 Solution

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 $e^{-r_{\tau }x}/4$.