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The scattering amplitude at zero deflection angle. Its real part changes the phase of a coherent beam in matter; its imaginary part is related by the optical theorem to removal from that beam. Different forward amplitudes for a kaon and a neutral antikaon allow neutral kaon regeneration without fundamental CP violation.

Put $M_f=m_p+m_{{\pi }^{+}}+m_{\Lambda }+m_{K^0}$. In natural units, a stationary target proton and a beam proton of total energy $E_b$ have the Mandelstam invariantAt fixed-target production threshold the final particles have zero relative momentum in their centre-of-mass frame, so $s=M_f^2$. Restoring $c$, the total beam energy and its kinetic energy areUsing rounded masses in $\mathrm{MeV}/c^2$, $(938.27,139.57,1115.68,497.61)$, gives $E_{b,\mathrm{thr}}\simeq 2.921\mathrm{GeV}$ and $T_{b,\mathrm{thr}}\simeq 1.983\mathrm{GeV}$.

One possible strong interaction quark-line flavour rearrangement isIt needs newly created $d\bar{d}$ and $s\bar{s}$ pairs, with no net change of any quark flavor. For example, keep the first $uud$ as the outgoing proton; allocate the second incoming $u,d$ and the new $s$ to ${\Lambda }^0$, the remaining incoming $u$ and new $\bar{d}$ to ${\pi }^{+}$, and the new $d,\bar{s}$ to $K^0$. Two gluon splittings illustrate this flow. This is a possible hadronization topology, not a unique perturbative description of the proton collision.

After several short mean lifetimes, the surviving coherent beam is approximately $|K_L⟩=(|K^0⟩-|{\bar{K}}^0⟩)/\sqrt{2}$. In ordinary matter the forward scattering amplitudes of $K^0$ and ${\bar{K}}^0$ differ. For example, the $s$ in a neutral antikaon can enter a strange baryon through strong absorption; the opposite strangeness assignment has different available channels. If their forward transmission amplitudes through a thin block are $t_0$ and ${\bar{t}}_0$, respectively, thenThus the transmitted forward beam contains a regenerated short-lived component, with prompt two-pion decays after the block. Differential absorption changes the intensities as well as the relative phase. Scattered particles outside the forward beam can provide an incoherent component. This neutral kaon regeneration does not require CP violation: the material itself distinguishes matter and antimatter.

Normalize the incident intensity to one. Since $|K^0⟩=(|K_S⟩+|K_L⟩)/\sqrt{2}$, the initial probability amplitudes are $A_S(0)=A_L(0)=1/\sqrt{2}$. Define $g_j=e^{-i{m}_{j}t}{e}^{-t/(2{\tau }_j)}$. The flavor probability amplitudes are $(g_S+g_L)/2$ and $(g_S-g_L)/2$. Their absolute squares arewhere $\Delta m=m_L-m_S$ in natural units. Thus $A=B=1/4$ for both, $\varphi =\Delta mt$, andMultiplying all three coefficients by the incident intensity gives any other normalization. Crucially, $C$ is time dependent. The printed template cannot hold with all of $A,B,C$ constant. For example, a nonzero constant $C$ would leave an undamped oscillation at arbitrarily late times, even after both unstable components have decayed.

In seconds, $\varphi =\Delta m{c}^{2}t/\hslash$. With $x=t/{\tau }_S$, the supplied mass difference and mean lifetimes giveInitially the intensities are one and zero. Their sum is $\frac{1}{2}(e^{-x}+e^{-r_{\tau }x})$, so these are unconditioned survival intensities, not probabilities conditioned on survival. The curves first meet at $x=\pi /(2\times 0.474975)\simeq 3.307$. Thereafter the neutral antikaon component briefly exceeds the $K^0$ component, before the interference dies away and both approach the surviving long-lived intensity $e^{-r_{\tau }x}/4$.

