= Solution
\b[Chirality and the component expansion.] A <chiral superfield> is constrained by
$$
\bar D_{\dot\alpha}\Phi=0.
$$
Use <left Grassmann derivatives>. The sign from differentiating an odd factor matters:
$$
\bar\partial_{\dot\alpha}(\theta\sigma^\mu\bar\theta)
=-(\theta\sigma^\mu)_{\dot\alpha},\qquad
\bar D_{\dot\alpha}y^\mu=0.
$$
For a general <superfield> written in coordinates $(y,\theta,\bar\theta)$, the odd chain rule consequently turns the given <superspace covariant derivative> into
$$
\bar D_{\dot\alpha}=-\bar\partial_{\dot\alpha}\big|_y.
$$
The chirality constraint removes the explicit $\bar\theta$ dependence at fixed $y$. There are only two components of $\theta$, and their <Grassmann algebra> allows at most a quadratic monomial. Its finite <chiral-superfield component expansion> is therefore
$$
\boxed{\Phi(y,\theta)=\varphi(y)+\sqrt2\theta\psi(y)+\theta^2F_{\mathrm{aux}}(y).}
$$
Here $\varphi$ is a <complex scalar field>, $\psi$ a <Weyl spinor>, and $F_{\mathrm{aux}}$ a complex <auxiliary field>; the last label avoids confusing it with the effective superpotential later. The factor $\sqrt2$ is the standard canonical component normalization. In ordinary coordinates the same statement is
$$
\Phi(x,\theta,\bar\theta)=
\exp\left(i\theta\sigma^\mu\bar\theta\,\partial_\mu\right)
\left[\varphi(x)+\sqrt2\theta\psi(x)+\theta^2F_{\mathrm{aux}}(x)\right].
$$
This translation exponential terminates because its shift is nilpotent; it displays the full component dependence without an unstated convention for the barred spinor square.
A <superspace covariant derivative> obeys the graded product rule. Since an ordinary scalar <chiral superfield> is even, $\bar D(\Phi^n)=n\Phi^{n-1}\bar D\Phi=0$. Linear combinations prove the polynomial claim. More generally, a nonsingular <holomorphic function> of chiral fields is chiral; inserting conjugate fields generally spoils this <holomorphic closure of chiral superfields>.
\b[The superspace action.] For a real <Kähler potential> and a <holomorphic superpotential>, the global chiral-field action has <Lagrangian density>
$$
\boxed{\mathcal L=\int d^2\theta\,d^2\bar\theta\,
K(\Phi^\dagger,\Phi)
+\left[\int d^2\theta\,W(\Phi)+\mathrm{h.c.}\right].}
$$
Full <superspace integration> gives a <D-term>, and chiral <superspace integration> an <F-term>. The action is real, and its supersymmetry variations are spacetime total derivatives. For one field, positive $K_{\varphi\bar\varphi}$ and algebraic elimination give $V=K^{\varphi\bar\varphi}|W_\varphi|^2$. The <Wess–Zumino model> uses the canonical choice $K=\Phi^\dagger\Phi$ at tree level.
\b[Scalar potential and the vertex.] In canonical normalization, the auxiliary part of the <Lagrangian density> is
$$
\mathcal L_{\mathrm{aux}}=|F_{\mathrm{aux}}|^2+
(F_{\mathrm{aux}}W_\varphi+\mathrm{h.c.}),\qquad
F_{\mathrm{aux}}=-\overline{W_\varphi}.
$$
Substitution leaves $\mathcal L_{\mathrm{aux}}=-|W_\varphi|^2$. Differentiating the given quadratic-plus-cubic <superpotential> therefore gives the tree-level <effective potential>
$$
\boxed{V_{\mathrm{tree}}=|m\varphi+g\varphi^2|^2
=|m|^2|\varphi|^2+m^*g\varphi^*\varphi^2
+mg^*\varphi\varphi^{*2}+|g|^2|\varphi|^4.}
$$
This applies for complex $m,g$. In terms of canonically normalized real fields $\varphi=(A+iB)/\sqrt2$, phases may be chosen to make $m,g$ real, in which case
$$
V_{\mathrm{tree}}=\frac{m^2}{2}(A^2+B^2)
+\frac{mg}{\sqrt2}A(A^2+B^2)
+\frac{g^2}{4}(A^2+B^2)^2.
$$
The complex-field quartic interaction is $\mathcal L_{\mathrm{int}}=-|g|^2\varphi^{*2}\varphi^2$. There are two identical external legs of each field type. Differentiating with respect to those four fields, or counting the Wick attachments to the vertex, produces the factor $2!2!$:
$$
\boxed{\text{two }\varphi\text{ and two }\varphi^*\text{ legs}:
\quad -i\,2!2!|g|^2=-4i|g|^2.}
$$
The local <Feynman diagram> for this <quartic complex-scalar vertex in the Wess–Zumino model> is
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-43-quartic.png]
{title=Quartic complex-scalar Wess–Zumino vertex with two legs of each field type and its factorial-normalized Feynman rule}
{height=420}
If real-field <Feynman rules> are preferred, the $AAAA$ and $BBBB$ vertices are $-6i|g|^2$, while the $AABB$ vertex is $-2i|g|^2$. These are the same interaction in a different component basis. They should not be confused with a convention that absorbs $2!2!$ into the coefficient of the complex quartic term.
