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The Goldstone theorem guarantees zero directions tangent to the vacuum orbit, not the absence of other zero eigenvalues. For $V(\phi )=\kappa ({\phi }^T\phi -v^2{)}^4$, $\kappa >0$, the minima form the same sphere as in ordinary $O(n)$ breaking, but the entire Hessian matrix vanishes there. Thus all n scalar quadratic masses vanish, although only n-1 are symmetry-protected Goldstone bosons. Exact Goldstone counting requires nonzero curvature transverse to the orbit.

For a gauged real scalar triplet with vacuum $v{e}_3$ and generators $(t^a{)}_{bc}=-i{ϵ}^{abc}$, the covariant kinetic term yields gauge-boson squared masses $g^2{v}^2\mathrm{diag}(1,1,0)$. Two Goldstone bosons become longitudinal vector modes through the Higgs mechanism. One radial scalar, two massive three-polarization vector fields and one massless two-polarization field remain. The physical count $3+3\cdot 2=1+2\cdot 3+2=9$ is unchanged. In unitary gauge, the scalar kinetic term includes $\frac{1}{2}{g}^2(v+h{)}^2[(B^1{)}^2+(B^2{)}^2]$.

A continuous symmetry belongs to a group whose parameters can vary continuously. Spontaneous breaking of such a symmetry can produce Goldstone bosons.

Canonical gauge interactions preserve the tree-level supertrace mass sum rule at a vacuum with vanishing D-terms. Put $C={\sum }_a{g}_a^2\Vert T_{a}v{\Vert }^2$ for the scalar expectation vector $v$. The extra real-scalar squared-mass trace is $2C$, the extra Weyl spinor squared-mass trace from gaugino mixing is $4C$, and the gauge boson squared-mass trace is $2C$. Their spin-weighted supertrace is $2C-2(4C)+3(2C)=0$. The scalar contribution comes from differentiating $\frac{1}{2}{\sum }_a{D}_a^2$; the fermionic one counts both blocks of the symmetric mixing matrix with entries $\sqrt{2}{g}_a{T}_{a}v$; the vector one follows from the covariant scalar kinetic term. Any zero-mass Goldstone bosons in the unreduced scalar Hessian do not change its trace.

For a real scalar triplet with $V=\frac{1}{2}{m}^2{\Phi }^2+\frac{\lambda }{4}({\Phi }^2{)}^2$, $\lambda >0$ and $m^2<0$, the vacuum sphere has radius $v^2=-m^2/\lambda$. At ${\Phi }_0=v{e}_3$, the Hessian is $\mathrm{diag}(0,0,2\lambda v^2)$. Global $O(3)\to O(2)$ breaking therefore gives two physical massless Goldstone bosons and one radial scalar with squared mass $2\lambda v^2$. The angular modes cannot be gauged away in a theory with only global rotations.

At tree level and energies much greater than a gauge-vector mass, an amplitude with longitudinal external gauge bosons equals the corresponding amplitude with their would-be Goldstone bosons, up to conventional external-state phases and relative $O(m_V/E)$ corrections for a nonzero leading amplitude. In a renormalizable gauge, the broken-symmetry identities relate $k_{\mu }{\mathcal{M}}^{\mu }$ to $m_V{\mathcal{M}}_{\chi }$; inserting ${\epsilon }_L^{\mu }=k^{\mu }/m_V+O(m_V/E)$ explains the result. Loop-level versions include field-normalization factors. This is different from an equivalence theorem for field redefinitions.

For ${\mu }_h^2>0$ and $\lambda >0$, the minimum has ${\varphi }^{†}\varphi =v^2/2$ with $v^2={\mu }_h^2/\lambda$. In unitary gauge, $\varphi =(0,v+H{)}^T/\sqrt{2}$ gives $V=\text{constant}+\lambda v^2{H}^2+\lambda v{H}^3+\lambda H^4/4$. The canonically normalized Higgs boson therefore has $m_H^2=2\lambda v^2$. The angular modes produce the Goldstone bosons absorbed in the Higgs mechanism.

A physical polarization of a massive spin-one gauge boson whose spatial polarization points along its spatial momentum. Together with two transverse polarizations it supplies the three physical spin states. In the Higgs mechanism, a broken-direction Goldstone boson supplies this extra state as the gauge boson acquires mass.

