A dynamically generated superpotential

Can a theory with no classical superpotential generate one through quantum effects? Perturbative nonrenormalization does not exclude an interaction proportional to an instanton weight. But a symmetry argument that permits an interaction does not show that nature generates it.

We will resolve that distinction in one-flavor $SU(2)$ SQCD, then calculate what happens to its vacua. The preparation is holomorphic scales and decoupling and the supersymmetric-vacuum test.

Use flat four-dimensional Minkowski spacetime with global $\mathcal N=1$ supersymmetry: four real supercharges, all manifest in $\mathcal N=1$ superspace. The gauge group is $SU(2)$, of rank one. There is one fundamental chiral $Q$ and one antifundamental chiral $\widetilde Q$, initially with $W_{\rm tree}=0$. Microscopic matter kinetic terms are canonical at the reference scale used for the semiclassical calculation. There is no FI term for this simple gauge group, no soft breaking, and no supergravity.

Identify the light variable

Write the same letters for the scalar components when discussing constant bosonic configurations. The gauge-invariant meson is

$$M=\widetilde Q_aQ^a,\qquad [M]=2.$$

There is no independent baryon: the antisymmetric contraction of two identical commuting $Q$ scalars vanishes. The two-color pseudoreality does not double the flavor count.

For $M\ne0$, a representative of each D-flat gauge orbit is

$$Q=\begin{pmatrix}v\\0\end{pmatrix},\qquad \widetilde Q=(v,0),\qquad M=v^2.$$

Indeed, $QQ^\dagger-\widetilde Q^\dagger\widetilde Q=0$, so every traceless D equation vanishes. The center of $SU(2)$ identifies $v$ and $-v$; $M$ is the physical coordinate. There are four elementary complex scalars and three complex gauge directions, leaving one complex modulus. At $v\ne0$ all three gauge bosons are massive.

When $|v|\gg|\Lambda|$, the Higgs scale is high enough that the gauge dynamics is weakly coupled. At energies well below the vector masses, the surviving light multiplet is the meson. This is the region in which we can calculate its metric and use a semiclassical instanton result. Tong, §6.2.4

Find the allowed interaction

The one-loop coefficient is $b=3(2)-1=5$. Keep the complex scale parameter

$$\mathcal L=\Lambda^5=\mu_{\rm R}^{\,5}e^{2\pi i\tau_{\rm h}(\mu_{\rm R})},$$

where $\mu_{\rm R}$ is a renormalization reference, not a quark mass. Two selection rules suffice.

The gauge-anomaly-free R assignment is $R(Q)=R(\widetilde Q)=-1$. Its mixed gauge anomaly vanishes because the adjoint gaugino contributes $2$ and the two fundamental Weyl fermions, each of R-charge $-2$, contribute $2(1/2)(-2)=-2$. Separately, rotate both chirals axially with charge $+1$ and compensate the anomaly by transforming $\mathcal L$. This axial transformation acts on a family of couplings; it is not a conserved symmetry at fixed theta.

Object	Engineering dimension	R-charge	Formal axial charge
$M$	$2$	$-2$	$2$
$\mathcal L$	$5$	$0$	$2$
$W$	$3$	$2$	$0$

These R-charges are not superconformal dimensions. For a candidate monomial $\mathcal L^aM^b$, the R and axial conditions require $-2b=2$ and $2a+2b=0$. Thus $b=-1$, $a=1$, and its dimension is $5-2=3$. No nonconstant combination of $M,\mathcal L$ is simultaneously dimensionless and neutral under both transformations. On the regular meson patch,

$$W_{\rm dyn}=C\,\frac{\mathcal L}{M}.$$

This argument fixes the form, including the inverse power of $M$. It permits $C=0$. The coefficient remains a dynamical question. The same distinction is explicit in Intriligator–Seiberg, (4.3).

Supply the nonzero dynamical input

An instanton is a finite-action Euclidean gauge configuration with nonzero topological charge. A charge-one configuration contributes the weight $e^{-8\pi^2/g_{\rm h}^2+i\theta}$. Here that is the coupling dependence of $\mathcal L$. Large squark expectation values suppress large instantons, making the Higgsed region suitable for a controlled calculation.

There is also a useful fermion count. The charge-one index gives four adjoint gaugino zero modes and one zero mode for each of the two fundamental Weyl fermions. The Higgs Yukawa coupling lifts two gaugino modes together with the two matter modes. Two supersymmetry modes remain, as required for a chiral $d^2\theta$ interaction. This explains why the calculation can produce a superpotential, but counting modes still does not evaluate its coefficient.

