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Count strangeness before drawing a weak interaction diagram. An Omega baryon has $S=-3$, a Lambda baryon has $S=-1$, a negatively charged kaon has $S=-1$, and a pion has $S=0$. Thus the two nonleptonic baryon decays requireThe strong interaction cannot supply either net change of strangeness.

In the upper Feynman diagram, one initial strange quark undergoes the weak charged current transitionThe remaining two strange quarks are spectator quarks. Combine the outgoing $u,d$ and one spectator $s$ into the Lambda baryon, and the remaining $s$ with $\bar{u}$ into the kaon. This uses one W boson propagator and two weak interaction vertices. Its Cabibbo-Kobayashi-Maskawa matrix factor has magnitude $|V_{us}{V}_{ud}|\simeq \sin {\theta }_C\cos {\theta }_C$.

For the lower Feynman diagram, let the upper initial $s$ first emit a $W^{-}$ and become a virtual $u$, then emit a $W^{+}$ and become a $d$. The $W^{+}$ is absorbed by a second initial $s$, turning it into a $u$; the third initial $s$ remains a spectator quark. The emitted $W^{-}$ produces $d\bar{u}$. The complete quark flavour flow is thereforeThe final $d,u,s$ form the Lambda baryon and the other $d\bar{u}$ form the pion. Electric charge conservation holds at every vertex. Internal lines are virtual; this is a contribution to one decay amplitude, not a sequence of freely propagating on-shell constituent-quark decays.

This is a possible lowest-weak-order diagram for a second-order strangeness-changing weak decay: it has two W boson propagators and four weak interaction vertices. Its Cabibbo-Kobayashi-Maskawa matrix factor has magnitude $|V_{us}{|}^2|V_{ud}{|}^2$ in the displayed light-quark topology. Other contributions can exist, but a single charged-current insertion cannot turn two net strange quarks into nonstrange ones. No extra loop Feynman diagram is compulsory: the displayed two-W tree already realizes the required quark flavour change.

At low energies, one-W exchange produces one power of the Fermi constant in an effective decay amplitude; the second process requires an additional weak insertion. Hadronic matrix elements and propagators supply the accompanying dimensions, so a numerical ratio cannot be obtained by simply squaring a dimensionful constant. The robust comparison isThe dominant reason is the extra weak order, with additional Cabibbo suppression in this representative topology. It is not a comparison between a strong decay and a weak decay.

Conservation laws. Use natural units $c=1$ and write $p_i^{\mu }=(E_i,{\mathbf{p}}_i)$ with the $+---$ metric. In the parent's rest frame, which is the final-state centre-of-momentum frame, $P^{\mu }=(M,0)$. Four-momentum conservation and the relativistic energy-momentum relation giveA real massless daughter has strictly positive energy. The kinematic threshold is $M\ge {\sum }_i{m}_i$; below, no decay satisfying these conditions exists. The endpoint discussion assumes open phase space, $M>{\sum }_i{m}_i$, to avoid collapsed threshold configurations.

Two-body energies. Momentum conservation requires ${\mathbf{p}}_2=-{\mathbf{p}}_1$, so both daughters have the same momentum magnitude $p$. Their on-shell relations implyand subtraction eliminates $p$:Since $E_1+E_2=M$, solving the sum and difference equations gives the relativistic two-body decay resultAs a check, their common momentum magnitude iswhere $\kappa$ is the Källén function. It vanishes at the massive two-body threshold and is real on the allowed side. If the symbols $M,m_i$ denote masses in kilograms rather than rest energies, multiply the displayed energy formulas by $c^2$ and the displayed momentum formula by $c$.

Minimum energy in three-body decay. The on-shell inequality already gives $E_1\ge m_1$. For $m_1>0$, the bound is attainable by putting daughter 1 at rest and letting daughters 2 and 3 recoil back to back:The last equation has a solution because $M-m_1>m_2+m_3$; its left side increases continuously from $m_2+m_3$ to infinity. ThusIf $m_1=0$, the same formula describes the lower boundary $E_1\to 0^{+}$. It is an infimum, not an attained physical zero-energy massless particle: this is a soft-particle endpoint.

