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For the longitudinal state,Insert this and $p^{\ast }=m_W/2$ into the phase-space formula:Rotational invariance makes the integrated decay probability independent of the orientation of the initial spin. Both transverse polarizations therefore have the same rate. An unpolarized sample averages these three equal rates; it does not add three copies. Hence its width into one specified charged-lepton flavour is also $g^2{m}_W/(48\pi )$.

At a hadron collider, select $W\to \ell \nu$ events and reconstruct the transverse massusing missing transverse momentum for the neutrino. Fit the shape, especially the region above the nominal Jacobian edge, with finite-width resonance templates convolved with detector resolution, recoil modelling and backgrounds. The Breit-Wigner width controls this off-shell tail. Calibrated momentum and missing-momentum resolutions must be separated from the physical width. This is a lineshape measurement; the lifetime is far too short for a direct flight-distance measurement.

Assume only Standard Model decay channels, negligible final-state masses, three lepton families, three quark colours and a unitary CKM matrix. The top-quark channels are kinematically closed. Summing the allowed quark channels gives $3{\sum }_{i=u,c}{\sum }_{j=d,s,b}|V_{ij}{|}^2\simeq 6$ times one leptonic width, while the leptons contribute three. Neglecting QCD and electroweak radiative corrections,The source's rounded CKM entries give almost the same factor; their tiny nonunitarity is not physically significant. ConsequentlyThe propagated uncertainty uses $\delta g/g=\frac{1}{2}\delta {\Gamma }_W/{\Gamma }_W$, taking $m_W$ as fixed. It is the measurement uncertainty within the stated tree-level approximation; omitted radiative corrections are an additional modelling limitation. In particular QCD increases the hadronic width and lowers the inferred coupling when included.

The printed electromagnetic comparison $4\pi /137$ is $e^2$, not the probability amplitude coupling $e$. Compare like quantities:Equivalently ${\alpha }_W=g^2/(4\pi )\simeq 0.0347$ exceeds ${\alpha }_{\mathrm{EM}}\simeq 1/137$. The weak interaction is weak at low energy primarily because of the massive propagator, rather than an exceptionally small fundamental coupling.

Use leptonic decays: an isolated high transverse momentum positive electron or muon, together with substantial missing transverse momentum from the neutrino, provides a clean $W^{+}$ signature. The lepton transverse momentum has a Jacobian scale near $m_W/2$, and the lepton–missing-momentum transverse mass has an edge near $m_W$, broadened by width, recoil and resolution. Lepton identification, isolation and a prompt primary-vertex origin suppress Jets faking leptons and heavy-flavour decays. Missing-momentum quality requirements suppress mismeasured multijet events; a veto on a second oppositely charged lepton suppresses $Z\to {\ell }^{+}{\ell }^{-}$ when both are reconstructed. Hadronic $W$ decays have a much larger ordinary-QCD background.

The flavour states $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$ have definite Strangeness and are distinguished by flavour-sensitive production or decay. The propagation states diagonalize the effective Hamiltonian $H_{\mathrm{eff}}=M-i\Gamma /2$; their complex eigenvalues give a pole mass and a decay width. The common flavour combinations specified in the source assume the usual CPT-symmetric two-state description and are normalized with $|p{|}^2+|q{|}^2=1$.

Choose the phase convention $CP|K^0⟩=|{\bar{K}}^0⟩$. If mixing conserves CP, $p=q$ can be chosen, and the symmetric and antisymmetric combinations are CP even and odd. More generally the equality must be understood with the corresponding flavour-state phase convention; $|p|=|q|$ alone is not a statement fixing an arbitrary written phase.

The almost CP-even state decays rapidly to two pions, whereas the almost CP-odd state is largely restricted to slower channels such as three pions. Their very different lifetimes are easier to distinguish than their tiny mass splitting, motivating the names $K_S^0$ and $K_L^0$. In the source's first notation the subscript $L$ means light, not long-lived: the light state is identified with the short-lived kaon, and the heavy state with the long-lived kaon. Below, $\ell ,h$ label light/heavy to avoid that collision.

