humanino said:
It seems to me confusing in the context of the original question to hide \hbar and c. Vanadium 50's and my result are on the first page of the particle data group booklet or review, so I think we were justified not to give further details (unless requested). I'd like to request a clarification about your formula. It seems to me, energy and length have inverse dimension for instance. Can you please re-establish the proper \hbar, c and (probably) G factors ?
Well, I agree that the OP really meant GeV^(-2) and agree with your answers. Now, if you put G = 1, then of course, any power of GeV could be a cross section (because you've made physics dimensionless).
Now, I don't work in particle physics so, I don't have the conversion factors in my head. So, what I always do is use a few well known formulae that contain hbar, c and G to do the conversion.
To convert GeV^2 to a cross section, you can use that in General Relativity, mass and length have the same dimensions (if you put c = G = 1). So, GeV^2 is already a cross section and no additional conversion using hbar needs to be performed.
To restore G and c, we just hijack the formula for gravitational potential energy, so:
m^2 G/r = energy = m c^2
this is a dimensionally correct expression, that doesn't need to make sense. So, we have:
m G/(c^2 r) = dimensionless
Or:
E G/(c^4) = length
where E is an energy. So, we see that:
cross section = E^2 G^2/c^8
If you know the formulas for Planck length, Planck energy etc. etc., you can do the conversion directly. To convert GeV^n to a cross section, you simply divide this by the Planck energy to the power n and multiply by the Planck length squared.