\b[Spurion symmetries.] Treat $m,g$ as chiral <spurions>. An ordinary $U(1)$ acts on $\Phi,m,g$ with charges $(1,-2,-3)$ and leaves $\theta$ neutral. An <R-symmetry> gives $\theta$ charge one and $\Phi,m,g$ charges $(1,0,-1)$. Thus
|| Quantity
|| Ordinary $U(1)$
|| $U(1)_R$
|| Mass dimension
| $\Phi$
| 1
| 1
| 1
| $m$
| $-2$
| 0
| 1
| $g$
| $-3$
| $-1$
| 0
| $\theta$
| 0
| 1
| $-1/2$
| $W$
| 0
| 2
| 3
Each superpotential term has ordinary charge zero and <R-charge> two; the chiral integration measure has <R-charge> minus two. In particular $m$ is neutral under the specified <R-symmetry>. These are formal transformations of fields and parameters together, not two exact symmetries of a theory with arbitrary fixed nontransforming numerical couplings. This <Wess–Zumino spurion charge assignment> is useful because it constrains possible quantum terms.
\b[The general holomorphic form.] Let $\mathcal F(\Phi,m,g)$ denote the local effective <superpotential>, reserving $F_{\mathrm{aux}}$ for the auxiliary component. The <holomorphy argument for superpotential non-renormalization> permits dependence on the chiral <spurions>, not on their conjugates. The dimensionless combination $z=g\Phi/m$ is neutral under both formal symmetries, whereas $m\Phi^2$ has the required dimension and charges. Hence, for $m\ne0$, their most general allowed form is
$$
\boxed{\mathcal F(\Phi,m,g)=m\Phi^2 f\left(\frac{g\Phi}{m}\right),}
$$
with a holomorphic function $f$ before perturbative regularity and matching conditions are imposed. Equivalently, a monomial $m^a g^b\Phi^c$ must obey
$$
-2a-3b+c=0,\qquad -b+c=2,\qquad a+c=3.
$$
Solving gives $a=1-b$, $c=b+2$. Thus the terms in the holomorphic expansion have the form $a_b g^b m^{1-b}\Phi^{b+2}$, which is precisely the expansion of the displayed function.
\b[What is and is not renormalized.] Apply the <non-renormalization theorem> to a local <Wilsonian effective action> retaining the elementary field and a nonzero infrared cutoff. Perturbative coefficients are regular as $g\to0$ and $m\to0$: no massless infrared modes have been integrated all the way to zero momentum. Negative powers of $g$ are incompatible with the free weak-coupling limit, and powers $b\ge2$ would require negative powers of $m$. Only the quadratic and cubic structures survive this regularity test. Their coefficients cannot acquire a loop correction here: a quadratic term with no $g$ is the free-theory mass term, while a cubic term only linear in $g$ is already the tree interaction. A loop renormalizing that cubic term requires additional interaction insertions. Such a dependence is excluded by the holomorphic charge constraints; dependence on $g^*$ cannot repair it in a <superpotential>. Matching to the specified tree action fixes
$$
f(z)=\frac12+\frac13z,\qquad
\boxed{\mathcal F_{\mathrm{pert}}=\frac12m\Phi^2+\frac13g\Phi^3.}
$$
The $m=0$ limit is taken in the final polynomial, not by evaluating the intermediate ratio $g\Phi/m$. \b[The holomorphic Wilsonian superpotential and its parameters $m,g$ receive no independent perturbative vertex renormalization.] Symmetries alone would only give the arbitrary function $f$; regularity and the free/tree matching are necessary to reach the stronger conclusion.
The <Kähler potential> is not protected by that theorem. In particular a corrected <kinetic term> $Z(\mu)\Phi^\dagger\Phi$ leads to <wave-function renormalization>. Writing $\Phi_c=Z^{1/2}\Phi$ in canonical normalization gives
$$
\boxed{m_c(\mu)=\frac{m}{Z(\mu)},\qquad
g_c(\mu)=\frac{g}{Z(\mu)^{3/2}}.}
$$
These <holomorphic and canonically normalized superpotential couplings> distinguish the two senses of “renormalized”: the physical/canonically normalized parameters can run, entirely through the common field normalization, even when the holomorphic coefficients do not. The scalar effective potential can consequently receive quantum corrections through the <Kähler potential>. Nor does the local perturbative statement automatically apply to infrared-singular one-particle-irreducible actions or to integrating out whole massive fields. \b[Non-renormalization protects the local holomorphic F-term, not the complete quantum action or every physically normalized mass and coupling.]
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