= Solution
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 invariant>
$$
s=(p_b+p_t)^2=2m_p^2+2m_pE_b.
$$
At <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> are
$$
\boxed{E_{b,\mathrm{thr}}=\frac{M_f^2-2m_p^2}{2m_p}c^2,\qquad
T_{b,\mathrm{thr}}=\frac{M_f^2-4m_p^2}{2m_p}c^2.}
$$
Using rounded <masses> in $\mathrm{MeV}/c^2$, $(938.27,139.57,1115.68,497.61)$, gives $E_{b,\mathrm{thr}}\simeq2.921\,\mathrm{GeV}$ and $T_{b,\mathrm{thr}}\simeq1.983\,\mathrm{GeV}$.
One possible <strong interaction> <quark-line flavour rearrangement> is
$$
(uud)+(uud)\longrightarrow
\underbrace{uud}_{p}+\underbrace{uds}_{\Lambda^0}
+\underbrace{u\bar d}_{\pi^+}+\underbrace{d\bar s}_{K^0}.
$$
It 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.
\Image[/past-exam-of-the-nuclear-and-particle-physics-course-of-the-university-of-oxford/2022-associated-kaon-production.png]
{title=One flavor-conserving quark-flow assignment for associated neutral-kaon production}
After several short <mean lifetimes>, the surviving coherent beam is approximately $|K_L\rangle=(|K^0\rangle-|\bar K^0\rangle)/\sqrt2$. 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, then
$$
\boxed{|K_L\rangle\longmapsto
\frac{t_0+\bar t_0}{2}|K_L\rangle+
\frac{t_0-\bar t_0}{2}|K_S\rangle.}
$$
Thus 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.
Back to article page
