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Take canonical kinetic terms, gauge coupling $g>0$, and define the Fayet–Iliopoulos term by $+\xi D$. If $\varphi$ is the scalar component of the charge-$q$ chiral superfield, the relevant auxiliary field Lagrangian isThe algebraic equation eliminates $D$ and gives the D-term scalar potentialSince the superpotential vanishes, $F=0$. In a homogeneous vacuum, the supersymmetry transformation of the gaugino includesThus $D_0\ne 0$ gives an inhomogeneous shift: no nonzero constant supersymmetry parameter leaves this vacuum invariant. The gaugino $\lambda$ is the goldstino, because it is the field shifting under the broken supercharge; there is no $F$ contribution here.

For $q\ne 0$, minimizing over $x=|\varphi {|}^2\ge 0$ separates three cases. If $q\xi <0$, the allowed value $x=-\xi /(gq)$ gives $D_0=V_0=0$: supersymmetry is unbroken while the internal gauge symmetry is Higgsed. If $q\xi >0$, the unconstrained zero of $D$ would have negative $x$, so the minimum is $\varphi =0$ with $D_0=-\xi$ and $V_0={\xi }^2/2$. If $\xi =0$, the minimum $\varphi =0$ is supersymmetric. Consequently, with this explicitly stated sign convention,Reversing the sign used to define the Fayet–Iliopoulos term reverses the printed sign criterion, not the physics. If $q=0$, the scalar is decoupled and flat: any $\xi \ne 0$ still breaks supersymmetry through $D_0=-\xi$, but the charged-field criterion does not apply.

On the broken branch, expand about $\varphi =0$:Writing $\varphi =({\varphi }_1+i{\varphi }_2)/\sqrt{2}$ gives both real scalar modes the squared mass $gq\xi$. There is no superpotential fermion mass and no vacuum expectation value to induce gaugino–matter-fermion mixing. The Abelian gauge symmetry is unbroken, so its vector is also massless. The complete mass spectrum of single charged-field D-term breaking isThe original chiral multiplet therefore has the squared-mass splittingThe original vector multiplet has zero mass splitting, although the nonzero $D_0$ means it is not an unbroken-supersymmetry representation. As a check, the tree-level supertrace mass sum rule gives $\mathrm{Str}{M}^2=2gq\xi$ when the two real scalar degrees of freedom are counted. On the $q=0$ broken branch all the propagating fields are massless.

There is a necessary scope qualification: one nonzero-charge Weyl fermion has an Abelian cubic gauge anomaly proportional to $q^3$ and a mixed gravitational gauge anomaly proportional to $q$. These are the classical vacuum and spectrum calculations requested by the model; a consistent quantum gauge theory needs an anomaly-cancelling completion, whose additional matter must also be included in its vacuum analysis.

Keep the PDF's hypercharge signs, which are the negatives of the frequently used convention. Write $U,D,E,N$ for the displayed $U^R,D^R,E^R,N^R$ chiral superfields. Their gauge group representations are conjugate right-handed-matter representations; their Weyl spinor components are left-chiral charge conjugates. Thus assign baryon number $B(Q)=\frac{1}{3}$, $B(U)=B(D)=-\frac{1}{3}$, and lepton number $L(L)=1$, $L(E)=L(N)=-1$, with both Higgs fields neutral under $B,L$. The corresponding electromagnetic convention is $Q_{\mathrm{em}}=T_3-Y$.

Let $X\cdot Y={ϵ}_{ab}{X}^a{Y}^b$ contract the two weak doublets, and use the color Levi-Civita symbol for three color antitriplets. The most general homogeneous cubic superpotential with right-handed-neutrino superfields separates as $W_3=W_{B,L}+W_{/B\text{or}/L}$, whereandRepeated generation and color indices are summed. The four Yukawa coupling matrices are arbitrary. The weak and color contractions imply ${\lambda }_{ijk}=-{\lambda }_{jik}$ and ${\lambda }_{ijk}^{\prime \prime }=-{\lambda }_{ikj}^{\prime \prime }$, while ${\rho }_{ijk}$ is completely symmetric. The $\lambda ,{\lambda }^{\prime }$ terms have $\Delta L=1$; ${\lambda }^{\prime \prime }$ has $\Delta B=-1$; $\kappa$ and $\rho$ have respectively $\Delta L=-1,-3$. Their conjugate interactions have the opposite changes. All four Yukawa terms preserve both numbers; all five remaining types violate at least one. In particular, the neutral $N$ fields printed in the PDF require both $N{H}_1{H}_2$ and $NNN$ terms.

