Superpotentials and a first interacting model

The symmetry lesson tested candidate interactions by their charges. We will now construct one and determine its dynamics. Start with one gauge-neutral chiral multiplet and the superpotential

$$W(\phi)=\frac m2\phi^2+\frac g3\phi^3,\qquad m\ne0,\quad g\ne0.$$

A superpotential is a holomorphic function: it depends on the chiral scalar $\phi$, not on the independent anti-chiral scalar $\widetilde\phi$. It organizes several component interactions. It is not itself the scalar potential energy.

The target is a three-dimensional Wess–Zumino model, meaning chiral matter with a superpotential and no gauge sector. We work in flat Euclidean space, with all background sources and the central charge set to zero. We preserve the full complexified $\mathcal N=2$ algebra, whose Lorentzian parent has four real supercharges. The kinetic normalization is that of the free chiral theory. All vacuum and mass statements below are classical, or equivalently tree level.

The dimensions are $[\phi]=1/2$, $[\psi]=1$, $[F_\Phi]=3/2$, and $[W]=2$. Thus $[m]=1$ and $[g]=1/2$. These are engineering dimensions, not a determination of an interacting scaling dimension. Aharony and collaborators describe this class of three-dimensional models in section 2.1, equation (2.2).

Find the component interaction

Keep tilded and untilded Euclidean fields independent. Write a second holomorphic function $\widetilde W(\widetilde\phi)$ for the anti-chiral partner. Its coefficients are initially independent too. Complex conjugation will be imposed only when discussing the physical scalar potential and an integration contour.

Primes mean derivatives with respect to the function’s own scalar argument. The interaction is

$$\boxed{ \mathcal L_W= W'F_\Phi-\frac12W''\psi\psi +\widetilde W'\widetilde F_\Phi +\frac12\widetilde W''\widetilde\psi\widetilde\psi . }$$

A term quadratic in fermions with a scalar-dependent coefficient is a Yukawa interaction. For our polynomial, the untilded terms are $(m\phi+g\phi^2)F_\Phi-\tfrac12(m+2g\phi)\psi\psi$. The relative factors and the different tilded Yukawa sign are essential.

We can derive them without a superspace prerequisite. Use the contractions of the earlier component lessons:

$$uv=u^TCv,\qquad C=\begin{pmatrix}0&-1\\1&0\end{pmatrix}, \qquad \psi\psi=-2\psi_1\psi_2.$$

For a commuting label $\zeta$ and an odd two-component spinor $\psi$,

$$(\zeta\psi)\psi=-\frac12\zeta(\psi\psi).$$

For example, the first component on either side is $\zeta_1\psi_1\psi_2$. This identity is where the factor $1/2$ enters.

The bottom component $W(\phi)$ transforms as $\delta_\zeta W=\sqrt2\zeta(W'\psi)$, so its composite fermion is $\psi_W=W'\psi$. Apply another transformation:

$$\begin{aligned} \delta_\zeta\psi_W &=W''(\delta_\zeta\phi)\psi+W'\delta_\zeta\psi\\ &=\sqrt2W''(\zeta\psi)\psi+\sqrt2W'\zeta F_\Phi\\ &=\sqrt2\zeta\left(W'F_\Phi-\frac12W''\psi\psi\right). \end{aligned}$$

The quantity in parentheses is the composite auxiliary component $F_W$. In the anti-chiral sector, $\delta_{\widetilde\zeta}\widetilde\phi=-\sqrt2\widetilde\zeta\widetilde\psi$ has an extra minus sign. The same calculation gives

$$\widetilde\psi_W=\widetilde W'\widetilde\psi,\qquad \widetilde F_W =\widetilde W'\widetilde F_\Phi +\frac12\widetilde W''\widetilde\psi\widetilde\psi.$$

Thus $\mathcal L_W=F_W+\widetilde F_W$. This is the F-term construction: use the highest component of a composite chiral multiplet and its anti-chiral partner as a Lagrangian. The signs agree with the chiral embeddings in P, (6.7) and (6.9) and the product rule (6.16). The paper’s (6.14) states the curved-background construction with its R-charge condition. Below we verify the flat interaction directly, including cases that do not have a continuous R-symmetry.

