Holomorphy and the quantum effective action

A heavy particle can disappear from the list of low-energy fields while leaving interactions behind. Supersymmetry makes part of that matching unusually simple: a holomorphic function determines the superpotential interactions. It does not determine every coefficient of the low-energy action.

Start from four-dimensional superspace and interactions. We use flat four-dimensional global $\mathcal N=1$ supersymmetry, with four real supercharges, all manifest in $\mathcal N=1$ superspace. The worked model has two gauge-neutral chiral multiplets and no gauge sector. Gravity and supersymmetry breaking are absent. We retain Lorentzian signature $(+---)$, the chiral expansion $\Phi=\phi+\sqrt2\theta\psi+\theta^2F$, and the superpotential component convention

$$\mathcal L_W=W_iF_i-\frac12W_{ij}\psi_i\psi_j+\text{h.c.}$$

The aim is to calculate a local effective interaction, then identify precisely what that calculation leaves undetermined.

Choose which fluctuations remain

Suppose a microscopic description is supplied at a cutoff $\Lambda_{\rm UV}$. Choose a lower scale $\mu$. A Wilsonian action $S_\mu$ incorporates fields heavier than $\mu$ and high-momentum fluctuations of the light fields. Light fluctuations below $\mu$ remain to be integrated over when computing an observable. Schematically, in Euclidean notation for this definition,

$$e^{-S_\mu[\Phi_<]} =\int\mathcal D\Phi_>\,e^{-S_{\Lambda_{\rm UV}}[\Phi_<,\Phi_>]}.$$

The split is an organizing definition; a supersymmetric calculation needs a regulator and counterterms consistent with the symmetries being used. It is not an instruction to impose an arbitrary sharp momentum cutoff separately on each component.

For external momenta much smaller than the eliminated scales, the action can be expanded in local operators and derivatives. Its coefficients depend on the matching prescription and on $\mu$. Changing $\mu$ while compensating those coefficients leaves observables unchanged. This is the renormalization-group viewpoint. For a dimensionless coupling $g_a(\mu)$, define

$$\beta_a=\mu\frac{d g_a}{d\mu}.$$

There is another effective action. With Euclidean sources, define $\mathcal W[J]=\log Z[J]$, $\phi_c=\delta\mathcal W/\delta J$, and

$$\Gamma[\phi_c]=\int J\phi_c-\mathcal W[J].$$

This Legendre transform is the one-particle-irreducible, or 1PI, effective action. Its vertices already include quantum fluctuations at all momenta. One uses those vertices to reconstruct connected functions without adding a second round of the loops already included in them. With massless particles, $\Gamma$ can contain nonlocal momentum dependence, such as logarithms and inverse differential operators. A zero-momentum limit then needs care.

Our holomorphy statements concern local Wilsonian superpotential terms at nonzero $\mu$. They do not assert that every zero-momentum 1PI expression is a local holomorphic function. This distinction is emphasized in Seiberg, §3 and Intriligator–Seiberg, §2.3.

Match a heavy multiplet

Call the heavy chiral superfield $H$ and the light one $X$, with scalar components $h$ and $x$. At the matching scale choose

$$K=H^\dagger H+X^\dagger X,\qquad W=\frac{M}{2}H^2+\frac{y}{2}HX^2.$$

Here $[H]=[X]=[M]=1$ and $[y]=0$. The complex parameter $M$ is nonzero. Around the supersymmetric origin, the heavy multiplet has mass $|M|$ and $X$ is massless. Work at weak coupling $|y|^2/(16\pi^2)\ll1$, with

$$E\ll\mu\ll |M|,\qquad u=\frac{|y|^2|x|^2}{|M|^2}\ll1.$$

The second condition keeps us near the vacuum where the heavy/light separation was chosen. In a superspace derivative expansion, auxiliary components must also be small on the heavy scale. This matters because a superspace derivative contains more than a spacetime gradient.

At tree level the leading algebraic part of the heavy superfield equation is

$$W_H=MH+\frac{y}{2}X^2=0,\qquad H_0=-\frac{y}{2M}X^2.$$

The full equation also contains superspace derivatives from the kinetic term. Thus $W_H=0$ is a controlled leading elimination, not an exact minimization of the full component potential at every fixed $x$.

