9
$\begingroup$

For gauge groups like $SU(2)$ and $SU(3)$ etc. we know that observable states such as mesons or baryons must be charge neutral. However, for a $U(1)$ gauge group we can have charged initial states in our scattering experiments. Why can we have states with an observable non-zero charge for the $U(1)$ but not for $SU(n)$?

Another way to state this question, to clarify, is the following: the global part of the $U(1)$ gauge group defines a good quantum number - the total charge - which allows us to split up our Hilbert space. Why does the global part of bigger gauge groups not supply a good quantum number? If it does supply such a number why must all states be charge neutral under it?


Having had a discussion with @user1504 and taking into account @FredericBrunner 's points I think I can answer this question to my satisfaction.

The summary is thus - it is possible in principle, regardless of the gauge group, to form localised lumps of the respective charges under the global part of the group. To me, this gives the global part of the group a meaning separate from the gauge part. For $U(1)$ the lumps can have positive and negative charge and for $SU(3)$ they can have each of the three colours. This means that in respect of the question, $U(1)$ and $SU(N)$ are not different. In a practical sense however, the $SU(N)$ groups are confining so one can only do this in principle at a temperature sufficiently high that the theory is deconfined and the notion of individual quarks makes sense. Similarly, $U(1)$ in 2+1 dimensions is confining and there the notion of lumps of positive and negative charge only make sense above the deconfinement transition.

$\endgroup$
4
$\begingroup$

The premise of this question is wrong: U(1) is not special. There are conserved charges associated to the global symmetry group $G$ of any gauge theory.

In the Standard Model, for example, we have the weak isospin charge, which is the conserved Noether charge of the electroweak $SU(2)$. (At low energies, this conservation is obscured by the Higgs mechanism, but at high enough energies, it becomes more plain.) There is also a conserved color charge associated to the $SU(3)$ color symmetry.

As Frederic Brunner correctly points out, in our world, the color force is confining, so the value of this conserved color charge in normal physics is always 'zero'. But this is not true above QCD's Hagedorn temperature, where the system deconfines. Nor is this entirely a theoretical issue: RHIC has produced collisions intense enough to reach this deconfined state.

$\endgroup$
7
  • $\begingroup$ Thank you for answering after so long. I am afraid that even now I am still a little unclear about what is going on. Perhaps you could explain why confinement necessarily means that a state of the system must be charge neutral? Thanks again. $\endgroup$ – Magician Apr 21 '20 at 19:09
  • 1
    $\begingroup$ @Magician It's essentially the definition of confinement. In confined systems, thanks to quantum effects, the interaction between objects with non-zero color are so strong that they always bind to each other forming objects with zero color. $\endgroup$ – user1504 Apr 21 '20 at 19:47
  • $\begingroup$ Ah - so in principle (if not practice) I could heat an SU(3) theory up and let my Maxwell demon select for me red quarks and put them in a box. Consequently I could construct a state of N red quarks at zero temperature? I think this is the core of my misunderstanding, as I can certainly construct states of positive and negative electron charge. $\endgroup$ – Magician Apr 21 '20 at 23:16
  • $\begingroup$ @Magician What would the box be made of? We're talking about temperatures not seen since the first second after the Big Bang? $\endgroup$ – user1504 Apr 22 '20 at 0:35
  • $\begingroup$ Sure, it's a hypothetical box (perhaps of the infinite potential well kind ;) ). My question is whether it is possible to prepare such a state in principle? This would seem to follow from what you have said. If the colour charge was pure gauge I would have said it is not possible to prepare such a state. $\endgroup$ – Magician Apr 22 '20 at 5:31
4
$\begingroup$

The reason is confinement. Yang Mills theories with $SU(2)$ and $SU(3)$ gauge groups exhibit confinement, while for example $U(1)$ electrodynamics does not. Whether a theory is confining or not can be found out by studying the properties of Wilson loops.

$\endgroup$
5
  • 1
    $\begingroup$ Hi Fredric. Could you elaborate on this a bit, I don't get the connection between confinement and the general spirit of the question. SU(2) charges (i.e. zeroth components of the Noether currents) certainly are globally conserved and do generate the chiral algebra, so they are useful too, but not as useful as their analogs for $U(1)$, which could lead to important conservation laws like the baryon number conservation. How does confinement figure in the difference, I did not understand. $\endgroup$ – 299792458 Jun 10 '15 at 13:39
  • $\begingroup$ Hi Fredric. Your answer seems to rely importantly on computations at one loop. It seems we could get around it by heating up the theory so that it deconfines or working only at tree level. Could you comment further? Many thanks. $\endgroup$ – Magician Jun 10 '15 at 14:31
  • $\begingroup$ @Magician: Why does it seem to be based on "one-loop"? The Wilson loops are not loops in a Feynman diagram, they are non-perturbative objects. $\endgroup$ – ACuriousMind Jun 10 '15 at 19:40
  • 1
    $\begingroup$ @ACuriousMind - You have a point. I was thinking about the beta function and not Wilson loops. Nonetheless you can still de-confine the theory so I was hoping for an argument that did not rely on confinement. Also, I seem to remember that even U(1) is confining in 2+1 dimensions. Do I need to clarify the question further somehow? $\endgroup$ – Magician Jun 11 '15 at 8:48
  • 1
    $\begingroup$ I think this answer is wrong. The SU(2) gauge theory in the Standard Model's electroweak sector is not confining. And we do see conservation of weak isospin in the Standard Model; this is the charge of the $SU(2)$. $\endgroup$ – user1504 Apr 20 '20 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.