Conservation of particle number corresponds to global circle group phase rotations of a nonrelativistic Bose field. A positive-density uniform mean-field saddle chooses a phase, with degenerate saddle manifold a circle. The associated Goldstone boson is the long-wavelength phase mode. This is a thermodynamic symmetry-selection statement; an exact finite-volume number eigenstate does not itself choose a phase.

The unbroken group is the stabilizer subgroup of the chosen vacuum expectation value:Its Lie algebra consists of generators $T$ with $T{\varphi }_0=0$. The vacuum manifold is the orbit $G{\varphi }_0\simeq G/H$ under the stated transitivity assumption. Its tangent space is the image of $T\mapsto T{\varphi }_0$, so its dimension is $r=\dim G-\dim H$.

Let $K_{ij}={\partial }_i{\partial }_{j}V({\varphi }_0)$ be the scalar mass matrix, with canonical scalar kinetic terms. Invariance of the scalar potential gives $({\partial }_{i}V)(T_a\varphi {)}_i=0$ for every $\varphi$. Differentiate with respect to ${\varphi }_j$ and evaluate at the minimum, where ${\partial }_{i}V=0$:Thus every independent tangent direction is a zero eigenvector of $K$. These are the Goldstone directions in the scalar mass matrix; for a global symmetry they are $r$ physical massless Goldstone bosons. For a generic nondegenerate minimum transverse to the orbit, they are precisely all the zero eigenvalues. Without that nondegeneracy assumption, additional massless scalars are possible.

For a local gauge group, use the orthogonal unitary-gauge slice near a scalar vacuum. Since ${\varphi }_0^T{T}_a{\varphi }_0=0$ by antisymmetry, the proposed condition is equivalent toIt sets the gauge-orbit components of the scalar fluctuation to zero. This is an admissible local unitary gauge: on a basis of independent broken tangent vectors $v_a=T_a{\varphi }_0$, the variation of these conditions under the broken gauge parameters is $v_a^T{v}_b$. This Gram matrix is invertible on that basis, so the implicit function theorem selects the required gauge transformation near the vacuum. Unbroken transformations remain as a residual $H$ gauge symmetry; no global gauge slice through zeros of the scalar field is being asserted.

Write $\varphi ={\varphi }_0+h$. To quadratic order, the covariant kinetic term containsThe middle term vanishes in this unitary gauge. The gauge-boson mass rank from a real scalar vacuum is therefore determined byThese are squared masses: the physical gauge boson masses are square roots of the matrix's nonzero eigenvalues after diagonalization. For real $c_a$,Its kernel consists exactly of the unbroken generators. With $g\ne 0$, the rank is $r$, so exactly $\dim G-\dim H$ gauge-field combinations become massive. Each removed scalar Goldstone mode supplies the extra longitudinal polarization of a massive gauge boson.

The remaining physical scalar fluctuations lie in the orthogonal complement of the orbit tangent space. They have no massless modes if $K$ is positive definite on that complement. The unconditional absence of massless scalars does not follow from the printed assumptions. For a concrete counterexample, take a real doublet, $G=SO(2)$ andIts entire minimum set is one circle orbit, as required, but its Hessian matrix vanishes at the minimum. Gauging the rotation gives one vector of squared mass $g^2{v}^2$ and removes the angular scalar. The radial scalar remains massless at quadratic order, since its potential starts at fourth order in the radial fluctuation. This is an accidental massless radial mode beyond Goldstone modes. Thus the intended no-massless-scalar conclusion needs nonzero transverse curvature.

In the Standard Model, the Higgs doublet contains four real components and breaks $SU(2{)}_L\times U(1{)}_Y$ to $U(1{)}_{\mathrm{em}}$. Three Goldstone bosons become the longitudinal components of $W^{\pm }$ and $Z$, while the photon remains massless. The remaining radial Higgs boson is massive for the ordinary quartic potential with positive curvature. With $⟨H⟩=(0,v_{\mathrm{EW}}/\sqrt{2}{)}^T$,The colour gauge group is unbroken. This is the Higgs mechanism, with the generic curvature condition realized by the standard Higgs potential.