A ground state spin-zero meson is a color singlet $q\bar{q}$ with $L=0$ and constituent spin zero, hence a pseudoscalar meson. The nine flavor combinations form $3\otimes \bar{3}=8\oplus 1$. Here $B=0$, so the vertical coordinate $B+S$ equals strangeness. The six peripheral weights in the right panel of the figure in part (a) are

At the center, $u\bar{u},d\bar{d},s\bar{s}$ all have $I_z=S=0$. A useful orthonormal flavor basis isThe neutral pion has $I=1$; ${\eta }_8$ and ${\eta }_1$ have $I=0$. The Eta octet state belongs to the octet, while ${\eta }_1$ is the flavor singlet. The physical eta and eta prime mesons are predominantly mixtures of these two isoscalar basis states,Unequal quark masses and the chiral anomaly prevent exact flavor symmetry and a degenerate nonet. The three central physical species are consequently ${\pi }^0,\eta ,{\eta }^{\prime }$, rather than three separately observable pure diagonal flavor pairs. Small isospin-breaking admixtures are neglected here.

The nuclear determination of $|V_{ud}|$. A $0^{+}\to 0^{+}$ beta decay transition has no nuclear parity change and can proceed with zero lepton orbital angular momentum. The allowed scalar operator gives a Fermi transition; the rank-one Gamow-Teller transition operator cannot connect total nuclear spins zero to zero. The strong isobaric-analog transitions used for precision measurements are superallowed beta decays. A generic $0^{+}\to 0^{+}$ assignment alone would not guarantee superallowed strength if the states were nonanalog.

At the quark level the weak charged current converts $d\leftrightarrow u$ and carries $V_{ud}$ in the amplitude. The nuclear Fermi operator is an isospin ladder operator, givingHere $f$ denotes the known phase space dependence, with the dimensional factors included in the proportionality constant. In the isospin-symmetric limit, $M_F$ between analog states is fixed by isospin algebra, for example $|M_F{|}^2=2$ for the usual $T=1$ analog transitions. A measured partial half-life and the computed phase space factor give the comparative half-life; after radiative corrections and isospin breaking corrections, its inverse determines $|V_{ud}{|}^2$. The decay rate measures a magnitude squared, not the complex phase of $V_{ud}$.

Ds decay diagrams and CKM comparisons. The original PDF has $K^{+}{\bar{K}}^0$, with ${\bar{K}}^0=s\bar{d}$. The local TeX drops that overbar; using $K^0=d\bar{s}$ instead would give a different flavour channel. Use the PDF's neutral antikaon throughout.

In the representative Ds nonleptonic tree amplitudes, the Ds meson spectator is $\bar{s}$. The elementary weak transitions are

In the first two channels, the produced $u$ joins the spectator $\bar{s}$ to form $K^{+}$; the other quark joins $\bar{d}$. For the first channel this uses the $-d\bar{d}/\sqrt{2}$ component of the neutral pion. In the third, $s\bar{s}$ forms the phi meson and $u\bar{d}$ forms ${\pi }^{+}$. The two possible final pairings have different colour contractions and hadronic matrix elements even when their weak CKM factors coincide.

The purely CKM rate suppression of $c\to d$ relative to $c\to s$ isThus $K^{+}{\pi }^0$ is singly Cabibbo suppressed, while $K^{+}{\bar{K}}^0$ and $\varphi {\pi }^{+}$ are Cabibbo-favored at this level. If reduced hadronic strengths and phase space effects were equal, including the neutral-pion component would giveThe supplied branching fractions instead giveThe first observed ratio is $0.814$ times the neutral-pion-corrected simple estimate, requiring only an order-one remaining hadronic and phase space factor. Relative to $\varphi {\pi }^{+}$, that remaining factor is $0.525$. The two Cabibbo-favored channels differ by a factor $0.644$ despite equal CKM magnitudes. The CKM hierarchy therefore explains why one channel is much rarer, but it does not calculate the three exclusive branching fractions; spin, phase space, color charge and strong interaction final-state effects also matter.

What spin and parity establish. In either pseudoscalar-pseudoscalar final state, the total intrinsic spin is zero. Conservation of angular momentum for a spin-zero parent forces $L=0$, so the final parity isopposite to the parent $0^{-}$ Ds meson. ConsequentlyFor a vector meson plus a pion, the intrinsic final spin is one. Coupling to total $J=0$ requires $L=1$, giving $P_f=(-1)(-1)(-1{)}^1=-1$, equal to the parent parity. Thus spin and parity alone permit a parity-conserving $\varphi {\pi }^{+}$ transition. That is the limited sense in which it is not necessarily weak from this information alone. In reality it is also weak: the initial charm flavour disappears, which the strong interaction and electromagnetic interaction conserve.

