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For an allowed decay channel in iii, $|\mathcal{M}{|}^2=h^2$. Integrating the spatial Dirac delta distribution in the printed Lorentz-invariant phase-space measure sets ${\mathbf{p}}_2=-{\mathbf{p}}_1$. With $p=|{\mathbf{p}}_1|$ and $E_p=\sqrt{p^2+m^2}$, its remaining value isThe positive root is $p_{\ast }=(M/2)\beta$, where $\beta =\sqrt{1-4m^2/M^2}$, and the Dirac delta distribution contributes the Jacobian determinant $E_{\ast }/(2p_{\ast })$. HenceIt follows that the literal printed integral evaluates toThe printed integration ranges count each final configuration twice, once as $({\mathbf{p}}_1,{\mathbf{p}}_2)$ and once in the opposite order. For two identical final bosons, the physical decay width therefore requires an additional identical final-state symmetry factor $1/2!$:There is one allowed decay channel per heavy particle, not an extra factor of two from summing the two charge-conjugate heavy species. Both expressions vanish below the two-body decay threshold $M=2m$, and the limiting value at threshold is zero. For the literally forbidden labels $\Phi \to \varphi \varphi$, the decay width is instead zero because its decay amplitude vanishes, as explained in iii.

For $g=0$, the invariant scattering amplitude from the tree-level Feynman diagram in i has a pole at $s=M^2$. The intermediate heavy antiparticle can go on shell and live for a time of order its mean lifetime. Near this scattering resonance, inserting a single self-energy is no longer a uniformly small correction. Dyson resummation gives, to leading accuracy in a narrow decay width,Here $M$ is understood as the renormalized resonance mass; the real self-energy shifts the original pole position. The imaginary term removes the unphysical divergence of the fixed-order invariant scattering amplitude.

For equal incoming and outgoing light masses, the two-body momentum ratio cancels in the cross section. The angle-independent invariant scattering amplitude, integrated over the full solid angle with the identical final-state symmetry factor $1/2!$, gives $\sigma =|\mathcal{M}{|}^2/(32\pi s)$. Thus the Breit-Wigner resonance cross section isAt $s\gg M^2$ it reproduces $h^4/(32\pi s^3)$. At the resonance peak,This is also $8\pi /k_{\ast }^2$ for $k_{\ast }=M\beta /2$, the elastic identical-boson S wave partial-wave unitarity limit when this is the only open decay channel. The approximation assumes ${\Gamma }_{\mathrm{phys}}\ll M$ and a resonance sufficiently far from threshold for the width to vary slowly across it.

If one mechanically uses the unsymmetrized ${\Gamma }_{\mathrm{printed}}=2{\Gamma }_{\mathrm{phys}}$ from d, the same displayed Breit-Wigner resonance cross section would have that doubled width and predict a peak $8\pi /(M^2{\beta }^2)$. This is the literal-width substitution, rather than the consistent physical result: the given high-energy cross section already includes the identical-final-particle factor that the printed decay width integral omits.

For exponential decay with mean lifetime $\tau$, the state amplitude evolves as $\exp [-i(E_0-i\Gamma /2)t/\hslash ]$ and its survival probability is $e^{-\Gamma t/\hslash }$. Hence the energy decay width isThe exceptionally small energy width expresses the long lifetime of this weak decay. Use $\hslash$, not $h$, because $\Gamma$ is an energy width in the complex-energy convention; changing to a frequency convention introduces the appropriate factor of $2\pi$. The quoted lifetime is treated as the mean lifetime. If a half-life were supplied instead, one would first use $\tau =t_{1/2}/\ln 2$. No extra logarithm is needed for the given mean-lifetime interpretation. This calculation uses the source's approximate 15-minute value rather than silently replacing it with a more precise contemporary neutron-lifetime measurement.

Evaluate the two-body Lorentz-invariant phase space in the parent's rest frame, $p=(m,0)$. The spatial Dirac delta function sets ${\mathbf{q}}_2=-{\mathbf{q}}_1$. For massless daughters $E_1=E_2=q$, soThe factor $1/2$ in the last step is the derivative of the delta-function argument. Substituting into the decay width gives, for labelled or distinct daughters,The angular domain is one sphere: $0\le \phi <2\pi$ and $-1\le \cos \theta \le 1$, equivalently $0\le \theta \le \pi$. The printed negative-to-positive polar-angle interval is not a second independent coverage of the sphere.

For identical daughters, integration over the full labelled two-body phase space counts each physical final state twice. One must therefore include an additional factor $1/2!$, or equivalently integrate over an unordered half-domain. This distinction is necessary for the two identical photons in solution.