For completeness of this list, a cubic color gauge-invariant can contain a $Q$–$U$ or $Q$–$D$ pair and one colorless doublet, or three color antitriplets. Vanishing hypercharge gives $Q{H}_{2}U$, $Q{H}_{1}D$, $LQD$ and $UDD$. Three $Q$ doublets have no weak singlet. Among colorless terms, two doublets and a singlet give $LLE$, $L{H}_{1}E$, $L{H}_{2}N$ and $H_1{H}_{2}N$; the apparent $H_1{H}_{1}E$ contraction vanishes since the identical bosonic superfields commute. Three singlets yield only $NNN$. This also explains why no omitted cubic monomial is available. The word cubic excludes the linear $N$ term and the quadratic $H_1{H}_2$, $L_i{H}_2$, $N_i{N}_j$ terms; these are separate possible terms in the full renormalizable superpotential.

To demonstrate squark-mediated proton decay, take ${\lambda }_{11k}^{\prime }$ and ${\lambda }_{11k}^{\prime \prime }$ with $k=2$ or $3$. Identical down-flavor indices would make ${\lambda }_{111}^{\prime \prime }$ vanish. Both interactions couple to the same scalar ${\tilde{D}}_k$ of mass $M$. Suppressing overall convention-dependent vertex signs, their relevant two-component fermion bilinears areThe color index on $A$ is contracted with that on ${\tilde{D}}_k$; Lorentz spinor indices within each bilinear are contracted too. Together with $-M^2|{\tilde{D}}_k{|}^2$, its algebraic equation at momenta small compared with $M$ gives ${\tilde{D}}_k=-A^{†}/M^2$. Substitution gives $A^{†}A/M^2$, whose cross term is the four-fermion matching for squark-mediated proton decayIts electron term contains three quark fields of flavors $uud$ and a lepton, and changes baryon number and lepton number together. The Hermitian-conjugate terms describe both orientations of the process. The hadronic matrix element of the three-quark proton-decay operator between a proton and a neutral pion has the right quantum numbers and is not forbidden by a remaining symmetry. Hence the operator permits $p\to e^{+}{\pi }^0$: the lepton field creates the positron and the three-quark matrix element connects the proton to the pion. This supplies the mechanism rather than merely naming the two violating interactions.

Each fermion field has mass dimension $\frac{3}{2}$, so this effective operator has dimension six and coefficient $C_6={\lambda }^{\prime }{\lambda }^{\prime \prime }{}^{\ast }/M^2$. By dimensional analysis its contribution to the decay rate has the formHere $c_h$ collects phase space, running and the dimensionless hadronic factor; a dimensional estimate takes it of order unity rather than pretending to calculate a strong-interaction matrix element. Assuming this exchange dominates and there is no tuned cancellation, the supplied lifetime gives the proton-lifetime bound on a product of R-parity-violating couplingsThe last estimate uses $m_p\simeq 1\mathrm{GeV}$ and ${\tau }_{\min }=2.4\times 10^{64}{\mathrm{GeV}}^{-1}$. Keeping $m_p=0.938\mathrm{GeV}$ instead changes $6.5$ to about $7.6$. A numerical limit independent of the squark mass or hadronic factor is not determined by the supplied data.

Finally, matter parity is $P_M=(-1{)}^{3(B-L)}$: all matter chiral superfields, including $N$, are odd, and the Higgs chiral superfields are even. R-parity of a component field is $R_p=P_M(-1{)}^{2s}$. In a superpotential, the spin factor is irrelevant to the allowed monomials: each of $Q{H}_{2}U,Q{H}_{1}D,L{H}_{1}E,L{H}_{2}N$ has two odd matter factors and is even. Each of $LLE,LQD,UDD,NNN$ has three odd matter factors, and $N{H}_1{H}_2$ has one, so all are odd and forbidden. Higgs vacuum expectation values are even; the allowed Yukawa interactions therefore still generate ordinary fermion masses.

