Nonrenormalization and a first quantum correction

If the superpotential receives no perturbative correction, where does an ordinary loop diagram go? We will calculate one rather than infer that all quantum effects cancel. Its answer is a nonzero kinetic coefficient. The same coefficient changes the cubic coupling after the fields are normalized.

Start from holomorphy and the effective action. We work in flat four-dimensional global $\mathcal N=1$ supersymmetry: four real supercharges, all manifest in superspace. There is one gauge-neutral chiral multiplet, no gauge sector, and no supersymmetry breaking. The Lorentzian signature is $+---$; Euclidean momentum integrals below are analytic continuations with an explicitly stated auxiliary contour.

Fix the vertex before calculating

At an ultraviolet scale $\Lambda$, take

$$K=\bar\Phi\Phi,\qquad W=\frac{m}{2}\Phi^2+\frac{y}{3!}\Phi^3.$$

Here $[\Phi]=[m]=1$ and $[y]=0$. The bar denotes the antichiral partner, whose Lorentzian components obey the usual conjugation condition. The factor $3!$ defines $y$. In the notation $W=h\Phi^3/3$ used in some references, $y=2h$.

The component convention from the prerequisite is

$$\mathcal L_W=W'F-\frac12W''\psi\psi+\text{h.c.}$$

For the loop, set $m=0$ and expand about $\phi=0$. The relevant Lorentzian terms are

$$\bar F F+\frac y2F\phi^2+\frac{\bar y}2\bar F\bar\phi^2 -\frac y2\phi\psi\psi-\frac{\bar y}2\bar\phi\bar\psi\bar\psi.$$

An external auxiliary field $F$ is a convenient probe of the kinetic multiplet. It need not satisfy its equation of motion during this calculation. At quadratic order, a local D-term $\delta Z\int d^4\theta\,\bar\Phi\Phi$ contains $\delta Z\,\bar F F$ in Lorentzian signature. Finding that coefficient therefore finds the common superfield normalization. It is not a new propagating auxiliary degree of freedom.

Regulate a whole supermultiplet

Keep fluctuations below a positive scale $\mu$ and integrate the higher ones. We require external momenta $E\ll\mu\ll\Lambda$ and weak $y$. Define the smooth scalar covariance

$$C_s(p)=\frac{e^{-p^2/s^2}}{p^2},\qquad G_>(p)=C_\Lambda(p)-C_\mu(p) =\frac{e^{-p^2/\Lambda^2}-e^{-p^2/\mu^2}}{p^2},$$

where $p$ is Euclidean. Use this split in the superfield Gaussian measure: multiply the free superfield covariance by the same momentum profile for all components, retaining their usual momentum numerators. Since functions of spacetime momentum commute with the supersymmetry generators and preserve chirality, the split preserves the linear off-shell supersymmetry. This neutral model needs no gauge regulator.

Equivalently,

$$G_>(p)=\int_{\Lambda^{-2}}^{\mu^{-2}}ds\,e^{-s p^2}.$$

The variable $s$ is a proper-time parameter, with dimension minus two. This representation shows that the propagator is finite at $p=0$ and suppressed at large $p$. The low covariance is $C_\mu$; their sum is exactly $C_\Lambda$. A smooth split weights a range of momenta instead of excising a sharp shell.

The resulting Wilsonian vertices have a derivative expansion around zero external momentum. We calculate its leading quadratic coefficient. Dependence on the smoothing profile, including finite constants near the endpoints, is part of the scheme; the logarithmic coefficient in the broad interval $\mu\ll p\ll\Lambda$ will have a universal one-loop interpretation.

