# I Dominant Feynman diagram for $b \to s ~l^+ l^−$

1. Feb 4, 2017

### Safinaz

Hi all,

It's written in QFT books, see for instance George_W._S. book "Flavor Physics and the TeV Scale" that the following Feynman diagram (1)

is the dominant Feynman diagram for $b \to s ~l^+l^−$ decay. Actually I compare this diagram via another possible diagram (2)

Both (1) and (2) are proportional to$G_F^2$, so they are on equal coupling order, also (1) proportional to

\begin{align}
\Big(\frac{1}{k\!\!/ - m_t}\Big)^2 ~ \frac{1}{k^2 - m_w^2},
\end{align}

while (2) proportional to

\begin{align}
\frac{1}{k\!\!/ - m_t} ~ \Big(\frac{1}{k^2 - m_w^2}\Big)^2 ,
\end{align}

I mean (1) has two heavy propagtors in the loop, while (2) has only one heavy propagator, so for the first sight, it seems (2) is the dominant diagram, but this is not the fact .. so any one can explain why (1) is dominant one ?

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2. Feb 4, 2017

### Dr.AbeNikIanEdL

Why do you assume the loop contains top quarks?

3. Feb 4, 2017

### Safinaz

because $\mathcal {M} \propto V_{ib} V_{is}$, so in the best cases the $V_{CKM}$ matrix elements are $V_{tb} V_{ts}$ or $V_{cb} V_{cs}$ . the first choice is the best because $\mathcal {M}$ also proportional to $m_i$

4. Feb 4, 2017

### ChrisVer

aren't you supposed to add them (each individual flavor within the loop)?

5. Feb 4, 2017

### Safinaz

Hi,

If you asking about $Z u_b \bar{u }_c$ vertex, there is $\delta_{bc}$, i.e., Z couples to the same quark flavor.

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