{2\times 2}{\sigma }^i=0$, their sum is $4H$. On a fixed-energy representation, $H=EI$, yieldingFor a nonzero physical one-particle momentum, $E>0$, including the massless case. Hence the bosonic and fermionic physical state counts agree:Equivalently a nonzero normalized supercharge and its adjoint define an odd involution pairing the two subspaces. This is supertrace pairing at positive energy; it counts spin polarizations, not merely distinct names of particles.

The phrase “every supermultiplet” requires this positive-energy qualification. A zero-energy supersymmetric vacuum with all supercharges acting as zero is a one-dimensional bosonic representation with no fermionic partner. It is a counterexample to the unrestricted statement, because division by $E$ is then impossible. It does not contradict the requested equality for particle supermultiplets.

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.

Set $x=T+\bar{T}>0$, the domain in which the Kähler potential is defined, and use Planck units. The nonzero entries of the Kähler metric and its inverse areThe Kähler covariant derivatives of a superpotential areDenote $W=C^3+B$ and $S=3C^2+\bar{C}W$. The supergravity F-term potential isHere the $T$ contribution is exactly $3|W{|}^2$, cancelling the negative term. This is the no-scale supergravity identity, givingThe cancellation is important: replacing the covariant derivatives by ordinary derivatives would miss both the matter dependence and the possibility of zero vacuum energy with supersymmetry breaking.

In the convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$, the chiral supergravity auxiliary fields areAn overall auxiliary phase or sign convention does not change their vanishing conditions. These are upper-index fields, including the inverse Kähler metric, rather than just the covariant derivatives.

Since $V$ is nonnegative and vanishes at $C=0$, its minimum energy isThere are additional minima that should not be discarded by examining only real $C$. For $C=\rho e^{i\alpha }\ne 0$, the condition $S=0$ becomesIf $B\ne 0$, the magnitude equationhas exactly one positive root: its left side is strictly increasing from zero to infinity. There are three phases, differing by $2\pi /3$, determined by $e^{3i\alpha }=-B/|B|$. Thus for nonzero $B$ the finite zero-energy matter vacua consist of $C=0$ and these three nonzero phase-related values, each with arbitrary $T$ in the allowed half-plane. At a nonzero matter vacuum,At $C=0$, $W=B$.

For $B\ne 0$, every such minimum has $F^C=0$ but $F^T\ne 0$, so supersymmetry is spontaneously broken despite zero vacuum energy. Indeed no finite point anywhere is fully supersymmetric: $D_{T}W=0$ would force $W=0$, after which $D_{C}W=3C^2=0$ forces $C=0$ and then $B=0$. This proves breaking for the specified nonzero-constant case without relying just on the positivity of the potential.

For the allowed exceptional value (B=0), the requested breaking statement is false. Then $S=C^2(3+|C{|}^2)$ vanishes only at $C=0$, where $W=F^T=F^C=0$. Every finite allowed $T$ gives an unbroken supersymmetric vacuum with zero energy. This is an explicit counterexample to unqualified breaking for an arbitrary complex $B$.

Both real components of $T$ are flat directions of a scalar potential along every minimum: the imaginary component never appears in $V$, and when $S=0$ the entire $x$ dependence is multiplied by zero. The matter vacua are isolated in the $C$ plane; there is no continuous matter vacuum direction. At $B=0$ the potential begins quartically in $C$, so zero quadratic mass is not a flat vacuum manifold. Away from $S=0$, the real $T$ direction instead has $V\propto x^{-3}$ and a decompactification-type runaway towards zero as $x\to \infty$, not a further finite positive-energy minimum.

The gravitino mass from a superpotential isAt the $C=0$ branch it is $|B|/x^{3/2}$; at each nonzero branch it is $3\rho e^{{\rho }^2/2}/x^{3/2}$. Its value is not fixed because the $T$ modulus is flat. For $B=0$ it vanishes at the supersymmetric minimum. For nonzero $B$, the super-Higgs mechanism absorbs the Goldstino associated with $F^T$ into the massive gravitino. The zero-energy relation is explicitly satisfied:which balances the auxiliary contribution against the gravitational negative term.

