Quantum-modified vacua and a confinement example

With one flavor, two-color SQCD has the runaway studied in the dynamical-superpotential lesson. Adding another massless flavor does not merely change a power in that answer. A continuous family of supersymmetric vacua survives, but its defining relation changes. Adding a third flavor gives a different phenomenon: a vacuum at the symmetric origin, described by extra light composite fields.

We will connect these statements by removing a flavor twice. The resulting chain tests signs and nonzero coefficients against physics already established, rather than treating a symmetry-allowed formula as proof that it occurs.

Keep the fields and scales fixed

Work in flat four-dimensional Lorentzian $\mathcal N=1$ SQCD, with four real supercharges, all manifest in N=1 notation. The gauge group is $SU(2)$, of rank one. There is one vector multiplet and $f$ fundamental/antifundamental chiral pairs, with no adjoint chiral. The unrescaled composites are

$$M^j{}_i=\widetilde Q^{\,j}_aQ_i^a,\qquad B=\det Q,\qquad \widetilde B=\det\widetilde Q \quad(f=2).$$

The row index of $M=\widetilde Q Q$ is $j$, and its column index is $i$. The tilde quarks and baryons are independent chiral fields, not complex conjugates. Retain the ordinary color trace $\operatorname{Tr}(t^At^B)=\delta^{AB}/2$ and the kinetic conventions of classical SQCD.

Use a subscript to distinguish the holomorphic scales of different theories:

$$\mathcal L_f=\Lambda_f^{\,6-f},\qquad \mathcal L_{f-1}=m\mathcal L_f . \tag{Q08.1}$$

The second equation applies when a pair has superpotential mass $m$ and is removed. These are complex parameters, including the mass phase. Use the unit-threshold convention of holomorphic decoupling and the absolute normalization in which the one-flavor result is $W=\mathcal L_1/M$. Thus $\mathcal L_2$ has dimension four, $\mathcal L_3$ dimension three. Changing the absolute scale convention would also change numerical coefficients below.

A quantum relation with the right charges

For general $N_f=N_c$, the classical relation is $\det M-B\widetilde B=0$. The quantum result in compatible scale and baryon conventions is

$$\det M-B\widetilde B=\Lambda^{2N_c}. \tag{Q08.2}$$

This is an established dynamical result, stated in Intriligator–Seiberg, (4.15). We will check its normalization explicitly for $N_c=N_f=2$.

The symmetric gauge-anomaly-free quark R-charge is $1-N_c/N_f=0$. The relevant charges are

Object	Engineering dimension	Baryon charge	R-charge	Formal axial charge
$M$	$2$	$0$	$0$	$2$
$B$	$N_c$	$N_c$	$0$	$N_c$
$\widetilde B$	$N_c$	$-N_c$	$0$	$N_c$
$\Lambda^{2N_c}$	$2N_c$	$0$	$0$	$2N_c$

The axial transformation is a selection rule that also transforms the theta angle; it is not a conserved symmetry with all parameters fixed. Both terms on the left of (Q08.2) have dimension $2N_c$ and axial charge $2N_c$. All terms are invariant under the non-Abelian flavor groups.

For two colors, the full flavor symmetry is enhanced to $SU(4)$, as the classical lesson demonstrated by converting all four chirals into doublets. Its six antisymmetric bilinears combine the four mesons and two baryons. Their quadratic invariant is, up to the already fixed orientation, $\det M-B\widetilde B$. A regular deformation with the same degree and charges can therefore add $C\mathcal L_2$. These checks permit a constant $C$; they do not rule out $C=0$.

The symbols in the quantum relation denote renormalized chiral operators and their expectation values. They are not products of numerical entries in one classical quark configuration. Relations among coincident composite operators require a quantum definition. There is therefore no instruction to find classical matrices whose determinants violate their ordinary matrix identity.