With outgoing physical photon polarization vectors, the Feynman rule givesThe overall phase does not affect the decay width. Replacing either polarization vector by its own four-momentum gives zero by antisymmetry. Thus the Ward identity permits replacement of each physical photon polarization sum by $-{\eta }^{\mu \nu }$ inside this amplitude. The printed sum is not an equality of free-standing physical two-polarization tensors: for a photon moving along $z$, its physical $00$ component is zero whereas $-{\eta }_{00}=-1$. Gauge-dependent longitudinal terms cancel only after contraction with the transverse amplitude.

The contraction of the two Levi-Civita symbols givessince $q_1^2=q_2^2=0$ and $2q_1\cdot q_2=m_a^2$. No initial spin average is needed for the spin-zero axion. The result is independent of direction in its rest frame. Including the identical-particle factor explained in solution therefore givesOmitting $1/2!$ would overestimate the physical rate by two. The normalization agrees with the standard axion-photon convention.^[ref]

The action is dimensionless and $[d^{4}x]=-4$, so $[\mathcal{L}]=4$. A canonical scalar kinetic term then requires $2+2[\varphi ]=4$, and the electromagnetic kinetic term requires $2[F_{\mu \nu }]=4$. ThusThe brackets here denote mass dimension. The Lagrangian has no explicit coordinate dependence, and all indices are contracted as Lorentz scalars. Its continuous spacetime symmetry is the Poincaré group. Noether's theorem gives conserved four-momentum from translations and conserved angular momentum, including the boost generators, from Lorentz transformations. The nonzero mass $M$ and dimensionful couplings prevent a dilation symmetry.

Sign qualification: the printed electromagnetic kinetic term has the opposite healthy relative sign to the scalar kinetic terms. With the stated mostly-minus Minkowski metric, a physical Maxwell Lagrangian is $-F_{\mu \nu }{F}^{\mu \nu }/4$. The printed plus sign instead gives photon Hamiltonian density $-({\mathbf{E}}^2+{\mathbf{B}}^2)/2$; a nonzero purely electric configuration already has negative energy, unbounded below as its amplitude grows. The physical decay widths below use the healthy Maxwell sign with the given interaction normalization. The dimensional and spacetime-symmetry conclusions themselves do not depend on this sign correction.

All daughter particles are massless, so the parent mass $M$ supplies the only kinematic scale at tree level. Each vertex has two derivatives, making its scattering amplitude scale as ${\lambda }_i{M}^2$. For the four-$b$ vertex, symmetrizing over the derivative-bearing legs yields a sum of pairwise four-momentum products, which is proportional to $M^2$ by four-momentum conservation.

For $n$ massless daughters, the Lorentz invariance Lorentz-invariant phase-space measure scales as $M^{2n-4}$: each $d^{3}q/(2E)$ contributes two powers and the four-dimensional Dirac delta removes four powers. Including the parent factor $1/(2M)$ gives $\Gamma \sim |{\lambda }_i{|}^2{M}^{2n-1}$. ThusThe unspecified factors are dimensionless angular integrals and identical-particle factors. Each result has mass dimension one, as a decay width must in natural units.

Counting the mass dimensions of each interaction operator givesIndeed $\Phi (\partial a{)}^2$ and $\Phi F^2$ have dimension five, whereas $\Phi b^2(\partial b{)}^2$ has dimension seven. These are interactions of an effective field theory with dimensionful couplings.

The direct one-vertex decay channels are $\Phi \to aa$, $\Phi \to bbbb$ and $\Phi \to \gamma \gamma$. Each leading scattering amplitude is linear in its corresponding coupling, and a decay width is quadratic in the amplitude. Therefore, holding the parent mass fixed,There are no tree-level interference terms between these different final states.

Every interaction contains an even number of $b$ fields, so the exact discrete symmetry $b\mapsto -b$ forbids $\Phi \to bbb$ to every perturbative order. There is no such prohibition for $\Phi \to bb$.

The latter decay channel is absent at tree level in the displayed interactions, but contracting two $b$ legs at a $\Phi b^2(\partial b{)}^2$ vertex produces a tadpole diagram with two external $b$ legs. For example, contracting the undifferentiated pair leaves the local operator $\Phi (\partial b{)}^2$ multiplied by an integral proportional to $\int d^4\ell /({\ell }^2+i0)$. Contracting the differentiated pair instead gives a local $\Phi b^2$ contribution. Both are allowed effective field theory operators. With a ultraviolet cutoff, these local terms generically carry quadratic or quartic cutoff dependence and require matching coefficients or counterterms.

