I'm trying to calculate the disintegration rate $\Gamma(t\rightarrow W^+b)$ from the Fermi Diagram:

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I know the lagrangian for the interaction would be: $$ -iF = \bar{u_b}\gamma^\mu P_Li\frac{g}{\sqrt2}U_{tb}\delta_{ij}u_t \epsilon^\mu $$ Hence, the $$|\bar{F}|^2 = \frac{g^2}{4}\sum \bar{u_b}\gamma^\mu P_Lu_t \bar{u_t} P_R\gamma^\nu u_b \epsilon^\mu \epsilon^{*\nu}$$

Then, I transform the sum into the trace, and substitute the $P_L = \frac{1-\gamma^5}{2}$, eliminating the $\gamma^5$ since it's asymmetrical. I don't want to make any approximation with the masses in order to reproduce the result in the PDG.

$$ |\bar{F}|^2 = \frac{g^2}{8}Tr[(P_b+m_b)\gamma^\mu(P_t+m_t)\gamma^\nu] \epsilon^\mu \epsilon^{*\nu}$$

This is where I get lost, not sure how to operate with the trace in order to obtain the momentum of each particle. Not sure either if my reasoning with the lagrangian is correct.

  • $\begingroup$ The trace of two gamma matrices is trivial, I hope you know, but your application of Casimir's trick for fermion bilinear and collapse of chiral projectors looks problematic, at the very least: Why don't you show your work on that? Have you worked out muon decay in class? $\endgroup$ Commented May 13, 2020 at 20:22
  • $\begingroup$ @CosmasZachos We haven't worked on that yet, but I'll do it as an exercise, why does the Casimir's trick seems problematic? I just skipped the steps on the indexes, what I did was "eliminating" the chiral projectors with $\gamma^5$ because it's asymetrical and the chiral projectors are symetrical $\endgroup$ Commented May 14, 2020 at 9:17

1 Answer 1


So, after a lot of algebra, I came to the solution as follows: $$ |\bar{F}|^2 = \frac{g^2}{8}4(P_{b\mu}P_{t\nu} + P_{b\nu}P_{t\mu} - P_bP_tg_{\mu\nu} + m_bm_tg_{\mu\nu})(-g^{\mu\nu}M_w^2+P^{\mu}_wP^{\nu}_w)$$ Contracting the indexes and asuming $m_b \approx 0$ in order to get the result of the PDG: $$ |\bar{F}|^2 = \frac{g^2}{8}4(P_tP_b+\frac{2P_tP_wP_bP_w}{M_w^2}) $$ Using $P_t = P_w+P_b \rightarrow P_t^2 = m_t^2=(P_w+P_b)^2$: $$ |\bar{F}|^2 = \frac{g^2}{8}4(\frac{m_t^2-m_w^2}{2}+\frac{(m_t^2-m_w^2)(m_t^2+m_w^2)}{2M_w^2}) $$ Knowing that $\Gamma(t\rightarrow W^+b) = \frac{\bar{F}|^2}{2m_t}\int dLips$ and $$\int dLips = \frac{m_t^2-m_w^2}{8\pi m_t^2}$$ for this specific case $$ \Gamma(t\rightarrow W^+b) = \frac{g^2}{64\pi}\frac{m_t^3}{m_w^2}(1-\frac{m_w^2}{m_t^2})^2(1+2\frac{m_w^2}{m_t^2}) $$


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