Fix the coefficient by removing a flavor

Write the meson as

$$M=\begin{pmatrix}X&Y\\ Z&T\end{pmatrix}$$

and add a mass for the second pair, $W_{\rm mass}=mT$. Implement the proposed constraint with a chiral Lagrange multiplier $\mathcal A$:

$$W=\mathcal A(XT-YZ-B\widetilde B-C\mathcal L_2)+mT . \tag{Q08.3}$$

$\mathcal A$ is a device for imposing a relation, with dimension $-1$ and R-charge two in the massless charge convention. We have not added a canonical particle with that engineering dimension.

The mass is a spurion of R-charge two in this charge assignment. Holding it fixed breaks that particular massless-theory R-symmetry; the surviving anomaly-free R-symmetry of the one-flavor theory uses a different assignment. Scale matching compares holomorphic parameters, not an unchanged R-charge table across theories.

Hold the light meson $X\ne0$ fixed and solve the equations of the fields being removed. First,

$$W_T=\mathcal A X+m=0 \quad\Longrightarrow\quad \mathcal A=-\frac mX\ne0 .$$

Then $W_Y=-\mathcal A Z$, $W_Z=-\mathcal A Y$, $W_B=-\mathcal A\widetilde B$ and $W_{\widetilde B}=-\mathcal A B$ set $Y,Z,B,\widetilde B$ to zero. Finally $W_{\mathcal A}=0$ gives $T=C\mathcal L_2/X$. Substituting into the full superpotential leaves

$$W_{\rm light}(X)=\frac{C m\mathcal L_2}{X} =C\,\frac{\mathcal L_1}{X}. \tag{Q08.4}$$

We have deliberately not imposed $W_X=0$: the remaining light theory will determine its own vacuum behavior. Its independently established ADS interaction has coefficient one, so compatibility requires $C=1$. A zero deformation would predict zero superpotential and fail this test.

This is the decoupling argument following Intriligator–Seiberg, (4.17), developed in Tong, (6.34)–(6.35) and the ensuing elimination. A weak heavy-flavor threshold supplies a controlled starting regime; holomorphy continues the relations in nonzero $m$. It does not make the strongly coupled Kähler metric calculable. The exact constraint and its continuation use the sourced dynamics; the charge table alone is not a derivation of the entire quantum vacuum space.

See what the deformation changes

For $f=2$ define

$$F=XT-YZ-B\widetilde B-\mathcal L_2.$$

Its gradient in the six composite coordinates is

$$dF=(T,-Z,-Y,X,-\widetilde B,-B). \tag{Q08.5}$$

All entries vanish only at the origin. When $\mathcal L_2\ne0$, that origin does not satisfy $F=0$. The quantum hypersurface is consequently smooth, with five complex dimensions. Locally one variable can be solved in terms of the other five. On the patch $X\ne0$, for example,

$$T=\frac{\mathcal L_2+YZ+B\widetilde B}{X}. \tag{Q08.6}$$

This is smoothing of the holomorphic vacuum space. It does not specify its exact kinetic metric or imply a mass gap for the entire theory: the moduli themselves are massless.

For a worked point choose $\mathcal L_2=\ell^4>0$ and $M=\ell^2\mathbf1_2$, $B=\widetilde B=0$. Here the tangent equation is $\delta X+\delta T=0$; $Y,Z,B,\widetilde B$ can also fluctuate. There are five complex tangent directions, all compatible with supersymmetric vacua.

The full $SU(4)$ flavor symmetry cannot remain unbroken at any point: a nonzero antisymmetric flavor tensor is required by the constraint, and that representation contains no $SU(4)$ singlet. At the displayed point the left/right flavor subgroup is broken to its diagonal subgroup. This is flavor-symmetry breaking, not supersymmetry breaking. No detailed statement about the full unbroken subgroup is needed to count the five complex moduli.