Maximum energy from the recoil mass. Group daughters 2 and 3 into a recoil system with four-momentum $Q=p_2+p_3=P-p_1$ and squared invariant mass $s_{23}=Q^2$. Directly in the parent rest frame,Maximizing $E_1$ therefore means minimizing the recoil invariant mass. This approach does not assume a recoil rest frame when that system is null.

If both recoil daughters have positive masses, use their pair rest frame. Their total energy there is at least $m_2+m_3$, with equality when both are at rest relative to one another. Equivalently, their future-directed four-velocities satisfy $p_2\cdot p_3\ge m_2{m}_3$, givingEquality means the two recoil daughters have a common velocity in the parent frame: their momenta are parallel and ${\mathbf{p}}_2/m_2={\mathbf{p}}_3/m_3$. Daughter 1 recoils in the opposite direction. HenceAt this endpoint their total energy and its sharing areThese explicitly exhibit a permitted configuration attaining the bound when both recoil masses are positive.

If exactly one recoil daughter is massless, say $m_3=0$ and $m_2>0$, work in the rest frame of daughter 2. For any positive photon-like daughter energy there,The lower recoil-mass bound is approached only as $E_3^{(2)}\to 0^{+}$. Thus the same general formula with $m_3=0$ is the supremum of $E_1$ for a genuine three-body event, rather than an attained maximum. Interchanging 2 and 3 covers the other mixed case. The distinction is important because the paper's category “at least one has mass” includes this soft-particle endpoint as well as the attainable two-massive-particle case.

If both recoil daughters are massless, instead computeIt can be exactly zero at nonzero $E_2,E_3$ when their momenta are parallel. Then $|\mathbf{Q}|=E_2+E_3$, and momentum conservation makes $|{\mathbf{p}}_1|=|\mathbf{Q}|$. Solving $E_1+|{\mathbf{p}}_1|=M$ and $E_1^2-|{\mathbf{p}}_1{|}^2=m_1^2$ givesBoth recoil daughters can have positive energies: set $E_2=x|{\mathbf{p}}_1|$ and $E_3=(1-x)|{\mathbf{p}}_1|$ for any $0<x<1$. This endpoint is attained without a soft daughter, even though their combined four-momentum is null and has no rest frame.

Combining the three-body decay energy endpoints, the closed kinematic range isFor masses in ordinary units, multiply both endpoint expressions by $c^2$. Open endpoints occur precisely in the soft limits described above. These bounds specify kinematic possibilities; the decay amplitude can still vanish at a boundary, so a possible endpoint need not have a nonzero event density.

Ignoring the requested final-mass difference and initially taking equal hadronic matrix elements, the common $W^{-}\to d\bar{u}$ factor cancels. A decay width is proportional to the squared decay amplitude, soWith the supplied Cabibbo angle,The quoted measurement, about $500\pm 10$, is larger by a factor $500/19.75\simeq 25.3$. If that whole difference were attributed to a ratio of other amplitudes, it would require a factor $\sqrt{25.3}\simeq 5.03$.

This does not invalidate quark mixing. The supplied Cabibbo-Kobayashi-Maskawa matrix is only a schematic hierarchy in powers of $\sin {\theta }_C$, particularly for $V_{ub}$; its omitted coefficients matter greatly after squaring. Equal phase space also does not imply equal hadronic decay amplitudes: the two transitions have different hadronic form factors and strong-interaction dynamics, and loop contributions can interfere with the charmless tree amplitude. Thus the calculation correctly predicts enhancement of the charmed channel, but the supplied CKM hierarchy alone does not quantitatively predict the measured ratio. It would be unjustified to treat $19.75$ as a precision prediction with only the quoted experimental error.

In the simple box-state counting used here, the final three-momentum lies in a spherical shell: $dn=C{p}^{2}dp$, where the common constant $C$ contains the normalization volume and angular factor. Momentum conservation fixes the other final particle's momentum. Hence the density of states isCombining parts (a) and (b) givesWith the paper's rounded masses,By Fermi's golden rule, equal squared matrix elements make the rate ratio equal to this density ratio. Thus the naive constant-matrix-element estimate is $R\simeq 0.287$, which favours the electronic decay by about $3.49$ to one.

