O’Raifeartaigh and Fayet–Iliopoulos Model Laboratories

O’Raifeartaigh and Fayet–Iliopoulos models are small enough that every claim about supersymmetry breaking can be checked directly. The first makes F-flatness equations incompatible and leaves a tree-level pseudomodulus; the second uses an Abelian D-term and exhibits an exact transition between gauge-preserving and Higgsed broken vacua. Solving both models exposes the extra work hidden behind the phrase “$F$-term” or “$D$-term breaking”: stationarity, physical masses, tachyon boundaries, gauge consistency, and the goldstino direction.

Required background. Use the order-parameter and fermion-zero-mode derivation on F- and D-Term Breaking, Vacuum Energy, and the Goldstino, together with the component action on Wess–Zumino Models.

Helpful background. The Abelian example uses the auxiliary and gauge conventions on Gauge–Matter Systems, F- and D-Term Potentials.

An O’Raifeartaigh F-term laboratory

Consider three canonical chiral multiplets $X,\phi_1,\phi_2$ with real positive parameters $f,h,m$ and

$$W=fX+\frac{h}{2}X\phi_1^2+m\phi_1\phi_2.$$

The continuous R-charges $R(X)=R(\phi_2)=2$ and $R(\phi_1)=0$ make $R(W)=2$. The auxiliary equations are

$$\begin{aligned} F_X^*&=-\left(f+\frac{h}{2}\phi_1^2\right),\\ F_1^*&=-(hX\phi_1+m\phi_2),\\ F_2^*&=-m\phi_1. \end{aligned}$$

For $m\neq0$, $F_2=0$ forces $\phi_1=0$; then $F_1=0$ forces $\phi_2=0$, while $F_X=-f$ remains. The F-flatness equations are incompatible. The scalar potential is

$$V=\left|f+\frac h2\phi_1^2\right|^2 +\left|hX\phi_1+m\phi_2\right|^2 +m^2\left|\phi_1\right|^2.$$

It has the stationary valley

$$\phi_1=\phi_2=0, \qquad X\ \text{arbitrary}, \qquad V=f^2.$$

The arbitrary complex $X$ is a pseudomodulus: flat at tree level but not protected from a quantum potential. For

$$y\equiv\frac{hf}{m^2}<1,$$

the valley is actually the global tree-level minimum. Indeed,

$$\left|f+\frac h2\phi_1^2\right|^2 \geq f^2-hf|\phi_1|^2+\frac{h^2}{4}|\phi_1|^4,$$

so $V\geq f^2+(m^2-hf)|\phi_1|^2+h^2|\phi_1|^4/4$. Equality requires $\phi_1=\phi_2=0$. This lower bound is the global check that the incompatible equations alone did not provide.

Spectrum at the symmetric point

Set $X=0$ on the valley. The chiral-fermion mass matrix is

$$W_{ij}= \begin{pmatrix} 0&0&0\\ 0&0&m\\ 0&m&0 \end{pmatrix}_{(X,\phi_1,\phi_2)}.$$

Thus $\psi_X$ is massless and is exactly the goldstino, while $\psi_1$ and $\psi_2$ form a Dirac fermion of mass $m$. Writing each complex heavy scalar as two real fields gives

$$m_B^2=\{m^2+hf,\ m^2-hf,\ m^2,\ m^2\}.$$

The real and imaginary parts of $X$ remain massless at tree level. The complete local diagnosis is therefore

Quantity	Result at $X=\phi_1=\phi_2=0$
Order parameter	$F_X=-f$
Vacuum energy	$f^2$
Goldstino	$G=\psi_X$ up to phase
Heavy fermions	two Weyl fermions with $m_F^2=m^2$
Heavy scalars	$m^2\pm hf,m^2,m^2$
Stability boundary	$hf=m^2$
Flat directions	one complex tree-level pseudomodulus $X$

At $hf=m^2$, an additional scalar becomes massless; for $hf>m^2$, the origin of the valley is tachyonic and cannot be used as a vacuum. The cancellation

$$\operatorname{STr}\mathcal M^2 =(m^2+hf)+(m^2-hf)+2m^2-2(2m^2)=0$$

is a useful normalization check, not a proof of stability. O’Raifeartaigh’s original construction established this chiral mechanism O’Raifeartaigh 1975, pp. 331–352; the relation between flat directions and the goldstino is treated systematically in Weinberg 2000, §26.5, pp. 83–86.

