= Solution
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 \b[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 $H=\{q,q^\dagger\}/2$, $q^2=0$, and let <fermion parity> anticommute with both charges. On a positive-energy eigenspace, $(q+q^\dagger)/\sqrt{2E}$ 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 index>
$$
I(\beta)=\operatorname{Tr}((-1)^F e^{-\beta H})
=n_B^{(0)}-n_F^{(0)}.
$$
For 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. \b[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. Define
$$
A=\frac d{dx}+w(x),\quad A^\dagger=-\frac d{dx}+w(x),\quad
H_-=\frac12A^\dagger A,\quad H_+=\frac12AA^\dagger.
$$
Positivity of these <Hamiltonian operators> shows that a zero mode must solve $A\psi_-=0$ or $A^\dagger\psi_+=0$. Thus the only candidates are
$$
\psi_-(x)\propto e^{-\int^x w(s)ds},\qquad
\psi_+(x)\propto e^{+\int^x w(s)ds}.
$$
For $w(x)=x$, 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 $[w^2\mp w']/2$ 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> gives
$$
F_i=-\overline{W_i},\qquad U=\sum_i|W_i|^2.
$$
A <supersymmetric vacuum> solves every $W_i=0$. At a stationary broken vacuum,
$$
0=\partial_iU=\sum_j W_{ij}\overline{W_j}.
$$
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 $\delta\psi_i=\sqrt2\epsilon F_i+\cdots$ independently identifies that direction through its inhomogeneous shift.
For one canonical chiral field, $W=fZ$, $f\ne0$, 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 $W'$ 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:
$$
W=fX+\frac h2X\Phi_1^2+m\Phi_1\Phi_2,\qquad fm\ne0.
$$
The auxiliary equations require $W_2=m\phi_1=0$, then $W_1=hX\phi_1+m\phi_2=0$, leaving $W_X=f\ne0$. They cannot vanish simultaneously. For real positive $f,h,m$ with $m^2>fh$, the potential obeys
$$
\begin{aligned}
U&=|f+h\phi_1^2/2|^2+|hX\phi_1+m\phi_2|^2+m^2|\phi_1|^2\\
&\geq f^2+(m^2-fh)|\phi_1|^2+\frac{h^2}{4}|\phi_1|^4.
\end{aligned}
$$
Therefore its global tree-level minimum has $\phi_1=\phi_2=0$, 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.
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