I currently read an article and they mentioned, that using the unitarity relation between the 1. and 3. column of the CKM matrix, one can easily show that the area spanned of the unitarity triangle is given by $2A = |Im(V_{ub}V^*_{ud}V^*_{cb}V_{cd})|$. So I tried to show that by myself. I have absolutely no idea if this is gonna work, but here is my attempt:

I started with the unitarity relation of the 1. and 3. column:

$V^*_{ub}V_{ud} + V^*_{cb}V_{cd} + V^*_{tb}V_{td} = 0$

This relation can be represented in the complex plain as a triangle, called unitarity trianlge. I made then the following to vectors:

$\vec{a} = \begin{pmatrix}Re(V^*_{ub}V_{ud})\\Im(V^*_{ub}V_{ud}) \\ 0\end{pmatrix}, \vec{b} = \begin{pmatrix}Re(V^*_{cb}V_{cd})\\Im(V^*_{cb}V_{cd}) \\ 0\end{pmatrix}$

Then taking the cross product of these two vectors should yield the mentioned area in the article:

$2A=|\vec{a}\times \vec{b}|= |\begin{pmatrix}0\\ 0 \\ Re(V^*_{ub}V_{ud})Im(V^*_{cb}V_{cd})-Im(V^*_{ub}V_{ud})Re(V^*_{cb}V_{cd})\end{pmatrix}|= |Re(V^*_{ub}V_{ud})Im(V^*_{cb}V_{cd})-Im(V^*_{ub}V_{ud})Re(V^*_{cb}V_{cd})|$

Now I dont see how that equals $|Im(V_{ub}V^*_{ud}V^*_{cb}V_{cd})|$?

Thank you.


You use that Im$(ab)= $Re$(a)$Im$(b) + $ Re$(b)$Im$(a)$, for $$a = V_{ub}V^*_{ud}\qquad\qquad b=V^*_{cb}V_{cd}$$ to get

\begin{align*} \text{Im}(V_{ub}V^*_{ud}V^*_{cb}V_{cd})&=\text{Re}(V_{ub}V^*_{ud})\text{Im}(V^*_{cb}V_{cd}) + \text{Re}(V^*_{cb}V_{cd})\text{Im}(V_{ub}V^*_{ud}) \\ &=\text{Re}(V_{ub}^*V_{ud})\text{Im}(V^*_{cb}V_{cd}) - \text{Re}(V^*_{cb}V_{cd})\text{Im}(V_{ub}^*V_{ud}) \end{align*} Where in the second equation I used \begin{align*} \text{Re}(a^*) = \text{Re}(a) \qquad\qquad \text{Im}(a^*) = -\text{Im}(a) \end{align*}


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