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Adding a holomorphic function $f$ and its conjugate does not change the Kähler metric. In global supersymmetry, these terms integrate to zero in the D-term action. In Planck-unit supergravity, the simultaneous transformation $W\mapsto e^{-f}W$ leaves the scalar potential invariant; shifting $K$ while holding $W$ fixed generally does not.

Use the canonical Kähler potential and normalize superspace integration by $\int d^2\theta {\theta }^2=1$. The superspace Lagrangian density isThe full superspace integration produces a D-term, while the chiral superfield integral produces an F-term. Work in chiral coordinates for the chiral-superfield component expansion, so the omitted spacetime-derivative terms do not enter the extraction of $F$.

Set $\delta \Phi =\sqrt{2}\theta \psi +{\theta }^{2}F$. A Taylor expansion of the superpotential givesHigher terms vanish because there are only two Grassmann variables ${\theta }^{\alpha }$. With $\theta \psi ={\theta }^{\alpha }{\psi }_{\alpha }$ and $\psi \psi ={\psi }^{\alpha }{\psi }_{\alpha }$, their anticommutation gives $(\theta \psi {)}^2=-\frac{1}{2}{\theta }^2\psi \psi$. ThusThe canonical Kähler potential contributes $F^{\ast }F$. Consequently the part of the Lagrangian density containing the auxiliary field isThe fermion terms coming from the superpotential areThe Weyl spinor kinetic term and the complex scalar field kinetic term are canonical because $K={\Phi }^{†}\Phi$.

Treat $F$ and $F^{\ast }$ as independent variables in their algebraic Euler-Lagrange equations. They giveEquivalently,After eliminating the auxiliary field, the Lagrangian density contains $-V$, so the F-term scalar potential isIt is nonnegative for arbitrary complex $m,g,\phi$, since it is a modulus squared. A supersymmetric vacuum satisfies F-flatness, so its zeros are:If $m=0$ and $g\ne 0$ these coincide; if $g=0$ and $m\ne 0$ only the origin remains; if $m=g=0$ every constant $\phi$ is a supersymmetric vacuum.

To compare masses, expand around a supersymmetric vacuum $v$ satisfying $W^{\prime }(v)=0$, writing $\phi =v+\eta$. With $M=W^{\prime \prime }(v)=m+2gv$,Writing $\eta =(A+iB)/\sqrt{2}$ gives the scalar field mass term $-\frac{1}{2}|M{|}^2(A^2+B^2)$; the Weyl spinor has physical mass $|M|$. Therefore the supersymmetric mass degeneracy isIn particular, at $v=0$ the boson and fermion masses are both $|m|$. At the second supersymmetric vacuum $v=-m/g$ the sign of $M$ reverses, but the physical mass is again $|m|$. Equality here concerns fluctuations about a supersymmetric vacuum, not arbitrary backgrounds with $W^{\prime }(v)\ne 0$.

In the convention $V\supset \lambda |\phi {|}^4$ and ${\mathcal{L}}_{\mathrm{Yukawa}}=-y\phi \psi \psi +h.c.$, the same superpotential coefficient gives $\lambda =|g{|}^2$ and $y=g$. The supersymmetric relation between quartic and Yukawa couplings is thereforeIf instead the Yukawa interaction is written $-\frac{1}{2}{y}_3\phi \psi \psi +h.c.$, then $y_3=2g$ and the same physical relation is $\lambda =|y_3{|}^2/4$. The numerical factor depends on the definition of the Yukawa coupling.

The perturbative non-renormalization theorem applies to the local holomorphic superpotential in the Wilsonian effective action, using regularization in quantum field theory that preserves supersymmetry. There are two complementary ways to see why it applies here.

First, in supergraph calculations in perturbative quantum field theory, the algebra of supersymmetric covariant derivatives puts any candidate local correction to the superpotential into a full $d^4\theta$ superspace integration. It therefore corrects a D-term, such as the Kähler potential, rather than a local chiral $d^2\theta$ F-term. Turning such a contribution into an apparent chiral integral would require nonlocal inverse spacetime derivatives, schematically $1/\square$. These are excluded from the local Wilsonian effective action with a fixed nonzero momentum cutoff. This gives the diagrammatic non-renormalization theorem.

Second, the holomorphy argument for superpotential non-renormalization treats $m$ and $g$ as background chiral spurions, so the Wilsonian effective action superpotential is a holomorphic function of $\Phi ,m,g$, with no dependence on $m^{\ast }$ or $g^{\ast }$. Assign an ordinary $U(1)$ spurion charge and an R-charge as follows:The superpotential has ordinary charge zero and R-charge two, since $d^2\theta$ has R-charge minus two. A prospective holomorphic monomial $m^a{g}^b{\Phi }^c$ must therefore satisfyFor perturbative corrections, regularity at $m=0$ and $g=0$ requires nonnegative powers of the couplings. Regularity in $m$ is justified by keeping a nonzero Wilsonian momentum cutoff and retaining $\Phi$ as a light field. Hence only $c=2$ and $c=3$ survive: the existing $m{\Phi }^2$ and $g{\Phi }^3$ structures. The quadratic coefficient is fixed by the free theory at $g=0$, where there are no interaction loops. The cubic coefficient is fixed at first order in $g$, where a connected three-field interaction has only its tree-level Feynman diagram; loop corrections would require further powers of the interaction couplings, which the holomorphy argument for superpotential non-renormalization charge constraints forbid. ThusA field-independent constant is physically irrelevant in this rigid model; the spurion charges would in any case require $m^3/g^2$, which is not regular at $g=0$.

The Kähler potential can still receive wave-function renormalization. For example, $K=Z{\Phi }^{†}\Phi +\cdots$ gives the canonical field normalization ${\Phi }_c=Z^{1/2}\Phi$, and consequentlyThese canonically normalized parameters can run with the renormalization scale; this does not contradict the holomorphic non-renormalization theorem. When massless modes are included in the full quantum effective action, nonlocal terms associated with infrared divergences can also imitate an F-term. Distinguishing that action from the local Wilsonian effective action is part of the theorem's convention.

Superspace integration uses the Berezin integral over Grassmann variables. A chiral $d^2\theta$ integral extracts an F-term, while a full $d^4\theta$ integral extracts a D-term.