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The electroweak interaction is the $SU(2{)}_L\times U(1{)}_Y$ gauge theory that unifies the weak and electromagnetic interactions.

An instructive tension motivates the unification: the photon is massless, while the W boson and Z boson are massive; electromagnetic interaction and weak interaction also have different observed low-energy strengths and chiral couplings. A gauge theory can relate their interactions while the Higgs field gives different masses after electroweak symmetry breaking. The resulting theory predicts relationships between processes, not merely a common name for observed phenomena.

Its useful organizing principle is the combination of weak isospin and hypercharge. The surviving electromagnetic generator is $Q=T_3+Y$ in the convention used here. The photon and Z boson are different mixtures of neutral gauge fields, while the charged current uses the charged combinations. The electroweak doublet gauge-boson mass matrix explains why one neutral mixture stays massless. At low energies, a massive W boson propagator instead produces the approximate Fermi interaction, withHere $g$ is the $SU(2{)}_L$ gauge coupling and $v$ the Higgs doublet vacuum expectation value. The inverse squared mediator mass suppresses low-energy weak rates. Unification does not require setting $g$ equal to the electromagnetic coupling $e$: their relation is $e=g\sin {\theta }_W$.

A striking experimental consequence is a neutral current in addition to the familiar charged current. The Gargamelle experiment tests this extra channel through visible charged secondaries, with a background problem fundamentally different from directly identifying a Z boson. The muon-neutrino electron recoil spectrum shows how the same Weinberg angle enters another process quantitatively.

The Standard Model Higgs field is a complex $SU(2{)}_L$ doublet of hypercharge $1/2$. Its nonzero vacuum expectation value triggers the Higgs mechanism.

The Higgs field is a scalar field: its components carry internal weak isospin and hypercharge, but no spacetime spin. A complex Higgs doublet contains four real field components. The vacuum expectation value is a property of the lowest-energy state, not a gas of Higgs bosons. A Higgs boson is an excitation about that state.

Why introduce this field? A bare mass term for a gauge field generally fails to respect the local gauge symmetry. A bare Dirac mass term for a charged Standard Model fermion also joins left- and right-handed fields with different Standard Model representations. A Higgs doublet permits gauge-invariant interactions whose expansion about a nonzero vacuum expectation value supplies both masses and definite Higgs boson interactions. The field therefore links the mass problem to experimentally testable coupling relations, as illustrated by the Higgs boson coupling to Z bosons.

Chirality distinguishes the two inequivalent Weyl spinor representations of the proper Lorentz group. For a Dirac spinor, define ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ and the chiral projectorsThey satisfy $P_L{P}_R=0$ and $P_L+P_R=I$. Proper Lorentz transformations preserve each chiral representation, whereas parity exchanges them. Helicity instead measures the projection of spin angular momentum along momentum. For a massive particle it can change under a boost that overtakes the particle, so it must not be identified with chirality. For massless positive-energy particles the left-chiral field annihilates negative-helicity particles and creates positive-helicity antiparticles.

The Standard Model is a chiral gauge theory: left- and right-chiral fields have different gauge group representations. In the convention $Q=T_3+Y$, one generation containsThe minimal field content has no right-handed neutrino. A sterile right-handed field is a possible extension, not a required member of that minimal list. The table displays weak isospin, hypercharge and color charge assignments, not assignments of massive-particle helicity.

The weak charged current contains $\bar{\nu }{\gamma }^{\mu }{P}_{L}e{W}_{\mu }^{+}$ and its Hermitian conjugate, with coupling $g/\sqrt{2}$. Only the left-chiral lepton doublet participates, giving parity violation. Both chiralities can couple to the Z boson, but their coefficients differ:Here $T_3$ is that of the left-chiral component. For the electron, it is $-1/2$. The photon and gluon couplings of a charged massive fermion are vectorlike; this does not make the full electroweak interaction vectorlike.

