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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.