# About the decay of Higgs into 2 Z bosons

I need to know if I understand this point.

The factor of HZZ vertex in the SM Lagrangian is half the value of the Feynman rule which appears in Diagrammatica.

When I calculate the decay of Higgs into 2 Z, I have to integrate over only half the phase space, because the two Z's are identical.

So, at the end, I get the same result as if I took the exact Lagrangian factor and integrated over the entire phase space, i.e. $2\cdot0.5=1$.

But, in the second way, I would in fact do two mistakes which cancel each other and accidentally get the correct answer. Is this correct?

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It's very strange how you're canceling mistakes against correct factors. The SM Lagrangian at two places is either equivalent up to a field redefinition or one of the places are wrong. Moreover, the decay width is proportional to the squared amplitude, so if the amplitude is doubled by a factor of 2 error at the beginning, then the probabilities and decay rates will be quadrupled ie changed by a factor of 4 which doesn't cancel against 2. – Luboš Motl Feb 7 '12 at 19:52
To add to what Lubos said, I think we might need to see the actual coefficients of the HZZ vertex and of the Feynman rule you're talking about in order to give a sensible answer to this question. – David Z Feb 9 '12 at 7:05
Thanks Lubos. It's 4*0.5 as you said, I didn't sqaure the amplitude. – user7126 Feb 9 '12 at 9:51