A bare Dirac mass term $m\bar{\psi }\psi =m({\bar{\psi }}_L{\psi }_R+{\bar{\psi }}_R{\psi }_L)$ joins the two chiral representations. Their different electroweak quantum numbers forbid that term before electroweak symmetry breaking. A Yukawa interaction, such as $-y_e{\bar{L}}_{L}H{e}_R+h.c.$, is gauge invariant; the Higgs doublet expectation value then gives $m_e=y_{e}v/\sqrt{2}$. Up-type quarks use the conjugate doublet. Thus chirality explains the weak charge assignments, their parity violation, and why masses arise through the Higgs mechanism. The consistency of these chiral assignments also requires Standard Model anomaly cancellation.

Use units in which the proton electric charge is one. The up quark and down quark both have spin angular momentum $1/2$ and form a flavour isospin doublet:Their antiquarks have the opposite charges and third components of isospin, with the same total isospin and spin. This is strong-interaction flavour isospin, not the chiral weak isospin assignment.

The Clebsch-Gordan decomposition for SU2 says that spins $j_1,j_2$ combine into every $j$ from $|j_1-j_2|$ to $j_1+j_2$ in unit steps. It applies independently to spin and isospin. ThusBefore exchange symmetry is imposed, three light quarks have total spin $S=1/2$ or $3/2$, and total isospin $I=1/2$ or $3/2$; the doublet occurs twice in the uncoupled tensor cube. With zero orbital angular momentum, $J=S$.

Physical baryons must also be colour singlets. The three-quark colour singlet is completely antisymmetric, so the Pauli exclusion principle requires symmetric spatial-spin-flavour symmetry. For a symmetric spatial S-wave state, the spin-flavour factor is symmetric. The $I=3/2$ flavour quartet is symmetric, whereas the $I=1/2$ flavour doublets have mixed permutation symmetry; the spin factors have the analogous symmetries. Symmetric flavour therefore pairs with symmetric spin, while the two mixed factors pair in their unique symmetric combination. This is the Pauli constraint on three-quark flavour multiplets; it leavesEquivalently the symmetric cube of the four-dimensional two-flavour spin space has dimension $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20=2\cdot 2+4\cdot 4$. The unmatched spin-isospin pairings in the bare tensor product do not describe the usual symmetric-spatial ground-state baryons.

For three light quarks, charge addition gives $Q=I_3+1/2$. The observed nucleon doublet and Delta baryon quartet have the following assignments; the listed contents specify flavour counts, not the complete spin-flavour wavefunction:The proton and neutron have $J=1/2$; all four Delta baryons have $J=3/2$. These states have positive parity in the zero-orbital ground-state model.

For a light quark-antiquark meson, both spin and flavour doublets combine into singlet plus triplet. In a conventional choice of phases the flavour states areEach row can have $J=0$ or $1$, since the two constituent spins couple to zero or one. At zero orbital angular momentum their parity is negative, following the meson parity and charge conjugation rule $P=(-1{)}^{L+1}$. The spin-zero isotriplet is the pion multiplet, and the spin-one isotriplet the rho meson multiplet. The nonstrange spin-one isosinglet is the idealized omega meson flavour direction. Its spin-zero counterpart is the nonstrange direction contributing to the eta and eta prime mesons; it is not automatically a physical mass eigenstate once other flavours and mixing are included.

Now introduce an isospin-zero quark $h$ of spin $1/2$ and charge $q_h$. For baryons with one isospin-singlet quark, the two light quarks have either the antisymmetric flavour singlet $[ud]=(ud-du)/\sqrt{2}$ or the symmetric triplet $uu,\{ud\},dd$. Their colour pair is antisymmetric. Hence light-diquark spin-isospin correlation gives $S_{ud}=0$ for $I=0$ and $S_{ud}=1$ for $I=1$. Coupling the $h$ spin producesThe three triplet rows have $I_3=1,0,-1$. For a strange-like charge $q_h=-1/3$, these are the neutral spin-one-half Lambda baryon, the spin-one-half Sigma baryon triplet of charges $1,0,-1$, and its spin-three-half starred partners. For a charm-like charge $q_h=2/3$, they are the spin-one-half Lambda c baryon of charge $1$, and the Sigma c baryon triplets of charges $2,1,0$ with spins $1/2$ and $3/2$. Both possible new charge assignments have thus been covered; charge-conjugate antibaryons have opposite charges and unchanged $I,J$.

