- #1
Valeriia Lukashenko
- 8
- 1
Homework Statement
I want to proof for $$V_{us}V^{*}_{ub}+V_{cs}V^{*}_{cb}+V_{ts}V^{*}_{tb}=0$$ unitarity triangle that left angle is $$\pi-\gamma$$ (see below picture from my lecture notes).
Homework Equations
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$$\gamma \approx - arg(V_{ub})$$
$$\beta_s \approx arg(V_{ts})+\pi$$
$$arg(V_{us}V_{ub}^{*})=arg(V_{ub}^{*})$$
since only Vub*has an imaginry part on this order of Wolfstein parametrization.
$$arg(V_{ts}V_{tb}^{*})=arg(V_{ts})$$
same reason
Vcs and Vcb* are real.
The Attempt at a Solution
Let's call the angle we want to know δ.
$$\delta=\pi+arg(V_{us}V_{ub}^{*})-arg(V_{ts}V_{tb}^{*})=\pi+arg(V_{ub}^{*})-arg(V_{ts})=\pi - arg(V_{ub})+\pi - \beta_s =2\pi -(\beta_s+arg(V_{ub}))=-(\beta_s-\gamma)=\gamma-\beta_s$$
and it is not equal to π-γ.
Could anyone, please, check my solution, because I don't get where my mistake is.