Initial-state radiation of gluons gives the W boson nonzero transverse momentum. Its transverse Lorentz boost then changes the charged-lepton $p_T$, washing out the sharp $M_W/2$ edge and permitting events above it. The recoil must be modelled with Quantum chromodynamics and constrained by measured hadronic activity. Independently, the finite W boson decay width spreads its actual invariant mass, also smearing the fixed-mass endpoint. Either is a theoretical broadening mechanism; neither is a detector-resolution effect.

In a relativistic two-body decay, four-momentum conservation in the parent's rest frame requires ${\mathbf{p}}_1^{\ast }=-{\mathbf{p}}_2^{\ast }$ andFor fixed daughter masses, this fixes a unique $p^{\ast }$, independent of direction. In the Lorentz-invariant phase space integral, the Dirac delta distribution enforcing conservation of energy therefore performs the radial momentum integration. Only the solid angle remains to be integrated; the resulting $p^{\ast }$ can be taken outside that integral, even if the matrix element depends strongly on direction.

For three or more final particles, four-momentum conservation generally leaves continuously variable daughter energies and momenta, so the decay rate includes integrations over them, or equivalently over independent invariant masses in a Dalitz plot. Integrating over a variable daughter invariant mass is also necessary when an unstable daughter is treated with a finite decay width rather than a fixed mass.

Useful examples with no net quark flavor are the eta c meson ${\eta }_c=c\bar{c}$ and the eta b meson ${\eta }_b=b\bar{b}$. Examples with nonzero quark flavor are the kaon $K^{+}=u\bar{s}$ and the D meson $D^{+}=c\bar{d}$. These are pseudoscalar mesons: in the lowest quark model configuration the constituent spins combine to $S=0$, the relative orbital angular momentum is $L=0$, and the total angular momentum and parity are $J^P=0^{-}$. For the self-conjugate examples, meson parity and charge conjugation also gives $C=(-1{)}^{L+S}=+1$. Electrical neutrality alone would not imply zero net quark flavor; for example $K^0=d\bar{s}$ has strangeness.

Restoring reduced Planck constant, decay width and mean lifetime satisfy $\Gamma \tau =\hslash$. Thus the representative numbers correspond toThe mean lifetimes differ by at least $3.0\times 10^{10}$. Hidden-heavy-flavor quarkonium can undergo quark-antiquark annihilation into gluons, followed by hadronization. A $0^{-+}$ color-singlet state can annihilate into two gluons. A lightest open-flavor meson cannot simply annihilate through the strong interaction, which conserves net quark flavor. With no lighter accessible final state carrying that flavor, it must undergo a weak interaction changing a constituent's flavor. The small weak interaction amplitude, CKM matrix factors and available Lorentz-invariant phase space produce a much smaller decay width.

Qualification. A decay width of order tens of MeV is not a universal property of flavor-neutral pseudoscalar mesons. The neutral pion has $\Gamma \simeq 7.8\mathrm{eV}$ and decays predominantly by the electromagnetic interaction, a concrete counterexample. Its mass lies below the lightest allowed hadronic final state. The examples above illustrate the intended annihilation-versus-open-flavor comparison; the numerical scale depends on thresholds and selection rules.^[ref]

Assume the charmonium state annihilates through a virtual photon, with lepton universality and negligible higher-order corrections. For a vector meson of mass $M$, the partial decay width into a charged lepton pair contains both the relativistic two-body decay factor ${\beta }_{\ell }$ and a vector-current spin factor:^[ref] For example, contracting the vector polarization sum with the Dirac trace for the two final fermions gives a factor proportional to $M^2+2m_{\ell }^2$, explaining why retaining only ${\beta }_{\ell }$ is insufficient. Using $m_{\tau }=1.777\mathrm{GeV}$, $m_{\mu }=0.1057\mathrm{GeV}$ and $M=3.686\mathrm{GeV}$ givesThe supplied total decay width also gives ${\Gamma }_{\mu \mu }=2.352\mathrm{keV}$ and ${\Gamma }_{\tau \tau }\simeq 0.91\mathrm{keV}$. It cancels from the branching fraction ratio.