R-parity forbids all the baryon/lepton-number-violating cubic terms above, not every possible violating operator. In particular, $\frac{1}{2}{M}_{ij}{N}_i{N}_j$ is even under matter parity but changes lepton number by two: matter parity allows Majorana neutrino masses. Thus the literal unrestricted claim would be false for these very fields; its valid cubic interpretation gives precisely the requested exclusion while preserving the Yukawa coupling mass terms.

Use a real scalar field and Minkowski action $S[\varphi ]$, with the vacuum selected by a Feynman propagator pole prescription or adiabatic vacuum boundary conditions. The normalized correlation functions areHere $|\Omega ⟩$ is the interacting vacuum and $\mathcal{T}$ denotes time ordering. These are the canonical vacuum expectation values, rather than unordered products of noncommuting fields. The functional integral formula assumes a regulated measure and the same vacuum prescription on both sides.

Couple an external source $J$ linearly to the scalar field and choose the normalized vacuum generating functionalThen functional derivatives insert the fields:In the convention used below, the connected generating functional isThe logarithm removes products of disconnected components. For example the connected correlation function $G_2^c=G_2-G_1{G}_1$; at three points one subtracts all three $G_2{G}_1$ products and adds $2G_1{G}_1{G}_1$. Normalization has already removed vacuum bubbles.

Assume $\lambda >0$ and $v>0$. The minima of the scalar potential have $|\varphi |=v$. Choose the vacuum expectation value ${\varphi }_0=v{e}_3$. Its stabilizer subgroup consists exactly of matrices $\mathrm{diag}(R_2,1)$ with $R_2\in O(2)$, soThe continuous spontaneous symmetry breaking is $SO(3)\to SO(2)$ and has two broken generators. For the full orthogonal group, the cross-product notation requires the gauge connection to transform in the adjoint, as an axial vector: $R[A{]}_{\times }{R}^{-1}=[(\det R)RA{]}_{\times }$. The scalar field transforms as the ordinary vector $R\varphi$. This axial transformation of orthogonal gauge connections makes the stated covariant derivative transform correctly even for $\det R=-1$; assigning an ordinary-vector transformation to the connection would fail.

Near this nonzero vacuum, unitary gauge removes the two angular Goldstone bosons and gives $\varphi =(0,0,v+h)$. PutThese are the physical charged gauge bosons, neutral massless gauge boson, and radial scalar field. Directly substituting into the original gauge field strengths givesThe scalar field kinetic term has no radial-vector cross term, because ${\partial }_{\mu }\varphi$ and ${\mathbf{A}}_{\mu }\times \varphi$ lie in perpendicular internal directions. The complete physical gauge-scalar Lagrangian becomesFor each real vector the mass term is $m^2{A}_{\mu }{A}^{\mu }/2$, while for the complex pair it is $m_W^2{W}_{\mu }^{+}{W}^{-\mu }$. Hence the adjoint triplet Higgs spectrum isThe radial scalar also has $m_h^2=\lambda v^2$ in the question's potential normalization. The two lost scalar modes supply the longitudinal polarizations of the massive vectors through the Higgs mechanism.

For the lepton kinetic terms let $P_L=(1-{\gamma }^5)/2$, $P_R=(1+{\gamma }^5)/2$, and denote the left neutral triplet entry by $n_L=c_{\alpha }{\nu }_L+s_{\alpha }{N}_L$. The missing singlet is the perpendicular combinationIts overall sign is immaterial. The real orthogonal matrix $\left(\begin{array}{cc}{c}_{\alpha }& s_{\alpha }\\ -s_{\alpha }& c_{\alpha }\end{array}\right)$ is orthogonal, so the sum of the kinetic terms of $n_L$ and $L^S$ is ${\bar{\nu }}_{L}i/\partial {\nu }_L+{\bar{N}}_{L}i/\partial N_L$ with no mixed terms. Combining with the right triplet gives ordinary Dirac field kinetic terms for $E^{+},N,e$, and only a left-handed kinetic term for the massless neutrino. This is orthogonal completion of mixed neutral lepton kinetic terms. The heavy charged field is $E^{+}$, as printed in the PDF, not the TeX's $N^{+}$.

