# Kinematics of 4 body decays for Higgs to 4 leptons

I would like to calculate the decay width of the Higgs boson $$h$$ to 4 leptons via $$ZZ^*$$ intermediate states. I just did a similar calculation for $$h$$ to $$W^{\pm}$$ and 2 fermions via $$WW^*$$, so I am trying to base this new calculation off of that.

I have constructed the matrix element squared, averaged over spins, $$|\mathcal{M}|^2$$. In the case of the 3-body decay, at this stage I simply used equation 49.22 from the kinematics document of the PDG guide:

$$\begin{equation} d\Gamma=\frac{1}{(2\pi)^3}\frac{1}{32 M^3}|\mathcal{M}|^2dm_{12}^2dm_{23}^2 \end{equation}$$

where $$M$$ is the mass of the decaying particle and $$m_{ij}^2=(p_i+p_j)^2$$ where $$i,j$$ label the outgoing three particles. Then the integration bounds are given by 49.23. For fixed $$m_{12}$$:

$$\begin{equation} (m_{23}^2)_{max}=(E_2^{*}+E_3^{*})^2-\left(\sqrt{(E_2^{*})^2-m_2^2}-\sqrt{(E_3^{*})^2-m_3^2}\right)^2 \end{equation}$$

$$\begin{equation} (m_{23}^2)_{min}=(E_2^{*}+E_3^{*})^2-\left(\sqrt{(E_2^{*})^2-m_2^2}+\sqrt{(E_3^{*})^2-m_3^2}\right)^2 \end{equation}$$

where $$E_i^{*}$$ indicates the energy of the particle in the $$m_{12}$$ rest frame. Then $$m_{12}^2$$ gets integrated from $$(m_1+m_2)^2$$ to $$(M-m_3)^2$$. (For simplicity I am approximating all the fermion masses as zero in the eventual numerical calculation.)

So, I clearly need to generalize this now that I have four final states. The next section of the PDG simply says to treat multiple final states as "effective states" thus allowing you to use previous formulas. You are supposed to make the generalization:

$$\begin{equation} m_{ijk...}=\sqrt{(p_i+p_j+p_k+...)^2} \end{equation}$$

in place of e.g. $$m_{23}$$. Naively I would start with this:

$$dm_{12}^2dm_{23}^2\to dm_{12}^2dm_{234}^2$$

But then I do not know how to get the bounds for the integration. Can you assist me?

I learned that for this particular decay we do not actually consider it as a 4-body problem in this way. Only one of the $$Z$$'s is off shell, $$Z^*$$, which decays right away to the 2 leptons. The other is on shell. So for the first step, we consider a 3-body decay using exactly the equations given in the question.
Then, you separately consider the decay of the on-shell $$Z$$ to two leptons. Combining the two steps gives the overall process $$h$$ to 4 leptons via $$ZZ^*$$.