The excluded origin is a major difference from the classical description. Near the strong scale the useful light fields are composite moduli, while far out on the same space the weakly Higgsed quark description is useful. Fundamental matter can screen electric sources; this interpolation does not require a sharp Higgs-versus-confinement phase boundary. Intriligator–Seiberg, sections 1.1 and 4.2

Add one flavor: fifteen light composites

Now take $N_c=2$, $f=3$, still with four supercharges. There are nine mesons and three baryons of each type. Define their signs using $\varepsilon_{123}=1$:

$$B_i=\frac12\varepsilon_{ikl}\,q_k^TJq_l,\qquad \widetilde B_j=\frac12\varepsilon_{jkl} \det\begin{pmatrix}\widetilde q^{\,k}\\ \widetilde q^{\,l}\end{pmatrix}, \qquad J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}. \tag{Q08.7}$$

Thus $B_3=\det(q_1,q_2)$, and $\widetilde B_3$ is the determinant of the first two tilde rows, agreeing with the two-flavor orientation. Regard $B,\widetilde B$ as three-component columns. The contraction is $\widetilde B^TMB=\sum_{i,j}B_i M^j{}_i\widetilde B_j$.

The established confining description has these fifteen unconstrained chiral fields, regular kinetic terms near the origin, and a superpotential of the form

$$W_3=\frac{c}{\mathcal L_3} \left(\widetilde B^TMB-\det M\right). \tag{Q08.8}$$

This infrared field content is dynamical input, not a conclusion from naming gauge invariants. The source is Intriligator–Seiberg, section 4.3, (4.18)–(4.20). We will verify its equations and fix $c$ by decoupling.

All fifteen fields have ultraviolet engineering dimension two. The quark R-charge is now $1/3$, so every composite in (Q08.8) has R-charge $2/3$. The numerator has dimension six, R-charge two and axial charge six; $\mathcal L_3$ has dimension three, R-charge zero and axial charge six. Baryon charges cancel in the first term. The full flavor symmetry is enhanced to $SU(6)$; the fifteen bilinears are its antisymmetric representation. The manifest left/right subgroup is enough for the index calculations below.

For $c\ne0$, the F equations are

$$MB=0,\qquad M^T\widetilde B=0,\qquad B_i\widetilde B_j=\operatorname{cof}_{ji}(M). \tag{Q08.9}$$

The cofactor $\operatorname{cof}_{ji}(M)$ is $(-1)^{j+i}$ times the determinant obtained by deleting row $j$ and column $i$; it equals $\partial(\det M)/\partial M^j{}_i$. This definition works even when $M$ is singular. Using $\det M\,M^{-1}$ there would be inappropriate.

These are precisely the classical polynomial relations. For actual two-color quarks, the signed-minor vector satisfies $QB=0$, the tilde vector satisfies $\widetilde B^T\widetilde Q=0$, and hence the first two equations follow. The two-by-two Cauchy–Binet identity gives the cofactor equation. Here the relations arise as F equations of an action for independent composites, rather than restrictions removing those fields before fluctuations are studied.

At the origin all F equations and the quadratic superpotential Hessian vanish. Thus all fifteen chiral multiplets are massless there, and flavor symmetry can remain unbroken. At a generic vacuum only nine complex moduli remain, matching the microscopic count $2N_cf-(N_c^2-1)=12-3=9$; the extra composite directions acquire holomorphic masses. A singular vacuum variety can therefore sit inside a regular low-energy action with additional fields.

We can see this count directly at $M=\operatorname{diag}(v^2,v^2,0)$, $B_3=\widetilde B_3=v^2\ne0$, with the other baryons zero. Define independent linear combinations of fluctuations

$$u_i=\delta B_i+\delta M^i{}_3,\quad \widetilde u_i=\delta\widetilde B_i+\delta M^3{}_i \quad(i=1,2),\qquad h=\delta B_3+\delta\widetilde B_3-\delta M^1{}_1-\delta M^2{}_2 .$$

Expanding (Q08.8) to quadratic order gives

$$W_{3,\mathrm{quad}}=\frac{c v^2}{\mathcal L_3} \left(\widetilde u_1u_1+\widetilde u_2u_2+\delta M^3{}_3\,h\right). \tag{Q08.9a}$$

For $c\ne0$ these are three bilinear pairs, so the holomorphic Hessian has rank six. Its nine-dimensional kernel is the tangent space of the vacuum variety at this point. At the origin the quadratic term instead vanishes entirely.