For massless up and down quarks, the Quantum chromodynamics kinetic term splits into two independent chirality sectors,The colour covariant derivative acts identically on both flavours and commutes with constant flavour matrices. Independent transformations $q_R\mapsto A{q}_R$, $q_L\mapsto B{q}_L$, with $A,B\in SU(2)$, therefore leave the action invariant. This is chiral symmetry. The non-singlet axial transformations have traceless flavour generators, so they are not broken by the colour axial anomaly; the anomalous axial $U(1)$ is not being included in this symmetry.

For $V_{ij}={\bar{q}}_{Lj}{q}_{Ri}$, transforming each index givesThe nonzero condensate $⟨V⟩=-vI$ is unchanged precisely when $A=B$. Thus chiral symmetry breaking leaves the diagonal vector subgroup $SU(2{)}_V$. Three of the six generators are broken, producing three massless Goldstone bosons in the exact massless limit. They are the pions, represented at low energy by the coset coordinate $U\in SU(2)$.

The uniqueness of the two-derivative two-flavour chiral action follows from the symmetry, rather than merely from guessing an invariant. The chiral group acts transitively on $U$, so every derivative-free invariant is constant. At $U=I$, the stabilizer acts on the three tangent directions by conjugation, the irreducible rotation representation. Its invariant symmetric quadratic form is a single multiple of ${\delta }_{ab}$. Transporting that form around the group gives a unique invariant two-derivative metric. Equivalently, the Maurer-Cartan form $U^{†}{\partial }_{\mu }U$ is traceless, so an independent double-trace term vanishes; the remaining contracted single trace is the same metric. Lorentz invariance supplies the spacetime contraction. Up to total derivatives, a constant term and its overall coefficient, the massless action is thereforeThe positive coefficient is conveniently expressed through the pion decay constant $F$.

Use the Pauli matrices with $\mathrm{tr}({\tau }_a{\tau }_b)=2{\delta }_{ab}$. Expanding the matrix exponential gives ${\partial }_{\mu }U=i{\tau }_a{\partial }_{\mu }{\pi }_a/F+\cdots$. Henceso the three fields have canonical kinetic terms. There is no nonconstant invariant potential in the exact chiral theory, consistent with their masslessness.

For real diagonal $M$, the two terms in the mass trace areTheir sum therefore gives the usual Dirac mass term,These terms explicitly break chiral symmetry. In the low-energy chiral perturbation theory prescription, $V\mapsto -vU$ yieldsThe Pauli matrix algebra implies $(\mathbf{\pi }\cdot \mathbf{\tau }{)}^2=|\mathbf{\pi }{|}^{2}I$. ThusandComparing with the canonically normalized scalar mass term gives the leading pion mass from a two-flavour condensate:Positive quark masses therefore give positive pion squared masses. Even if $m_u\ne m_d$, this leading action depends only on their sum and gives equal masses for all three pions. Strong isospin-breaking effects beyond this order and electromagnetic effects can split them. In the condensate convention used here, $⟨\bar{u}u⟩=⟨\bar{d}d⟩=-2v$, which also explains the factor of two in the mass relation. The massive pions are pseudo-Goldstone bosons, becoming massless as both quark masses tend to zero.

Let $t^A$ be Hermitian color generators with $\mathrm{Tr}(t^A{t}^B)={\delta }^{AB}/2$, and take $D_{\mu }={\partial }_{\mu }-i{g}_s{G}_{\mu }^A{t}^A$. In explicit spinor, color and flavor indices the massless kinetic term isThe full Quantum chromodynamics Lagrangian also has $-G_{\mu \nu }^A{G}^{A\mu \nu }/4$. A local color gauge transformation acts asThus $D_{\mu }q\mapsto V{D}_{\mu }q$ and the kinetic term is gauge-invariant.