The auxiliary two-point diagram

After continuation, the auxiliary and its interactions enter the Euclidean density as

$$\mathcal L_{E,F}=-\bar F F-\frac y2F\phi^2 -\frac{\bar y}2\bar F\bar\phi^2.$$

The negative free term is not a prescription to integrate over $\bar F=F^*$. Treat $F,\bar F$ as independent complex variables and choose the convergent free contour $\bar F=-F^*$, so $-\bar F F=|F|^2$. The interactions are then imaginary for $\bar\phi=\phi^*$. Their Gaussian continuation reproduces the positive scalar potential after eliminating the auxiliary. In computing an off-shell vertex we first retain independent external $F,\bar F$ and then apply this contour if they are integrated.

Only the two scalar vertices attach to an external $F,\bar F$ pair at order $|y|^2$. Write their integrated interaction as $S_{\rm int}$. The connected Gaussian expansion is

$$\Delta S_E=-\log\langle e^{-S_{\rm int}}\rangle_> =\langle S_{\rm int}\rangle_> -\frac12\langle S_{\rm int}^2\rangle_{>,c}+\cdots.$$

The first term vanishes. There are two choices of which vertex is first in the square, and two Wick pairings of the identical scalars:

$$\langle\phi^2(x)\bar\phi^2(z)\rangle_>=2G_>(x-z)^2.$$

Thus the coefficient is

$$-\frac12\times2\times\frac{|y|^2}{4}\times2 =-\frac{|y|^2}{2},$$

and

$$\Delta S_E^{(2)} =-\frac{|y|^2}{2} \int d^4x\,d^4z\,F(x)\bar F(z)G_>(x-z)^2.$$

For slowly varying external fields, the leading local term is $-\delta Z\bar F F$, where

$$\delta Z=\frac{|y|^2}{2}I_{\Lambda,\mu},\qquad I_{\Lambda,\mu}=\int\frac{d^4p}{(2\pi)^4}G_>(p)^2.$$

The two minus signs now have a transparent meaning: the connected expansion produces a negative coefficient of $\bar F F$, while a positive Euclidean kinetic normalization multiplies $-\bar F F$. Hence $\delta Z$ is positive. There is no closed fermion loop in this auxiliary diagram and no missing fermion-loop sign.

To evaluate the integral, use the area $2\pi^2$ of the unit three-sphere and set $t=p^2$, $a=\Lambda^{-2}$, $b=\mu^{-2}$:

$$I_{\Lambda,\mu} =\frac1{16\pi^2}\int_0^\infty\frac{dt}{t} \left[e^{-2at}+e^{-2bt}-2e^{-(a+b)t}\right].$$

For positive $A,B$, differentiating with respect to $B$ and fixing the value at $A=B$ gives $\int_0^\infty dt\,(e^{-At}-e^{-Bt})/t=\log(B/A)$. Apply this to two convergent differences:

$$I_{\Lambda,\mu} =\frac1{16\pi^2} \log\frac{(a+b)^2}{4ab} =\frac1{16\pi^2} \log\frac{(\Lambda^2+\mu^2)^2}{4\Lambda^2\mu^2}.$$

The completed one-loop result in this scheme is

$$\boxed{ Z(\mu)=1+\frac{|y|^2}{32\pi^2} \log\frac{(\Lambda^2+\mu^2)^2}{4\Lambda^2\mu^2} +O(|y|^4). }$$

It vanishes as a correction when $\mu=\Lambda$, is positive for $0<\mu<\Lambda$, and is dimensionless. The positive integral and $(a+b)^2\geq4ab$ independently verify its sign. The local action contains $Z\bar\Phi\Phi$; supersymmetry supplies the same $Z$ for the scalar, fermion and auxiliary terms. Tong, §3.3.2, (3.34) describes this common normalization; the calculation above fixes its coefficient for our vertex and regulator.

In the wide scale interval,

$$\delta Z=\frac{|y|^2}{16\pi^2}\log\frac{\Lambda}{\mu} -\frac{|y|^2}{32\pi^2}\log4 +O\!\left(|y|^2\frac{\mu^2}{\Lambda^2},\,|y|^4\right).$$

The finite $-\log4$ is specific to the smooth covariance split. Keeping only the logarithm would not correctly describe the endpoint $\mu=\Lambda$.

