F- and D-Term Breaking, Vacuum Energy, and the Goldstino

In rigid four-dimensional $\mathcal N=1$ theory, spontaneous supersymmetry breaking has three equivalent local signatures under standard Goldstone-theorem assumptions: positive vacuum-energy density, a nonzero auxiliary-field expectation value, and a massless fermion with a nonzero supercurrent matrix element. This page derives those equivalences for a homogeneous Lorentz-invariant vacuum, identifies the goldstino in a mixed chiral–vector theory, and marks the points at which stationarity, gauge invariance, boundaries, or supergravity change the conclusion.

Required background. Use the component normalizations and auxiliary equations from Gauge–Matter Systems, F- and D-Term Potentials, and the positive-energy argument from Four-Dimensional N=1 Supersymmetry.

Helpful background. The current-algebra step uses the charge and surface-term conditions developed in Spacetime Currents, Stress Tensors, and Charge Algebras.

Auxiliary fields and the positive potential

Take chiral multiplets $(\phi^i,\psi^i,F^i)$ in a unitary representation of a compact gauge group and vector multiplets $(A_\mu^a,\lambda^a,D^a)$. For the explicit component derivation, assume canonical Kähler metric, constant gauge couplings $g_a$, no boundary, and a gauge-invariant superpotential $W(\phi)$. An Abelian factor may have a rigid Fayet–Iliopoulos parameter $\xi^a$. The algebraic part of the Lagrangian is

$$\begin{aligned} \mathcal L_{\rm aux} ={}&F_i^*F^i+W_iF^i+\overline W_{\bar i}F^{*\bar i}\\ &+\frac12D^aD^a +g_aD^a\!\left(\phi^\dagger T^a\phi+\xi^a\right), \end{aligned}$$

where $W_i=\partial W/\partial\phi^i$ and the normalization of $\xi^a$ is chosen to place it inside the parentheses. Completing squares gives

$$F^i=-\overline W_{\bar i}, \qquad D^a=-g_a\!\left(\phi^\dagger T^a\phi+\xi^a\right),$$

and

$$V(\phi,\bar\phi) =\sum_i|F^i|^2+\frac12\sum_a(D^a)^2\geq0.$$

For a noncanonical Kähler potential, the first term is $W_iK^{i\bar j}\overline W_{\bar j}$; positivity requires a positive Kähler metric. In either case a field configuration is supersymmetric precisely when all auxiliary transformations vanish,

$$F^i=0, \qquad D^a=0,$$

provided the configuration is Lorentz invariant so that derivative and field-strength terms in the fermion variations vanish. Therefore incompatible $F$- and $D$-flatness equations prove that no supersymmetric configuration exists in the stated field domain. They do not yet prove that the potential has a stationary, stable, or normalizable vacuum.

The operator algebra gives the same criterion without weak coupling. If the global supercharges are well defined and surface terms vanish,

$$\langle\Omega|H|\Omega\rangle =\frac14\sum_{\alpha=1}^{2} \left( \lVert Q_\alpha|\Omega\rangle\rVert^2 +\lVert Q_\alpha^\dagger|\Omega\rangle\rVert^2 \right).$$

The normalization follows from $\{Q_\alpha,\bar Q_{\dot\alpha}\}=2\sigma^\mu_{\alpha\dot\alpha}P_\mu$. Thus a normalizable zero-energy ground state is supersymmetric, while a ground state of strictly positive energy breaks supersymmetry. For an infinite homogeneous system, both sides are understood through energy density and smeared local charges before the thermodynamic limit. Weinberg 2000, §§29.1–29.2, pp. 248–263 develops both the algebraic and current forms of this statement.

