Dynamical Supersymmetry Breaking and Calculability

Dynamical supersymmetry breaking (DSB) means that a theory with no explicit supersymmetry-breaking parameter generates a strictly positive vacuum energy through quantum gauge dynamics. The strongest demonstrations combine an exact holomorphic obstruction to solving all $F$- and $D$-term equations with a tunable limit in which the vacuum lies at weak coupling. The classic $SU(3)\times SU(2)$ “3–2 model” has both features.

Required background. Chiral-theory anomaly constraints supplies the microscopic consistency checks; Nelson–Seiberg-type criteria supplies the $R$-symmetry hypotheses; and quantum chiral rings and Konishi anomalies explains exact operator relations. Helpful background. The Witten index and its failure modes explains why a vanishing or ill-defined index is not a proof of breaking.

What a convincing DSB argument must exclude

A positive-energy stationary point is not enough. A stable DSB conclusion must address:

• all gauge and global anomalies, including discrete or global gauge anomalies;
• every invariant branch and every supersymmetric solution at finite field values;
• runaways on which $V\to0$ only at infinite distance;
• the Kähler metric in the region where the vacuum is claimed;
• extra vacua that enter from infinity when a deformation is varied;
• whether an $R$ symmetry is exact, accidental, explicitly broken, or spontaneously broken.

Holomorphy can establish an incompatible set of $F$-term equations, but a vacuum energy depends on the nonholomorphic Kähler metric. “Calculable DSB” therefore requires the vacuum to sit where the gauge couplings and Kähler corrections are parametrically controlled.

This separation between exact structural evidence, controlled dynamical evidence, and a broader infrared proposal is the same one used in the SQCD regimes, exact results, and evidence limits table. Here the exact layer rules out finite solutions of the holomorphic vacuum equations; the controlled weak-coupling analysis excludes the would-be runaway and locates the positive-energy minimum.

Theory card for the 3–2 model

Take the simply connected product gauge group

$$G=SU(3)\times SU(2)$$

and chiral fields

$$Q:({\bf3},{\bf2}),\quad \overline U:(\overline{\bf3},{\bf1}),\quad \overline D:(\overline{\bf3},{\bf1}),\quad L:({\bf1},{\bf2}),$$

with

$$W_{\mathrm{tree}}=\lambda\,\epsilon^{\alpha\beta} Q^a_{\alpha}\overline D_aL_\beta.$$

This is the generic nonzero renormalizable interaction, not a special deletion of a second coupling. Before choosing a flavor basis one may write $\lambda_i\epsilon^{\alpha\beta}Q^a_\alpha\overline Q_{ia}L_\beta$ for the two antifundamentals. A unitary rotation of $(\overline U,\overline D)$ aligns the coupling vector along one combination, which we call $\overline D$; Shacham 2012, § 2.1, p. 2, arXiv PDF makes this basis choice explicit.

No nontrivial subgroup of the $\mathbb Z_3\times\mathbb Z_2$ gauge center acts trivially on all four fields, so there is no quotient ambiguity for this matter content and no surviving electric one-form symmetry.

The $SU(3)^3$ anomaly cancels because the two $SU(2)$ components of $Q$ give two fundamentals against the two antifundamentals. The $SU(2)$ factor has four left-handed doublets—three colors of $Q$ plus $L$—so its mod-two global gauge anomaly also vanishes. The one-loop coefficients are

$$\begin{aligned} b_3&=3(3)-\left[2\left(\frac12\right) +\frac12+\frac12\right]=7,\\ b_2&=3(2)-\left[3\left(\frac12\right)+\frac12\right]=4. \end{aligned}$$

Thus the two holomorphic scales are $\Lambda_3^7$ and $\Lambda_2^4$.

One anomaly-free ordinary symmetry and one anomaly-free $R$ symmetry can be represented by the scalar charges

$$\begin{array}{c|rrrr} &Q&\overline U&\overline D&L\\ \hline X&1&-4&2&-3\\ R&-1&0&0&3 \end{array}$$

The tree term has $(X,R)=(0,2)$. Direct substitution gives zero for both $SU(3)^2X$ and $SU(2)^2X$, and for each mixed gauge–$R$ anomaly after including the charge-one gaugino. An $X$ rotation by $\alpha=\pi/3$ acts exactly like the gauge-center element $(\omega_3^2,-1)$ on every matter field. The faithful ordinary symmetry is therefore

$$U(1)_X/\mathbb Z_6.$$

As on the preceding chiral-theory page, we record the spin-$R$ extension separately from the ordinary flavor quotient. The mixed generator $R_0=R+X$ has even scalar charges $(0,-4,2,0)$, so $e^{i\pi R_0}=(-1)^F$. This makes the fermion-parity identification explicit without obscuring the $\mathbb Z_6$ gauge-center overlap of $X$.