To compute the charge commutator, the canonical anticommutation relations implyThe quartic terms cancel on reordering the fermions. Thus well-defined normal-ordered integrated bilinears obey fermionic bilinear charge algebra, $[Q_C,Q_D]=Q_{[C,D]}$. Left/right mixed commutators vanish since $P_L{P}_R=0$.

In either triplet, writeThe displayed current has a factor two relative to the projected bilinears, since $1\mp {\gamma }^5=2P_{L,R}$. Consequently its charges are $T^{+}=2(Q_K^L+Q_K^R)$ and $T^{-}=(T^{+}{)}^{†}$. The neutral combinations have canonical anticommutators because $c_{\alpha }^2+s_{\alpha }^2=1$. Applying the matrix commutator giveswhere $Q_{\ell }=e\int :(E^{+†}{E}^{+}-e^{†}e):$ is the electromagnetic charge in the lepton sector. The neutrino, heavy neutral lepton and singlet contribute zero. This establishes triplet charged-current closure on electromagnetic charge, with the requested proportionality independent of normalization choices for the weak Noether charges. The PDF correctly writes $J^{-}=(J^{+}{)}^{†}$; the TeX repeats $J^{+}$ incorrectly. Equal-time algebra does not require conservation of the separate fermionic weak currents after symmetry breaking. For full gauge-theory charges, the corresponding charged-boson contributions must also be included.

Work with canonically normalized real components of the scalar field; a complex representation can first be written in terms of its real and imaginary components. At a vacuum expectation value ${\phi }_0$, expansion of the scalar potential givesThe linear term vanishes by stationarity. Thus the scalar mass matrix is the Hessian matrix of the scalar potential. Invariance under each Lie algebra generator gives the identity ${\partial }_{r}V(\phi )(i{T}_a\phi {)}_r=0$. Differentiating this identity with respect to ${\phi }_s$ and evaluating at the stationary point removes the term containing ${\partial }_{r}V$, leavingThe kernel of the map $T\mapsto iT{\phi }_0$ is precisely the Lie algebra of the stabilizer subgroup H. Consequently its restriction to a complementary space of broken Lie algebra generators is injective: a linear combination that annihilates ${\phi }_0$ would belong both to the stabilizer algebra and its complement, and hence would vanish. There are therefore $\dim G-\dim H$ independent Goldstone directions in the scalar mass matrix, tangent to the vacuum manifold. There are $\dim G-\dim H$ symmetry-protected massless scalars.

This Goldstone theorem guarantees at least that many zero quadratic masses. The claim that these are all the massless scalars additionally requires nonzero curvature transverse to the vacuum orbit. For example, $V=\kappa ({\phi }^T\phi -v^2{)}^4$, with $\kappa >0$, has an ordinary spherical vacuum manifold but its entire Hessian matrix vanishes there. These accidental massless radial modes beyond Goldstone modes are not excluded by the hypotheses. For the indicated orthogonal group breaking, the protected number is

For the Higgs doublet, put $r=\sqrt{2{\varphi }^{†}\varphi }=v+H$ and $z=\sqrt{2}\varphi /r$ wherever $r\ne 0$. If $z=(z_1,z_2{)}^T$, thenThis explicitly exhibits the three angular coordinates as an SU(2) transformation of a real radial field. In particular, the polar coordinates for the electroweak Higgs doublet can locally be chosen in the formAt the vacuum the three variations generated by $-i{\sigma }_1$, $-i{\sigma }_2$, and $i(I-{\sigma }_3)$ acting on $(0,1{)}^T$ are independent over the reals. They span the three broken directions; the remaining generator $Q=({\sigma }_3+I)/2$ annihilates the vacuum. This proves that the exponential supplies local angular coordinates. An inverse local gauge transformation, applied also to the gauge fields, removes these three fields in unitary gauge. They supply the longitudinal polarizations of the massive W bosons and Z boson through the Higgs mechanism. The parameterization is local about a nonzero vacuum, rather than a coordinate system through every possible zero of the Higgs doublet.