Since the gauge coupling is independent of flavor and chirality, the classical internal global transformations are $q_L\mapsto U_L{q}_L$, $q_R\mapsto U_R{q}_R$, with constant flavor matrices $U_L,U_R\in U(2)$. Equivalently these consist of $SU(2{)}_L\times SU(2{)}_R$, a common vector phase, and a singlet axial phase $q\mapsto e^{i\eta {\gamma }_5}q$, modulo shared discrete factors. The vector phase measures quark number, or baryon number after dividing its generator by three. The nonsinglet axial rotations may be written $q\mapsto e^{-i{\theta }_a{\gamma }_5{\tau }_a/2}q$. At the quantum level the singlet axial phase is anomalous; the exact continuous internal symmetry is $SU(2{)}_L\times SU(2{)}_R\times U(1{)}_V$ up to those discrete identifications. The usual spacetime Poincaré transformations also leave the theory invariant; the massless classical theory has dilation symmetry, which is broken by quantum running. These distinctions are massless two-flavor QCD symmetries.

At equal time the canonical relations, including all indices, areNo constraint removes any of the four spinor components of the Dirac field in these equal-time relations.

To evaluate the axial charges, first prove the elementary bilinear identity. For any constant matrix $K$ on the spinor/flavor space,For example $[q_I^{†}{q}_J,q_K]=q_I^{†}\{q_J,q_K\}-\{q_I^{†},q_K\}{q}_J$; the spatial delta function collapses the integral. This proves the charge action directly from the canonical anticommutation relations.

For $K_a={\gamma }_5{\tau }_a/2$, anticommutation of ${\gamma }^0$ and ${\gamma }_5$ then givesThe charge is an even operator, so its commutator obeys the ordinary product rule. ConsequentlySimilarly, using ${\gamma }_5^2=1$ and the Pauli matrices identity $\{{\tau }_a,{\tau }_b\}=2{\delta }_{ab}I$,Composite fields can be defined with a common symmetry-preserving regulator; these nonsinglet identities have no color axial anomaly. This is the nonsinglet axial charge algebra of quark densities.

If every conserved axial charge left the vacuum invariant, the vacuum expectation of each such commutator would vanish. A nonzero quark chiral condensate instead makes $⟨[Q_{5,a},P_b]⟩=-i{\delta }_{ab}⟨S⟩\ne 0$ for $a=b$. It therefore diagnoses spontaneous breaking of the nonsinglet chiral symmetry. In the massless two-flavor limit, the observed hadronic pattern is described by $SU(2{)}_L\times SU(2{)}_R\to SU(2{)}_V$, with three Goldstone boson modes identified with the pions. The vacuum condensate is a dynamical property of QCD, not something derivable from the canonical algebra alone. The light pion multiplet and absence of parity-degenerate light hadrons support this pattern; the singlet axial anomaly prevents interpreting an additional singlet mode as a fourth Goldstone boson.

For equal nonzero masses, add ${\mathcal{L}}_m=-m\bar{q}q$. It is invariant under local color gauge transformations and common vector flavor rotations, but not under independent left/right rotations. The exact continuous flavor symmetry becomes $SU(2{)}_V\times U(1{)}_V$. The kinetic term and canonical anticommutators are unchanged, and the two proved equal-time commutators still hold. The charges, however, are no longer conserved:The first equation follows from the massive Dirac equations, or the last from $H_m=m\int S$ and $\dot{Q}=i[H,Q]$. This is the flavor-nonsinglet axial Ward identity with equal quark masses. The pions become pseudo-Goldstone bosons with $m_{\pi }^2=O(m)$, rather than exact massless modes. To leading order, in the convention $S=\bar{u}u+\bar{d}d$, $f_{\pi }^2{m}_{\pi }^2=-m⟨S⟩+O(m^2)$. The condensate persists for small masses, but the symmetry is now explicitly broken, so the exact-symmetry Goldstone argument is replaced by an approximate one. The singlet anomaly and gauge symmetry are unaffected.

In the supersymmetric Higgs mechanism, an internal gauge symmetry is broken by a charged scalar field expectation value while supersymmetry can remain unbroken. A massless vector multiplet absorbs a chiral multiplet to form a massive N=1 vector multiplet: the vector absorbs a scalar Goldstone boson, and the gaugino and matter fermion acquire the matching masses. In the super-Higgs mechanism, local supersymmetry itself is spontaneously broken: the gravitino absorbs the spin-$1/2$ Goldstino and becomes a massive spin-$3/2$ particle. In a flat vacuum this increases its physical states from two to four.

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.

