 Solution

ID: past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2022/2/b/solution

CP ∣ K^{0} ⟩ = ∣ \overset{ˉ}{K}^{0} ⟩

. Neglecting CP violation, the kaon CP eigenstates can then be written

∣ K_{S} ⟩ = \frac{∣ K ^{0} ⟩ + ∣ K ˉ ^{0} ⟩}{2}, ∣ K_{L} ⟩ = \frac{∣ K ^{0} ⟩ - ∣ K ˉ ^{0} ⟩}{2} .

A = A (\overset{ˉ}{D}^{0} \to K^{0} π^{0})

and

B = A (\overset{ˉ}{D}^{0} \to \overset{ˉ}{K}^{0} π^{0})

, projection on these coherent final states gives

A_{S} = \frac{A + B}{2} = \frac{A}{2} (1 + B / A), A_{L} = \frac{A - B}{2} = \frac{A}{2} (1 - B / A) .

This 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.

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