First calculation: eliminate the auxiliaries and find the vacua

Add the interaction to the normalized free action:

$$\begin{aligned} \mathcal L={}& \partial_\mu\widetilde\phi\,\partial_\mu\phi -i\widetilde\psi\gamma^\mu\partial_\mu\psi -\widetilde F_\Phi F_\Phi\\ &+W'F_\Phi+\widetilde W'\widetilde F_\Phi -\frac12W''\psi\psi +\frac12\widetilde W''\widetilde\psi\widetilde\psi. \end{aligned}$$

The auxiliary fields have no derivatives. Vary them independently:

$$\frac{\partial\mathcal L}{\partial F_\Phi} =-\widetilde F_\Phi+W'=0,\qquad \frac{\partial\mathcal L}{\partial\widetilde F_\Phi} =-F_\Phi+\widetilde W'=0.$$

Therefore

$$F_\Phi=\widetilde W',\qquad \widetilde F_\Phi=W'.$$

Substitution gives the scalar term $+W'\widetilde W'$, not its negative. A useful way to see all signs at once is

$$-\widetilde F_\Phi F_\Phi+W'F_\Phi+\widetilde W'\widetilde F_\Phi =-(\widetilde F_\Phi-W')(F_\Phi-\widetilde W') +W'\widetilde W'.$$

Now choose $\widetilde\phi=\phi^*$ and $\widetilde W(\widetilde\phi)=\overline{W(\overline{\widetilde\phi})}$, so the tilded coefficients are the complex conjugates of $m,g$. The scalar potential is

$$\boxed{ V(\phi,\phi^*)=|W'(\phi)|^2 =|m\phi+g\phi^2|^2 =|\phi|^2|m+g\phi|^2 . }$$

It appears with a plus sign in the Euclidean action. After continuation to Lorentzian signature it is the nonnegative potential energy with the canonical scalar kinetic normalization.

A constant bosonic configuration preserves all flat supercharges if its fermion variations vanish. With vanishing gradients and fermions, $\delta\psi=\sqrt2\zeta F_\Phi$ and $\delta\widetilde\psi=\sqrt2\widetilde\zeta\widetilde F_\Phi$. Hence both auxiliaries must vanish. After elimination this becomes the F-term vacuum equation

$$W'(\phi)=0,$$

together with its complex conjugate. For the stated nonzero couplings, factorization gives two distinct classical supersymmetric vacua:

$$v_1=0,\qquad v_2=-\frac m g,\qquad V(v_1)=V(v_2)=0.$$

These are actual zero-energy minima of the nonnegative potential. Solving the weaker stationarity equation $\partial V=0$ can also find nonsupersymmetric saddles; exercise 2 checks one explicitly.

Check the small fluctuations

Around either vacuum write $\phi=v+h$. Define $\mu_v=W''(v)$; it is a mass parameter for fluctuations, not a new coupling. Then

$$W'(v+h)=\mu_v h+gh^2,\qquad \mu_{v_1}=m,\quad \mu_{v_2}=-m.$$

The quadratic scalar potential is $|\mu_v|^2|h|^2$. Writing $h=(h_1+ih_2)/\sqrt2$ gives two canonically normalized real scalar modes, each with mass squared $|m|^2$. The quadratic Yukawa terms have coefficients $\mu_v,\mu_v^*$:

$$\mathcal L_{\rm fermion}^{(2)} =-i\widetilde\psi\gamma^\mu\partial_\mu\psi -\frac12\mu_v\psi\psi +\frac12\mu_v^*\widetilde\psi\widetilde\psi.$$

Using left Grassmann variations gives

$$i\gamma^\mu\partial_\mu\psi=\mu_v^*\widetilde\psi,\qquad i\gamma^\mu\partial_\mu\widetilde\psi=-\mu_v\psi.$$

Apply the Dirac operator again and use $(\gamma^\mu\partial_\mu)^2=\partial^2$:

$$(-\partial^2+|\mu_v|^2)\psi=0.$$

The same equation holds for $\widetilde\psi$. The Euclidean inverse propagator has the mass scale $|\mu_v|=|m|$, agreeing with the scalar modes. This is a direct tree-level mass check at both vacua; it is not a calculation of quantum mass renormalization.

