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For $\varphi =\kappa (v+h+ib)$, the real fluctuation kinetic terms are $-{\kappa }^2[(\partial h{)}^2+(\partial b{)}^2]$. The same $v$ labels a Minimum of $V=\lambda (|\varphi {|}^2-v^2/2{)}^2/4$ and gives canonically normalized real fields exactly when ${\kappa }^2=1/2$. This condition also makes $G=\partial \cdot A-\xi gvb$ cancel the gauge-scalar derivative mixing: its coefficient is $(2{\kappa }^2-1)gvA\cdot \partial b$. Changing the normalization of the fields without changing the gauge function need not be a conventional diagonal $R_{\xi }$ gauge.

For a real scalar field with canonical field normalization, impose $[\dot{\varphi }(t,\mathbf{x}),\varphi (t,\mathbf{y})]=-i{\delta }^3(\mathbf{x}-\mathbf{y})$. The Källén–Lehmann spectral representation also represents its vacuum commutator as the weighted sum of free commutators. The equal-time derivative of each free commutator is $-i{\delta }^3(\mathbf{x}-\mathbf{y})$ independently of mass. Comparison therefore gives the displayed sum rule, whenever the canonical commutator and spectral integral admit this distributional limit. Since the measure is nonnegative, an isolated one-particle weight obeys $0\le Z\le 1$, and is strictly positive if the field couples to that particle. Arbitrary rescaling of the field rescales the total spectral weight, so positivity alone does not imply this normalization or upper bound.

For the ghost coupling in a linear Cartesian Abelian Higgs gauge, the common factor in the scalar parametrization cancels from the gauge transformation. Separating real and imaginary parts givesConsequentlyFor the variables and gauge function actually printed, the ghost mass and Higgs-ghost vertex are thereforeFor the usual nonnegative gauge parameter, $m_c=\sqrt{\xi }g|v|$. The momentum-space ghost propagator is $-i/(k^2+m_c^2-i0)$ in this kinetic convention. There is no direct ghost-photon vertex or ghost-$b$ vertex. At $\xi =0$ the ghost becomes massless and decouples.

The scalar prefactor $1/4$ is incompatible with the stated canonical broken-vacuum interpretation. At $h=b=0$ it gives $|\varphi {|}^2=v^2/16$, whereas the potential is minimized at $|\varphi {|}^2=v^2/2$. In fact the printed parametrization givesIts $h,b$ kinetic terms have coefficient $-1/16$, not $-1/2$, and its vector mass parameter is $g^2{v}^2/8$. The given gauge function consequently does not cancel the vector-scalar mixing in those literal variables. The ghost operator computed above nevertheless follows algebraically from that gauge function without changing the printed fields.

With a canonical Abelian Higgs shift $\varphi =(v+h+ib)/\sqrt{2}$, the expansion is about a true minimum, the vector mass is $m_A=g|v|$, the gauge fixing cancels the mixing, and the same ghost formulas acquire the usual interpretation $m_c^2=\xi m_A^2$. For the literal $1/4$ parametrization, a canonical fluctuation is $h_{\mathrm{can}}=h/\sqrt{8}$, so its ghost vertex would instead be $-i\sqrt{8}\xi g^{2}v$; the non-vacuum background problem still remains. These two interpretations must not be silently mixed.

Since the hypercharge is zero, electric neutrality means $T^{3}v=0$. Choose the real-triplet vacuum $v_i=v{\delta }_{i3}$, with $v^2=m^2/(4\lambda )$. With canonical real-scalar kinetic density $-\frac{1}{2}(D_{\mu }\Phi {)}^T{D}^{\mu }\Phi$, insert this vacuum into the covariant derivative. The hypercharge field $B_{\mu }$ has no coupling to this scalar, while only $T^{1}v,T^{2}v$ are nonzero. ThusThe charged combinations $W^{\pm }=(W^1\mp i{W}^2)/\sqrt{2}$ haveBoth $T^3$ and the independent hypercharge generator annihilate the vacuum. The connected unbroken subgroup is thereforeThe electromagnetic generator $Q=T^3+Y/2$ belongs to it, but it is only one linear combination of two surviving neutral generators. This model has two massless neutral gauge fields, rather than the single massless photon and massive neutral vector of the usual scalar-doublet electroweak model. A neutral-field rotation does not create a massive eigenvalue from this zero block.

The two transverse triplet components are eaten, leaving one real, electrically neutral radial scalar $H={\varphi }_3-v$. The radial eigenvalue of the Hessian found in part i isIndeed the radial potential is a constant plus $4\lambda v^2{H}^2+4\lambda v{H}^3+\lambda H^4$, fixing the factor of two between its quadratic coefficient and its mass squared. The physical degree-of-freedom count is preserved: three real scalars plus four massless vectors give $3+4\times 2=11$, while one scalar, two massive vectors and two massless vectors give $1+2\times 3+2\times 2=11$.