Why the superpotential does not acquire this correction

Now restore a small $m$ while keeping $\Phi$ as a light field, with $|m|\ll\mu$. Promote $m,y$ to constant supersymmetric chiral spurions. In a regulator respecting these selection rules, the local derivative-free Wilsonian superpotential is holomorphic in them. Its perturbative expansion at fixed $\mu>0$ is regular near $m=y=0$: we have not integrated away the light field or its infrared modes.

Assign formal flavor and R-charges:

Object	Flavor	R
$\Phi$	$1$	$1$
$m$	$-2$	$0$
$y$	$-3$	$-1$
$W$	$0$	$2$

These transformations include the parameters; they need not be symmetries of the fixed massive theory. The cutoffs have neither charge. A holomorphic monomial $m^a y^b\Phi^n$ therefore obeys

$$n-2a-3b=0,\qquad n-b=2, \qquad\Longrightarrow\qquad n=b+2,\quad a=1-b.$$

Analytic perturbation theory has nonnegative integers $a,b$. Only $m\Phi^2$ and $y\Phi^3$ survive. Neutral functions of $\Lambda/\mu$ are not excluded by this charge argument alone; we must still fix their coefficients.

At $y=0$ the theory is Gaussian, so no interaction loop changes the quadratic coefficient. At order $y$ a cubic vertex with three external legs has no internal line and hence no loop. Higher powers of $y$ cannot correct these two monomials without violating the holomorphic charge conditions. Their coefficients consequently retain the matched tree values:

$$W_\mu=\frac m2\Phi^2+\frac y{3!}\Phi^3 \qquad\text{in the fixed holomorphic field coordinate.}$$

This is the perturbative nonrenormalization reasoning of Seiberg, §§3–4.1, (4.1)–(4.3), applied with an explicit infrared scale and our factorial convention. Holomorphy does real work: $|y|^2\bar\Phi\Phi$ is allowed in $K$ but cannot be put into a holomorphic superpotential coefficient.

The regularity assumption also matters. A term $y^2\Phi^4/m$ has the formal charges, but is singular at $m=0$ and would arise from exchanging the very field we are retaining, not a local interaction generated by its high fluctuations. By contrast, integrating out a distinct heavy field can generate inverse heavy masses, as the prerequisite demonstrated. Intriligator–Seiberg, §2.2 distinguishes those tree matching contributions. This argument neither excludes nonperturbative superpotentials in other theories nor assumes a nonperturbative ultraviolet completion of the four-dimensional Wess–Zumino model.

Running in canonical units

Return to $m=0$. At scale $\mu$, define $\Phi_c=\sqrt{Z}\Phi$ and

$$y_c=yZ^{-3/2},\qquad \eta=\mu\frac{d\log Z}{d\mu},\qquad \gamma=-\frac{\eta}{2}.$$

The last equation defines our anomalous-dimension convention. The prerequisite defined $\eta$ without naming it $\gamma$. Holding the holomorphic $y$ and ultraviolet input fixed, our finite-regulator answer gives

$$\eta=-\frac{|y|^2}{16\pi^2} \frac{\Lambda^2-\mu^2}{\Lambda^2+\mu^2} +O(|y|^4).$$

In the scaling window $\mu/\Lambda\to0$, replace $y$ by $y_c$ in the one-loop coefficient; their difference affects higher orders. Then

$$\gamma=\frac{|y_c|^2}{32\pi^2}+O(|y_c|^4),\qquad \beta_{y_c}\equiv\mu\frac{dy_c}{d\mu} =-\frac32\eta y_c =\frac{3y_c|y_c|^2}{32\pi^2}+O(y_c|y_c|^4).$$

These are running local-action parameters in canonical coordinates, not a claim that a scattering amplitude equals a superpotential coefficient. The remaining low fluctuations must still be included in an observable.