Use the metric $(+---)$ and define the charged W bosons and neutral gauge fields bywhere $s_W=\sin {\theta }_W$, $c_W=\cos {\theta }_W$, and $\tan {\theta }_W=g^{\prime }/g$. Direct substitution into the gauge-covariant kinetic term givesand henceThe electroweak doublet gauge-boson mass matrix in the $(A^3,B)$ basis isIts normalized zero direction is $(s_W,c_W)$ and its massive direction is $(c_W,-s_W)$. Thus the scalar-vector terms in physical fields arewith the tree-level electroweak gauge-boson massesThese are also charge eigenstates. The unbroken U(1) gauge symmetry has generator $Q=T_3+Y$; on the Higgs doublet, $Y=I/2$. Since $[Q,T_{\pm }]=\pm T_{\pm }$, the combinations $W^{\pm }$ have charges $\pm 1$, while Z, the photon, and H are neutral. The photon disappears from the displayed scalar-vector terms because $g{A}^3-g^{\prime }B=\sqrt{g^2+g^{\prime 2}}Z$. The photon is massless and has no direct tree-level coupling to the remaining radial scalar in this sector. This does not remove its interactions with charged W bosons, or exclude a scalar-to-photon interaction generated by loops.

Finally, retain the printed normalization of the Higgs field potential:The kinetic term is $\frac{1}{2}(\partial H{)}^2$, so the Higgs radial mass for a half-normalized quartic potential is $\boxed{m_H^2=\lambda v^2}$. The extra factor of two familiar from a potential with prefactor $\lambda$, rather than $\lambda /2$, does not apply here.

A useful model of spontaneous breaking of a Z2 scalar symmetry is the four-dimensional real scalar theoryThe action is invariant under the Z2 symmetry $\varphi \mapsto -\varphi$. Its classical vacuum manifold consists of the two values $\varphi =\pm v$. In the broken quantum phase, choose a pure vacuum with $⟨\varphi ⟩=v$; the symmetry maps it to a different vacuum with opposite expectation value, although it leaves the action unchanged. To distinguish quantum symmetry breaking from simply minimizing a classical potential, introduce a small source $J\varphi$ and take the infinite-volume limit before $J\to 0^{+}$. A finite-volume symmetry-preserving ground state can remain an even superposition because of tunneling, whereas the selected infinite-volume pure vacua have nonzero vacuum expectation values. The parameter $v$ below is the tree-level expectation, with its quantum value determined by renormalized parameters.

Writing $\varphi =v+h$ givesThe original sign symmetry relates expansions about the two vacua; in one expansion it acts as $h\mapsto -2v-h$. It is not a symmetry that fixes the chosen vacuum. Since the broken group is discrete, there is no continuous broken generator and no required Goldstone boson.

With relativistically normalized external states, the general two-body partial decay width isHere the bar sums final spins and averages initial spins only if the initial ensemble is unpolarized. The identical final-state symmetry factor is $\mathcal{S}=2!$ for two identical daughters and one for distinguishable daughters. Integrating the three-momentum delta function in the parent rest frame leaves ${\mathbf{p}}_2=-{\mathbf{p}}_1$. The energy delta function fixes $k=|{\mathbf{p}}_1|$ and has radial Jacobian $k(1/E_1+1/E_2)$. Hence the two-body decay phase space isConsequentlywhen the spin-summed amplitude is angle independent.

For Higgs decay to two W bosons, the scalar parent has no spin average and the amplitude, up to an irrelevant phase, is $\mathcal{M}=g{M}_W{\epsilon }_1^{\ast }\cdot {\epsilon }_2^{\ast }$. Use the massive vector polarization sum with $p_i^2=M_W^2$:Momentum conservation gives $p_1\cdot p_2=(M_H^2-2M_W^2)/2$. With the PDF's ratio $x=M_W/M_H$, this becomesThe charged daughters are distinguishable. Combining the last two equations with $g^2/M_W^2=4\sqrt{2}{G}_F$ givesFor Higgs decay to two Z bosons, the Higgs boson coupling to Z bosons is $2i{M}_Z^2{g}_{\mu \nu }/v=ig{M}_Z^2{g}_{\mu \nu }/M_W$, using $v=2M_W/g$. This compensates the changed powers of $M_Z$ in the polarization contraction, giving the same normalization of the squared amplitude with $x$ replaced by $y=M_Z/M_H$. The two Z bosons are identical, so the phase-space symmetry factor supplies an additional half:Both expressions have mass dimension one, vanish at their two-body thresholds and approach the ratio $2:1$ for $M_H\gg M_Z$. The cubic large-mass behavior comes from longitudinal-vector polarizations. These are on-shell tree-level widths under the stated heavy-Higgs hypothesis; below threshold the physical off-shell multi-particle decays require a different calculation.