Why the interaction is supersymmetric

The auxiliary elimination used equations of motion. To prove the action’s symmetry, return to the full off-shell fields. Constant commuting labels generate odd operations. Keep the graded product rule $\delta(XY)=(\delta X)Y+(-1)^{|X|}X\delta Y$. The relevant untilded transformations are

$$\begin{aligned} \delta_\zeta\phi&=\sqrt2\zeta\psi,& \delta_\zeta\psi&=\sqrt2\zeta F_\Phi,& \delta_\zeta F_\Phi&=0,\\ \delta_{\widetilde\zeta}\phi&=0,& \delta_{\widetilde\zeta}\psi&=-\sqrt2i\gamma^\mu\widetilde\zeta\,\partial_\mu\phi,& \delta_{\widetilde\zeta}F_\Phi&=-\sqrt2i\widetilde\zeta\gamma^\mu\partial_\mu\psi. \end{aligned}$$

First, $\delta_\zeta F_W=0$. Indeed,

$$\delta_\zeta(\psi\psi)=2\sqrt2F_\Phi\zeta\psi,$$

so the variation of $W'F_\Phi$ cancels that of the quadratic fermion term. The remaining candidate term is proportional to $W'''(\zeta\psi)(\psi\psi)$ and vanishes: it contains three factors chosen from the same two odd components. This is a local classical Grassmann identity; no fermion equation is involved.

For the opposite family, use $(\gamma^\mu\widetilde\zeta)\psi=-\widetilde\zeta\gamma^\mu\psi$. The two contributions are

$$\begin{aligned} \delta_{\widetilde\zeta}F_W &=-\sqrt2i\,W'\widetilde\zeta\gamma^\mu\partial_\mu\psi -\sqrt2i\,W''(\partial_\mu\phi)\widetilde\zeta\gamma^\mu\psi\\ &=-\sqrt2i\,\partial_\mu \left(\widetilde\zeta\gamma^\mu W'\psi\right). \end{aligned}$$

The anti-chiral calculation must respect its different Yukawa sign. Its rules are

$$\begin{aligned} \delta_{\widetilde\zeta}\widetilde\phi&=-\sqrt2\widetilde\zeta\widetilde\psi,& \delta_{\widetilde\zeta}\widetilde\psi&=\sqrt2\widetilde\zeta\widetilde F_\Phi,& \delta_{\widetilde\zeta}\widetilde F_\Phi&=0,\\ \delta_\zeta\widetilde\phi&=0,& \delta_\zeta\widetilde\psi&=\sqrt2i\gamma^\mu\zeta\,\partial_\mu\widetilde\phi,& \delta_\zeta\widetilde F_\Phi&=-\sqrt2i\zeta\gamma^\mu\partial_\mu\widetilde\psi . \end{aligned}$$

Under $\delta_{\widetilde\zeta}$, the variation $-\sqrt2\widetilde W''\widetilde F_\Phi \widetilde\zeta\widetilde\psi$ from $\widetilde W'\widetilde F_\Phi$ cancels $+\sqrt2\widetilde W''\widetilde F_\Phi \widetilde\zeta\widetilde\psi$ from the Yukawa term. Again the triple-fermion term vanishes. Under $\delta_\zeta$ the two terms combine as

$$\begin{aligned} \delta_\zeta\widetilde F_W &=-\sqrt2i\,\widetilde W'\zeta\gamma^\mu\partial_\mu\widetilde\psi -\sqrt2i\,\widetilde W''(\partial_\mu\widetilde\phi)\zeta\gamma^\mu\widetilde\psi\\ &=-\sqrt2i\,\partial_\mu \left(\zeta\gamma^\mu\widetilde W'\widetilde\psi\right). \end{aligned}$$

Together,

$$\delta\mathcal L_W=\partial_\mu B_W^\mu,\qquad B_W^\mu=-\sqrt2i\left( \widetilde\zeta\gamma^\mu W'\psi +\zeta\gamma^\mu\widetilde W'\widetilde\psi \right).$$

The kinetic action already has the boundary expressions derived in the free-theory lesson. Add $B_W^\mu$ to them. Periodic fields on a flat torus, or sufficiently decaying fields on $\mathbb R^3$, give zero integrated boundary flux. This proves off-shell supersymmetry of the complete interacting action without requiring $m$ or $g$ to transform.

