The Affleck–Dine–Seiberg Superpotential

For massless $SU(N_c)$ SQCD with $0<N_f<N_c$ and $W_{\rm tree}=0$ , the exact Wilsonian superpotential is

W_{\rm ADS} =(N_c-N_f) \left(\frac{\Lambda^{3N_c-N_f}}{\det M}\right)^{1/(N_c-N_f)}.

Symmetry, holomorphy, and dimension fix the functional form; a controlled $N_f=N_c-1$ instanton fixes the initial coefficient; holomorphic decoupling fixes the coefficient for every lower $N_f$ . The massless theory has a runaway rather than a finite supersymmetric vacuum.

Required background. The SQCD theory card fixes $M$ , baryon number, R-charge, global form, and $\Lambda$ . Nonperturbative superpotentials supplies the general holomorphy and spurion logic.

Helpful background. Instantons, zero modes, and condensates gives the controlled calculation at $N_f=N_c-1$ .

Precise domain of the formula

Assume:

four-dimensional rigid $\mathcal N=1$ supersymmetry;
gauge group $SU(N_c)$ , not only gauge algebra $\mathfrak{su}(N_c)$ ;
$0<N_f<N_c$ pairs $Q,\widetilde Q$ in fundamental and antifundamental representations;
vanishing tree superpotential and no additional singlets;
meson $M_i{}^{\tilde\jmath}=\widetilde Q_a{}^{\tilde\jmath}Q^a{}_i$ with no powers of $\Lambda$ absorbed;
$B(Q)=+1$ , $B(\widetilde Q)=-1$ ; and
$\Lambda^{3N_c-N_f}=\mu^{3N_c-N_f}\exp[-8\pi^2/g_h^2+i\theta]$ .

For $N_f<N_c$ , no baryon can be formed and $M$ is the only polynomial gauge-invariant coordinate. At a generic point the unbroken gauge group is $SU(N_c-N_f)$ , so the fractional power below also records the vacuum branches of that strong sector.

Symmetry and dimension leave one candidate

Let $k=N_c-N_f>0$ and $b_0=3N_c-N_f$ . The exact nonanomalous R-charge is

R(M)=2\left(1-\frac{N_c}{N_f}\right) =-\frac{2k}{N_f}.

Therefore

R(\det M)=N_fR(M)=-2k.

A superpotential has R-charge two, so $(\det M)^{-1/k}$ has exactly the required charge. Engineering dimensions give the same exponent:

\dim\Lambda^{b_0}=3N_c-N_f, \qquad \dim\det M=2N_f,

and their difference is $3(N_c-N_f)=3k$ . Taking the $k$ th root produces dimension three. Flavor and baryon symmetries allow no additional tensor. Thus

W=C_k \left(\frac{\Lambda^{b_0}}{\det M}\right)^{1/k},

where holomorphy alone leaves the number $C_k$ undetermined. This is the important logical pause: symmetry finds a one-dimensional space of candidates, not a nonzero coefficient.

Fixing the coefficient

The controlled anchor

At $N_f=N_c-1$ , a generic large meson expectation value completely Higgses $SU(N_c)$ . A one-instanton calculation is weakly coupled, the Higgs expectation value cuts off the size integral, and Yukawa interactions lift every zero mode except the universal two. With the conventions above it gives

W=\frac{\Lambda^{2N_c+1}}{\det M},

so $C_1=1$ . This is the controlled dynamical input of Affleck, Dine, and Seiberg 1984, pp. 493–534.

The decoupling recursion

Suppose the answer for $N_f$ flavors is $W=kY$ , with

Y=\left(\frac{\Lambda^{3N_c-N_f}} {x\det M'}\right)^{1/k},

where $x=M_{N_f}{}^{\widetilde N_f}$ and $M'$ is the light-flavor block. Add $W_{\rm tree}=mx$ . The $x$ equation is

0=\frac{\partial W}{\partial x} =-\frac{Y}{x}+m, \qquad x=\frac{Y}{m}.

Substitution gives

Y^{k+1} =\frac{m\Lambda^{3N_c-N_f}}{\det M'} =\frac{\Lambda_L^{3N_c-(N_f-1)}}{\det M'},

where $\Lambda_L^{3N_c-(N_f-1)}=m\Lambda^{3N_c-N_f}$ . The low-energy superpotential becomes

W_L=(k+1)Y.