The decisive change at $3.770\mathrm{GeV}$ is crossing the open-charm D meson-pair threshold, approximately $3.73\mathrm{GeV}$. The strong interaction can now create a light quark-antiquark pair and rearrange $c\bar{c}$ into $D\bar{D}$, without annihilating the heavy pair. This increases the total decay width from hundreds of keV to tens of MeV. The leptonic decay partial decay width is additionally smaller because $\psi (3770)$ is predominantly a $1{}^3{D}_1$ state, whereas $\psi (3686)$ is predominantly $2{}^3{S}_1$; the short-distance annihilation overlap is suppressed for nonzero orbital angular momentum. A roughly hundredfold increase in total decay width together with an order-ten reduction in leptonic partial decay width explains the order-thousand reduction in branching fraction. Attributing all three orders of magnitude to the threshold alone would miss the second effect.^[ref]

The three structures are the radiative cascadesIn the quark model, the J/psi meson and $\psi (2S)$ are respectively $1{}^3{S}_1$ and $2{}^3{S}_1$ charmonium states. Thus $L=0$, $S=1$, $J=1$, and meson parity and charge conjugation gives $J^{PC}=1^{--}$. The intermediate chi c mesons are $1{}^3{P}_J$: $L=1$, $S=1$, and addition of angular momentum permits $J=0,1,2$. ConsequentlyThese are electric dipole transitions, changing $L$ by one and preserving $S$ to leading order. A singlet $1{}^1{P}_1$ state has $J^{PC}=1^{+-}$ and is not another member of these spin-triplet cascades.

The primary photon energy is $(M_{2S}^2-M_{{\chi }_J}^2)/(2M_{2S})$, about $261$, $171$ and $128\mathrm{MeV}$ for $J=0,1,2$. The secondary photon remains more energetic even after its boost into the parent rest frame. Hence $m^2(J/\psi {\gamma }_h)=M_{{\chi }_J}^2$: with this quantity on the vertical axis, each resonance gives a horizontal band. A finite decay width and detector measurement resolution broaden each band; photon ordering removes the mirrored duplication of an unordered Dalitz plot.

The faint band beside the negative-slope diagonal boundary is $\psi (2S)\to J/\psi {\pi }^0$, followed by ${\pi }^0\to \gamma \gamma$. The neutral pion is predominantly $(u\bar{u}-d\bar{d})/\sqrt{2}$, a ${}^1{S}_0$ pseudoscalar meson with $J^{PC}=0^{-+}$. This production transition violates isospin, accounting for its small rate. To locate the band, set $X=m^2(J/\psi {\gamma }_l)$ and $Y=m^2(J/\psi {\gamma }_h)$. Four-momentum conservation givesFor the neutral pion, $m_{\gamma \gamma }^2=m_{{\pi }^0}^2$ is small, putting the line just inside the $m_{\gamma \gamma }^2=0$ boundary. The allowed $J/\psi \eta$ channel, with the eta meson also $0^{-+}$, produces a parallel line further inside, at $m_{\gamma \gamma }^2=m_{\eta }^2$. The positive-slope edge $X=Y$ instead comes from ordering the two photon energies; proximity to that edge alone would not identify the two-photon state.

Use a clean Z boson sample with final-state radiation, $Z\to e^{+}{e}^{-}\gamma$, and independently calibrated electron four-momenta. For measured $P=p_{e^{+}}+p_{e^{-}}$ and measured photon direction ${\hat{n}}_{\gamma }$, write $p_{\gamma }=E_{\gamma }(1,{\hat{n}}_{\gamma })$. The known Z boson mass suppliesComparing the reconstructed photon energy with this prediction constrains its scale versus energy, detector position and conversion status. In practice fit the $e^{+}{e}^{-}\gamma$ invariant mass distribution, allowing for the physical Z boson decay width, detector response and backgrounds, instead of assigning every event exactly $M_Z$. Select final-state-radiation topologies: an initial-state-radiation photon is not part of the Z boson decay and generally does not satisfy the same mass constraint. The calibration applies over the populated photon energy range; extrapolation to still higher energies needs a validated response model.