The meson doublets with one isospin-singlet quark are $h\overline{u},h\overline{d}$, withFor $q_h=-1/3$ their charges are $-1,0$, giving $K^{-},{\overline{K}}^0$-like kaons and their vector partners; the conjugate doublet has $u\overline{h},d\overline{h}$ and charges $1,0$. For $q_h=2/3$ the charges are $0,1$, giving $D^0,D^{+}$-like D mesons and their vector partners, with charge-conjugate partners as well. Finally a hidden-flavour pair $h\overline{h}$, if included, is neutral with $I=0$ and $J=0$ or $1$. These are quantum-number classifications; charge and isospin alone do not determine particle masses or mixing angles.

The derivative convention uses hypercharge Y with $Q=T_3+Y$. For each lepton family $l=e,\mu ,\tau$, the left-handed fields form an SU(2) doublet $L_l=({\nu }_l,l{)}_L^T$, with $Y=-1/2$. Their weak isospin values are $T_3=+1/2,-1/2$ and their electric charges are $0,-1$. The right-handed charged lepton $l_R$ is an SU(2) singlet, with $T_3=0$, $Y=-1$ and charge $-1$. The minimal model used here has no right-handed neutrino. A sterile right-handed neutrino, if added as a singlet with $Y=0$, would have no charged weak current interaction. Antiparticles carry the conjugate representations and opposite electric charges.

Expanding ${\bar{L}}_{l}i{\gamma }^{\mu }{D}_{\mu }{L}_l$ supplies $-g{\bar{L}}_l{\gamma }^{\mu }{A}_{\mu }^a{\sigma }_a{L}_l/2$. The off-diagonal entries of $A^a{\sigma }_a$ are $\sqrt{2}{W}^{+}$ and $\sqrt{2}{W}^{-}$. With the chiral projector $P_L=(1-{\gamma }^5)/2$, one has ${\bar{\nu }}_L{\gamma }^{\mu }{l}_L=\bar{\nu }{\gamma }^{\mu }{P}_{L}l$. Thus the two W boson interactions areFor quarks, separate rotations diagonalize the up-type and down-type mass terms. If their left-handed rotations are $U_{uL}$ and $U_{dL}$, the charged interaction contains the CKM matrix $V=U_{uL}^{†}{U}_{dL}$:There is an implicit sum over color charge. This quark mixing permits transitions between different families, rather than just the paired fields in a single doublet.

Let $J^{\mu }={\sum }_l{\bar{\nu }}_l{\gamma }^{\mu }(1-{\gamma }^5)l+J_{\mathrm{had}}^{\mu }$ and $\kappa =g/(2\sqrt{2})$. At momenta much smaller than $m_W$, neglect derivatives in the heavy W boson equation. Its mass and source terms can be completed to a square:This explicitly performs integrating out a charged vector boson. The resulting Fermi interaction isThe omitted terms contain derivatives and are relatively suppressed by momentum squared divided by $m_W^2$.

