A If there is a particle at 750 GeV, what is it? a quark?

  • A
  • Thread starter Thread starter edguy99
  • Start date Start date
  • Tags Tags
    Particle Quark
edguy99
Gold Member
Messages
449
Reaction score
28
A couple of quotes from papers that have shown up recently:

from: http://arxiv.org/pdf/1512.05327.pdf

We consider several models for these fermions, including a single vector-like charge 2/3 T quark, a doublet of vector-like quarks (T, B), and a vectorlike generation including leptons that also contribute to the X → γγ decay amplitude.

from: http://arxiv.org/pdf/1512.05326.pdf

Models of a scalar resonance which explain the excess can come from sectors with a wide variety of field content and quantum number assignments [3]. The simplest possibility which avoids many correlated bounds is a resonance that is a singlet under the Standard Model gauge group. This implies that the coupling to protons and photons is generated by loops of new non-Standard Model particles that are colored and charged ... The most economical model consists of adding a single pair of colored and hypercharged fermions, thus providing for loop-level couplings to gluons and photons.
 
Physics news on Phys.org
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
5
Views
124
Replies
10
Views
3K
Replies
35
Views
8K
Replies
7
Views
6K
Replies
4
Views
1K
Replies
19
Views
3K
Replies
2
Views
3K
Back
Top