Fit radiatively corrected visible cross sections across the Z boson resonance. The line shape determines $M_Z$, the total decay width ${\Gamma }_Z$, and pole normalizationsThe first extraction uses the fitted total and visible partial decay widths. Assuming lepton universality, ${\Gamma }_e={\Gamma }_{\mu }={\Gamma }_{\tau }={\Gamma }_{\ell }$, an absolute leptonic pole cross section gives ${\Gamma }_{\ell }={\Gamma }_Z\sqrt{M_Z^2{\sigma }_{\ell }^0/(12\pi )}$. The hadronic-to-leptonic ratio gives $R_{\ell }={\Gamma }_{\mathrm{had}}/{\Gamma }_{\ell }$. Subtracting every visible channel then yieldsWithout lepton universality, separately fitted visible leptonic widths replace $3{\Gamma }_{\ell }$, with sufficient independent information to disentangle their products with ${\Gamma }_e$.

A second, useful peak-normalization extraction first determines the invisible-to-leptonic ratio. Becauseone obtainsThis ratio extraction does not require the measured line width. Multiplication by an independently predicted leptonic partial decay width, or by ${\Gamma }_Z/(R_{\mathrm{inv}}+R_{\ell }+3)$ from the scan, converts it to ${\Gamma }_{\mathrm{inv}}$. Alternatively dividing $R_{\mathrm{inv}}$ by the predicted single-species ratio ${\Gamma }_{\nu }/{\Gamma }_{\ell }$ gives the number of light active neutrinos. These are two standard uses of line-shape information, with different choices of theoretical input; if both use the same fitted ${\Gamma }_{\ell }$, they are algebraically equivalent and statistically correlated, not independent determinations.

Statistical uncertainties in the reference-process counts give fluctuations in the collider luminosity at individual scan points, propagating into normalization and fitted line width. They decrease with additional reference events. A common multiplicative luminosity-scale systematic largely cancels in $R_{\ell }$ and the normalized line shape, but shifts every absolute peak cross section. It therefore does not cancel from either invisible-width extraction. For a fractional upward luminosity shift $\epsilon$, ${\sigma }_{\mathrm{had}}^0$ shifts by $-\epsilon$, and at fixed $R_{\ell },M_Z,{\Gamma }_Z$,Thus overestimating luminosity increases the inferred invisible contribution. Energy-dependent systematic errors can also distort the line shape and its total decay width. The covariance matrix must retain correlations between scan points and experiments, including common theoretical uncertainty in the Bhabha scattering calculation.

Tag an energetic initial-state radiation photon in $e^{+}{e}^{-}\to \gamma Z^{(\ast )}\to \gamma +\text{invisible}$, observing a single-photon signature with missing energy and missing momentum. At known squared center-of-mass energy $s$, the leading single-photon recoil invariant mass isAbove the Z boson mass, radiative-return events cluster at $m_{\mathrm{recoil}}\simeq M_Z$. At or near the pole the full off-shell recoil spectrum is fitted instead. Folding the Breit-Wigner resonance cross section with the calculable QED radiation kernel gives a predicted tagged rate proportional to ${\Gamma }_e{\Gamma }_{\mathrm{inv}}$ for fixed total decay width. A fit with the known luminosity and visible line-shape parameters therefore determines ${\Gamma }_{\mathrm{inv}}$ directly from invisible events, rather than subtracting visible widths. Comparing identically selected radiative-return events with visible Z boson decays can instead constrain ${\Gamma }_{\mathrm{inv}}/{\Gamma }_{\mathrm{vis}}$ while cancelling much of the luminosity normalization.