This is the particular noncovariant state-counting convention in the question. A calculation with the Lorentz-invariant phase-space measure places factors $1/(2E)$ in the measure and correspondingly normalizes its decay amplitude differently. Only the complete product of phase space and squared matrix element is convention-independent; mixing the two conventions would give a wrong rate.

The charged pion has valence content ${\pi }^{-}=d\bar{u}$. Its constituents annihilate into a virtual W boson, followed by the lepton pair:

The quark vertex carries the $u$--$d$ Cabibbo-Kobayashi-Maskawa matrix element, and the leptonic vertex is the weak charged current. The pion's nonperturbative spatial structure is encoded in the pion decay constant when evaluating the decay amplitude.

The weak charged current creates a negative charged lepton in a left-chiral state. For a massless particle, that would have negative helicity; part (e) instead requires positive helicity. A nonzero mass supplies the necessary wrong-helicity component. This helicity suppression is extremely strong for the fast electron, but much weaker for the slow muon.

For a positive-helicity charged lepton with spinors normalized to unit probability, the left-chiral weight is $(1-{\beta }_{\ell })/2$. With the box-state normalization of part (c), this gives the useful consistent rate ratioIndeed $1-{\beta }_{\ell }=2m_{\ell }^2/(m_{\pi }^2+m_{\ell }^2)$ cancels the extra energy-dependent factor in that density ratio, leaving the standard pion electron-to-muon decay ratio in inverted form:Using the rounded masses in this paper gives $R\simeq 8185$, close to the stated order of $8000$. Precision predictions require more accurate masses and electromagnetic radiative corrections. The essential mechanism is the charged-lepton-mass dependence of the decay amplitude; simply counting available final momenta misses it.

The charged pion has spin zero. In its rest frame, the outgoing muon or Positron and the neutrino travel in opposite directions. With a nearly massless left-helicity neutrino, total angular momentum zero requires the positive charged antilepton to have negative helicity. The weak charged current instead creates a massless positive antilepton with positive helicity. A nonzero charged-lepton mass supplies the required opposite-helicity component.

Thus the decay amplitude is proportional to $m_{\ell }$: this is helicity suppression. Explicitly, contracting the pion's axial-current matrix element, proportional to its four-momentum $p_{\pi }^{\mu }$, with the leptonic current and applying the Dirac equation leaves a factor $m_{\ell }$. The leptonic pseudoscalar decay width is, in the $f_{\pi }$ convention,The phase space advantage of the lighter Positron cannot offset its much stronger helicity suppression. The common factors cancel in the pion electron-to-muon decay ratio:This tree-level estimate explains why the muonic branch dominates by roughly four orders of magnitude. The paper's table gives an electronic branching fraction $1.230\times 10^{-4}$; electromagnetic corrections change the simple ratio by a few percent without changing the mechanism.

Put $\tau =\Gamma t$, $r=r_D$ and $\delta ={\delta }_D$. Two indistinguishable paths reach the wrong-sign kaon-pion final state: a direct doubly Cabibbo suppressed decay amplitude $r{e}^{-i\delta }{g}_{+}$ and a neutral meson mixing amplitude followed by the favoured decay, $g_{-}$. They must be added before taking the absolute square:Direct complex-trigonometric multiplication givesThereforeThe three terms are direct decay, quantum interference and decay after mixing. In particular, the minus sign multiplying $\sin \delta \sin x\tau$ follows from the source's $+i\sin$ convention and its definition $r{e}^{-i\delta }$; importing another convention without transforming $\delta$ would change a physical prediction incorrectly. The common exponential accounts for survival before decay. The expression is a decay-rate density up to an overall normalization, rather than a normalized cumulative probability.

