A neutral kaon of isospin decays to two pions

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Particle Physics & Quantum Fields · Symmetry, angular momentum and spin

Symmetry and Flavour, Ch Neutral kaons: mixing, strangeness oscillation and CP violation

Problem

A neutral kaon of isospin I=12I=\tfrac12 decays to two pions. Because the two pions are identical bosons in a relative SS-wave, Bose symmetry restricts the final state to total isospin I=0I=0 or I=2I=2, and the Clebsch-Gordan decomposition (as constructed for two-pion states) gives ∣π+π−⟩=23 ∣I=0⟩+13 ∣I=2⟩,∣π0π0⟩=13 ∣I=0⟩−23 ∣I=2⟩.\begin{aligned}|\pi^+\pi^-\rangle &= \sqrt{\tfrac23}\,|I=0\rangle + \sqrt{\tfrac13}\,|I=2\rangle,\\ |\pi^0\pi^0\rangle &= \sqrt{\tfrac13}\,|I=0\rangle - \sqrt{\tfrac23}\,|I=2\rangle.\end{aligned} Write the two isospin decay amplitudes as A0eiδ0A_0e^{i\delta_0} and A2eiδ2A_2e^{i\delta_2}, with δ0\delta_0 and δ2\delta_2 the pion-pion final-state strong-interaction phase shifts. Take ∣A0/A2∣=22|A_0/A_2| = 22 and δ0−δ2=45∘\delta_0 - \delta_2 = 45^\circ (a standard low-energy value of the strong rescattering phase difference), and neglect any difference in two-body phase space between the two channels (the momenta released are almost equal). ((a)) Write ∣A(π+π−)∣2|A(\pi^+\pi^-)|^2 and ∣A(π0π0)∣2|A(\pi^0\pi^0)|^2 in terms of A0A_0, A2A_2, and cos⁡(δ0−δ2)\cos(\delta_0-\delta_2). ((b)) Evaluate the ratio ∣A(π+π−)∣2/∣A(π0π0)∣2|A(\pi^+\pi^-)|^2/|A(\pi^0\pi^0)|^2 numerically and compare it with the measured ratio of KSK_S branching fractions, B(KS→π+π−)/B(KS→π0π0)=69.20/30.69\mathcal{B}(K_S\to\pi^+\pi^-)/\mathcal{B}(K_S\to\pi^0\pi^0) = 69.20/30.69 (PDG 2024). ((c)) State briefly why this calculation, on its own, says nothing about CP violation in the decay (the parameter that would).

Related theory

Browse the problem library