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Vector boson decay

 Starting from the Feynman rule for the HZZ vertex [7] in fig. 2.7 the partial Higgs width $\Gamma_{H \rightarrow ZZ}^{}$ can be calculated. The matrix element is

 
i M H $\scriptstyle\rightarrow$ ZZ = $\displaystyle\sum_{\lambda,\rho}^{}$ig$\displaystyle{\frac{m_{Z}}{\cos\theta_{W}}}$g$\scriptstyle\mu$$\scriptstyle\nu$$\displaystyle\epsilon^{\ast\mu}_{1\lambda}$$\displaystyle\epsilon^{\ast\nu}_{2\rho}$ (27)

with $\epsilon_{1}^{}$,$\epsilon_{2}^{}$ the polarisation vectors of the two Z bosons with polarisation indices $\lambda$,$\rho$ .
  
Figure 2.7: The decay of a Higgs boson to two Z 0 bosons with four momentum p , q and the Feynman rule for the vertex.
\begin{figure}
 \begin{center}
 \leavevmode
 
\epsfig {file=hzz_decay.eps,width=\singlefig}
 \end{center}\end{figure}

For the two-body decay to two equal mass particles the differential decay rate [16] is given as
   \begin{align}
 \frac{{d}\Gamma}{{d}\Omega} &= \frac{\sqrt{\lambda(m_{H}^2,m_{Z}^...
 ...m_{H}}\right)^2} \\ \intertext{and}
 {S} &= \prod_k n_k^{-1} = 2^{-1}\end{align}
where n is the number of identical final state particles of type k . The squared matrix element can from (2.41) be written as

 
|M |2 = $\displaystyle\left(\frac{g\,m_{Z}}{\cos\theta_{W}}\right)^$2$\displaystyle\sum_{\lambda,\rho}^{}$g$\scriptstyle\mu$$\scriptstyle\nu$$\displaystyle\epsilon^{\ast\mu}_{1\lambda}$$\displaystyle\epsilon^{\ast\nu}_{2\rho}$g$\scriptstyle\alpha$$\scriptstyle\beta$$\displaystyle\epsilon^{\alpha}_{1\lambda}$$\displaystyle\epsilon^{\beta}_{2\rho}$ (28)

The summation over the three polarisation states of the massive Z bosons [16] is

 
$\displaystyle\sum_{\lambda}^{}$$\displaystyle\epsilon^{\mu}_{\lambda}$$\displaystyle\epsilon^{\ast\nu}_{\lambda}$ = - g $\scriptstyle\mu$$\scriptstyle\nu$ + $\displaystyle{\frac{p^\mu p^\nu}{m_{Z}^2}}$, (29)

which is just the Lorentz covariant generalisation of having three orthonormal polarisation vectors e $\scriptstyle\lambda$ in the rest frame of the Z where

 
$\displaystyle\sum_{\lambda}^{}$e i$\scriptstyle\lambda$e j$\scriptstyle\lambda$ = $\displaystyle\delta^{ij}_{}$. (30)

The squared matrix element (2.45) can now be simplified as
  \begin{align}
\vert{M}\vert^2 &= \left(\frac{g\,m_{Z}}{\cos\theta_{W}}\right)^2
...
 ...(1
 - \frac{4m_{Z}^2}{m_{H}^2}
 + \frac{12m_{Z}^4}{m_{H}^4}
 \right).\end{align}
The decay rate can be written as
\begin{align}
\Gamma_{H \rightarrow ZZ}&= \int{d}\Omega 
 \frac{{d}\Gamma}{{d}\O...
 ...(1
 - \frac{4m_{Z}^2}{m_{H}^2}
 + \frac{12m_{Z}^4}{m_{H}^4}
 \right).\end{align}
The width of the Higgs decaying to W +W - is the same as the calculation above, except that the two decay particles are not equal leading to a factor two larger result from (2.44). In a notation with scaling variables the results can be given as
    \begin{align}
 \Gamma_{H \rightarrow ZZ} &= \frac{g^2}{128\pi}\frac{m_{H}^3}{m_{...
 ...W}^2}
 \sqrt{1-x_{W}}
 \left( 1 - x_{W} + \frac{3x_{W}^2}{4}
 \right)\end{align}
where

xZ = 4$\displaystyle{\frac{m_{Z}^2}{m_{H}^2}}$xW = 4$\displaystyle{\frac{m_{W}^2}{m_{H}^2}}$. (31)

It is the longitudinal polarisation that is responsible for the second part of the sum in (2.46) and by following this part onwards to (2.60) it is seen that the transverse polarisation only enters with powers of xZ/xW or higher. At high Higgs masses the Higgs coupling to vector mesons is thus totally dominated by the longitudinal vector boson states.


next up previous contents
Next: Fermionic decays Up: Higgs decays Previous: Higgs decays
Ulrik Egede
1/8/1998