There is a relevant qualification to a claim that the one-loop decay width must be nonzero. Since $b$ is exactly massless here, the one-vertex tadpole integrals have no external scale and are scaleless integrals. They vanish in dimensional regularization. Thus the correct conclusion is allowed, and generically generated by quantum corrections or matching, not a regulator-independent nonzero one-loop prediction from ${\lambda }_b$ alone. No exact symmetry protects the two-$b$ channel; its numerical rate requires the renormalized coefficients of the allowed operators.

Work in the parent rest frame, $p=(M,0)$. For massless daughters, the spatial Dirac delta fixes ${\mathbf{q}}_2=-{\mathbf{q}}_1$ and $E_1=E_2=q$. The Lorentz invariance two-body Lorentz-invariant phase-space measure becomesThe factor $1/2$ in the last step is the Jacobian of the energy-conservation Dirac delta. Consequently the labelled two-body decay width isFor two identical final particles, the full angular integral counts each unordered final state twice, so the physical decay width includes an additional factor $1/2!$. This factor is needed for both channels in part e; it is separate from any vertex combinatorics already included in the Feynman rule.

The massless two-body kinematics gives $q_1\cdot q_2=M^2/2$. The scalar scattering amplitude therefore has constant magnitude $|{\mathcal{A}}_{aa}|=|{\lambda }_a|M^2/2$. Use $\int d\Omega =4\pi$ and the identical-particle factor from part d:For the photons, the two-photon vertex of a scalar field-strength interaction following from $\Phi F_{\mu \nu }{F}^{\mu \nu }$ is, up to an irrelevant overall phase,Vertex qualification: the source prints same-index polarization contractions in the second product, whereas the contractions must be crossed as above. The printed expression fails the Ward identity. For a concrete counterexample take $q_1=(E,0,0,E)$, $q_2=(E,0,0,-E)$ and ${ϵ}_2=(1,1,0,-1)$. This is a transverse unit spacelike photon polarization vector, since $q_2\cdot {ϵ}_2=0$ and ${ϵ}_2^2=-1$. Replacing ${ϵ}_1$ by $q_1$ in the printed expression leaves $4{\lambda }_{\gamma }{E}^3$ instead of zero. The corrected crossed expression vanishes under either such replacement.

Choose physical spatial transverse photon polarizations along $x$ and $y$ in the parent rest frame. Then both crossed products vanish; the two equal-polarization combinations have $|{ϵ}_1\cdot {ϵ}_2{|}^2=1$ and the two orthogonal combinations give zero. There is no initial-spin average for a spin-zero parent. HenceThe calculation in this spatial transverse gauge also yields the intended result from the misprinted expression, but only the corrected vertex makes its gauge invariance explicit. For equal-magnitude couplings, the photon decay width is twice the two-scalar decay width because of its two nonzero polarization combinations.

At $E=E_R$, the Breit-Wigner resonance cross section loses its explicit dependence on the total decay width after writing ${\Gamma }_{ab}=B_{ab}\Gamma$ and ${\Gamma }_X=B_X\Gamma$. For an unpolarized ultrarelativistic electron-Positron initial state and a spin-one resonance, $g=3/4$ and $p_i=E_R/2$, giving ${\sigma }_{\mathrm{peak}}=12\pi B_e{B}_X/E_R^2$ in natural units. An elastic resonant channel therefore measures the square $B_e^2$, rather than one power of the electronic branching fraction.

Section 62.1 distinguishes inner bremsstrahlung from structure-dependent radiative pion decay and their interference. Equations (62.3)–(62.4) give the double differential decay width in terms of the scaled photon and charged-lepton energies and the squared lepton-to-meson mass ratio.

At SPEAR, counterrotating electron and Positron beams collided inside a magnetic particle detector. Finer steps in stored-beam energy resolved a sharp production peak. A magnetometer and measured orbit positions monitored the energy settings; Bhabha scattering normalized the cross sections. The paper attributed the observed width to beam-energy spread and radiation effects, reporting an upper limit on intrinsic width. Its muon identification system was not functioning: the non-electron two-body excess was not conclusively a muon pair.^[ref]

A scan must compare normalized production rates, since more selected events can result from more beam encounters or different detection efficiency. A proposed diagnostic is to revisit energies on either side of the peak while monitoring the energy calibration and event selection. Such neighboring settings test the local production background; their event totals alone are not an independent measurement of the intrinsic decay width.