The goldstino direction

At a constant vacuum, the inhomogeneous parts of the fermion transformations are, in these conventions,

$$\delta\psi^i=\sqrt2\,\epsilon F^i, \qquad \delta\lambda^a=i\epsilon D^a.$$

Define the breaking scale by the vacuum energy density,

$$f^2\equiv V_{\rm vac} =F_i^*F^i+\frac12D^aD^a.$$

If $f\neq0$, the normalized fermionic direction

$$G =\frac{1}{f} \left(F_i^*\psi^i-\frac{i}{\sqrt2}D^a\lambda^a\right)$$

shifts as $\delta G=\sqrt2f\,\epsilon+\text{derivative terms}$. Orthogonal fermion combinations have no constant shift. This identifies the goldstino without diagonalizing a mass matrix; it also shows how F- and D-term order parameters mix.

The shift must be compatible with a zero eigenvalue of the fermion mass matrix at a stationary point. For canonical matter, the relevant matrix in the basis $(\psi^i,\lambda^a)$ is, up to field-redefinition phases,

$$\mathcal M_F= \begin{pmatrix} W_{ij} & \sqrt2g_a(\phi^\dagger T^a)_i\\ \sqrt2g_b(\phi^\dagger T^b)_j & 0 \end{pmatrix}.$$

Differentiating the potential gives

$$0=\partial_iV =-W_{ij}F^j-g_aD^a(\phi^\dagger T^a)_i.$$

Gauge invariance, $W_i(T^a\phi)^i=0$, gives the conjugate relation $(\phi^\dagger T^a)_iF^i=0$. Consequently,

$$\mathcal M_F \begin{pmatrix}F^j\\D^a/\sqrt2\end{pmatrix}=0.$$

The null vector is a consequence of stationarity plus gauge invariance, not merely of $V>0$ at an arbitrary field point. In a pure Wess–Zumino model this collapses to $W_{ij}F^j=0$. A nonzero F-term therefore supplies an exact tree-level massless fermion at every stationary point, even when the scalar Hessian later reveals a saddle.

With a curved Kähler metric, ordinary derivatives are replaced by Kähler-covariant ones and the normalized direction uses the metric. With field-dependent gauge kinetic functions, gaugino masses and additional mixing enter, but the full stationary mass matrix still annihilates the supersymmetry-variation direction when the action is exactly supersymmetric.

The supercurrent pole

The massless fermion is not only a tree-level accident. Let $S^\mu_\alpha$ be a conserved supercurrent whose charge generates supersymmetry. Spontaneous breaking means that some local operator has a nonzero vacuum variation, equivalently that the charge does not annihilate the vacuum. The supersymmetric Ward identity then requires a zero-mass singularity in the current correlator. Lorentz covariance permits

$$\langle0|S^\mu_\alpha(0)|G(p)\rangle =i\sqrt2f\,(\sigma^\mu\bar u(p))_\alpha \quad\text{up to improvement and phase conventions}.$$

Insertion into the spectral representation produces a $1/p^2$ pole with residue fixed by $f^2$. Interactions may rotate an elementary combination into a composite one, but cannot gap the goldstino while global supersymmetry remains exact and spontaneously broken. This is the fermionic Goldstone theorem first formulated in current-algebra language by Salam and Strathdee 1974, pp. 465–467; a component derivation and normalization check appear in Weinberg 2000, §29.2, pp. 256–263.

The improvement ambiguity $S^\mu\mapsto S^\mu+\partial_\nu B^{[\nu\mu]}$ changes contact terms and the representative current, not the conserved charge or the physical pole when surface terms vanish. If a boundary supports flux, however, the improvement term can change the charge and the boundary degrees of freedom must be included.

A reliable breaking test

For a proposed weakly coupled vacuum, perform the following checks in order.

1. Specify the field domain. Include gauge identifications, singular loci, and allowed behavior at infinity.
2. Solve auxiliary equations. Show whether $F_i=D^a=0$ can be satisfied anywhere, including runaway limits.
3. Find stationary points. Solve $\partial V=0$; do not infer one from incompatible flatness equations.
4. Remove redundancies and test stability. Quotient gauge orbits, separate Goldstone directions, and diagonalize the physical scalar Hessian.
5. Identify the goldstino. Compute $(F^i,D^a/\sqrt2)$ and verify the full fermion mass matrix has the corresponding null vector.
6. State global status. Distinguish a global vacuum, a local minimum with a specified lower endpoint, a saddle, and a runaway.
7. Name the regime. Record loop expansion parameters, cutoff, and whether the theory is rigid or coupled to gravity.