Use signature $(+---)$ and covariant one-particle state normalization $⟨p^{\prime }|p⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-{\mathbf{p}}^{\prime })$. The future scattering condition uses the outgoing asymptotic scalar field:As with the incoming condition, this is understood in suitably smeared scattering matrix elements, not as unrestricted pointwise strong convergence of interacting and free local operators.

The free incoming field has expansionThe creation operator defines $|p⟩=a_{\mathrm{in}}^{†}(p)|0⟩$, with $a_{\mathrm{in}}|0⟩=0$, so $⟨p|{\varphi }_{\mathrm{in}}(x)|0⟩=e^{ipx}$. On the other hand, translation invariance givesMatching the stable one-particle state component in the asymptotic condition fixes the constant overlap. Choose its state phase so that it is real and nonnegative. ThenThe same stable particle appears in the outgoing description, with compatible one-particle phases. Thus $Z$ is the squared overlap measured by wave-function renormalization.

Insert a complete set of four-momentum eigenstates between the fields. For $x^0>0$, translation invariance and Hermiticity giveFor $x^0<0$, reversing the time-ordered product gives the same sum with $e^{i{p}_{\alpha }x}$. Therefore, with continuous-state integrals implicit in the completeness sum,Here $\theta$ is the Heaviside step function.

For the usual spectral representation, take a Lorentz-invariant vacuum, a positive-norm Hilbert space and the future-cone spectrum condition. Also take $⟨0|\varphi |0⟩=0$, as for the unbroken real scalar field, or subtract this constant and use the connected correlation function. Otherwise the vacuum intermediate state adds a disconnected constant not represented by the ordinary mass-shell measure below.

Let $W(x)=⟨0|\varphi (x)\varphi (0)|0⟩$ for that centered field, and use the Fourier transform convention $W(x)=\int d^{4}p(2\pi {)}^{-4}{e}^{-ipx}\tilde{W}(p)$. Define the Källén–Lehmann spectral density by the distributional equalityLorentz invariance makes the weight on each future mass shell a function or measure of $p^2$; the spectrum condition gives $p^2\ge 0$. Its nonnegativity follows from the displayed squared matrix elements. ThusApplying time ordering to this expression mass shell by mass shell proves the Källén–Lehmann spectral representation:The factor $i$ is outside ${\Delta }_F$ under the convention in the question.

The stable one-particle state part of completeness is $\int d^3\mathbf{p}[(2\pi {)}^{3}2E_p{]}^{-1}|p⟩⟨p|$. Its overlap is $\sqrt{Z}$, so its contribution to $W$ is $Z$ times the free mass-$m$ two-point function. HenceOther isolated particles as well as multiparticle states can belong to ${\rho }_{\mathrm{rest}}$. The full scalar propagator accordingly has isolated pole $Z/(p^2-m^2+i0)$, making $Z$ its pole residue.

Finally use canonical field normalization: the canonical momentum is $\dot{\varphi }$ and the equal-time canonical commutation relation is $[\dot{\varphi }(t,\mathbf{x}),\varphi (t,\mathbf{y})]=-i{\delta }^3(\mathbf{x}-\mathbf{y})$. The vacuum commutator $C(x)=W(x)-W(-x)$ has the same spectral decomposition. For each free mass shell,independently of $\sigma$. Comparing with the interacting equal-time commutator gives the canonical scalar spectral sum rule:For the particle genuinely interpolated by the stated asymptotic field, $Z>0$; the general nonnegative bound allows a field with zero overlap. The total weight and the upper bound require this canonical normalization and a well-defined distributional equal-time limit. They do not follow from the asymptotic condition alone: rescaling $\varphi$ rescales both $Z$ and the entire spectral measure.

After integrating out the shell, let a chosen operator's coefficient be $\kappa +\Delta \kappa$ and write the two-derivative quadratic term as $\frac{1}{2}(1+\Delta Z)(\partial \varphi {)}^2$. Here $\Delta \kappa$ is the shell-induced change before rescaling, and $\Delta Z$ is the shell correction to the wavefunction renormalization. Work in an operator basis in which that quadratic term has this form; the other generated operators remain in the Wilsonian effective action.

