= Neutral-kaon CP impurity and two-pion decay
{title2=$|K_L\rangle=\frac{|K_2\rangle+\epsilon|K_1\rangle}{\sqrt{1+|\epsilon|^2}}$}
Use the <kaon CP eigenstates> convention $CP|K^0\rangle=-|\bar K^0\rangle$, matching <kaon CP mixing parameter>. Then $|K_1\rangle=(|K^0\rangle-|\bar K^0\rangle)/\sqrt2$ is <CP> even, while $|K_2\rangle=(|K^0\rangle+|\bar K^0\rangle)/\sqrt2$ is <CP> odd. With <CPT symmetry> and a common small mixing parameter, the physical states can be written
$$
|K_S\rangle=\frac{|K_1\rangle+\epsilon|K_2\rangle}{\sqrt{1+|\epsilon|^2}},\qquad
|K_L\rangle=\frac{|K_2\rangle+\epsilon|K_1\rangle}{\sqrt{1+|\epsilon|^2}}.
$$
Here $S,L$ mean short and long <mean lifetimes>; the <kaon CP mixing parameter> $\epsilon$ depends on the flavor-state phase convention. A positive flavor-exchange convention instead makes the symmetric flavor combination <CP> even, as explained in <exchange-positive CP convention for neutral kaon mixing>. Mixing these conventions would reverse the interpretation of the written sums.
Why does the two-<pion> final state test <CP symmetry>? A spin-zero <kaon> decaying into two spin-zero <pions> has relative <orbital angular momentum> $\ell=0$ by <angular momentum conservation>. Two negative <intrinsic parities> multiply to plus, and $(-1)^\ell=1$. <Charge conjugation> exchanges $\pi^+$ and $\pi^-$; the $\ell=0$ state is even under this exchange. Hence this two-<pion> state has <CP> <eigenvalue> $+1$. If a decay transition operator $T$ preserves <CP symmetry>, inserting <CP> on both sides of $\langle\pi\pi|T|K_2\rangle$ changes its sign without changing its value: that <decay amplitude> must vanish.
Assume for illustration that <CP violation> occurs only through the state admixture and the decay transition preserves <CP symmetry>. Put $A_1=\langle\pi\pi|T|K_1\rangle$. Then
$$
A_L=\frac{\epsilon A_1}{\sqrt{1+|\epsilon|^2}},\qquad
A_S=\frac{A_1}{\sqrt{1+|\epsilon|^2}},\qquad
\eta_{\pi\pi}=\frac{A_L}{A_S}=\epsilon.
$$
Neglecting the tiny difference in two-body <Lorentz-invariant phase space> between the physical <masses> gives $\Gamma_L^{\pi\pi}/\Gamma_S^{\pi\pi}\simeq|\epsilon|^2$. A <branching fraction>, however, divides a partial <decay width> by the total <decay width>. With $\hbar=1$, $\tau_i=1/\Gamma_i$, so
$$
\frac{B_L^{\pi\pi}}{B_S^{\pi\pi}}
\simeq |\epsilon|^2\frac{\Gamma_S}{\Gamma_L}
=|\epsilon|^2\frac{\tau_L}{\tau_S}.
$$
For chosen values $|\epsilon|=0.01$ and $\tau_L/\tau_S=100$, the <branching fraction> ratio is $0.01$, whereas the partial <decay width> ratio is $0.0001$. This is original arithmetic, not a historical fit. The longer observation time amplifies the integrated fraction relative to the instantaneous partial <decay rate>.
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