The right-sign denominator has its own small mixing contribution. In the same CP symmetry convention its decay amplitude is ${\mathcal{A}}_{\mathrm{RS}}=g_{+}+r{e}^{-i\delta }{g}_{-}$. After cancelling $e^{-\tau }$, the exact numerator and denominator areDefine $u=y\cos \delta$ and $v=x\sin \delta$. Their Taylor expansions are $N=r^2+r(u-v)\tau +\{x^2+y^2+r^2(y^2-x^2)\}{\tau }^2/4+O({\tau }^3)$ and $D=1+r(u+v)\tau +\{y^2-x^2+r^2(x^2+y^2)\}{\tau }^2/4+O({\tau }^3)$. Dividing the series gives the strict second-order answerThe usually quoted approximation additionally drops terms suppressed by powers of $r$ in the right-sign denominator:For $r=0.1$ and $x=y=0.01$, this is $0.01+0.001\tau +0.00005{\tau }^2$ for $\delta =0$, and $0.01-0.001\tau +0.00005{\tau }^2$ for $\delta =\pi /2$. The first increases; the second initially decreases, reaches approximately $0.005$ at $\tau =10$, then rises. The strict series instead has $(C_1,C_2)=(0.00099,0.000049005)$ and $(-0.00101,0.000051005)$. At $\tau =20$ the exact ratios are approximately $0.04913$ and $0.009934$; the leading approximations give $0.05$ and $0.01$.

With no neutral meson mixing, $x=y=0$ and $g_{-}=0$, so $\boxed{R=r^2=0.01}$ at all decay times. The wrong-sign signal then remains through direct doubly Cabibbo suppressed decay.

Use $B^0=d\bar{b}$ and ${\bar{B}}^0=\bar{d}b$. A box Feynman diagram with two W bosons and internal up-type quarks changes one neutral flavour into the other. The top quark contribution dominates the dispersive mixing amplitude and carries the squared CKM combination involving $V_{td},V_{tb}$. With a consistent meson phase convention,The illustration in b shows this box and the dominant decay topology. A suitable common final state is $f=J/\psi K_S$, reconstructed for example through $J/\psi \to {\mu }^{+}{\mu }^{-}$ and $K_S\to {\pi }^{+}{\pi }^{-}$. The leading $\bar{b}\to \bar{c}c\bar{s}$ tree transition gives $B^0\to J/\psi K^0$, while the conjugate gives ${\bar{B}}^0\to J/\psi {\bar{K}}^0$. Neutral kaon mixing projects both kaon flavours onto the observed $K_S$.

The final state is approximately a CP eigenstate with ${\eta }_f=-1$: $J/\psi$ and $K_S$ each have CP approximately $+1$, and their relative orbital angular momentum is $L=1$ to couple the vector charmonium to the spin-zero parent. Neglecting the small kaon CP impurity and subleading decay amplitudes,There are two indistinguishable paths, direct decay and mixing followed by decay, so their amplitudes interfere. For $\Delta {\Gamma }_d\simeq 0$ and negligible direct CP violation, the time-dependent neutral-B CP asymmetry isMeasure the production flavour and proper decay time, fit the oscillation coefficient, and extract $\sin 2\beta$. This determines $\beta$ up to the usual trigonometric ambiguities unless further information is used. The cancellation of the dominant decay phase is why this channel isolates the angle involving $V_{td}$ without needing $V_{ub}$.

For $\Psi (p_1)\to \chi (p_2)+\varphi (p_3)$, the sole tree-level Feynman diagram is a single flavour-changing Yukawa coupling interaction vertex, with $p_1=p_2+p_3$. Its invariant decay amplitude isThe solid arrow passes from the incoming $\Psi$ to the outgoing $\chi$; the dashed outgoing line carries $p_3$.

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 $1/2!$.

In the convention of iii, the invariant decay amplitude is ${\mathcal{M}}_{s_1{s}_2}=-g{\bar{u}}_{\chi }^{s_2}(p_2)u_{\Psi }^{s_1}(p_1)$. Including the common overall $i$ in the definition changes no squared decay amplitude. An unpolarized ensemble is an incoherent equal mixture of the two initial spin states; hence take a spin average over $s_1$ and a spin sum over the unobserved final $s_2$. Different final spin states are orthogonal outcomes, so their probabilities, rather than their decay amplitudes, add.