Substitute $H_0$ into both terms of $W$:

$$\begin{aligned} W_{\rm eff} &=\frac{M}{2}\frac{y^2X^4}{4M^2} +\frac{y}{2}\left(-\frac{yX^2}{2M}\right)X^2\\ &=\frac{y^2X^4}{8M}-\frac{y^2X^4}{4M} =-\frac{y^2}{8M}X^4. \end{aligned}$$

The relative sign is essential. Merely deleting $H$ would miss the interaction. An equivalent check completes the holomorphic square:

$$W=\frac{M}{2}\left(H+\frac{yX^2}{2M}\right)^2 -\frac{y^2X^4}{8M}.$$

This is tree exchange of a heavy multiplet, not a light-field loop correction. The heavy-field example appears in Tong, §3.3.3, (3.35)–(3.36); the calculation here fixes its coefficient and keeps track of the kinetic terms as well.

Several checks are immediate. The result has dimension three. It vanishes as $y\to0$ or $|M|\to\infty$ at fixed $X,y$. The coefficient contains $1/M$, not $1/|M|$: it retains the complex phase of a chiral coupling.

The pole at $M=0$ is meaningful. The field we removed is no longer heavy there, so this description cannot be continued through that point at fixed $E,\mu$. The expression is holomorphic on the patch $M\ne0$, and meromorphic as a function of $M$ across the excluded point. A singular reduced description need not mean that the original two-field Lagrangian is singular.

Recover kinetic data and the scalar potential

An effective superpotential is only part of an effective action. The same leading substitution in $K$ gives

$$K_{\rm eff}=X^\dagger X+ \frac{|y|^2}{4|M|^2}(X^\dagger X)^2+\cdots.$$

The omitted terms include superspace derivative corrections and quantum corrections. A general real Kähler potential defines the scalar kinetic metric

$$g_{x\bar x}=\frac{\partial^2K}{\partial x\,\partial\bar x}.$$

Treat $x,\bar x$ as independent coordinates while differentiating, then impose the Lorentzian conjugation condition. Differentiating the displayed quartic term twice gives a factor four:

$$g_{x\bar x}=1+\frac{|y|^2|x|^2}{|M|^2}+\cdots=1+u+\cdots.$$

The coefficient of $\partial_\mu x\,\partial^\mu\bar x$ has changed. The correction depends on $y$ and $\bar y$, and is fully compatible with supersymmetry.

The same metric multiplies $F\bar F$ in the bosonic auxiliary Lagrangian:

$$\mathcal L_F=g_{x\bar x}F\bar F+W_xF+\overline{W_x}\bar F.$$

Varying with respect to $\bar F$ gives $F=-\overline{W_x}/g_{x\bar x}$. Substitution produces $\mathcal L_F=-V$, where

$$V=g_{x\bar x}^{-1}|W_x|^2.$$

This is the one-field version of the inverse-metric potential in Tong, (3.28)–(3.29). A superpotential by itself determines neither this metric nor the resulting potential away from its supersymmetric zeros.

Keeping the leading small-field term gives

$$W_x=-\frac{y^2}{2M}x^3,\qquad V=\frac{|y|^4|x|^6}{4|M|^2}\,[1+O(u)].$$

The interaction is sextic in the scalar even though $W_{\rm eff}$ is quartic in the superfield. Its coefficient has dimension minus two, so $V$ has dimension four. It is nonnegative and has its only zero at $x=0$ when $y\ne0$. Around that zero the light field still has no quadratic mass term.

There is a useful independent check before making a derivative expansion. The full two-field scalar potential is

$$V_{\rm full}(h,x)=\left|Mh+\frac{y}{2}x^2\right|^2 +|yhx|^2.$$

At fixed constant $x$, set $A=|M|^2+|y|^2|x|^2>0$. Completing an ordinary positive square gives

$$\begin{aligned} h_{\min}(x)&=-\frac{\bar M\,y\,x^2}{2A} =\frac{h_0(x)}{1+u},\\ V_{\rm full}(h,x)&=A|h-h_{\min}|^2+ \frac{|y|^4|x|^6}{4A}. \end{aligned}$$

Consequently the leading small-$u$ term in the minimum agrees with the effective potential. The displacement $h_{\min}-h_0$ is relative order $u$; $W_H=0$ was not the exact fixed-$x$ minimization. Substituting only $h_0$ into the original canonical kinetic terms does correctly recover the displayed leading metric correction through $|dh_0/dx|^2=u$.

For this simple model, using precisely the displayed $g=1+u$ in $g^{-1}|W_x|^2$ even reproduces the rational expression for the constant-field minimum. That agreement does not establish an exact action for arbitrary fields: the full heavy equation still generates derivative and auxiliary corrections, and quantum corrections have not been computed. The approximation is a statement about an action and its regime, not just one successful function comparison.

Let the couplings carry charges

Why was $1/M$ permitted while $1/|M|$ was not? A superpotential integrand must be chiral. Promote $M$ and $y$ temporarily to nondynamical chiral superfields. The local superpotential is then a holomorphic function of $X,M,y$; dependence on their conjugates belongs to other structures, such as D-terms, rather than an ordinary derivative-free superpotential.