The first two steps diagnose supersymmetric configurations; all seven are needed for a controlled vacuum claim.

Where the theorem changes

Explicit breaking. If the Lagrangian contains nonsupersymmetric terms, the supercurrent is not conserved and no exactly massless goldstino is required. A spurion can keep bookkeeping supersymmetric, but freezing its F-component is still explicit breaking in the visible theory.

Supergravity. Local supersymmetry gauges the current. The goldstino becomes the longitudinal helicity-$1/2$ component of a massive gravitino, and the scalar potential contains negative terms proportional to the superpotential. Neither $V=\sum|F|^2+D^2/2$ nor a physical massless goldstino survives unchanged.

Boundaries and central charges. Boundaries, defects, or extended objects can add surface or central terms to the superalgebra. A positive energy may then saturate a BPS bound rather than signal complete breaking. One must test the preserved linear combinations of supercharges.

No vacuum. A potential whose infimum occurs only at infinite field distance may have arbitrarily small energy without a normalizable ground state. “No finite solution of $F=0$” and “spontaneous breaking in a vacuum” are different conclusions.

Index information. A nonzero well-defined Witten index obstructs breaking, but a zero index is compatible with either a broken vacuum or paired supersymmetric vacua. The index cannot replace the local and global analysis above.

Common pitfalls

Evaluating $V$ at a convenient point. Positive $V$ diagnoses breaking only if the point is a physical state of interest, normally a stationary vacuum. At a generic point it is simply potential energy.

Calling every massless fermion a goldstino. The goldstino is selected by the inhomogeneous supersymmetry transformation and the supercurrent residue. An accidental zero eigenvalue with zero current overlap is not enough.

Ignoring gauge directions in the Hessian. A zero scalar eigenvalue tangent to a gauge orbit is not a pseudomodulus, and a negative eigenvalue in a gauge-fixed but unphysical direction is not a tachyon. Test the quadratic form on gauge-invariant fluctuations.

Exercises

1. The chiral null vector. For canonical chiral fields with $V=W_i\overline W_i$, show that every stationary point with some $F^i\neq0$ has a massless Weyl fermion at tree level.

Solution

Since $F^i=-\overline W_i$,

$$0=\partial_iV=W_{ij}\overline W_j=-W_{ij}F^j.$$

The chiral-fermion mass matrix is $W_{ij}$, so $F^j$ is a nonzero null vector. The corresponding normalized fermion is $G=F_i^*\psi^i/(F_j^*F^j)^{1/2}$. This proves a massless fermion at a stationary point; it says nothing about scalar stability.

2. Mixed breaking. Suppose $|F|^2=3D^2/2$ for one chiral and one Abelian vector multiplet. What fraction of the normalized goldstino norm lies in the gaugino?

Solution

The norm is $f^2=|F|^2+D^2/2=2D^2$. The gaugino contribution is $D^2/2$, so its fraction is $(D^2/2)/(2D^2)=1/4$. The remaining $3/4$ lies in the chiral fermion.

3. Diagnose the claim. A potential has no finite solution of $F_i=0$, but along $\phi=t$ one finds $V(t)\to0$ as $t\to\infty$. Is supersymmetry spontaneously broken?

Solution

Not enough information is given. There is no finite supersymmetric configuration, but the infimum is zero at infinite field distance. One must determine whether a normalizable ground state exists, whether the limit is a physical vacuum, and whether the global charges are defined. The appropriate conclusion is a supersymmetric runaway or no vacuum, not automatically spontaneous breaking in a stable state.

References

• Salam, A., and J. Strathdee. “On Goldstone Fermions.” Physics Letters B 49 (1974): 465–467. DOI.
• Weinberg, S. The Quantum Theory of Fields, Volume III: Supersymmetry. Cambridge: Cambridge University Press, 2000, §§29.1–29.2. DOI.