We now use an established result, rather than attempt the full instanton integral here. Finnell–Pouliot, §3.1, (3.1)–(3.9) evaluates the weakly Higgsed amplitude, including the measure normalization, and obtains a nonzero coefficient equal to one in its DR-matched scale convention. Its comparison includes the light-fermion bilinear coefficient $3\Lambda^5/v^4$.

Adopt that absolute scale normalization, with our meson normalized as $M=\widetilde Q Q$. In the source’s two-fundamental notation, choose the invariant determinant $X=M$. We then have the Affleck–Dine–Seiberg, or ADS, result

$$W_{\rm ADS}=\frac{\Lambda^5}{M}.$$

The threshold convention of the preceding lesson and this absolute choice are distinct. A common finite rescaling of the high and low scale parameters could preserve unit threshold matching while changing $C$. For example, if $\mathcal L'=a\mathcal L$ with $M$ fixed, the same interaction has coefficient $C'=1/a$. The source fixes the additional normalization we use here. Finnell–Pouliot, (2.1)–(2.7)

Holomorphy and the absence of further invariant parameter combinations extend this superpotential from the calculable Higgsed region across its regular meson patch. They do not make the Kähler potential exact. The singularity at $M=0$ warns against treating a meson-only weak-coupling action as a complete local description there.

Derive the metric before the potential

Substitution of a D-flat representative into the canonical microscopic Kähler potential gives

$$K_{\rm cl}=Q^\dagger Q+\widetilde Q\widetilde Q^\dagger =2|v|^2=2\sqrt{\overline M M}.$$

The metric is a derivative of this Kähler potential, not the Kähler potential itself:

$$G_{M\overline M}=\partial_M\partial_{\overline M}K_{\rm cl} =\frac1{2|M|},\qquad G^{M\overline M}=2|M|.$$

Consequently the kinetic density is $|\partial M|^2/(2|M|)$. Its dimension is four because $[M]=2$ and $[G]=-2$. Assigning a dimensionless unit metric to $M$ would already fail this check.

In the site’s auxiliary convention, the effective bosonic F density is

$$\mathcal L_F=G_{M\overline M}F_M\overline F_M +W_MF_M+\overline{W_M}\,\overline F_M.$$

Completing its square gives $F_M=-G^{M\overline M}\overline{W_M}$ and the positive potential

$$V=G^{M\overline M}|W_M|^2 =\frac{2|\Lambda^5|^2}{|M|^3} =\frac{2|\Lambda|^{10}}{|v|^6}.$$

There is a separate microscopic check. Since $\partial W/\partial Q^a=W_M\widetilde Q_a$ and $\partial W/\partial\widetilde Q_a=W_MQ^a$, the canonical sum of their F contributions on the D-flat orbit is $(|Q|^2+|\widetilde Q|^2)|W_M|^2=2|M||W_M|^2$, the same answer.

This potential uses the leading classical metric at large Higgs expectation value. The Kähler metric receives quantum corrections; choosing a reference near the Higgs scale controls large logarithms, and perturbative corrections become small as the gauge coupling decreases. The exact holomorphic superpotential and this leading metric are different levels of information.

For every finite regular $M$, $W_M=-\Lambda^5/M^2\ne0$. Far out on the branch, $V>0$ but approaches zero. Along a positive real $v$, $dV/dv=-12|\Lambda|^{10}/v^7<0$. The potential therefore drives the field outward instead of selecting a finite vacuum. The asymptotic metric places infinity at infinite field distance, so it is not a missing finite point in a poor coordinate.

This is the ADS runaway. It is not a finite positive-energy vacuum that has been shown to break supersymmetry, nor a normalizable vacuum at infinity. Our controlled calculation exhibits the runaway and excludes supersymmetric vacua at finite regular points. It does not establish the shape of the strongly coupled potential or exclude every possible nonsupersymmetric local extremum there. Tong, §6.2.2 makes this Kähler-potential qualification explicit.

Stabilize the runaway with a mass

Add the microscopic superpotential $m\widetilde Q Q=mM$, with $m\ne0$ and $[m]=1$. As a spurion, $m$ has R-charge $4$ and axial charge $-2$. The exact holomorphic meson interaction becomes

$$W(M;m)=\frac{\mathcal L}{M}+mM.$$

It is worth understanding why we may use this sum. Symmetry alone would allow $(\mathcal L/M)f(t)$ with $t=mM^2/\mathcal L$. At fixed $t$, take the weakly Higgsed limit while sending $m$ to zero accordingly. The massless instanton term and the tree mass term then give $f(t)=1+t$. Higher corrections do not survive this limit; no inverse-mass threshold is introduced because the light meson is retained. Holomorphic continuation gives the displayed result. This is the argument of Intriligator–Seiberg, (4.4)–(4.5), not a claim that symmetry alone fixed an arbitrary function.