A constant supersymmetric coupling means a constant lowest component with zero background fermion and auxiliary component. A nonzero background auxiliary would change the problem. The background-field argument and its regulator assumptions are explained in Seiberg, §3. Here we use it as a selection rule and still perform matching to determine a coefficient.

Assign three formal charges:

Object	$F_X$	$F_H$	$R$
$X$	$1$	$0$	$1$
$H$	$0$	$1$	$0$
$M$	$0$	$-2$	$2$
$y$	$-2$	$-1$	$0$
$W$	$0$	$0$	$2$

Every microscopic term has the required charge. The R column describes the superfield’s lowest component; $\theta$ has R-charge one and $d^2\theta$ has R-charge minus two. The first two columns are ordinary flavor charges.

These are transformations of a family of theories, allowing the parameters to change. Each column separately need not generate a symmetry of a theory with $M,y$ fixed. In this example a suitable combination does survive: $R-\tfrac12F_X+F_H$ assigns $R(X)=1/2$, $R(H)=1$, and leaves both parameters neutral. Formal charges are useful without confusing them with separately conserved currents.

Consider a derivative-free monomial $y^aM^bX^n$ in a perturbative local expansion, regular in $X$ near zero and in $y$ near zero, but allowing inverse powers of the nonzero heavy mass. Charge conservation requires

$$n-2a=0,\qquad -a-2b=0,\qquad n+2b=2.$$

Solving gives $a=2$, $b=-1$, $n=4$. Thus the allowed field-dependent form is

$$W_{\rm eff}=c\,\frac{y^2X^4}{M}.$$

Charge and holomorphy arguments have not fixed $c$, or shown by themselves that it is nonzero. The tree calculation gave $c=-1/8$. Conversely, the real combination $|y|^2(X^\dagger X)^2/|M|^2$ is neutral in every column and is allowed in $K$. The two selections have different content.

This illustrates the method of Intriligator–Seiberg, §2.1: combine holomorphy and symmetry with controlled limits or a matching calculation. It is not a rule that every symmetry-allowed term must appear.

Canonical normalization can still run

At some scale the quadratic kinetic term might be $Z(\mu)X^\dagger X$, with real $Z>0$. Define $X_c=\sqrt Z\,X$ to make it canonical at that scale. Even if a holomorphic coefficient $\kappa=-y^2/(8M)$ is unchanged in the original field coordinate,

$$W_{\rm eff}=\kappa X^4 =\frac{\kappa}{Z^2}X_c^4.$$

The canonically normalized coefficient is $\kappa_c=\kappa/Z^2$. It depends on kinetic data. It is not a holomorphic function of spurions in general, since $Z$ need not be one.

For clarity, define $\eta=\mu\,d\log Z/d\mu$; conventions for a symbol called an anomalous dimension vary. If $\kappa$ is held fixed in this holomorphic coordinate, then

$$\mu\frac{d\kappa_c}{d\mu}=-2\eta\,\kappa_c,\qquad \widehat\kappa=\mu\kappa_c,\qquad \mu\frac{d\widehat\kappa}{d\mu}=(1-2\eta)\widehat\kappa.$$

The one in the last expression is the engineering scaling of a coefficient of dimension minus one. These formulas specify how a given $Z$ affects normalization; they do not calculate $Z$ or a loop coefficient. Nor is a coefficient in a local action automatically an on-shell measured amplitude. That requires the appropriate kinematics, external-state normalization, and remaining light-field fluctuations.

The next quantum calculation must therefore ask two questions separately: what constrains the Wilsonian superpotential, and what nonzero correction appears in the kinetic term?

Exercises

1. Test the constant-field approximation

Take real $M>0$, $y=1/2$, and a real light field $x=M/2$. Compute $u$, $h_0$, $h_{\min}$, and $V_{\min}$. Compare the leading sextic potential with $V_{\min}$. Explain why the discrepancy is consistent with the approximation.

Hint 1

Use the dimensionless ratio $x/M$ first. The heavy-field displacement is governed by $1/(1+u)$.

Hint 2

Express $V_{\min}$ in units of $M^4$. Compare the leading value with the exact fixed-$x$ minimum, not with a new vacuum.

Solution

The field amplitude gives $u=(1/4)(1/4)=1/16$. Hence

$$h_0=-\frac{M}{16},\qquad h_{\min}=-\frac{M}{17}.$$

The leading sextic value and the exact constant-field minimum are

$$V_0=\frac{M^4}{4096},\qquad V_{\min}=\frac{16}{17}V_0=\frac{M^4}{4352}.$$

Their relative difference, measured against $V_{\min}$, is $1/16=u$. The leading approximation has exactly the expected relative error. The positive square $A|h-h_{\min}|^2$ also shows that the fixed-$x$ minimum is stable in the heavy directions. It is not a vacuum of the full theory: varying $x$ lowers the potential toward the supersymmetric origin.