After eliminating the auxiliaries, the reduced action remains supersymmetric: the extra terms from varying their field-dependent solutions are multiplied by their vanishing algebraic Euler derivatives. The reduced transformation algebra, however, closes using the interacting equations of motion. Keeping an invariant action and keeping off-shell closure are distinct statements.

Check the Euclidean auxiliary integral

The algebraic saddle is not a reason to impose $\widetilde F_\Phi=F_\Phi^*$ in the off-shell Euclidean action; that would leave a growing Gaussian from $-\widetilde F_\Phi F_\Phi$.

At fixed scalar, set $P=W'=A+iB$, with real $A,B$ and $\widetilde W'=P^*$. On the free auxiliary contour

$$F_\Phi=u+iv,\qquad \widetilde F_\Phi=-u+iv,\qquad u,v\in\mathbb R,$$

the auxiliary Lagrangian becomes

$$\mathcal L_{\rm aux}=u^2+v^2+2i(Bu+Av).$$

Its real part is positive. For one finite-dimensional mode with positive Gaussian coefficient $a$, direct Fourier Gaussian integration gives

$$\int_{\mathbb R^2}\!du\,dv\, e^{-a[u^2+v^2+2i(Bu+Av)]} =\frac\pi a\,e^{-a(A^2+B^2)} =\frac\pi a\,e^{-a|W'|^2}.$$

This independently checks the sign of the effective potential. Equivalently, deform to the saddle-centered contour $F_\Phi=P^*+f$, $\widetilde F_\Phi=P-f^*$. The completed expression is then $|f|^2+|P|^2$. For this finite Gaussian the integrand is entire, and shifting the contours is justified by their Gaussian decay. The interacting saddle generally does not lie on the original contour $\widetilde F_\Phi=-F_\Phi^*$.

This check concerns auxiliary integration with the scalar held fixed. A full continuum path integral additionally requires a regulator and a consistent prescription for the remaining fields; we have not calculated its partition function.

Supersymmetry and R-symmetry impose different tests

For fixed, neutral $m,g$, a continuous phase R-symmetry of the form taught in the preceding lesson would require

$$2r=2,\qquad 3r=2.$$

There is no common solution. The flat component proof nevertheless went through. The supercharge algebra can act on an action that lacks this continuous R-symmetry.

With $g=0$, the quadratic term permits $r=1$. With $m=0$, the cubic permits $r=2/3$. Treating $m,g$ as transforming spurions can organize formal covariance, but it does not create a symmetry at fixed nonzero couplings. The dimensions $[m]=1$, $[g]=1/2$ are consistent with each term of $W$ having dimension two and each component term having dimension three.

This distinction matters for curved backgrounds constructed from a conserved R-current. The generic fixed-coupling quadratic-plus-cubic model cannot simply be placed into that R-symmetric construction using one phase assignment for $\phi$. Our result is the explicitly stated flat theory; a curved completion would require its own symmetry and background analysis.

Exercises

1. Determine the Yukawa signs instead of remembering them

Replace the two composite auxiliary expressions by

$$F_W^{(c)}=W'F_\Phi+cW''\psi\psi,\qquad \widetilde F_W^{(\widetilde c)} =\widetilde W'\widetilde F_\Phi +\widetilde c\,\widetilde W''\widetilde\psi\widetilde\psi .$$

Use the same-family transformations to determine $c,\widetilde c$. Explain why choosing the same sign for both fails in these conventions.

First hint

The three-fermion term vanishes. Compare the coefficient of $W''F_\Phi\zeta\psi$ and its tilded counterpart.

Second hint

The variation of the untilded $W'$ has a plus sign, while that of $\widetilde W'$ has a minus sign. The variation of either spinor square contains twice its auxiliary-spinor contraction.

