# Cross section for two identical particles in the final state

1. Nov 11, 2009

### krishna mohan

I was calculating the cross section for a particular decay and saw that I was off by a factor of half from the known value..

Now, the decay has two identical particles in the final state and I seem to remember having read somewhere that this requires a factor of half....but I am not able to find the statement in standard books like Peskin(P93-symmetry factors)...maybe it is given somewhere else in Peskin...

Could anyone give me a reference for the above statement?

2. Nov 12, 2009

### hamster143

The rule is that you don't sum over diagrams that differ only by vertex permutation. In your case, two vertices corresponding to final state particles are identical.

3. Nov 13, 2009

### krishna mohan

No...in this case there is only one vertex....I was considering the decay of the Higgs into two Z bosons...

But I know now why there is a factor of two....

In calculating the Cross section, we integrate over all the momenta of the two Z particles, say Z_1 and Z_2, with momenta denoted by p_1 and p_2.....

But in doing so we count the points

( p_1 = x , p_2 = y ) and (p_1 = y, p_2 = x )

separately.....although they are infact one and the same point in phase space since Z_1 and Z_2 are identical particles.....