The scalar charges in the displayed card must not be inserted directly into fermion anomalies. The left-handed matter fermions have $R=(-2,-1,-1,2)$, while the eleven gauginos have $R=1$. Including gauge multiplicities gives

$$\begin{aligned} \operatorname{Tr}X&=-6,& \operatorname{Tr}X^3&=-216,\\ \operatorname{Tr}R&=-3,& \operatorname{Tr}R^3&=-27,\\ \operatorname{Tr}X^2R&=-36,& \operatorname{Tr}XR^2&=-6. \end{aligned}$$

Together with the gauge-anomaly checks, this is the complete continuous Abelian anomaly record in the chosen basis. The selection rules permit the nonperturbative term below; they do not generate it by themselves.

The exact obstruction from the SU(3) node

Choose a hierarchy $\Lambda_3\gg\Lambda_2$ and eventually take $\lambda\ll1$. Viewed as an $SU(3)$ theory, $Q$ provides two fundamentals and $(\overline U,\overline D)$ two antifundamentals, so $N_f=2=N_c-1$. The exact ADS term is

$$W_3=\frac{\Lambda_3^7} {\det(Q\overline Q)}, \qquad \overline Q=(\overline D,\overline U).$$

The full superpotential in this regime is

$$W_{\mathrm{eff}} =\lambda Q\overline D L +\frac{\Lambda_3^7}{\det(Q\overline Q)}.$$

If all $F$ terms vanished at a finite point, $F_L=0$ would require $Q\overline D=0$. That makes the first column of the $2\times2$ meson matrix vanish in the ordering $\overline Q=(\overline D,\overline U)$, hence $\det(Q\overline Q)=0$, precisely where the ADS term is singular. The equations therefore have no simultaneous finite solution. This is an exact holomorphic incompatibility, first exploited in Affleck, Dine, and Seiberg 1984, pp. 1678–1680.

The argument would be incomplete if the fields could run to infinity with every auxiliary field tending to zero. The stabilization can be exhibited rather than asserted. Choose $\epsilon^{12}=+1$ and $t^3=\operatorname{diag}(1,-1)/2$ for the displayed $SU(2)$ doublets. Up to gauge transformations and harmless phases, a $D$-flat representative is

$$\begin{aligned} Q&=v_0 \begin{pmatrix} a&0\\ 0&b\\ 0&0 \end{pmatrix},& \overline D&=v_0(a,0,0),\\ \overline U&=v_0(0,b,0),& L&=v_0\begin{pmatrix}0\\ \sqrt{a^2-b^2}\end{pmatrix}, \end{aligned} \qquad a>b>0.$$

The $SU(3)$ $D$ terms cancel because $QQ^\dagger$ equals the sum of the two antifundamental outer products. The $SU(2)$ diagonal $D$ term is proportional to $a^2-b^2-\lvert L\rvert^2$, and all off-diagonal terms vanish. With this alignment,

$$\epsilon^{\alpha\beta}Q^a_\alpha\overline D_aL_\beta =a^2\sqrt{a^2-b^2}\,v_0^3,$$

so the stabilizing tree term is visibly nonzero.

Why the vacuum is calculable

Let $g_2(v)$ be the running $SU(2)$ coupling at the Higgsing scale and define

$$\Delta=\left(\frac{g_2(v)}{\lambda}\right)^2.$$

The standard calculable minimum is the regime $\Delta\gg1$, with $g_3(v)\gg\lambda$ as well. Gauge fluctuations transverse to the $D$-flat manifold are then heavier than the supersymmetry-breaking curvature, so one first minimizes the $F$-term potential on $D=0$. Let $v$ denote the common magnitude of the Higgsing expectation values. Dimensional analysis gives

$$W_{\mathrm{tree}}\sim\lambda v^3, \qquad W_3\sim\frac{\Lambda_3^7}{v^4}.$$

Balancing their $F$ terms,

$$\lambda v^2\sim\frac{\Lambda_3^7}{v^5},$$

yields

$$v\sim\Lambda_3\lambda^{-1/7}, \qquad F\sim\Lambda_3^2\lambda^{5/7}, \qquad V_{\min}\sim\Lambda_3^4\lambda^{10/7}.$$