Use natural units and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. A massive spin-one particle has three physical particle polarizations, whereas a massless gauge boson has two. Giving a vector a mass must account for this extra longitudinal state while preserving a positive physical state space, controlled high-energy scattering and, if a fundamental perturbative theory is sought, renormalizability.

At the free-field level there is no inconsistency. The Proca actiongives the Proca equation ${\partial }_{\nu }{F}^{\nu \mu }+m^2{A}^{\mu }=0$. Taking its divergence gives $m^2{\partial }_{\mu }{A}^{\mu }=0$, and then $(\square +m^2)A^{\mu }=0$. For $m>0$ the divergence constraint removes one of the four components, leaving three positive-norm physical modes. The time component is constrained, rather than an independent negative-norm propagating oscillator. The mass term does, however, change under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\omega$, so the ordinary massless gauge invariance is no longer manifest.

The difficulty becomes acute for generic interactions. Inverting the quadratic operator gives the Proca propagatorThe longitudinal part approaches order $1/m^2$ at large momentum, rather than falling as $1/k^2$. Thus the usual ultraviolet power counting deteriorates. There is a parallel external-state problem: for $p^{\mu }=(E,\mathbf{p})$,Longitudinal external legs can generate powers of $E/m$ in a scattering amplitude. These are the high-energy obstruction for a hard vector mass. Conserved Abelian currents can remove dangerous contractions, so one should not conclude that every massive-vector interaction is impossible.

For a massive non-Abelian vector with gauge-like cubic and quartic couplings, cancellations among diagrams remove the largest powers, but without a suitable additional sector longitudinal scattering still grows like $s/v^2$, where $v$ is the symmetry-breaking scale. A partial-wave coefficient is then of order $s/(16\pi v^2)$. Partial-wave unitarity requires its real part to remain bounded, so perturbation theory fails at energies of order the electroweak scale times a modest loop factor, around the TeV scale for electroweak vectors. This is a limit on the perturbative description, not a proof that a low-energy massive-vector effective field theory is inconsistent.

The Higgs mechanism supplies a weakly coupled completion. For an Abelian example, take a complex scalar field withLocally write $\varphi =(v+h)e^{i\chi /v}/\sqrt{2}$. Its gauge-covariant kinetic term becomesThe combination is invariant under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\omega$, $\chi \mapsto \chi +gv\omega$. In unitary gauge $\chi =0$, it contains $m^2{A}_{\mu }{A}^{\mu }/2$ with $m=gv$, as well as the correlated $hAA$ and $hhAA$ interactions. The two original gauge polarizations and two real scalar components become three massive-vector polarizations and one radial Higgs boson. The would-be Goldstone boson supplies the longitudinal state. This counting explains why the Goldstone theorem for spontaneously broken global symmetries does not imply an extra physical massless particle here. Local gauge symmetry is a redundancy; choosing a Higgs background after gauge fixing does not explicitly break the gauge invariance of the action.

The extra scalar interactions also repair high-energy scattering. By the Goldstone-boson equivalence theorem, the leading longitudinal-vector amplitude can be computed using the would-be Goldstone bosons. For a charged-to-neutral channel in the linear scalar model, the scalar contact and radial-exchange terms giveThis Higgs cancellation in longitudinal vector scattering removes the uncontrolled $s/v^2$ growth. It also shows why a very large scalar self-coupling would itself make perturbation theory unreliable; introducing a scalar is not a license to ignore partial-wave unitarity.

Although the unitary gauge propagator looks like the badly behaved Proca propagator, renormalizability is conveniently established in an R-xi gauge. Using Cartesian scalar fluctuations, the quadratic mixing is $-m{A}_{\mu }{\partial }^{\mu }\chi$. The gauge-fixing term $-(\partial \cdot A+\xi m\chi {)}^2/(2\xi )$ cancels it and givesAt fixed finite $\xi$ this falls as $1/k^2$. The would-be scalar modes and Faddeev-Popov ghosts remain in intermediate calculations. Gauge identities, organized through BRST symmetry, make unphysical states cancel from physical amplitudes and constrain counterterms to the gauge-invariant renormalizable form. With a renormalizable scalar sector and cancellation of gauge anomalies, spontaneously broken gauge theory is renormalizable and unitary on its physical states; the bosonic construction is demonstrated in the original massive Yang-Mills renormalizability proof. The ultraviolet limit and $\xi \to \infty$ limit do not commute, so the unitary-gauge numerator alone is not a valid disproof of this result.