For the leptonic pion decay, Lorentz covariance restricts a vacuum-to-spin-zero current matrix element to a constant times $p^{\alpha }$. In the rest frame, rotations eliminate the spatial components. Under parity, the temporal vector current is even, whereas the vacuum is even and the pion is odd, so its temporal matrix element equals its negative and vanishes. The temporal axial current is odd and is allowed. This proves the vacuum-to-pseudoscalar current selection rule, rather than merely assuming an axial current. Define the pion decay constant byThen the vector-minus-axial charged weak current has matrix element $-i{f}_{\pi }{V}_{ud}{p}^{\alpha }$, up to an irrelevant overall phase convention. Consequently the scattering amplitude for the decay can be writtenUse $p=k+q$, the momentum-space Dirac equation ${\bar{u}}_e/k=m_e{\bar{u}}_e$, and $/q{v}_{\nu }=0$ for a massless neutrino. Since $/q(1-{\gamma }^5)=(1+{\gamma }^5)/q$, the neutrino contribution vanishes, leavingThe fermion spin sum is, using the supplied gamma matrix trace identities,Here $(k+q{)}^2=m_{\pi }^2$, $k^2=m_e^2$, and $q^2=0$. The mass term in the trace vanishes; anticommuting ${\gamma }^5$ reduces the remaining trace to twice $\mathrm{tr}(/k/q)$. It follows thatUsing two-body decay phase space, with final momentum magnitude $(m_{\pi }^2-m_e^2)/(2m_{\pi })$, gives the leptonic pseudoscalar decay widthThus the requested $m_e^2(m_{\pi }^2-m_e^2)$ factor is already present in the squared amplitude, and the phase space supplies one more threshold factor.

The physical explanation is helicity suppression. In the pion rest frame the two momenta are opposite. The massless antineutrino has positive helicity; zero total angular momentum would require the electron to have positive helicity too. The charged weak current instead creates a left-chiral electron, whose helicity is negative when its mass vanishes. A nonzero electron mass admits the required opposite helicity with an amplitude proportional to $m_e$. The decay amplitude vanishes for $m_e=0$, and the rate is suppressed by $m_e^2$.

For a purely electromagnetic interaction continuum with five active quark flavors, the usual estimate is $N_{e^{+}{e}^{-}}/N_{\mathrm{had}}=3/11$. The unspecified particle resonance and the word “total” require qualifications before this can be treated as a unique physical prediction.

In natural units, neglecting final masses, single-photon annihilation givesHere $\alpha$ is the fine-structure constant, $Q_f$ the quark electric charge in elementary-charge units, and the factor three counts orthogonal colour final states. Consequently the inverse hadronic R ratio isWith $u,d,s,c,b$ accessible, the denominator is $3[2(2/3{)}^2+3(1/3{)}^2]=11/3$. The corresponding values for $u,d,s$ and $u,d,s,c$ are $1/2$ and $3/10$. Inclusive Quantum chromodynamics corrections increase the hadronic denominator, approximately by $1+{\alpha }_s/\pi$ away from thresholds and particle resonances. These are annihilation ratios before selection and detection efficiency corrections, not ratios of uncorrected detector counts.

A future muon-collider Higgs resonance is a plausible reading of part (c), since its narrow decay width makes a detuning of a few $\mathrm{MeV}$ relevant. Even then, two further assumptions are necessary. First, being a few $\mathrm{MeV}$ below the Higgs boson peak does not eliminate it: for detuning $\Delta =\sqrt{s}-m_H$, the ideal narrow Breit-Wigner resonance cross section retains the relative factorFor ${\Gamma }_H=4\mathrm{MeV}$ and $\Delta =-4\mathrm{MeV}$ this is $1/5$, not zero. Resonant hadronic production can still change the requested denominator substantially.

Second, at $\sqrt{s}\simeq 125\mathrm{GeV}$ the Z boson continuum is not negligible. A more complete annihilation estimate including both tree-level Feynman diagrams usesHere $s_W^2$ is the squared sine of the weak mixing angle, $T_3^f$ is weak isospin, and $N_c^f$ is one for a lepton and three for a quark. Thus the continuum ratio is $F_e/[3(2F_u+3F_d)]$, rather than necessarily $3/11$. For example, $s_W^2=0.231$, $m_Z=91.1876\mathrm{GeV}$ and ${\Gamma }_Z=2.4952\mathrm{GeV}$ give approximately $0.102$ at $125\mathrm{GeV}$ before Quantum chromodynamics corrections.

