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I have been trying to calculate the decay width for the $t \rightarrow bW^+$ decay. After having done this numerous times, I can not figure out how the result

$$\Gamma = \frac{g^2m_t^3}{64\pi m_W^2}\bigg(1-\frac{m_W^2}{m_t^2}\bigg)^2\bigg(1+2\frac{m_W^2}{m_t^2}\bigg)$$ is obtained.

Every time I do this calculation, I find the reduced matrix element to be

$$\frac{1}{2}\sum|\mathcal{M}|^2 = \frac{g^2}{4}\frac{m_t^4}{m_W^2}\bigg(1-\frac{m_W^2}{m_t^2}\bigg)^2$$

which gives $$\Gamma = \frac{g^2m_t^3}{64\pi m_W^2}\bigg(1-\frac{m_W^2}{m_t^2}\bigg)^3$$

What I can't seem to figure out is how the $$\bigg(1+2\frac{m_W^2}{m_t^2}\bigg)$$ part comes about.

I know this is a somewhat tedious calculation so if someone were able to just point me in the direction of some resources that may be helpful it would be much appreciated. As of yet, I can not find anything that helps clear up this issue.

Just for reference, I was starting with the following interaction Lagrangian

$$\mathcal{L} = -\frac{g}{\sqrt{2}}W_\mu^-\big(\bar{b}\gamma^{\mu}P_Lt\big) + c.c.$$

and was assuming the bottom quark to be massless.

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    $\begingroup$ What is the density of states factor? $\endgroup$ – G. Smith Apr 7 at 17:44
  • $\begingroup$ For the Lorentz invariant phase space factor I was using $$d\phi_2 = \frac{K}{16\pi^2E_{cm}}d\phi_1d(cos\theta_1)$$ K is the three momentum of the final state particles in the COM frame which I had as $K=\frac{m_t}{2}\big(1-\frac{m_W^2}{m_t^2}\big)$ $\endgroup$ – Baugh_Mania Apr 7 at 17:47
  • $\begingroup$ In the COM frame there seem to be just two types of helicity amplitudes: The "massless" left-handed b going off oppositely to the t spin or else colinearly with its spin, the W spin reversing it. Somehow, your - here should be +2 ... $\endgroup$ – Cosmas Zachos Apr 7 at 21:55

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