For $X\neq0$, the heavy masses depend on $X$ even though the tree potential does not. Their exact eigenvalues and the resulting one-loop lifting are calculated on Pseudomoduli, Quantum Lifting, and Metastability.

An anomaly-free Fayet–Iliopoulos laboratory

Now take a rigid $U(1)$ gauge theory with canonical chiral fields $\Phi_+$ and $\Phi_-$ of charges $+1$ and $-1$, superpotential

$$W=m\Phi_+\Phi_-,$$

and a positive FI parameter $\xi$. The opposite charges cancel both the cubic gauge anomaly and the mixed gauge–gravitational anomaly. With

$$x=|\phi_+|^2, \qquad z=|\phi_-|^2,$$

the potential is

$$V=m^2(x+z)+\frac{g^2}{2}(x-z+\xi)^2, \qquad D=-g(x-z+\xi).$$

There is no supersymmetric configuration when $m\xi\neq0$: F-flatness sets both scalars to zero, where $D=-g\xi$. Unlike the F-term example, however, the identity of the vacuum changes at a calculable tachyon boundary.

Gauge-preserving branch

At the origin,

$$m_+^2=m^2+g^2\xi, \qquad m_-^2=m^2-g^2\xi.$$

For $g^2\xi<m^2$, both are positive and the origin is the global minimum:

$$\phi_+=\phi_-=0, \qquad V_0=\frac{g^2\xi^2}{2}, \qquad D=-g\xi.$$

The gauge symmetry is unbroken. The two matter fermions have mass $m$; the massless gaugino shifts under supersymmetry and is the goldstino. At $g^2\xi=m^2$, $\phi_-$ becomes massless, so an expansion that assumes the origin is isolated fails.

Higgsed branch

For $g^2\xi>m^2$, the origin has a $\phi_-$ tachyon. Minimization instead gives

$$\phi_+=0, \qquad |\phi_-|^2=v^2=\xi-\frac{m^2}{g^2},$$

with

$$D=-\frac{m^2}{g}, \qquad F_+=-m\phi_-^*, \qquad V_0=m^2\xi-\frac{m^4}{2g^2}.$$

The $U(1)$ is Higgsed. In the normalization $\phi_-=v+(h+ia)/\sqrt2$, the gauge boson and radial scalar have

$$m_A^2=m_h^2=2g^2v^2,$$

while the two real components of $\phi_+$ have $m^2=2m^2$. The phase $a$ is the gauge Goldstone mode. Up to phases, the fermion matrix in the basis $(\lambda,\psi_-,\psi_+)$ is

$$\mathcal M_F= \begin{pmatrix} 0&\sqrt2gv&0\\ \sqrt2gv&0&m\\ 0&m&0 \end{pmatrix}.$$

It has one zero eigenvalue and two eigenvalues with squared mass $m^2+2g^2v^2$. The zero mode is precisely

$$G=\frac{1}{\sqrt{V_0}} \left(F_+^*\psi_+-\frac{iD}{\sqrt2}\lambda\right).$$

The bosonic and fermionic weighted mass-squared sums agree:

$$4m_A^2+4m^2 =4(m_A^2+m^2),$$

another useful check on the quadratic expansion. Fayet and Iliopoulos introduced the Abelian D-term mechanism in Fayet and Iliopoulos 1974, pp. 461–464; the two branches and their mixed goldstino are derived in Weinberg 2000, §§27.2 and 27.5, pp. 122–128 and 144–148.

What the two models teach

Question	O’Raifeartaigh model	FI model
Why are flatness equations incompatible?	$F_2=F_1=0$ leaves $F_X=-f$	F-flatness fixes the origin, where $D=-g\xi$
Primary order parameter	F-term	D-term, or mixed F/D after Higgsing
Gauge symmetry	absent	preserved below and Higgsed above $g^2\xi=m^2$
Tree-level danger	pseudomodulus and $hf=m^2$ tachyon boundary	charged-scalar tachyon at $g^2\xi=m^2$
Goldstino	$\psi_X$	gaugino below the boundary; mixed above it
Next calculation	loop-lift $X$	include gauge thresholds and check FI consistency

Both examples are renormalizable rigid theories. A constant FI term exists only for an Abelian factor. Its coupling to supergravity and to a UV-complete theory imposes additional gauge-invariance and global-structure conditions; the rigid calculation does not settle those questions. Likewise, a field-content choice with uncancelled gauge anomalies would invalidate the FI model before any vacuum analysis.