The order-one ratios follow from an explicit dimensionless problem. Set $v_0=\Lambda_3\lambda^{-1/7}$ in the $D$-flat representative and choose the relative phase that permits destructive interference between the tree and ADS derivatives. The nonzero dimensionless auxiliary fields are

$$\frac{F_Q}{\Lambda_3^2\lambda^{5/7}} =\frac{F_{\overline Q}^{\mathsf T}}{\Lambda_3^2\lambda^{5/7}} = \begin{pmatrix} a\sqrt{a^2-b^2}-\dfrac{1}{a^3b^2}&0\\ 0&\dfrac{1}{a^2b^3}\\ 0&0 \end{pmatrix}, \qquad \frac{F_L}{\Lambda_3^2\lambda^{5/7}} =\begin{pmatrix}0\\a^2\end{pmatrix}.$$

Thus

$$\frac{V_F}{\Lambda_3^4\lambda^{10/7}} =\mathcal V(a,b) =2\left(a\sqrt{a^2-b^2}-\frac{1}{a^3b^2}\right)^2 +\frac{2}{a^4b^6}+a^4.$$

Minimizing for $a>b>0$ gives $a\simeq1.164$ and $b\simeq1.132$, a finite point with $\mathcal V>0$. Corrections from departing the $D$-flat manifold are suppressed by $\Delta^{-1}$ and $\lambda^2/g_3(v)^2$. For $\lambda\ll1$ at fixed separated strong scales, $v\gg\Lambda_3\gg\Lambda_2$, the gauge couplings at $v$ are weak, and eventually $g_{2,3}(v)\gg\lambda$. Kähler corrections are suppressed by powers of $\Lambda_3/v\sim\lambda^{1/7}$ and weak logarithms. The scaling, finite minimum, and positive energy are then controlled, not merely dimensional estimates. The minimization regimes and their control parameters are analyzed in Shacham 2012, §§ 2.1–2.2, pp. 2–5, arXiv PDF.

The order parameters also make the symmetry statement concrete. At the representative above,

$$\left\langle\epsilon^{\alpha\beta}Q^a_\alpha\overline D_aL_\beta\right\rangle\neq0$$

has $(X,R)=(0,2)$, so every $R+tX$ combination is spontaneously broken while $U(1)_X/\mathbb Z_6$ remains unbroken. The phase therefore contains an $R$ axion when the continuous $R$ symmetry is exact. Gauging additional interactions or adding higher-dimension terms can explicitly reduce that symmetry and change the light spectrum; those are deformations, not properties of the minimal model.

Other exact mechanisms and their limits

A different route couples singlets to composites on a quantum-modified moduli space. If the quantum constraint forbids all composites from vanishing while singlet $F$ terms demand that they do, the equations have a rank-condition obstruction. The $SU(2)$ model with four doublets and singlets developed by Intriligator and Thomas 1996, §§ 2–3, pp. 126–132 is the canonical example. Such a proof still needs Kähler control to compute the vacuum quantitatively; symmetry can sometimes locate it near a point where qualitative breaking is robust.

Two popular shortcuts are unsafe:

• A zero Witten index permits breaking but does not require it. Noncompact directions and vacua arriving from infinity can invalidate deformation arguments.
• The Nelson–Seiberg criterion assumes a generic superpotential in a specified set of infrared fields. Strong dynamics may add fields, constraints, or accidental symmetries, so the theory card and exact chiral relations must come first; the theorem’s hypotheses and exceptions are stated in Nelson and Seiberg 1994, pp. 46–62.

Order of limits

The controlled statement takes $0<\lambda\ll1$ at fixed nonzero $\Lambda_3$, with $\Lambda_2/\Lambda_3$ sufficiently small and $\Delta\gg1$. Setting $\lambda=0$ first restores flat directions and sends the stabilized point to $v=\infty$. Raising $\Lambda_2$ to compete with $\Lambda_3$ removes the sequential-node derivation, even though holomorphy may continue some protected information. At $\Delta\ll1$ a different controlled scaling can arise, while part of the interpolation is not under quantitative control; it must not be inferred from the boxed $\Delta\gg1$ result. None of these altered limits disproves the finite-small-$\lambda$ vacuum; each changes the available description.

Common pitfalls

Stopping after incompatible F terms. One must also exclude $D$-flat runaways and show that the potential has a finite minimum. The 3–2 tree interaction and ADS singularity do this together.