The F equation is

$$W_M=m-\frac{\mathcal L}{M^2}=0, \qquad \boxed{M_\pm=\pm\sqrt{\frac{\mathcal L}{m}}.}$$

These are two different meson expectation values. They are not the gauge-equivalent choices $v$ and $-v$ for one value of $M$. At either root, $F_M=0$ and the D-flat representative still has $D^a=0$. Both vacua preserve all four real supercharges.

Using the leading metric, the full potential on the regular branch is

$$V(M,\overline M;m)=2|M| \left|m-\frac{\mathcal L}{M^2}\right|^2.$$

It is nonnegative and vanishes at each root. At large $|M|$, its leading behavior is $2|m|^2|M|$, so the nonzero mass prevents escape to infinity.

The controlled regime follows directly:

$$\frac{|M_\pm|}{|\Lambda|^2} =\left(\frac{|\Lambda|}{|m|}\right)^{1/2}, \qquad \frac{|v_\pm|}{|\Lambda|} =\left(\frac{|\Lambda|}{|m|}\right)^{1/4}.$$

Thus $|m|\ll|\Lambda|$ puts the vacua far out in the weakly Higgsed region. For a concrete example choose $\mathcal L=L_0^5>0$ and $m=\epsilon L_0>0$. Then

$$M_\pm=\pm L_0^2\epsilon^{-1/2},\qquad |v_\pm|=L_0\epsilon^{-1/4},\qquad W_\pm=\pm2L_0^3\sqrt\epsilon.$$

For $\epsilon=10^{-8}$, the meson magnitude is $10^4L_0^2$ and the Higgs expectation value is $100L_0$. The physical meson multiplet has leading mass $4|m|$, as derived in Exercise 2; the vector multiplets have masses of order $g|v|$. The separation becomes parametrically large as $\epsilon\to0$.

At fixed $\mathcal L$, sending $m\to0$ moves both vacua to infinity. They do not merge into a finite vacuum of the massless theory.

Relate the answer to decoupling

For a sufficiently heavy flavor, the low theory is pure $\mathcal N=1$ $SU(2)$. In our compatible scale conventions, the previous lesson gives

$$\Lambda_{\rm pure}^{\,6}=m\Lambda^5.$$

Evaluation of the superpotential in the two vacua therefore yields

$$W_\pm=\pm2\sqrt{m\Lambda^5} =\pm2\sqrt{\Lambda_{\rm pure}^{\,6}}.$$

This is holomorphic vacuum data, not a claim that the classical meson metric remains reliable during decoupling. The small-mass Higgs calculation and the large-mass pure-gauge description have different useful degrees of freedom. Continuing the holomorphic vacuum branches relates them; extrapolating the mass $4|m|$ to an elementary meson particle in pure Yang–Mills would not.

A constant $W_\pm$ is not a vacuum energy in global supersymmetry. Both energies are zero. Its dependence on the background mass remains meaningful: on a chosen branch, $dW_\pm/dm=M_\pm$, the holomorphic source-response relation. Intriligator–Seiberg, (2.8)

Exercises

1. Use a locally canonical coordinate

On a simply connected patch with $M\ne0$, define $X=\sqrt{2M}$. Rewrite the massless Kähler potential, superpotential and scalar potential in $X$. Find the fermion bilinear coefficient $W_{XX}$ at $M=v^2>0$, and compare with the instanton amplitude quoted above. Does this coefficient describe a particle mass about a stationary vacuum?

Hint 1

Choose one square-root branch on the patch. Then $M=X^2/2$ and $K_{\rm cl}=|X|^2$.

Hint 2

Differentiate $W=2\mathcal L/X^2$ twice. For the scalar potential use its first derivative and the canonical $X$ metric.

Solution

The leading action has

$$K_{\rm cl}=|X|^2,\qquad W=\frac{2\mathcal L}{X^2}, \qquad V=|W_X|^2=\frac{16|\mathcal L|^2}{|X|^6}.$$

Since $|X|^6=8|M|^3$, the potential is again $2|\mathcal L|^2/|M|^3$. Its dimension is four, with $[X]=1$.