Neither example has a supersymmetric vacuum in its finite canonical field space for the stated nonzero parameters. More elaborate O’Raifeartaigh models can possess runaways or distant supersymmetric vacua; a local spectrum must therefore never be promoted to a global statement without inspecting the full potential.

Common pitfalls

Rank condition without a minimum. Incompatible F equations prove the absence of an F-flat point, not the existence of a stable nonsupersymmetric vacuum. The scalar Hessian and behavior at infinity remain mandatory.

A zero scalar mass labeled “Goldstone.” The O’Raifeartaigh field $X$ is a pseudomodulus, not a symmetry Goldstone mode. In the FI Higgs phase, the phase of $\phi_-$ is a gauge direction and is removed from the physical spectrum.

Pure D-breaking assumed on both FI branches. Above the Higgsing boundary, $F_+=-m\phi_-^*$ is also nonzero. The goldstino is consequently a chiral–gaugino mixture.

Exercises

1. Prove the O’Raifeartaigh lower bound. Show that $hf<m^2$ makes $\phi_1=\phi_2=0$ the global tree-level minimum for every $X$.

Solution

Use $\operatorname{Re}(\phi_1^2)\geq-|\phi_1|^2$ to obtain

$$\left|f+\frac h2\phi_1^2\right|^2 \geq f^2-hf|\phi_1|^2+\frac{h^2}{4}|\phi_1|^4.$$

The other two terms in $V$ are nonnegative, and $m^2|\phi_1|^2$ combines with the negative term to give $(m^2-hf)|\phi_1|^2>0$. Equality requires $\phi_1=0$, then $|m\phi_2|^2=0$. The value of $X$ is unrestricted, so the minimum is a valley with energy $f^2$.

2. Derive the FI branches. Starting from $V=m^2(x+z)+g^2(x-z+\xi)^2/2$, minimize over nonnegative $x,z$ and recover the boundary $g^2\xi=m^2$.

Solution

The stationary conditions before imposing nonnegativity are

$$\frac{\partial V}{\partial x}=m^2+g^2(x-z+\xi), \qquad \frac{\partial V}{\partial z}=m^2-g^2(x-z+\xi).$$

They cannot vanish simultaneously for $m\neq0$. On the boundary $x=0$, minimizing in $z$ gives $z=\xi-m^2/g^2$ when this is nonnegative. Otherwise the constrained minimum is $x=z=0$. Thus the origin applies for $g^2\xi\leq m^2$ and the Higgsed solution for $g^2\xi\geq m^2$, with a massless charged scalar at equality.

3. Verify the Higgsed goldstino. Find the null vector of the displayed fermion matrix and compare the absolute chiral-to-gaugino ratio with the auxiliary-field order parameters.

Solution

For the real matrix shown, a null vector is proportional to

$$\begin{pmatrix}1\\0\\-\sqrt2gv/m\end{pmatrix}.$$

Hence the absolute chiral-to-gaugino ratio is $\sqrt2gv/m$. The auxiliary direction gives

$$\frac{|F_+|}{|D|/\sqrt2} =\frac{mv}{m^2/(\sqrt2g)} =\frac{\sqrt2gv}{m}.$$

The relative phase depends on the gaugino and vacuum conventions, while the ratio and norm $|F_+|^2+D^2/2=V_0$ are invariant.

References

• Fayet, P., and J. Iliopoulos. “Spontaneously Broken Supergauge Symmetries and Goldstone Spinors.” Physics Letters B 51 (1974): 461–464. DOI.
• O’Raifeartaigh, L. “Spontaneous Symmetry Breaking for Chiral Scalar Superfields.” Nuclear Physics B 96 (1975): 331–352. DOI.
• Weinberg, S. The Quantum Theory of Fields, Volume III: Supersymmetry. Cambridge: Cambridge University Press, 2000, §§26.5, 27.2, and 27.5. DOI.