The normalized Fock vacuum obeys $a_{\mathbf{p}}|0⟩=0$ for every momentum and $⟨0|0⟩=1$. Taking adjoints gives $⟨0|a_{\mathbf{p}}^{†}=0$. Of the four products in the mode expansion of $\varphi (x)\varphi (y)$, only $a_{\mathbf{p}}{a}_{\mathbf{q}}^{†}$ has a nonzero vacuum expectation value. The commutator impliesConsequentlyThis unordered Wightman function differs from the time-ordered Feynman propagator when the temporal ordering is reversed.

In the incoming Bunch-Davies vacuum, only $⟨a_{\mathbf{k}}{a}_{\mathbf{q}}^{†}⟩={\delta }_{\mathbf{kq}}$ contributes to the coincident vacuum expectation value. ThusReplacing the box sum by $\mathcal{V}\int d^{3}k/(2\pi {)}^3$ givesThe subhorizon vacuum contribution is the flat-space zero-point term; at fixed comoving $k$ it redshifts as $a^{-2}$. The other term gives frozen scalar power on superhorizon scales. Specifically, the dimensionless cosmological power spectrum per logarithmic interval isIts late-time value is independent of $k$, producing a scale-invariant inflationary power spectrum. Each fixed Fourier mode freezes separately as $\tau \to 0$.

The coincident integral is formal: the vacuum part diverges at large $k$, and an unlimited scale-invariant part diverges at small $k$. The conclusion concerns the power in each finite logarithmic band. A finite inflationary history supplies a largest generated wavelength, as described by the infrared qualification of a scale-invariant covariance; taking the late-time limit inside an unregulated integral would not be justified.

For a single chiral superfield, the supergravity F-term potential, with $\kappa =m_p^{-1}$ restored, isThe canonical Kähler potential has $K_{z\bar{z}}=K^{z\bar{z}}=1$, so the Kähler covariant derivative of a superpotential isConsequently the Polonyi model has the scalar potentialThe negative term is essential; retaining only the square of the supergravity auxiliary field would give the global F-term scalar potential rather than the supergravity F-term potential.

Set $x=\kappa z=a+iy$ and $b=\kappa \beta =2-\sqrt{3}$. In this notation $V=|\mu {|}^2{m}_p^2{e}^{|x{|}^2}U$, whereAt $a_0=\sqrt{3}-1$, $y=0$, one has $a_0+b=1$ and $1+a_0(a_0+b)=\sqrt{3}$, so $U=0$. To prove that this is a local minimum in both real directions, rather than checking only the real axis, put $u=a-a_0$. Direct expansion gives the stable zero-energy Polonyi vacuum identityEvery coefficient outside the square is strictly positive, and the exponential factor is positive. Thus $V\ge 0$ everywhere and equality requires $u=y=0$, provided $\mu \ne 0$. This proves the stronger conclusiona unique global minimum of this Polonyi model. If $\mu =0$, the scalar potential vanishes identically and this uniqueness and supersymmetry breaking conclusion do not hold.

For an independent local stability check, the Hessian matrix of $U$ at this global minimum is diagonal:Since $U$ and its first derivatives vanish there, differentiating the exponential introduces no additional Hessian matrix terms at the global minimum. For canonically normalized real fluctuations $z=z_0+(s+it)/\sqrt{2}$, the squared scalar masses are thereforewhich are both positive.

Although the vacuum energy is zero, the supergravity auxiliary field is not. In the convention $F^z=-e^{{\kappa }^{2}K/2}\overline{D_{z}W}$, its vacuum expectation value obeysThe matter fermion transformation contains a term proportional to $F^zϵ$, so no nonzero constant supersymmetry parameter leaves this vacuum invariant. Hence supersymmetry is spontaneously broken even though the vacuum energy vanishes: the positive auxiliary field contribution cancels the negative gravitational term. The matter fermion supplies the goldstino, which is absorbed by the gravitino in the super-Higgs mechanism.