In the Standard Model, a Higgs doublet breaks the manifest electroweak group from $SU(2{)}_L\times U(1{)}_Y$ to $U(1{)}_{\mathrm{em}}$. The scalar kinetic term yieldsThree scalar directions supply the longitudinal polarizations of $W^{\pm }$ and $Z$, while the unbroken electromagnetic generator leaves the photon massless. The radial Higgs boson remains physical.

There is an important Abelian alternative. The Stueckelberg mechanism introduces a scalar with ${\mathcal{L}}_{\mathrm{mass}}=(m{A}_{\mu }-{\partial }_{\mu }\chi {)}^2/2$, invariant under $A\mapsto A+\partial \omega$, $\chi \mapsto \chi +m\omega$. With suitable conserved-current couplings this can describe a renormalizable massive Abelian gauge theory without a radial Higgs particle. A naive non-Abelian analogue involves a nonlinear group-valued scalar and derivative interactions suppressed by inverse powers of the symmetry-breaking scale; it is generally an effective field theory, not a power-counting-renormalizable substitute. Strongly coupled or composite sectors can provide other completions. A free vector mass is consistent; the central problem is obtaining the longitudinal state and its interactions in a theory with controlled ultraviolet behavior and physical unitarity.

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.

Adopt the phase convention $CP|K^0⟩=-|{\bar{K}}^0⟩$ and $CP|{\bar{K}}^0⟩=-|K^0⟩$. Changing both phases changes the convention for the mixing parameter, not physical predictions. The flavor contents are $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$. A pair of charged weak currents changes strangeness by two units through the following box contractions, with $u_i,u_j=u,c,t$ summed over.

The drawing uses unitary gauge: the W boson gauge-boson propagator contains its longitudinal part. In a gauge of a renormalizable quantum field theory, charged Goldstone boson boxes must also be included to obtain the same gauge-independent result. The external lines represent the valence quarks; the meson matrix element also contains strong interaction binding effects.

Put ${\lambda }_i=V_{is}^{\ast }{V}_{id}$, $x_i=m_i^2/m_W^2$ and $P_L=(1-{\gamma }^5)/2$. The charged weak box contribution to kaon mixing has the effective four-fermion interaction structureHere $G_F$ is the Fermi constant, $V$ is the CKM matrix, and $S$ is a dimensionless loop function from a Feynman integral; the estimate suppresses convention-dependent numerical factors. Four weak interaction Feynman vertices and a one-loop Feynman integral give the coupling and loop order suppression. Unitarity of the CKM matrix gives ${\sum }_i{\lambda }_i=0$, so the GIM mechanism cancels the mass-independent part. With a negligible up quark mass, the leading charm quark term has $S(x_c,x_c)\sim x_c$. Define $f_K$ by $⟨0|\bar{s}{\gamma }_{\mu }{\gamma }^{5}d|K^0(p)⟩=i{f}_K{p}_{\mu }$. Estimating the hadronic matrix element by $f_K^2{m}_K{B}_K$, with the dimensionless kaon bag parameter $B_K$ of order unity, givesThe hadronic normalization includes division by $2m_K$ when relativistically normalized states are converted to the effective Hamiltonian for neutral-meson mixing. Since $|V_{cs}^{\ast }{V}_{cd}|\simeq \sin {\theta }_C\cos {\theta }_C$, the Cabibbo suppression is explicit. Illustrative scales $G_F\sim 10^{-5}{\mathrm{GeV}}^{-2}$, $m_c\sim 1\mathrm{GeV}$, $f_K\sim 0.1\mathrm{GeV}$ and $m_K\sim 0.5\mathrm{GeV}$ give an order $10^{-16}$ to $10^{-15}\mathrm{GeV}$ estimate. The full sum includes charm quark-top quark and top quark-top quark terms, important for CP violation; long-distance strong interaction contributions prevent this dimensional estimate from being a precision prediction. A meson matrix element necessarily needs hadronic input in addition to electroweak parameters.