The small-field check does not test a momentum expansion. One must additionally keep $E/|M|$ small for time-dependent configurations.

2. Change the normalization

Consider one light chiral field with

$$K=Z X^\dagger X,\qquad W=\frac{m}{2}X^2+\frac{\lambda}{3!}X^3,\qquad Z>0.$$

Find the canonically normalized mass and cubic coefficient. Derive the scalar mass squared at $x=0$ by eliminating $F$ and rescaling the scalar. Explain why a fixed holomorphic $m$ would not imply a fixed mass in canonical units.

Hint 1

Rescale the whole superfield, including its auxiliary field.

Hint 2

Before rescaling, the potential is $|m x+\lambda x^2/2|^2/Z$, while the kinetic term has coefficient $Z$.

Solution

With $X_c=\sqrt Z\,X$,

$$W=\frac{m}{2Z}X_c^2+ \frac{\lambda}{3!Z^{3/2}}X_c^3.$$

Thus the complex canonical mass parameter is $m_c=m/Z$, the physical free quadratic mass is $|m|/Z$, and the cubic coefficient is $\lambda_c=\lambda/Z^{3/2}$. Eliminating $F$ gives

$$F=-\frac{\bar m\bar x+\bar\lambda\bar x^2/2}{Z}, \qquad V=\frac{|m|^2}{Z}|x|^2+O(|x|^3) =\frac{|m|^2}{Z^2}|x_c|^2+O(|x_c|^3).$$

The scalar mass squared is $|m|^2/Z^2$, agreeing with the squared fermion mass from the canonically normalized superpotential. A scale-dependent $Z$ changes these parameters even if $m,\lambda$ are constant in the original coordinate. Further momentum-dependent corrections are needed to identify an exact interacting pole mass. For $Z=1$ all expressions reduce to the original canonical formulas.

3. Distinguish a heavy threshold from an infrared limit

For positive $a,b$, evaluate the convergent Euclidean integral

$$I(a,b)=\int\frac{d^4p}{(2\pi)^4} \left[\frac1{(p^2+a^2)^2}-\frac1{(p^2+b^2)^2}\right].$$

Explain what happens as $a\to0$ at fixed $b$. Then examine the two-field superpotential at $M=0$, $y\ne0$: solve its supersymmetric vacuum equations. How does this help interpret the pole of $W_{\rm eff}$? The integral is a diagnostic of infrared behavior, not a claimed loop correction with the model’s vertices or multiplicities.

Hint 1

The unit three-sphere has area $2\pi^2$. Set $t=p^2$ in the radial integral, and subtract before removing the upper limit.

Hint 2

At $M=0$, use both $W_H=yX^2/2$ and $W_X=yHX$. Determine which field must be kept light.

Solution

Angular integration gives

$$I(a,b)=\frac1{16\pi^2}\int_0^\infty dt\,t \left[\frac1{(t+a^2)^2}-\frac1{(t+b^2)^2}\right].$$

An antiderivative of $t/(t+c)^2$ is $\log(t+c)+c/(t+c)$. The upper-end difference vanishes; the lower-end difference is $\log(a^2/b^2)$. Therefore

$$I(a,b)=\frac1{16\pi^2}\log\frac{b^2}{a^2}.$$

The difference is ultraviolet convergent. As $a\to0$ the first term behaves like $dp/p$ at small momentum, causing an infrared logarithm. Holding a nonzero lower momentum boundary would remove that endpoint. This demonstrates why integrating fluctuations all the way to zero momentum can differ from keeping a Wilsonian scale.

At $M=0$, the equations are

$$\frac y2x^2=0,\qquad yhx=0.$$

Thus $x=0$ and $h$ is arbitrary. The original theory has a supersymmetric flat direction with $H$ retained. Its microscopic polynomial action has no $1/M$ singularity. Integrating out $H$ near the origin is invalid at this point: precisely the degrees of freedom omitted by the reduced description must return. The infrared logarithm and the heavy-field pole arise in different calculations, but both warn against taking a singular limit after removing necessary low-energy information.

You can now obtain a local superpotential by matching and test its charges, dimensions, and domain. To predict a potential or a normalized interaction, also specify the Kähler metric and which fluctuations remain. Nonrenormalization and a first quantum correction makes that distinction quantitative: calculate a nonzero loop correction, then determine which interaction coefficients it can change.