If the unspecified particle resonance instead means the Z boson, detuning by a few $\mathrm{MeV}$ is tiny compared with its $2.5\mathrm{GeV}$ decay width. The resonant ratio remains approximately $\Gamma (Z\to e^{+}{e}^{-})/\Gamma (Z\to \mathrm{hadrons})=1/20.767\simeq 0.0482$.^[ref] This is a concrete counterexample to treating any slight downward detuning as permission to discard the particle resonance. The source supplies neither its identity nor the detuning and selection needed for a unique all-process ratio. The boxed continuum formula is the charge-counting answer under its stated approximation; a literal total must also include remaining resonant and other accepted hadronic channels.

Let $z=\cos {\theta }^{\star }$. Rotating the spin-one intermediate state between initial and final axes gives probability amplitudes proportional to $d_{\pm 1,\pm 1}^1$ for equal fermion helicities and $d_{\mp 1,\pm 1}^1$ for opposite ones. Thus, apart from a common positive normalization,

The common factor $1/4$ from the rotation elements is absorbed into the normalization. Distinct external helicities are orthogonal, so an unpolarized rate adds their squared probability amplitudes incoherently.

The source interchanges the conventional left and right electron coupling labels. With $g_L=T_3-Q{\sin }^2{\theta }_W$ and $g_R=-Q{\sin }^2{\theta }_W$, the physical electron values areAt ${\sin }^2{\theta }_W=0$, for example, the left electron still carries weak isospin $-1/2$, while the right electron is a weak singlet; the printed assignment gives the opposite result. This fixes how the helicity-channel table should be interpreted.^[ref]

For a muon final state, lepton universality gives the same pair of couplings. Crucially, swapping both labels does not change the summed distribution. Put $L=g_L$, $R=g_R$; thenHere $\mathcal{N}$ includes the resonance propagator, initial spin average and common coupling normalization. Equivalently,The source's simultaneous label interchange reverses ${\mathcal{A}}_{\ell }$ but leaves its square unchanged.

The covariant derivative $D_{\mu }$ differentiates a field while preserving its transformation law under the local electroweak gauge symmetry $SU(2{)}_L\times U(1{)}_Y$. Here ${\partial }_{\mu }$ is the ordinary spacetime derivative and $\mu$ a Lorentz index. The gauge fields $W_{\mu }^j$, $j=1,2,3$, are the three SU(2) fields; $B_{\mu }$ is the U(1) field. The corresponding dimensionless coupling constant is $g$ for SU(2) and $g^{\prime }$ for U(1).

The ${\tau }_j$ are weak isospin generators in the representation of the differentiated field, with ${\tau }_j={\sigma }_j/2$ on a doublet and ${\tau }_j=0$ on an SU(2) singlet. The repeated index $j$ is summed. The generator $Y/2$ supplies the field's weak hypercharge in the convention $Q={\tau }_3+Y/2$, where $Q$ is its fractional electric charge. The factor $-i$ follows the convention of Hermitian generators. The two connection terms compensate the derivatives of spacetime-dependent gauge transformations.

Write $f=(v+h)/\sqrt{2}$. The Pauli matrices give the weak isospin operatorsFor the Higgs doublet $\Phi =(0,f{)}^T$ with weak hypercharge $Y=1$,The Higgs field direction, including its vacuum expectation value, is electrically neutral. Consequently the electromagnetic U(1) leaves the vacuum invariant: electroweak symmetry breaking preserves electromagnetism and the photon remains massless. The result concerns this neutral vacuum direction; it does not say that both components of the full Higgs doublet have zero electric charge.

Weak isospin labels representations of the SU(2) factor in the electroweak gauge group. A left-handed lepton doublet has $T=1/2$ with third-component eigenvalues $+1/2,-1/2$; a right-handed charged-lepton singlet has $T=0$. Along with electric charge, the third component determines the weak neutral current through $T^3-Q{\sin }^2{\theta }_W$.