Timeline for Why is $SU(3)$ chosen as the gauge group in QCD?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 7, 2014 at 7:47 | vote | accept | dan-ros | ||
| Apr 16, 2014 at 18:19 | comment | added | innisfree | Cheng and Li, gauge theory of elementary particles @JSchwinger | |
| Apr 16, 2014 at 17:35 | comment | added | Melquíades | @innisfree Can you give a reference explaining why only $SU(N)$ exhibit asymptotic freedom? | |
| Apr 16, 2014 at 13:32 | comment | added | Flint72 | @innisfree Ah yes, only SU(N) gives asymptotic freedom. that had been annoying me. Thank you! | |
| Apr 16, 2014 at 13:31 | comment | added | Qmechanic♦ | Suggestion to the answer (v2): Mention $SU(N_1)\times\ldots \times SU(N_r)$. | |
| Apr 16, 2014 at 13:19 | comment | added | innisfree | No, for a good symmetry, the probability of seeing something should be invariant under the action of tht symmetry. If the symmetry were not unitary, probabilities could be made arbitrarily big or small by successive symmetry transformations. unbounded probabilities greater than 1 make no sense. | |
| Apr 16, 2014 at 13:14 | comment | added | dan-ros | Ok, but why do we want to conserve probability. i thought that only has to hold in a non-relativistic theory | |
| Apr 16, 2014 at 13:12 | history | edited | innisfree | CC BY-SA 3.0 |
deleted 11 characters in body
|
| Apr 16, 2014 at 13:06 | history | answered | innisfree | CC BY-SA 3.0 |