The detector needs efficient single-photon triggering, good photon energy and angular resolution, reliable timing, and nearly hermetic charged-particle and particle calorimeter coverage. Forward vetoes reject radiative Bhabha scattering with escaped electrons; additional-photon vetoes reject $\gamma \gamma$ events with a lost photon. Require sufficient transverse photon momentum that a recoiling visible system could not all disappear down a beam hole. Timing rejects cosmic rays and beam backgrounds. Detector inefficiencies and beam holes must be measured, since visible final states with lost particles imitate the signal. The irreducible $e^{+}{e}^{-}\to {\nu }_e{\bar{\nu }}_e\gamma$ contribution from $t$-channel W boson exchange and its interference must be included when isolating the Z boson invisible decay width.

The incident particle obeys the energy–momentum relation $E^2=p^2{c}^2+m_1^2{c}^4$. The stationary target contributes energy $m_2{c}^2$ and zero momentum. A single product inherits the total four-momentum, so its invariant mass satisfiesHenceHere $E$ is the incident total energy, including rest energy. In a one-particle final state, exact production of a specified mass fixes the incident energy; there is no adjustable final relative motion to absorb excess energy. A resonance with a finite decay width relaxes that ideal exact equality over its line shape.

For $\Psi (p_1)\to \chi (p_2)+\varphi (k_1)+\varphi (k_2)$, the sole tree-level Feynman diagram has a Yukawa coupling interaction vertex producing an off-shell scalar field particle of four-momentum $q=p_1-p_2=k_1+k_2$, followed by its cubic scalar field interaction vertex. Its invariant decay amplitude isTwo successive Yukawa coupling interaction vertices would change the fermion species twice and leave a $\Psi$, so they cannot produce this final state. Interchanging the two identical outgoing bosons does not create another Feynman diagram; the cubic factorial is already included in the Feynman rule. The integrated decay width instead includes the identical final-state symmetry factor $1/2!$.

The printed integration labels need correction. The parent's four-momentum is fixed, so the two Lorentz-invariant phase-space measures must integrate $p_2$ and $p_3$, rather than the printed $p_1$ and $p_2$. Write $P=p_1=(M,0)$ and $p=p_2$, $k=p_3$. The physical decay width isThe spatial Dirac delta function sets $\mathbf{k}=-\mathbf{p}$. Performing the angular integral in spherical coordinates, with $E=\sqrt{p^2+m^2}$, gives the integrated two-body decay phase spaceThe positive root is $p_{\ast }=\sqrt{M^2/4-m^2}$. Since $d(2E)/dp=2p/E$, the Dirac delta function integration yields ${\Phi }_2=p_{\ast }/(4\pi M)$. ThusThe two final particles have different species, so there is no identical final-state symmetry factor. The result tends to $g^{2}M/(16\pi )$ as $m\to 0$ and vanishes at the two-particle decay threshold.

Put $Q=M-3m$. For $Q<0$ the decay width is zero by conservation of energy. Just above the three-particle decay threshold, all three daughters are nonrelativistic. The decay amplitude of iv remains finite: the internal scalar propagator denominator approaches $(2m{)}^2-m^2=3m^2$, and the Dirac scalar bilinear does not vanish at rest. Thus, for nonzero fixed $g$ and $\lambda$, the leading suppression comes entirely from three-body threshold phase space.

Use the spatial Dirac delta function to eliminate one daughter momentum. The remaining integral has six independent momentum coordinates and one conservation of energy constraint,Rescaling both momenta by $\sqrt{mQ}$ gives $Q^3$ from the measure and $Q^{-1}$ from the Dirac delta function. The smooth factors $1/(2E)$ do not change the power. Consequentlywith a finite positive leading coefficient when $g\lambda \ne 0$; $\Gamma =0$ for $m\ge M/3$. The $1/2!$ for the identical bosons changes only this coefficient. Equivalently, at fixed $M$ the opening is quadratic in $M/3-m$. If $g\lambda =0$, this leading tree-level Feynman diagram vanishes, so the stated leading coefficient is zero.