Starting from $C_1=1$ , this recursion yields $C_k=k=N_c-N_f$ . It simultaneously checks the exponent, threshold power, and branch count. The same chain is summarized in Intriligator and Seiberg 1996, § 4.1, pp. 12–15.

Fractional powers and branches

The shorthand

\left(\frac{\Lambda^{b_0}}{\det M}\right)^{1/k}

has $k$ local branches. They are the gaugino-condensate branches of the unbroken $SU(k)$ sector at a generic meson point. A path around $\det M=0$ permutes them. The expression is therefore not a globally single-valued ordinary function on the punctured meson space; it is branch data for the effective theory.

The singularity at $\det M=0$ is physical. There the chosen description has integrated out degrees of freedom that become important, and the unbroken gauge group is larger. One should not smooth the singularity by choosing an arbitrary root or by treating $M$ as the only weakly coupled field there.

Mass deformation and the $N_c$ vacua

Add a nonsingular mass matrix,

W_{\rm tot}=W_{\rm ADS}+\operatorname{tr}(mM).

Define

S\equiv \left(\frac{\Lambda^{3N_c-N_f}}{\det M}\right)^{1/(N_c-N_f)}.

The matrix F-term is

m=S M^{-1}, \qquad M=Sm^{-1}.

Taking determinants and using the definition of $S$ gives

S^{N_c}=\Lambda^{3N_c-N_f}\det m.

Thus there are $N_c$ supersymmetric solutions,

S_k= \left(\Lambda^{3N_c-N_f}\det m\right)^{1/N_c} e^{2\pi ik/N_c},

and at each solution

W_k=N_cS_k.

This is precisely pure-SYM gaugino condensation after the scale match

\Lambda_{\rm SYM}^{3N_c} =\det m\,\Lambda^{3N_c-N_f}.

The calculation is an independent consistency loop: ADS plus masses reproduces the pure-gauge vacuum count and superpotential. The mass-deformed vacuum analysis and its decoupling interpretation are developed in Seiberg 1994, pp. 6857–6863.

The massless theory runs away

For $m=0$ , differentiating $W_{\rm ADS}$ gives an inverse power of $M$ that cannot vanish at any finite nonsingular meson. Along $M=t^2\mathbf1$ with $t\to\infty$ ,

W_{\rm ADS}\sim \Lambda^{(3N_c-N_f)/k}t^{-2N_f/k}\longrightarrow0,

and the F-term potential approaches zero. The theory therefore has a supersymmetric runaway at infinity, not a normalizable finite vacuum on this branch.

This distinction matters for index arguments. Sending the mass to zero moves the $N_c$ massive-theory vacua to infinite field values, so the asymptotic behavior of field space changes and the naive finite-volume index-continuity slogan does not apply.

The ADS term is an exact Wilsonian F-term. It does not determine the Kähler metric near strong coupling, and a potential estimate that uses a canonical metric outside the large-field regime is not exact.

Worked example: $SU(3)$ with one flavor

Here $b_0=8$ , $k=2$ , and $M$ is one complex field:

W=2\left(\frac{\Lambda^8}{M}\right)^{1/2}+mM.

The F-term gives $M=S/m$ and

S^3=m\Lambda^8.

There are three massive vacua, as required after the flavor decouples to pure $SU(3)$ . When $m\to0$ , $|M|\sim|\Lambda|^{8/3}|m|^{-2/3}\to\infty$ : the vacua do not remain at the origin but escape along the ADS runaway.

Common pitfalls

Using the formula at $N_f=N_c$ . The exponent diverges because the correct object changes: the next regime has a quantum-modified constraint, not an ADS superpotential.

Calling symmetry a coefficient calculation. Symmetry and holomorphy fix the monomial. The nonzero coefficient requires an instanton anchor, decoupling, or another dynamical input.

Forgetting the branch. The fractional root represents the vacua of an unbroken gauge sector. A principal-value root erases physical monodromy.

Exercises

Check that the ADS superpotential has baryon number zero and R-charge two.

Solution

$M$ has baryon number zero, so both $\det M$ and the full expression do. The exact R-symmetry leaves $\Lambda^{b_0}$ invariant, while $R(\det M)=-2(N_c-N_f)$ . Raising its inverse to $1/(N_c-N_f)$ gives R-charge two.

Decouple one flavor from the $N_f=N_c-1$ result and recover the coefficient two for $N_f=N_c-2$ .