Under $x^{\prime }=bx$, the volume element changes by $d^{4}x=b^{-4}{d}^4{x}^{\prime }$ and each derivative by ${\partial }_x=b{\partial }_{x^{\prime }}$. Canonical field normalization is restored by definingIndeed, the volume factor, two derivatives and two field factors cancel in the kinetic term. A term with $n$ fields and $m$ derivatives consequently obtains the factor $b^{-4}{b}^m{b}^n(1+\Delta Z{)}^{-n/2}$. HenceThis is Wilsonian rescaling of a scalar coupling. The exponent is minus the coupling's engineering mass dimension, $[\kappa ]=4-n-m$. The rescaling also returns the low-momentum cutoff to $\Lambda$, because $p^{\prime }=p/b$.

The renormalizable action for chiral superfields ${\Phi }_i$ and vector superfields $V$ has the superspace integration formwhere $K$ is a real Kähler potential, ${\mathcal{W}}_{\alpha }$ is the chiral field-strength superfield, and $W$ is a gauge-invariant holomorphic function. For the renormalizable theory choose canonical positive kinetic terms, a polynomial $W$ of degree at most three, and field-independent gauge kinetic coefficients $f_{ab}$. We prove the non-renormalization theorem for the local Wilsonian effective action, with a nonzero infrared cutoff and a regulator preserving supersymmetry. The elementary chiral superfields are retained; this is not a claim that eliminating whole massive fields leaves the same polynomial.

Write $W={\sum }_A{\lambda }_A{\mathcal{O}}_A(\Phi )$, treating masses and interaction coefficients alike as background chiral spurions. A local F-term is chiral even when these backgrounds vary, so its coefficient depends holomorphically on $\Phi ,{\lambda }_A$ and the holomorphic gauge couplings. Dependence on ${\bar{\lambda }}_A$ would not be chiral. Such dependence can enter D-terms or terms with supercovariant derivatives instead. With the infrared cutoff kept fixed, the perturbative local coefficients are regular power series around vanishing masses and interactions. This removes the inverse-mass and infrared singular terms that would spoil the following spurion selection rule for perturbative non-renormalization.

Assign every elementary ${\Phi }_i$ R-charge zero, every ${\lambda }_A$ R-charge two, and $\theta$ R-charge one. The superpotential must have R-charge two because $d^2\theta$ has R-charge minus two. Holomorphy excludes the conjugate spurions, so regularity and this formal R-symmetry require exactly one factor of a superpotential spurion in any perturbative candidate. Thus $W_{\mathrm{eff}}$ is linear in the ${\lambda }_A$, with the ordinary flavor spurion charges fixing the permitted field monomials. An anomalous formal R-symmetry is implemented by also transforming the holomorphic gauge coupling; its anomaly must not simply be ignored.

Gauge interactions cannot provide a perturbative coefficient modifying this conclusion. With $\tau =\vartheta /(2\pi )+4\pi i/g^2$, perturbative diagrams are independent of the topological angle $\vartheta$. A holomorphic coefficient invariant under continuous real shifts of $\tau$ is constant, by the Cauchy-Riemann equations. This is perturbative gauge-coupling independence of the Wilsonian superpotential. Contributions involving $e^{2\pi i\tau }$ are nonperturbative and are outside this argument. For an Abelian factor the same angle-independent holomorphy argument applies. Hence a putative perturbative coefficient linear in ${\lambda }_A$ is independent of the gauge coupling and can be evaluated as that coupling tends to zero.

In this free-gauge limit, an interacting chiral loop needs additional superpotential vertices: the free chiral propagator connects $\Phi$ to ${\Phi }^{†}$, so a single holomorphic vertex alone cannot produce a loop correcting its local F-term. Extra vertices introduce extra spurions or their conjugates, both excluded by the preceding charge and holomorphy argument. The surviving coefficient is its tree value. Therefore

For an explicit charge check, take $W=m{\Phi }^2/2+g{\Phi }^3/3$ in the Wess–Zumino model. Give $(\Phi ,m,g)$ ordinary charges $(1,-2,-3)$ and R-charges $(1,0,-1)$. A holomorphic monomial $m^a{g}^b{\Phi }^n$ of the required charges satisfies $n-2a-3b=0$ and $n-b=2$, giving $a=1-b$, $n=b+2$. Perturbative regularity requires $a,b\ge 0$, hence only $(a,b,n)=(1,0,2)$ and $(0,1,3)$ survive. The free limit fixes their coefficients to $1/2$ and $1/3$; there is no room for a coupling-dependent loop correction. This illustrates the holomorphy argument for superpotential non-renormalization explicitly.

The Kähler potential can acquire wave-function renormalization. Restoring canonical field normalization changes physical masses and Yukawa couplings, so their running does not contradict the boxed statement about holomorphic coordinates. Gauge kinetic F-terms have their own renormalization and are not the superpotential. Infrared effects in the one-particle-irreducible quantum effective action, tree-level elimination of retained fields, and genuine nonperturbative effects must likewise be distinguished from this perturbative Wilsonian effective action theorem.

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.