Calling a strong-coupling estimate calculable. Calculability here comes from $v/\Lambda_3\sim\lambda^{-1/7}\gg1$. Without a tunable hierarchy, exact holomorphy does not determine the Kähler metric or vacuum energy.

Taking a deformation to zero in the wrong order. At $\lambda=0$ the vacuum escapes to infinity. Continuity at a fixed compact field point is therefore not a valid argument across that limit.

Exercises

1. Check both gauge factors

Verify the perturbative $SU(3)^3$ anomaly, the $SU(2)$ mod-two anomaly, and the two one-loop coefficients.

Solution

For $SU(3)^3$, $Q$ is two fundamentals and $\overline U,\overline D$ are two antifundamentals, giving $2-1-1=0$. The $SU(2)$ factor sees $3+1=4$ doublets, an even number. The index sums are $2$ for each gauge factor, so $b_3=9-2=7$ and $b_2=6-2=4$.

2. Derive the small-coupling exponents

Starting only from $W_{\mathrm{tree}}\sim\lambda v^3$ and $W_3\sim\Lambda_3^7/v^4$, recover the powers of $\lambda$ in $v,F$, and $V$.

Solution

$\partial_vW$ balances as $\lambda v^2\sim\Lambda_3^7/v^5$, so $v^7\sim\Lambda_3^7/\lambda$ and $v\sim\Lambda_3\lambda^{-1/7}$. Then $F\sim\lambda v^2\sim\Lambda_3^2\lambda^{5/7}$ and $V\sim\lvert F\rvert^2\sim\Lambda_3^4\lambda^{10/7}$.

3. Find the faithful ordinary symmetry

Show that an $X$ rotation by $\pi/3$ is gauge-equivalent to the identity and compute $\operatorname{Tr}X$, $\operatorname{Tr}X^3$, $\operatorname{Tr}R$, and $\operatorname{Tr}R^3$.

Solution

With $\omega_3=e^{2\pi i/3}$, the center element $(\omega_3^2,-1)$ acts on $(Q,\overline U,\overline D,L)$ as $(e^{i\pi/3},e^{2\pi i/3},e^{2\pi i/3},-1)$, exactly the phases from $e^{i\pi X/3}$. Hence the faithful action is $U(1)_X/\mathbb Z_6$.

Using multiplicities $(6,3,3,2)$ gives $\operatorname{Tr}X=-6$ and $\operatorname{Tr}X^3=-216$. The matter-fermion $R$ charges are $(-2,-1,-1,2)$, and the $8+3$ gauginos have charge one, giving $\operatorname{Tr}R=-3$ and $\operatorname{Tr}R^3=-27$.

4. Verify the finite minimum

Check that the displayed representative is $D$-flat, derive $\mathcal V(a,b)$ from the nonzero auxiliary fields, and verify that $a\simeq1.164$, $b\simeq1.132$ lies at a positive local minimum.

Solution

The two antifundamental outer products reproduce $QQ^\dagger$, while $\lvert L_2\rvert^2=a^2-b^2$ cancels the diagonal $SU(2)$ moment map. Differentiating the tree and ADS terms gives the auxiliary fields displayed above and therefore

$$\mathcal V(a,b)=2\left(a\sqrt{a^2-b^2}-\frac{1}{a^3b^2}\right)^2 +\frac{2}{a^4b^6}+a^4.$$

Substitution gives finite positive energy. Differentiating numerically yields a stationary point near $(1.164,1.132)$ with positive Hessian eigenvalues on the domain $a>b>0$. This verifies stabilization within the $D$-flat approximation; the approximation itself requires $g_{2,3}(v)\gg\lambda$.

References

• Affleck, Ian, Michael Dine, and Nathan Seiberg. “Calculable Nonperturbative Supersymmetry Breaking.” Physical Review Letters 52 (1984): 1677–1680. DOI.
• Intriligator, Kenneth, and Scott Thomas. “Dynamical Supersymmetry Breaking on Quantum Moduli Spaces.” Nuclear Physics B 473 (1996): 121–142. DOI; arXiv.
• Nelson, Ann E., and Nathan Seiberg. “R Symmetry Breaking versus Supersymmetry Breaking.” Nuclear Physics B 416 (1994): 46–62. DOI; arXiv.
• Shacham, Tomer. “On the Vacuum Structure of the 3–2 Model.” Journal of High Energy Physics 2012, no. 5 (2012): 087. DOI; arXiv.

Further reading

• Skiba, Witold. “Dynamical Supersymmetry Breaking.” Modern Physics Letters A 12 (1997): 737–750. DOI; arXiv.