Regular kinetic terms do not mean known canonical normalization. Near the origin their leading coefficients multiply quantities such as $|M|^2/|\Lambda_3|^2$, with flavor symmetry relating the composite terms. The positive wavefunction coefficient and its quantum running are not fixed by (Q08.8). We cannot read physical masses directly from a dimension-two composite’s superpotential Hessian.

Check the second decoupling and its sign

Give the third flavor mass $m$ and focus on the branch containing the retained two-flavor fields:

$$M=\begin{pmatrix}\widehat M&0\\0&T\end{pmatrix}, \qquad B=\begin{pmatrix}0\\0\\B_3\end{pmatrix},\qquad \widetilde B=\begin{pmatrix}0\\0\\\widetilde B_3\end{pmatrix}.$$

The F equations of the mixed components vanish on this restriction. The remaining superpotential, including the mass, is

$$W=T\left[ \frac{c}{\mathcal L_3} (B_3\widetilde B_3-\det\widehat M)+m \right]. \tag{Q08.10}$$

In the three-flavor R-charge assignment, $m$ is now a spurion of R-charge $4/3$, since $T$ has charge $2/3$. As in the first decoupling, the fixed-mass low-energy R-symmetry must be identified afresh.

Its $T$ equation yields

$$\det\widehat M-B_3\widetilde B_3 =\frac{m\mathcal L_3}{c}.$$

The scale relation is $\mathcal L_2=m\mathcal L_3$. Agreement with the already normalized two-flavor constraint fixes $c=1$, including the relative sign in (Q08.8). If $c=0$, the $T$ equation would instead be the impossible condition $m=0$.

On the deformed hypersurface its gradient never vanishes, so the remaining F equations set $T=0$ in a supersymmetric vacuum. Its role in (Q08.10) becomes the constraint multiplier: $\mathcal A=-T/\mathcal L_3$ has the same dimension as in (Q08.3). This matches the calculation following Tong, (6.37)–(6.38).

This three-flavor phenomenon is called s-confinement: a description in gauge-invariant composites covers the origin without additional light gauge fields, and confinement need not force flavor-symmetry breaking there. It does not mean that every excitation is massive. Nor does it assert an indefinitely growing potential between fundamental test charges; dynamical fundamental pairs can screen them. The two- and three-flavor theories have different vacuum geometry and different symmetry realizations despite belonging to one decoupling chain.

Exercises

1. Are two square roots two isolated vacua?

Set $\mathcal L_2=\ell^4>0$. Show that $M=\ell^2\mathbf1_2$ and $M=-\ell^2\mathbf1_2$, with both baryons zero, lie on the quantum space. Connect them by a continuous path of supersymmetric vacua with zero baryons. Can the first point be obtained from numerical classical quark matrices satisfying these same composite definitions?

Hint 1

Keep the product of the two diagonal meson entries equal to $\ell^4$ while varying their phases in opposite directions.

Hint 2

For ordinary square matrices the determinant of a product is the product of the determinants, without a quantum constant.

Solution

Both determinants equal $\ell^4$. The path

$$M(t)=\ell^2\operatorname{diag}(e^{it},e^{-it}), \qquad 0\le t\le\pi,\qquad B=\widetilde B=0$$

stays on the constraint and connects the two points. They are different points on a continuous moduli space, not two isolated vacua counted by choosing a root of $\mathcal L_2$.

