Past exam of the particle physics course of the University of Oxford / 2016 / 4 / 2

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- Solution 2

Solution

Put

τ = Γ t

,

r = r_{D}

and

δ = δ_{D}

. Two indistinguishable paths reach the wrong-sign kaon-pion final state: a direct doubly Cabibbo suppressed decay amplitude

r e^{- i δ} g_{+}

and a neutral meson mixing amplitude followed by the favoured decay,

g_{-}

. They must be added before taking the absolute square:

A_{WS} = r e^{- i δ} g_{+} + g_{-} .

Direct complex-trigonometric multiplication gives

∣ g_{+} ∣^{2} = \frac{e ^{- τ}}{2} (cosh y τ + cos xτ), ∣ g_{-} ∣^{2} = \frac{e ^{- τ}}{2} (cosh y τ - cos xτ), g_{+}^{*} g_{-} = \frac{e ^{- τ}}{2} (sinh y τ + i sin xτ) .

Therefore

∣ A_{WS} ∣^{2} = e^{- τ} [\frac{r ^{2}}{2} (cosh y τ + cos xτ) + r (cos δ sinh y τ - sin δ sin xτ) + \frac{1}{2} (cosh y τ - cos xτ)] .

The three terms are direct decay, quantum interference and decay after mixing. In particular, the minus sign multiplying

sin δ sin xτ

follows from the source's

+ i sin

convention and its definition

r e^{- i δ}

; importing another convention without transforming

δ

would change a physical prediction incorrectly. The common exponential accounts for survival before decay. The expression is a decay-rate density up to an overall normalization, rather than a normalized cumulative probability.

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