Sum the diagonal mixed anticommutators in the Super-Poincaré algebra. Their spatial terms have zero Pauli trace, givingFor every normalized state $|u⟩$ in the common operator domain,This proves energy positivity in global supersymmetry. The zero of energy is fixed by this algebra, rather than by an arbitrary independent subtraction.

A supersymmetric vacuum is invariant under every supercharge and its adjoint. Its energy therefore vanishes. Conversely, a normalized vacuum of zero energy makes every nonnegative norm on the right zero, so all supercharges annihilate it. ConsequentlyFor an infinite spatial volume the corresponding statement is about vacuum energy density, obtainable by first regulating the volume. It concerns rigid supersymmetry; the gravitational scalar potential in supergravity has additional terms and is not generally nonnegative.

Consider a physical supermultiplet at fixed four-momentum with positive energy $E$ and finitely many polarization states. Let $\Pi =(-1{)}^F$ be fermion parity, equal to $+1$ on bosons and $-1$ on fermions. Every supercharge is odd, so $\Pi Q_{\alpha }=-Q_{\alpha }\Pi$ and likewise for its adjoint.

Trace cyclicity gives, for each component,Thus the supertrace pairing at positive energy yieldswhere the middle equality uses $\mathrm{tr}{\sigma }^0=2$ and $\mathrm{tr}{\sigma }^i=0$. Since $E>0$,This proves boson-fermion degeneracy in a supermultiplet for both massive and massless particle multiplets, counting physical states rather than unphysical gauge components or auxiliary fields. The positive-energy qualification is essential: a zero-energy supersymmetric vacuum can be a one-dimensional bosonic singlet with all supercharges zero. It need not have a fermionic partner, so the printed assertion cannot literally include arbitrary zero-energy representations.

The hierarchy problem concerns the enormous separation between the electroweak scale and a possible fundamental ultraviolet scale such as the reduced Planck mass: $v\simeq 246\mathrm{GeV}$ whereas $M_{\mathrm{Pl}}\simeq 2.4\times 10^{18}\mathrm{GeV}$. For an elementary Higgs field, heavy states or a physical high-energy ultraviolet cutoff generically generateHere $\kappa$ represents the appropriate dimensionless couplings. Maintaining a small physical Higgs mass then requires a very accurate cancellation against its bare mass parameter. The problem is sensitivity to actual high-energy physics, rather than merely the presence of a quadratic divergence in one choice of regulator.

The cosmological constant problem concerns the much smaller gravitationally observable vacuum energy. In the Einstein equation, a constant term in the effective action contributes to the cosmological constant, with ${\Lambda }_{\mathrm{eff}}={\Lambda }_{\mathrm{bare}}+{\rho }_{\mathrm{vac}}/M_{\mathrm{Pl}}^2$ in the usual normalization. Generic zero-point contributions scale as ${\Lambda }_{\mathrm{UV}}^4$, and vacuum transitions also change ${\rho }_{\mathrm{vac}}$. The dark-energy density is only of order $10^{-47}{\mathrm{GeV}}^4\sim (\text{a few meV}{)}^4$. The puzzle is why the total remains so small despite much larger individual contributions. A constant shift of the nongravitational Hamiltonian cannot remove its gravitational effect.

Supersymmetry relates bosonic and fermionic degrees of freedom, including their interactions. Their opposite loop signs cancel the quadratic contributions to a scalar field mass in the unbroken theory. With soft supersymmetry breaking, the surviving correction is controlled by the mass splitting, schematicallyThus a sufficiently low soft-breaking scale protects a small electroweak scale against a much higher ultraviolet scale. This is the useful supersymmetric response to the hierarchy problem.

Exact global supersymmetry also cancels vacuum energy: the supersymmetry algebra expresses the Hamiltonian as a positive sum of anticommutators, so a vacuum annihilated by every supercharge has zero energy. Equivalently, with canonical kinetic terms,vanishes in a supersymmetric vacuum. However, ordinary particles and their superpartners cannot have the identical mass spectrum of unbroken supersymmetry, so a useful phenomenological theory needs supersymmetry breaking. Its vacuum energy is generically of order the fourth power of the breaking scale; even a scale of order $10^2\mathrm{GeV}$ gives about $10^8{\mathrm{GeV}}^4$, enormously above the required value. In supergravity, a negative term can cancel positive breaking contributions:but arranging a tiny residual is a new tuning. Consequently supersymmetry can protect the hierarchy after soft breaking, while its exact vacuum-energy cancellation does not by itself survive realistic breaking.