No ordinary classical $Q,\widetilde Q$ can realize the first point with both baryons zero: $\det(\widetilde Q Q)=B\widetilde B=0$ would contradict $\det M=\ell^4$. The quantum coordinates are expectation values of renormalized composite operators. Their relation is not a new determinant rule for numerical matrices, and their nonzero values do not imply supersymmetry breaking.

2. Give both flavors a mass

In the two-flavor theory add $\mu_1X+\mu_2T$ to the constraint-multiplier superpotential, with $\mu_1,\mu_2\ne0$. Find all its supersymmetric vacua and the values of $W$. Compare with the pure-theory scale $\mathcal L_0=\mu_1\mu_2\mathcal L_2$.

Hint 1

The equations $\mathcal A T+\mu_1=0$ and $\mathcal A X+\mu_2=0$ imply $\mathcal A\ne0$; use that in the other equations.

Hint 2

The remaining equations are $XT=\mathcal L_2$ and $\mu_1X=\mu_2T$. The two sign choices must be correlated.

Solution

$Y=Z=B=\widetilde B=0$, and

$$X_\pm=\pm\sqrt{\frac{\mu_2\mathcal L_2}{\mu_1}}, \qquad T_\pm=\pm\sqrt{\frac{\mu_1\mathcal L_2}{\mu_2}}, \qquad \mathcal A_\pm=-\frac{\mu_2}{X_\pm}.$$

Choose square-root branches so that $X_\pm T_\pm=\mathcal L_2$. These give the two solutions; independent sign choices would fail the constraint. The multiplier term vanishes on either solution, leaving

$$W_\pm=\mu_1X_\pm+\mu_2T_\pm =\pm2\sqrt{\mu_1\mu_2\mathcal L_2} =\pm2\sqrt{\mathcal L_0}.$$

This agrees with the holomorphic vacuum data of the previous mass-deformed ADS calculation after two thresholds. The two roots are now isolated because the mass terms lift the continuous family. Their energies are zero in global supersymmetry. The unknown Kähler metric prevents extracting a quantitative strongly coupled particle spectrum from these equations alone.

3. Test the massless composites with anomalies

For the three-flavor theory at the origin, match $\operatorname{Tr}R$, $\operatorname{Tr}R^3$ and the cubic $SU(3)_L$ flavor anomaly between the microscopic fields and the fifteen composite chiral multiplets. Normalize a left-flavor fundamental Weyl fermion’s cubic anomaly to $+1$ and its antifundamental’s to $-1$. What does agreement establish?

Hint 1

There is one adjoint gaugino with three color components and twelve quark Weyl components. Their R-charges are $1$ and $-2/3$. Each composite Weyl component has R-charge $-1/3$.

Hint 2

The mesons provide three left-flavor fundamentals. The baryons $B_i$ form one antifundamental; the tilde baryons are left-flavor singlets.

Solution

The microscopic traces are

$$\operatorname{Tr}R=3+12\left(-\frac23\right)=-5, \qquad \operatorname{Tr}R^3=3+12\left(-\frac23\right)^3=-\frac59.$$

The composite traces give $15(-1/3)=-5$ and $15(-1/3)^3=-5/9$. For the left-flavor cubic anomaly, the two microscopic colors give $2$. In the composite description the result is $3-1=2$. We have tested a subgroup of the enhanced $SU(6)$ flavor symmetry; calling it the full flavor group would miss that enhancement.

All these are fermion traces; using scalar R-charges would fail. The agreement is a necessary consistency check of the proposed light spectrum and its symmetry realization. It does not prove confinement or determine the Kähler metric. The independent mass-deformation check provides different evidence. Intriligator–Seiberg, section 4.3

We can now follow one fixed gauge group through a runaway, a smooth quantum-modified family, and a symmetric vacuum with massless composites. Build a Seiberg-duality dictionary takes the next step: an infrared description with a new gauge group and a precise map of mesons and baryons.