The bilinear coefficient is $W_{XX}=12\mathcal L/X^4=3\mathcal L/v^4$ at $X=\sqrt2v$. The canonical tangent fermion there is the equal combination of the two quark fluctuations, agreeing with Finnell–Pouliot’s normalization check. This differentiates a sourced superpotential; it does not independently evaluate the instanton measure.

There is no stationary vacuum at finite $v$ in the massless theory. The coefficient is a local term in the effective action about a chosen background, not an on-shell mass spectrum of a stable vacuum. Finally $X$ is only a local coordinate: changing the square-root branch sends $X\to-X$ and leaves $M$ unchanged.

2. Check both physical scalar fluctuations

For the mass-deformed theory, expand near either root $M_0^2=\mathcal L/m$ in the weakly Higgsed regime. Normalize the complex fluctuation using $G_0=1/(2|M_0|)$. Find the two real scalar masses squared and the magnitude of the fermion mass. Explain why a positive result here is stronger than inspecting the potential along only real $M$.

Hint 1

At the root $W_M=0$ and $W_{MM}=2m/M_0$. Derivatives of the metric do not contribute to the quadratic potential multiplied by a nonzero constant $W_M$, because there is no such constant.

Hint 2

Use the canonical fluctuation $\eta=\sqrt{G_0}\,\delta M=(u+iw)/\sqrt2$. Its quadratic superpotential coefficient is $W_{MM}/G_0$.

Solution

To quadratic order,

$$V=G_0^{-1}|W_{MM}(M_0)|^2|\delta M|^2 =\frac{|W_{MM}(M_0)|^2}{G_0^2}|\eta|^2 =16|m|^2|\eta|^2.$$

Thus each canonically normalized real field $u,w$ has mass squared $16|m|^2$. The fermion coefficient has magnitude

$$\left|\frac{W_{MM}(M_0)}{G_0}\right| =\left|\frac{2m}{M_0}\,2|M_0|\right|=4|m|.$$

The scalar and fermion masses agree as required in a supersymmetric vacuum. Checking both real directions excludes an overlooked angular instability; a minimum along one line would not suffice. Gauge directions have already been removed in the meson description. In the microscopic weak-coupling description the other physical fields belong to three massive vector multiplets with mass of order $g|v|$.

These masses use the leading classical metric. Holomorphy protects the vacuum equation but does not fix quantum wavefunction corrections to the physical masses.

3. Follow a phase and distinguish two limits

Fix nonzero $\mathcal L$ and continue $m=m_0e^{i\alpha}$ once around the origin. What happens to a continuously followed vacuum and to the unordered pair of vacua? Then take $|m|\to\infty$ while holding $P=m\mathcal L=\Lambda_{\rm pure}^{\,6}\ne0$ fixed. Find the limits of $M_\pm$ and $W_\pm$. Which metric and mass claims from the weak Higgs regime can be carried into this limit?

Hint 1

Continue a square root continuously rather than resetting it to a principal branch at each value of $\alpha$.

Hint 2

At fixed $P$, write $\mathcal L=P/m$ before taking the limit. Contrast this with $m\to0$ at fixed $\mathcal L$.

Solution

A chosen meson branch changes as $M(\alpha)=M(0)e^{-i\alpha/2}$. At $\alpha=2\pi$ it reaches the other root. The unordered pair returns to itself. Two different roots of the vacuum equation are physical vacua; this is different from assigning a physical multiplicity to roots of a conventionally defined strong scale.

At fixed $P$ the branch equations are

$$M_\pm=\pm\frac{\sqrt P}{m}\longrightarrow0,\qquad W_\pm=\pm2\sqrt P.$$

The holomorphic source check still gives $dW_\pm/dm=M_\pm$ when differentiating at fixed microscopic $\mathcal L$; it is not a derivative along the path that holds $P$ fixed.

The meson expectation values enter the region where its classical quotient metric is unreliable. Neither $G=1/(2|M|)$ nor the weak-region particle mass $4|m|$ should be continued as a quantitative pure-gauge spectrum. The two holomorphic vacuum branches and scale matching do continue. The opposite limit $m\to0$ at fixed $\mathcal L$ sends $|M_\pm|\to\infty$ and recovers the runaway.

The calculation has used three different kinds of input: symmetry fixes an allowed function, an instanton calculation establishes its nonzero coefficient in a specified normalization, and a controlled Kähler metric converts that function into a potential. Keeping those roles distinct lets the mass deformation provide a genuine dynamical check.

Continue with Quantum-modified vacua and a confinement example to add massless flavors. The same scale matching will connect the runaway to a smooth quantum vacuum space and then to a description with additional massless composites.