Complete solution

The two variations are

$$\begin{aligned} \delta_\zeta F_W^{(c)} &=\sqrt2(1+2c)W''F_\Phi\zeta\psi,\\ \delta_{\widetilde\zeta}\widetilde F_W^{(\widetilde c)} &=\sqrt2(-1+2\widetilde c) \widetilde W''\widetilde F_\Phi \widetilde\zeta\widetilde\psi . \end{aligned}$$

They vanish for arbitrary off-shell fields only if $c=-1/2$ and $\widetilde c=+1/2$. This remains necessary for the quadratic example, where $W''=m\ne0$. For constant fields the incorrect residual contains no spacetime derivative and can have nonzero integral on a periodic box. A field redefinition could change a whole convention package; changing only one Yukawa sign does not.

2. A stationary point need not be a supersymmetric vacuum

Take real positive $m,g$ and write $\phi=x+iy$, with $x,y$ real. Show that $\phi_*=-m/(2g)$ is a stationary point of $V=|m\phi+g\phi^2|^2$. Compute its energy and the quadratic terms in real displacements about it. Is it a stable vacuum? Does it preserve supersymmetry?

First hint

At a constant configuration, $\partial_{\phi^*}V=W'(\phi)\overline{W''(\phi)}$. Stationarity can follow from $W''=0$ even when $W'\ne0$.

Second hint

Write $\phi=-m/(2g)+\xi+i\eta$. The derivative becomes $W'=-m^2/(4g)+g(\xi+i\eta)^2$.

Complete solution

At this point $W''=m+2g\phi_*=0$, whereas $W'(\phi_*)=-m^2/(4g)$. Its energy is

$$V(\phi_*)=\frac{m^4}{16g^2}>0.$$

The full expansion is

$$V=\frac{m^4}{16g^2} -\frac{m^2}{2}\xi^2+\frac{m^2}{2}\eta^2 +g^2(\xi^2+\eta^2)^2.$$

The Hessian in the coordinates $(\xi,\eta)$ has eigenvalues $-m^2,+m^2$. The point is a saddle, with an unstable real displacement. Its nonzero $W'$ also gives nonzero auxiliary expectation values, so it fails the constant-configuration supersymmetry test. The two zero-energy minima found in the main calculation are the stable classical supersymmetric vacua; the intervening stationary point is neither another minimum nor another supersymmetric vacuum.

3. Follow the vacua when a coupling vanishes

For real positive couplings, compare two limits: $g\to0$ at fixed $m>0$, and $m\to0$ at fixed $g>0$. Track both roots, the quadratic fluctuation masses, and the limiting potential. In each case, explain whether the two distinct roots are lost through escape or merging, and whether the limiting theory is free.

First hint

Track the root $-m/g$ before setting either coupling to zero. Roots can escape to infinity as well as merge.

Second hint

At $m=0$, use $V=g^2|\phi|^4$. A zero quadratic term does not mean the interaction has vanished.

Complete solution

For $g\to0$ at fixed $m$, the root at zero remains and the other root tends to negative infinity. In every fixed bounded region of field space,

$$V\longrightarrow m^2|\phi|^2.$$

The remaining vacuum has scalar and fermion mass $m$. The distant vacuum had the same local mass for every nonzero $g$, but it leaves the bounded region. The potential limit is not uniform over the entire unbounded field space.

For $m\to0$ at fixed $g$, both roots merge at zero and

$$V\longrightarrow g^2|\phi|^4,\qquad W''(0)=0.$$

The quadratic masses vanish at the merged classical vacuum, but a quartic scalar interaction and a scalar-dependent Yukawa interaction remain. This is not a free theory. The first limit loses a root to infinity; the second merges two roots. These are different mechanisms. Neither classical calculation determines a quantum index or a quantum scaling dimension.

Checkpoint

Given $W$, you can now construct the Yukawa and auxiliary terms, verify their boundary variation, eliminate the auxiliaries, and distinguish F-term vacua from other stationary points. For the quadratic-plus-cubic example, the same derivative $W''$ controls the Yukawa coefficient and the bosonic fluctuation mass around each supersymmetric vacuum.

Before extending this construction to gauge theories, explain why a charge test alone would not have fixed the Yukawa signs, and why off-shell invariance survives even though the generic fixed-coupling model has no continuous phase R-symmetry. Then construct the vector multiplet and test which matter interactions can consistently be gauged.