Casimir Trick in \( e^+ e^- \to H \to f \bar{f} \)?

[dingo_d]

Casimir trick in e+e-->H->ffbar

Homework Statement

I have the process:

$e^+e^-\to H\to f\bar{f}$

I have calculated the amplitude and it's conjugate, and now I want to find the averaged, unpolarized square of the invariant amplitude $\langle|M|^2\rangle$.

I average over the initial spins and sum over the final and usually in some simple processes like Moller scattering, I would play with Casimir trick and traces. But here I have:

$\langle|M|^2\rangle=\frac{1}{2}\frac{1}{2}\left( \frac{g_w^2}{4m_w^2} m_e m_f\right)^2\sum_{spins} \bar{u}_4v_2\bar{v}_1u_3\bar{v}_2u_4\bar{u}_3v_1$

Where $\bar{v}_1$ is the incoming positron with impulse p_1 and spin s_1, $u_3$ is the incoming electron, $v_2$ is the outgoing anti fermion, and $\bar{u}_4$ is the outgoing fermion.

If I look at the spinor components, I can arrange them into pairs and use the relations:

$\sum_{s_1}u_{1\delta}\bar{u}_{1\alpha}=({\not} p_1+m_1)_{\delta\alpha}$ and $\sum_{s_2}v_{2\beta}\bar{v}_{2\gamma}=({\not} p_2-m_2)_{\beta\gamma}$

But I'm not getting any trace out of this :\

What am I doing wrong?

 

[member 553083]

Homework Equations \langle|M|^2\rangle=\frac{1}{2}\frac{1}{2}\left( \frac{g_w^2}{4m_w^2} m_e m_f\right)^2\sum_{spins} \bar{u}_4v_2\bar{v}_1u_3\bar{v}_2u_4\bar{u}_3v_1\sum_{s_1}u_{1\delta}\bar{u}_{1\alpha}=({\not} p_1+m_1)_{\delta\alpha} and \sum_{s_2}v_{2\beta}\bar{v}_{2\gamma}=({\not} p_2-m_2)_{\beta\gamma}The Attempt at a Solution To solve this problem, you need to use the Casimir trick. This is an algebraic trick which allows you to calculate the unpolarized, averaged square of the amplitude by summing over the initial spins and then taking the trace of the matrix. First, you need to express the spinors in terms of the momentum 4-vectors and the mass of the particle. Then, you can rearrange the spinors so that you have pairs of spinors which can be summed over using the Casimir trick. For example, you can rearrange the spinors so that you have: \langle|M|^2\rangle=\frac{1}{2}\frac{1}{2}\left( \frac{g_w^2}{4m_w^2} m_e m_f\right)^2\sum_{spins} Tr[\bar{v}_1({\not} p_1+m_1)v_2\bar{u}_3({\not} p_2-m_2)u_4] Now, you can take the trace of the matrix to get: \langle|M|^2\rangle=\frac{1}{2