As a coefficient check, specialize Martin–Vaughn, (1.1), (2.7), (2.10) to one neutral chiral field and no soft breaking. Their cubic tensor is $Y^{111}=y_c$, and their one-loop anomalous dimension is $(|y_c|^2/2)/(16\pi^2)$, agreeing with ours. This comparison concerns the universal leading coefficient, not the finite smoothing constant.

For a concrete check, let $|y|=1/2$ and $\Lambda/\mu=10$, with $E/\mu$ negligible. The finite expression gives

$$\delta Z=\frac1{128\pi^2}\log\frac{10201}{400} \approx0.002564,\qquad \frac{y_c}{y}=1-\frac32\delta Z+O(|y|^4) \approx0.99615.$$

The phase does not change at this order; the magnitude decreases toward lower scales. The expansion requires $\delta Z\ll1$. Replacing the truncated expression for $Z$ in an exact fractional power does not turn it into an all-orders answer.

The infrared endpoint changes the question

The full one-particle-irreducible auxiliary kernel instead includes all momenta. At nonzero Euclidean external momentum $k$, its scalar loop contains

$$I_{\rm 1PI}(k)= \int\frac{d^4p}{(2\pi)^4}\frac1{p^2(p+k)^2}.$$

Subtract the same expression at $k_0^2>0$. Feynman parameterization shifts the denominator to $[\ell^2+x(1-x)k^2]^2$ for $0<x<1$. The convergent subtracted radial integral from the prerequisite gives

$$I_{\rm 1PI}(k)-I_{\rm 1PI}(k_0) =\frac1{16\pi^2}\int_0^1dx\, \log\frac{x(1-x)k_0^2}{x(1-x)k^2} =-\frac1{16\pi^2}\log\frac{k^2}{k_0^2}.$$

Use a common translation-invariant ultraviolet regularization before making the subtraction and momentum shifts. The integrable parameter endpoints are understood by their limits. The resulting two-point coefficient has a logarithm of external momentum, and no Taylor expansion at $k=0$. It is not the local Wilsonian $Z(\mu)$ with $E\ll\mu$.

This is why taking $\mu\to0$ at fixed external momentum is not a permitted limit of the local calculation. Massless 1PI functions can have infrared singular and nonlocal structures; the Wilsonian superpotential statement is not a statement about every way of representing them. Seiberg, §3 explicitly identifies this distinction. Our one-loop example exhibits a kinetic logarithm; it does not claim a newly generated local F-term.

Exercises

1. Translate the cubic normalization

A calculation uses $W=h\Phi^3/3$ instead of $y\Phi^3/3!$. Find the coefficient of the auxiliary vertex, its one-loop $\delta Z$ in terms of $I_{\Lambda,\mu}$, and the scaling-window $\gamma$ and $\beta_{h_c}$. Explain the error in transferring the numerical coefficient of $\beta_y$ without translating the coupling.

Hint 1

Differentiate the superpotential before drawing the diagram. Each external auxiliary vertex couples to two identical scalars.

Hint 2

Use $y=2h$ and $h_c=hZ^{-3/2}$. The definition $\gamma=-\eta/2$ remains unchanged.

Solution

The vertex is $hF\phi^2+\text{h.c.}$ in the Lorentzian action. The same two pairings now give

$$\delta Z=2|h|^2I_{\Lambda,\mu}.$$

This equals $|y|^2I_{\Lambda,\mu}/2$ when $y=2h$. In the wide scale interval,

$$\gamma=\frac{|h_c|^2}{8\pi^2}+O(|h_c|^4),\qquad \beta_{h_c}=3h_c\gamma =\frac{3h_c|h_c|^2}{8\pi^2}+O(h_c|h_c|^4).$$

The factor four reflects the square of the rescaling of the cubic coefficient. It is not a change of theory or a disagreement about diagrams. Both conventions give the same normalized Yukawa term, since $-y\phi\psi\psi/2=-h\phi\psi\psi$.