W branching fractions. In the massless-daughter, tree-level estimate, each leptonic channel contributes one common unit of decay width. Each accessible quark pair contributes $3|V_{ij}{|}^2$ because of the three final colours. The charged-current W branching-fraction counting denominator isOnly $u$ and $c$ are accessible up-type quarks; the large $V_{tb}$ is irrelevant because a W boson cannot decay into the heavier top quark. HenceA final bottom quark arises in $W^{-}\to b\bar{u}$ or $b\bar{c}$, soThis counts the primary decay channels; finite daughter masses and QCD corrections refine the approximation.

The requested top-lifetime estimate and its limitation. Under the instructed equal-coefficient Sargent's rule scaling, $\Gamma \propto m^5$ and the mean lifetime is inversely proportional to that fifth power. Converting both masses to GeV givesThis is the requested order-of-magnitude estimate, assuming the same effective coefficient for the two total decay rates. It corresponds to $c\tau \simeq 54\mathrm{fm}$ and is longer than the hadronization time, roughly a few $10^{-24}\mathrm{s}$. Taken literally, it would allow top-flavored hadrons to form; it does not establish the physical statement that the top decays before hadronizing.

The extrapolation fails because Sargent's rule uses a low-energy Fermi interaction approximation, whereas $m_t>m_W+m_b$ permits $t\to bW$ with an on-shell W boson. The physical leading decay width instead scales asUsing $m_W\simeq 80.4\mathrm{GeV}$, the Fermi constant $G_F=1.16638\times 10^{-5}\mathrm{Ge}{\mathrm{V}}^{-2}$ and $|V_{tb}|=0.999$ gives about $1.55\mathrm{GeV}$ at the supplied top mass. Then ${\tau }_t=\hslash /{\Gamma }_t$, with $\hslash =6.58212\times 10^{-25}\mathrm{GeV}\mathrm{s}$, gives ${\tau }_t\simeq 4.2\times 10^{-25}\mathrm{s}$. This corrected regime explains top-quark decay before hadronization.^[ref]

Starting with an antikaon instead givesThusThis factored form assumes $A_f,{\bar{A}}_f\ne 0$; otherwise use the preceding probability amplitude. The three mechanisms are separated below. Comparisons for distinct final states must pair $f$ with its CP conjugate $\bar{f}$.

The favoured weak charged current transition is $\bar{c}\to \bar{s}{W}^{-}$ followed by $W^{-}\to d\bar{u}$. The spectator quark $u$ combines with $\bar{u}$ to form the $u\bar{u}/\sqrt{2}$ component of the neutral pion, while $d\bar{s}$ forms $K^0$. The product of CKM matrix factors is $V_{cs}{V}_{ud}^{\ast }$. The doubly Cabibbo suppressed transition is $\bar{c}\to \bar{d}{W}^{-}$ followed by $W^{-}\to s\bar{u}$, with factors $V_{cd}{V}_{us}^{\ast }$; the final $s\bar{d}$ is the neutral antikaon. Both are colour-suppressed internal-emission Feynman diagrams, with the same spectator routing.

In the two-generation approximation to the CKM matrix, $V_{ud}=V_{cs}=\cos {\theta }_C$, $V_{us}=\sin {\theta }_C$ and $V_{cd}=-\sin {\theta }_C$. For equal magnitudes of the hadronic matrix elements,The rate ratio is about $2.8\times 10^{-3}$ with the supplied CKM matrix magnitudes. The minus sign of a particular product of CKM matrix elements must be combined with the phases of the hadronic matrix elements and kaon basis states; it is not by itself the observable interference sign.

Choose the exchange-positive CP convention for neutral kaon mixing, $CP|K^0⟩=|{\bar{K}}^0⟩$. Neglecting CP violation, the kaon CP eigenstates can then be writtenWriting $A=A({\bar{D}}^0\to K^0{\pi }^0)$ and $B=A({\bar{D}}^0\to {\bar{K}}^0{\pi }^0)$, projection on these coherent final states givesThis is the convention implicit in the printed sum. Changing the sign convention for the neutral antikaon changes both its coefficient in the kaon CP eigenstates and $B$; it cannot change either physical decay rate. The extremely small $K_S$–$K_L$ mass difference can be ignored in the two-body phase space.