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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.

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.

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 $\alpha$.