2. Test three proposed corrections

At fixed $\mu>0$, with the light field retained and $|m|\ll\mu$, consider

$$\delta W_1=c_1|y|^2y\Phi^3,\qquad \delta W_2=c_2\frac{y^2}{m}\Phi^4,\qquad \delta K_3=c_3|y|^2\bar\Phi\Phi.$$

The $c_i$ are neutral constants. Which obstruction applies to each proposed term? Would replacing $m$ by $\mu$ in the second expression repair it? State how the answer changes if a distinct heavy chiral field has actually been integrated out.

Hint 1

“Chiral for a constant number” is weaker than holomorphic dependence on a background chiral superfield. Check charges and analyticity separately.

Hint 2

The Wilsonian scale has dimension one but no flavor charge. A light-mass pole and a heavy-threshold pole need not have the same interpretation.

Solution

The first term has appropriate dimensions and constant-parameter charges, but depends on $\bar y$. It fails the holomorphic spurion condition for a local superpotential. The second is holomorphic away from $m=0$ and has flavor charge $-6+2+4=0$ and R-charge $-2+4=2$. Its pole violates the perturbative regularity required with this field still light at a nonzero Wilsonian scale.

Replacing $m$ with $\mu$ removes that pole but loses its flavor charge: $y^2\Phi^4/\mu$ has flavor charge $-2$, so it is not allowed by the spurion selection rule. Dimensional analysis alone missed this.

The third term is real for real $c_3$, neutral, and of dimension two, as a Kähler term must be. It is allowed. The loop actually generates it with positive scale-dependent coefficient $I_{\Lambda,\mu}/2$ at this order; symmetry alone would not fix that number or require it to be nonzero.

For a distinct heavy field of nonzero mass $M$, an inverse $M$ can be allowed in the theory with that field removed. Its validity is restricted to energies below $|M|$; the field must return when $M$ becomes light. This is the tree matching situation of the prerequisite, not an exception to the present loop argument.

3. Sum the leading logarithms

In the scaling window, set $u=|y_c|^2$ and use the leading beta function with initial value $u(\mu_0)=u_0>0$. Solve for $u(\mu)$. Determine the direction of running, the formal pole, and why this pole is not a controlled prediction of new physics. For $\log(\mu/\mu_0)=-1$ and $u_0=1/4$, estimate the change.

Hint 1

Differentiate $y_c\bar y_c$. The phase is constant at this order. The equation for $1/u$ is simpler than the equation for $u$.

Hint 2

The initial condition is specified at a scale already far below $\Lambda$. Continue only while the running coupling stays weak and both scales remain in that interval.

Solution

Both conjugate terms contribute:

$$\frac{du}{d\log\mu}=\frac{3u^2}{16\pi^2}, \qquad \frac{d(1/u)}{d\log\mu}=-\frac3{16\pi^2}.$$

Therefore the one-loop running solution is

$$u(\mu)= \frac{u_0}{1-\dfrac{3u_0}{16\pi^2}\log(\mu/\mu_0)}.$$

Lower scales increase the denominator and decrease $u$. For the specified change,

$$u(\mu)=\frac{1/4}{1+3/(64\pi^2)} \approx0.24882.$$

This running solution sums the logarithms implied by the leading differential equation; it does not include higher-loop beta coefficients. Its formal ultraviolet pole is $\mu_{\rm pole}=\mu_0\exp[16\pi^2/(3u_0)]$. The weak-coupling approximation fails before that pole, and the finite ultraviolet cutoff may be reached much earlier. The formula diagnoses growth toward the ultraviolet, not a reliable particle threshold or a demonstrated ultraviolet completion.

The loop and the selection rule answer different questions. The fixed-coordinate Wilsonian superpotential retains its perturbative coefficients, while the Kähler term changes and canonical couplings run. Before applying nonrenormalization, specify the action, field coordinate, infrared scale, and which fields remain in the theory.

For an application to a broken vacuum, An O’Raifeartaigh model and its Goldstino calculates a loop-generated scalar mass while the superpotential stays fixed. The scalar and its massless fermionic partner need not remain degenerate.