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 classical 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

$$y\equiv\frac{hf}{m^2}\leq1,$$

the valley is 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 $V\geq f^2+(m^2-hf)|\phi_1|^2+h^2|\phi_1|^4/4$. Its minimum requires $\phi_1=\phi_2=0$. For $y<1$ the valley is strictly stable in the transverse directions; at $y=1$ one additional real scalar is massless at quadratic order but is stabilized by the displayed quartic. 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 symmetric valley is tachyonic and the global minimum moves to a displaced branch. 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. 83–85.

The displaced global branch for y > 1

The same model remains exactly solvable beyond the tachyon boundary. For fixed $X$ and $\phi_1$, minimizing the nonnegative middle term sets

$$\phi_2=-\frac{hX}{m}\phi_1.$$

The phase that lowers the first term obeys $\phi_1^2=-|\phi_1|^2$. Writing $\Delta=hf-m^2>0$, the remaining one-variable polynomial is minimized at

$$|\phi_1|^2=r^2=\frac{2\Delta}{h^2}, \qquad \phi_1=\pm ir, \qquad X\ \text{arbitrary},$$

with

$$V_{\min} =f^2-\frac{\Delta^2}{h^2} =\frac{2fm^2}{h}-\frac{m^4}{h^2}.$$

The auxiliary order parameters are

$$F_X=-\frac{m^2}{h}, \qquad F_1=0, \qquad F_2=-m\phi_1^*,$$

so the normalized goldstino is

$$G=\frac{F_X^*\psi_X+F_2^*\psi_2}{\sqrt{V_{\min}}} =-\frac{(m^2/h)\psi_X+m\phi_1\psi_2}{\sqrt{V_{\min}}}.$$

At $X=0$, the two massive Weyl fermions have

$$m_F^2=2hf-m^2,$$

while the third Weyl fermion is the goldstino. The six real scalar squared masses are

$$m_B^2= \left\{ 0,0, 2hf-m^2,2hf-m^2, 2hf, 2(hf-m^2) \right\}.$$

The two zeros are tangent to the complex tree-level valley; every transverse eigenvalue is positive for $y>1$. Their weighted sum again gives $\operatorname{STr}\mathcal M^2=0$. Thus $y=1$ is the classical branch-changing boundary, not the end of the model’s vacuum analysis.

For every $y>0$, the lower-bound polynomial reaches the stated minimum, so there is no lower-energy classical runaway. The arbitrary $X$ direction is instead a noncompact flat valley. Until quantum lifting or another controlled deformation is included, the tree approximation alone does not supply a normalizable ground-state wavefunction along that direction.

For $X\neq0$ on the symmetric branch, the heavy masses depend on $X$ even though the tree potential does not. Their exact eigenvalues and the resulting one-loop lifting for $0<y<1$ 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

$$V=m^2(x+z)+\frac12\left[g(x-z)+\xi\right]^2, \qquad D=g(x-z)+\xi.$$

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

Gauge-preserving branch

At the origin,

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

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

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

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

Higgsed branch

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

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

with

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

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 $m_{\phi_+}^2=2m^2$. The phase $a$ is the gauge Goldstone mode. Keeping the gaugino phase fixed by the site gauge Yukawa interaction, the fermion matrix in the basis $(\lambda,\psi_-,\psi_+)$ is

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

It has one zero singular value and two singular values with squared mass $m^2+2g^2v^2$. Its null vector is proportional to $(iD/\sqrt2,0,F_+)$, so the massless field is precisely

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

The bosonic and fermionic weighted mass-squared sums cancel, with the three vector polarizations counted explicitly:

$$\operatorname{STr}\mathcal M^2 =\left[3m_A^2+m_h^2+2(2m^2)\right] -2\left[2(m^2+m_A^2)\right] =0,$$

This is 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, pp. 122–127; §27.5, pp. 144–148.

What the two models teach

For the hypothesis gate that turns these branch calculations into a breaking claim, and for the common fermion-null-vector construction, see From auxiliary order parameters to the Goldstino.

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 $D=\xi$
Primary order parameter	F-term	D-term, or mixed F/D after Higgsing
Gauge symmetry	absent	preserved below and Higgsed above $g\xi=m^2$
Tree-level branches	symmetric valley for $hf\leq m^2$; displaced valley for $hf>m^2$	origin for $g\xi\leq m^2$; Higgsed branch for $g\xi>m^2$
Goldstino	$\psi_X$ on the symmetric branch; $\psi_X$–$\psi_2$ mixture on the displaced branch	gaugino below the boundary; chiral–gaugino mixture above it
Next calculation	loop-lift $X$	include gauge thresholds and check FI consistency

Both examples are renormalizable rigid theories. A standard field-independent FI term exists only for an Abelian factor. Its supercurrent and stress tensor have important gauge-improvement subtleties, and coupling it to supergravity or embedding it in quantum gravity imposes conditions that the rigid scalar potential cannot settle Komargodski and Seiberg 2009, abstract and §§2–3. 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. In the FI model, $V\geq m^2(x+z)$ for $m>0$, so no large-field direction can lower the energy. The O’Raifeartaigh model has the noncompact flat valleys identified above but no lower-energy classical runaway. 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. Continue through the O’Raifeartaigh boundary. For $hf>m^2$, minimize over $\phi_2$ and the phase of $\phi_1$, then derive the displaced branch and its vacuum energy.

Solution

First set $\phi_2=-hX\phi_1/m$ so that $F_1=0$. The minimizing phase has $\phi_1^2=-|\phi_1|^2$. With $s=|\phi_1|^2$,

$$V_{\rm red}(s) =\left(f-\frac h2s\right)^2+m^2s =f^2+(m^2-hf)s+\frac{h^2}{4}s^2.$$

For $\Delta=hf-m^2>0$, $V_{\rm red}'(s)=0$ gives $s=2\Delta/h^2$. Substitution yields

$$V_{\min}=f^2-\frac{\Delta^2}{h^2}.$$

The two phase choices are $\phi_1=\pm i\sqrt{2\Delta}/h$, while $X$ remains arbitrary and $\phi_2=-hX\phi_1/m$. The nonzero $F_X$ and $F_2$ make the goldstino a $\psi_X$–$\psi_2$ mixture.

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

Solution

The stationary conditions before imposing nonnegativity are

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

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

4. 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 phase-aware matrix shown, a null vector is proportional to

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

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.
• Komargodski, Z., and N. Seiberg. “Comments on the Fayet–Iliopoulos Term in Field Theory and Supergravity.” Journal of High Energy Physics 2009, no. 06 (2009): 007. DOI. Open preprint.
• 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.