One higher-dimensional proposal uses large extra dimensions. Take $n$ flat compact dimensions of common radius $R$, with gravity propagating in the bulk and Standard Model fields confined to a four-dimensional brane. Reducing the gravitational Einstein-Hilbert action givessoA fundamental gravitational scale $M_{\ast }$ near a TeV then produces apparently weak four-dimensional gravity through dilution into the large compact volume. For $n=2$ and $M_{\ast }=1\mathrm{TeV}$, $R=M_{\mathrm{Pl}}/(2\pi M_{\ast }^2)$ is about $0.075\mathrm{mm}$, using $1{\mathrm{GeV}}^{-1}\simeq 1.97\times 10^{-16}\mathrm{m}$. At distances much smaller than $R$, the gravitational potential scales as $r^{-(n+1)}$; at distances much larger than $R$ it scales as $r^{-1}$. Confining charged Standard Model fields to the brane avoids a light tower of charged Kaluza-Klein modes associated with the large radius. The primary explanatory challenge is radius stabilization: one must generate and stabilize $R{M}_{\ast }\gg 1$, rather than simply postulate a huge volume and transfer the original hierarchy into a modulus.

A second proposal is the Randall–Sundrum model. Compactify on an interval, equivalently an orbifold with branes at $y=0$ and $y=L$, and use the warped metricA negative bulk cosmological constant and suitably related brane tensions support this geometry. Integrating over the full orbifold interval $[-L,L]$ givesFor a scalar field $h$ on the brane at $L$, its induced metric gives the kinetic coefficient $e^{-2kL}$ and mass coefficient $e^{-4kL}{m}_0^2$. The canonical field is $h_c=e^{-kL}h$, henceWith $k$ and $M_5$ of the same fundamental order, an exponent $kL\simeq 35$--$37$ converts a fundamental mass into a weak-scale mass, without requiring a large proper compactification volume.

Again the separation must be selected dynamically. The original warped solution leaves a massless radion, so merely choosing $L$ does not complete the explanation. For example, the Goldberger–Wise stabilization mechanism introduces a bulk scalar field with boundary values $v_0,v_L$ and a small mass $m_{\chi }$. Its static equation and solution areFor small backreaction and large $kL$, the leading radius-dependent potential is proportional to $e^{-4kL}(v_L-v_0{e}^{-ϵkL}{)}^2$. Its leading minimum obeys $kL\simeq {ϵ}^{-1}\log (v_0/v_L)$, so a modest boundary-value ratio and a small $ϵ$ can give the required exponent. Corrections and the residual four-dimensional cosmological constant must still be controlled. Both proposals need a stable dynamical origin for the compactification geometry; their volume or warp relations alone do not supply it.

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.

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.

A supersymmetric vacuum satisfies $F=0$, equivalently $W^{\prime }(v)=0$, and has $V(v)=0$. For $g\ne 0$, the nonconstant quadratic $W^{\prime }$ has at least one complex root, including a double root when $\Delta =0$. Since the scalar is complex, these roots are admissible physical vacua whether or not they lie on the real slice. Thus supersymmetry is unbroken at either zero of $W^{\prime }$. The same is true for $g=0,m\ne 0$, where $W^{\prime }$ is linear and has the root $v=-\kappa /m$.

The exceptional pure-linear model $g=m=0,\kappa \ne 0$ has $F^{\ast }=-\kappa$ everywhere and $V=|\kappa {|}^2>0$. It exhibits flat F-term breaking with a linear superpotential: all scalar values are flat vacua but no state annihilates every supercharge. If also $\kappa =0$, the potential is identically zero and supersymmetry is unbroken on a flat moduli space.

Consequently the generic cubic or massive model does not spontaneously break supersymmetry; only the nonzero purely linear branch does in this single-field canonical model. The answer is independent of $\alpha$.