Take $\varphi (p_1)+\bar{\varphi }(p_2)\to \varphi (p_3)+\bar{\varphi }(p_4)$ with the Mandelstam variables of i. The allowed tree-level Feynman diagram is the crossed $u$-channel $\Phi$ exchange in panel (c) of b's figure. At one end the incoming particle and outgoing antiparticle transfer charge $+2$; the other end absorbs that charge. The zero-charge $s$- and $t$-channel assignments are forbidden by the interaction vertices. ThusIn the center-of-mass frame with $|\mathbf{p}|\ll m$, $s=4m^2+O(|\mathbf{p}{|}^2)$ and $u=O(|\mathbf{p}{|}^2)$. For $M\gg m$, both scattering amplitudes approach $g^2/M^2$. The elastic two-body Lorentz-invariant phase space and incident flux agree for the two experiments. Usingfor distinguishable final particles, and an extra factor $1/2!$ when integrating the identical-particle final state over the whole sphere, givesThere is no additional division for identical incoming particles in the definition of a beam scattering cross-section. Before taking $M/m\to \infty$, the zero-momentum ratio is $\frac{1}{2}[M^2/(M^2-4m^2){]}^2$, provided the threshold is away from the heavy pole.

Near $s=M^2$, the bare $s$-channel scalar propagator makes fixed-order perturbation theory fail: repeated self-energy insertions contain powers of ${\Sigma }_{\Phi }(s)/(s-M^2)$. The $\Phi \to \varphi \varphi$ channel is open for $M>2m$, giving the leading decay widthResumming the self-energy gives a finite Breit-Wigner resonance cross section, with denominator $s-M_{\mathrm{pole}}^2+i{M}_{\mathrm{pole}}{\Gamma }_{\Phi }$ near the pole, rather than an infinite scattering cross-section. Also, $s\simeq M^2$ with $M\gg m$ lies outside the assumed nonrelativistic incoming kinematics; the ratio $1/2$ is not a resonance prediction.

Interpret the repeated species label in the Yukawa interaction as summed over $i=1,2$: each Feynman vertex is diagonal in species and equals $-ig$. Define the invariant scattering amplitude by $S-1=i(2\pi {)}^4{\delta }^{(4)}(P_f-P_i)\mathcal{M}$. There is one tree-level Feynman diagram, an $s$-channel scalar exchange. In the left panel, $j=2$ for this part; the right panel only occurs in d. Arrows show fermion-number flow, so an antifermion arrow opposes its physical momentum.

Writing $k=p_1+p_2=p_3+p_4$ and $s=k^2$, the momentum-space Feynman rules giveThe scalar carries no spin label. Average over the $2\times 2$ incoming spin states and sum the four outgoing combinations. For the real scalar Yukawa interaction, the conjugate of ${\bar{v}}_2{u}_1$ is ${\bar{u}}_1{v}_2$. The fermion spin sums then yieldThe bar denotes the initial spin average, not a Dirac adjoint. The gamma matrix trace identities giveand the second trace has the same value because $p_3+p_4=p_1+p_2$ and all external masses are $m$. ThereforeThis is the fixed-order expression away from $s=M^2$. If that pole is accessible, a physical scattering resonance requires a scalar self-energy and decay width; the tree formula alone is not finite there.

With the conventional particle labels, the allowed cubic decay channels areEach has one tree-level Feynman diagram, consisting of one cubic Feynman vertex. The final identical-leg contractions supply $2!$ and cancel the $1/2!$ in the Lagrangian density.

Taken literally with these labels, the printed $\Phi \to \varphi \varphi$ channel is forbidden: its initial Noether charge is $-2$ and its final Noether charge is $+2$. Its decay amplitude is zero, at every order preserving the global phase symmetry. If the source uses “$\Phi$” to name the heavy antiparticle instead, its intended answer is the nonzero decay amplitude $-h$ for $\bar{\Phi }\to \varphi \varphi$. The following decay width calculation concerns either allowed channel, whose widths are equal by charge conjugation.