Using the Dirac adjoint and the fermion spin sum givesThe gamma-matrix trace identities used here are $\mathrm{tr}{\gamma }^{\mu }=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4{\eta }^{\mu \nu }$. The mass-shell conditions $p_1^2=M^2$ and $p_2^2=p_3^2=m^2$, together with four-momentum conservation, implyThereforeIn the parent's rest frame this also says $E_2=E_3=M/2$.

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.

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.

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.

A decay amplitude changing strangeness by two units requires at least two insertions of the low-energy weak interaction in the Standard Model. Each charged-current insertion can remove only one net strange quark in such a process. Both two-W boson tree topologies and loop contributions can occur; a loop is not required for every exclusive final state. At hadronic energies, the extra power of the Fermi constant, with the needed powers of an energy scale to balance units, strongly suppresses a second-order amplitude relative to a first-order one. This is distinct from Cabibbo suppression within a fixed weak order.

If semileptonic decay tags the decaying flavour, wrong-sign decays require neutral meson mixing. With no direct CP violation and no wrong-sign direct decay amplitudes, their rate asymmetry is $(|p/q{|}^2-|q/p{|}^2)/(|p/q{|}^2+|q/p{|}^2)=(1-|q/p{|}^4)/(1+|q/p{|}^4)$. The common time dependence cancels; contaminating direct amplitudes invalidate that isolation.

Choose $m_{\tau }=2\mathrm{GeV}$ and $\sqrt{s}=6\mathrm{GeV}$ purely for illustration. Then each parent has $E_{\tau }=3\mathrm{GeV}$, $|{\mathbf{p}}_{\tau }|=\sqrt{5}\mathrm{GeV}$ and $\beta =\sqrt{5}/3$. Treat all six daughters as massless, including the charged electron and antimuon, and generate each three-body particle decay with constant squared decay amplitude. Choose the parent direction isotropically and the two decays independently; this omits the physical production angular law and spin correlations.

The Lorentz-invariant phase space factorizes through $t=q^2$ into a two-body decay to $\ell +q$, followed by a two-body decay of $q$:After angular integration the first factor contributes $1-t/m_{\tau }^2=x$, while the massless second factor is constant. Since $|dt|=m_{\tau }^{2}dx$, the normalized chosen density is $f(x)=2x$. Drawing $x=\sqrt{U}$ with $U$ uniform on $[0,1]$ implements this density. Isotropic directions, followed by a Lorentz boost for the invisible pair and another for the parent, conserve the complete four-momentum of each generated event.

Define transverse acoplanarity by $\alpha =\pi -\Delta \varphi$, where $\Delta \varphi \in [0,\pi ]$ is the angle between the two visible momentum projections perpendicular to the beam. A fully visible two-body event with zero initial transverse momentum has $\alpha =0$ and $M_{\mathrm{miss}}^2=0$. The toy tau lepton events spread away from that point. These observables are complementary: an angular imbalance asks about directions, while the missing invariant mass also uses the visible energies and longitudinal momenta.

A Lorentz-invariant phase space density supplies allowed configurations and their geometric measure. A physical weak interaction decay amplitude reweights them and correlates spins. Consequently the plotted density is a lesson about conservation and available configurations, not a predicted experimental rate or a fit to the discovery sample. The massless charged-daughter approximation is also unsuitable near the muon rest-energy threshold.

Combine the two massless neutrinos in ${\tau }^{-}\to {\ell }^{-}{\bar{\nu }}_{\ell }{\nu }_{\tau }$ into $q$. In the parent rest frame, four-momentum conservation givesThe lower bound follows from the nonnegative Minkowski norm of the sum of two future-directed null four-vectors; the upper bound follows from $E_{\ell }^{\ast }\ge m_{\ell }$. Substituting gives the displayed energy bounds. They are kinematic: they do not specify the weak interaction decay amplitude or the probability density function inside the interval.

If the charged daughter is also massless, put $x=2E_{\ell }^{\ast }/m_{\tau }\in [0,1]$. A parent with Lorentz factor $\gamma$ and speed $\beta$ in natural units producesHere ${\theta }^{\ast }$ is the daughter angle relative to the boost direction in the parent rest frame. Two opposite parent momenta therefore need not produce two opposite visible daughter momenta: the invisible system can recoil differently in each particle decay.