Lattice field theories and the continuum limit

jfy4
Messages
645
Reaction score
3
I have heard generally that it is possible to put different physical theories on a lattice and after renormalization get the same continuum theory. I mean, different lattice theories that lead to the same continuum theory. Is this true for, say, qcd, or other particle theories? Are there different varieties of lattice qcd that give the same continuum qcd theory? And if so, is it known how these lattice theories are related, besides their obvious shared macroscopic limit?

Thanks.
 
Physics news on Phys.org
Yes, there are many different "lattice actions" for QCD that all give the same continuum limit. In renormalization group language, the different actions differ by irrelevant terms (terms down by powers of the lattice spacing) that vanish as the lattice spacing goes to zero.
 
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}...
Back
Top