Kln theorem and initial-state singularities

eoghan
Messages
201
Reaction score
7
Hi!
If I have a pair q\bar q g in a final state, I know that the gluon has a IR singularity. But KLN theorem rescues me: if I sum over all degenerate states the IR singularity cancels away.
Otherwise, if the emission of the soft gluon is in an initial state, then the IR divergence cannot be canceled. Why KLN doesn't work for initial-state singularities?
I'm thinking of the hard scattering of a parton from an hadron with a target. If the parton radiates a gluon before it hits the target, then I can't cancel the IR singularity, unless I "renormalize" the parton distribution function. Why can't I sum over all initial degenerate states to cancel the singularity?
 
Physics news on Phys.org
This is a post from 2012 but I was wondering the same question ^^ - the remnant IR divergences coming from collinear/soft gluons in the initial state are 'put under the rug' into the PDF associated with the mother hadron from which the parton came from. This is the treatment I have seen everywhere but if I did an inclusive summation over all initial degenerate states would I get no IR divergences as the KLN theorem perhaps tells me and, if so, what would such initial states look like?
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Replies
1
Views
3K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
2
Views
2K
Back
Top