The tau is the charged lepton of the third fermion generation. In the massless approximation its charged-current W boson partial decay width equals those of the electron and muon, reflecting lepton universality.

The tau lepton has electric charge $-1$ in units of the positive elementary electric charge; ${\tau }^{+}$ is its antiparticle. Its weak charged current decays make the tau lepton useful for testing lepton universality. They also illustrate how particle detection can establish a short-lived parent through visible daughters and a consistent invisible recoil.

For three massive daughters with a smooth nonzero decay amplitude at the decay threshold, the integrated Lorentz-invariant phase space vanishes as $Q^2$, where $Q$ is the parent's mass minus the sum of daughter masses. The spatial Dirac delta function leaves six independent momentum coordinates. Rescaling each by $Q^{1/2}$ contributes $Q^3$, while the conservation of energy Dirac delta function contributes $Q^{-1}$. The factors $1/(2E)$ are smooth nonzero constants at the decay threshold. Thus the decay width also scales as $Q^2$ when the decay amplitude has a nonzero constant term. A vanishing decay amplitude, a massless daughter, or a threshold singularity can change this law. In three spatial dimensions the same argument gives $Q^{(3N-5)/2}$ for $N$ massive daughters.

With CPT symmetry, write the propagating states of neutral meson mixing as $p|B^0⟩\pm q|{\bar{B}}^0⟩$, defining the coefficient ratio $q/p$. Let $\Delta m$ be their mass splitting and use units $\hslash =1$.

For rates $f_{\pm }$ tagged as $B^0,{\bar{B}}^0$ and $\Delta t=t_f-t_{\mathrm{tag}}$, neglect the decay width difference and take $|q/p|=1$. DefineHere $A_f,{\bar{A}}_f$ are decay amplitudes. The tagged asymmetry is $(f_{+}-f_{-})/(f_{+}+f_{-})=S\sin (\Delta m\Delta t)-C\cos (\Delta m\Delta t)$ in this convention.^[ref]

For a single weak phase in a charmonium-kaon CP eigenstate with eigenvalue ${\eta }_f$, ${\lambda }_f\simeq {\eta }_f{e}^{-2i\beta }$ gives $S=-{\eta }_f\sin 2\beta$, $C\simeq 0$, withThis connects the CKM unitarity triangle to quantum interference between direct decay and decay after neutral meson mixing.^[ref]

For an independent teaching example choose $x=\Delta m\Delta t$, ${\lambda }_f=4i/3$ and an arbitrary common positive envelope $F(\Delta t)$. Then $S=24/25$, $C=-7/25$ andBecause $(24/25{)}^2+(7/25{)}^2=1$, both rates remain nonnegative. At $x=0$, the asymmetry is $7/25$; at $x=\pi /2$, it is $24/25$. These chosen values illustrate the sign convention and are not experimental estimates. More generally $S^2+C^2=1-[2\mathrm{Re}{\lambda }_f/(1+|{\lambda }_f{|}^2){]}^2\le 1$.

If a tag flips independently with probability $w$, exchanging $f_{+}$ and $f_{-}$ for that fraction of events gives $A_{\mathrm{observed}}=(1-2w)A_{\mathrm{tag}}$ by direct subtraction. The plotted toy choice $w=1/5$ reduces the amplitude to $3/5$ of its ideal value; at $w=1/2$ the tag contains no flavour information. Finite time resolution additionally averages over nearby oscillation phases. Extracting a weak phase consequently requires calibrated readout as well as a CKM matrix prediction.

Here $x=m_W^2/m_t^2$, and the tree-level decay width neglects the bottom quark mass. The open two-body W channel gives a top quark lifetime of a few $10^{-25}\mathrm{s}$, shorter than the hadronization time. A fifth-power Sargent's rule extrapolation misses this physical regime.^[ref]

With three massless lepton families, two accessible up-type quarks, three colors, and no Quantum chromodynamics radiative correction, a W boson has total decay width $9{\Gamma }_{\ell }$. With an identity CKM matrix, each leptonic channel has branching fraction $1/9$ and each of the two allowed quark-flavour channels has $1/3$. With mixing, each accessible quark channel is weighted by $|V_{ij}{|}^2$, while row unitarity preserves the total at this order if all down-type channels are accessible.

In the rest frame of the parent particle, four-momentum conservation requires $M=E_1+E_2\ge m_1+m_2$. For two equal masses the daughter momentum magnitude is $\frac{1}{2}\sqrt{M^2-4m^2}$; the two-body decay phase space closes at $M=2m$. For a finite constant decay amplitude, the decay width tends to zero at this threshold.