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For positive real $M,\mu ,\lambda$ with $M^2>2{\lambda }^2{\mu }^2$, minimizing $V={\lambda }^2|z^2-{\mu }^2{|}^2+M^2|z{|}^2+|2\lambda xz+My{|}^2$ gives $z=y=0$ with arbitrary $x$. Around $x=0$, the real scalar squared masses are $0,0,M^2,M^2,M^2-2{\lambda }^2{\mu }^2,M^2+2{\lambda }^2{\mu }^2$, while the three Weyl spinor masses are $0,M,M$. Their weighted supertrace vanishes. The massless ${\psi }_X$ is the goldstino; the two real $x$ modes form a tree-level pseudomodulus. The stability bound follows from $|z^2-{\mu }^2{|}^2\ge (|z{|}^2-{\mu }^2{)}^2$, after minimizing the final square over $y$.

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.

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.

For direct breaking in the Minimal supersymmetric Standard Model, the difficulty is both field content and the tree-level spectrum. A linear gauge-invariant superpotential term needs a gauge-singlet chiral superfield, which the minimal model does not contain. Moreover, with canonical kinetic terms and purely neutral F-term breaking, without D-term mass shifts, the MSSM tree-level sfermion mass constraint applies separately to conserved charge sectors. For one electron pair it gives $m_{{\tilde{e}}_1}^2+m_{{\tilde{e}}_2}^2=2m_e^2$: both scalar partners cannot be heavy. This is an illustrative charge-sector consequence under those assumptions, not an unrestricted inference from the total supertrace alone. A separate hidden supersymmetry-breaking sector communicating effective soft supersymmetry breaking through loops or suppressed operators avoids the direct canonical tree-level obstruction.

First assume global four-dimensional N=1 supersymmetry, canonical positive Kähler potential, and tree level. Write $W_i={\partial }_{i}W$ and $m_{ij}=W_{ij}$ at the stationary vacuum. The chiral-superfield fermion mass matrix is the symmetric matrix $m$, and the sum of squared Weyl spinor masses is $\mathrm{tr}(m^{†}m)$. From the F-term scalar potential $V={\sum }_k|W_k{|}^2$, its mixed and holomorphic Hessian matrix blocks areFor canonically normalized real and imaginary parts of the scalar fields, the real scalar mass matrix has trace twice the mixed trace. Its holomorphic blocks can split the two real scalar masses but do not change their sum. HenceA Weyl spinor has two spin states, so this proves the tree-level supertrace mass sum rule,The paper defines its supertrace with $(-1{)}^{2j+1}$, the negative of the conventional boson-minus-fermion supertrace used here. The asserted zero is the same in either convention. The result requires the stated canonical tree-level hypotheses; a noncanonical Kähler metric, supergravity, or radiative corrections can change the sum rule.

If vector multiplets are also present, the gauge contributions to the F-term supertrace cancel rather than being omitted. At $D_a=0$, define $C={\sum }_a{g}_a^2\Vert T_{a}v{\Vert }^2$ for the scalar expectation vector $v$. Differentiating $\frac{1}{2}{\sum }_a{D}_a^2$ adds $2C$ to the real-scalar trace. The symmetric gaugino-matter mass matrix has off-diagonal entries $\sqrt{2}{g}_a{T}_{a}v$, giving an extra $4C$ in its fermion squared-mass trace. The covariant scalar kinetic term gives vector squared-mass trace $2C$. Thus their contribution is $2C-2(4C)+3(2C)=0$. The gauge-theory result uses all spin states, including the vector weight three; it is not obtained by dropping massive gauge partners.

The MSSM tree-level sfermion mass constraint explains the phenomenological difficulty with direct visible-sector breaking. If F-term breaking is neutral under an unbroken electric and color gauge symmetry, and no D-term shifts are present, the same trace argument applies within each conserved-charge fermion block. Its corresponding scalar partners cannot all have squared masses above the mean of the light fermion squared masses. Thus canonical tree-level visible-sector supersymmetry breaking alone cannot make every squark and other scalar partner heavy while leaving the observed fermions light. The usual effective soft supersymmetry breaking terms arise after communicating breaking from a separate sector; integrating out that sector, noncanonical interactions and radiative effects evade the hypotheses. The global trace identity alone would not identify a particular light squark without the conserved-charge block argument.

For the specified O'Raifeartaigh model, phases may be chosen so that $\lambda ,\mu ,M$ are real and positive. Its superpotential derivatives areand its F-term scalar potential isFor fixed $x,z$, choose $y=-2\lambda xz/M$ to minimize the final square. Since $|z^2-{\mu }^2{|}^2\ge (|z{|}^2-{\mu }^2{)}^2$,Thus $M^2>2{\lambda }^2{\mu }^2$ makes $z=y=0$ a global minimum with arbitrary $x$. The hierarchy $M\gg \mu$ ensures this for fixed perturbative $\lambda$, rather than for arbitrarily large $\lambda$. The origin is one member of this classically flat family, with $V_0={\lambda }^2{\mu }^4>0$ and $F_X=\lambda {\mu }^2$. The complex field $x$ is a pseudomodulus.

At the origin the chiral-superfield fermion mass matrix, in the $({\psi }_X,{\psi }_Y,{\psi }_Z)$ basis, isIts physical fermion masses are $0,M,M$; the two massive Weyl spinors form one massive four-component fermion. The massless ${\psi }_X$ is the goldstino. More generally, stationarity gives $W_{ij}\overline{W_j}=0$, so the nonzero auxiliary field direction is a null vector of $m$, proving the goldstino zero mode from vacuum stationarity.

Writing $z=(z_R+i{z}_I)/\sqrt{2}$ and similarly for $x,y$, the quadratic scalar potential isThe mass spectrum of the quadratic-cubic O'Raifeartaigh model is thereforeThe scalar squared-mass sum is $4M^2$, while twice the fermion squared-mass sum is also $4M^2$, so the supertrace is zero, as required. The two massless real $x$ modes are tree-level pseudomoduli, which can be lifted by a quantum effective potential; their tree-level masslessness is not the exact symmetry protection enjoyed by the goldstino.

Forthe three F-terms areThe equation $W_Z=0$ forces $X=0$, after which $W_Y=\alpha \ne 0$; hence no supersymmetric vacuum exists. Minimizing over $Z$ sets $Z=-2\beta XY/\gamma$, leavingThe two real quadratic eigenvalues about $X=0$ are $|\gamma {|}^2\pm 2|\alpha \beta |$, so the stated hierarchy makes $X=0$ stable. The vacua areThis is an O'Raifeartaigh model: supersymmetry is spontaneously broken and $Y$ is a classical pseudomodulus.