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Yang Mills SU(N) charge couplings 
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#1
Dec313, 09:51 AM

P: 632

After playing around with possible charges for new fermionic fields, I came to the conclusion that ONLY the U(1) gauge group allows different particles to have different charge. This seems to be purely due to its commutative properties and since SU(N) gauge groups are not commutative, all particles must have the same charge with respect to that gauge field. This is supported by the standard model, where all couplings to the SU(2) field are g_{2}/2 and all couplings to the SU(3) field are g (including selfcoupling terms). So in reality it seems the only free parameters for the charge of a new fermionic field are its U(1) coupling strength, and whether or not it couples to the other two fields (2 discrete possibilities per field instead of an infinite number).
Is this conclusion correct? And can anyone explain it qualitatively with some higherlevel reason? I understand mathematically why its not possible (it breaks gauge symmetry), but I have no idea why this makes sense physically... Also take a meson for example. It contains two tightly bound quarks, and can be described by an effective field theory that treats the meson as a single field. We already know the U(1) charges add, but what would be the SU(2) charge of this meson in the effective theory?? If it's not g_{2}/2 I dont see how gauge invariance can be recovered.. *edit* I just realized that the meson would necessarily have 0 SU(2) charge if the charges of the quarks just add. So lets deal with baryons instead, where you have 3 quarks (no antiquarks) with the same SU(2) charge coupling. 


#2
Dec313, 10:48 AM

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P: 1,948

In a nonAbelian theory the coupling constant g is unique, as you said. What can be different is the group representation the particle belongs to. In a SU(2) group this representations combine exactly the same way that spin adds up  hence the name isospin. So, for instance two doublets can combine to form an isospin singlet  as you stated, but they may also combine to form a isospin triplet. SU(3) and higher groups will have their own (more complicated) rules for combining multiplets. This is what is called multiplication of representations. You get lots of coefficients analogous to the ClebschGordan coefficients. It gets complicated.



#3
Dec313, 11:05 AM

P: 883

In general, though, as dauto says you "add" representations of nonabelian groups in the same general way as you "add" spins, by working out ClebschGordan coefficients. 


#4
Dec313, 12:05 PM

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PF Gold
P: 16,471

Yang Mills SU(N) charge couplings
You get one coupling constant, but you can still have different charges: as a physical example, the quarks and gluons carry different color charges.



#5
Dec313, 12:37 PM

P: 1,059

ehmm... Sorry to put a question here, but it came in my mind.... and so I am not sure...
Why do you say that the charge is due only to [itex]U(1)[/itex]? Of course [itex]SU(2)_{Left} × U(1)_{Hypercharge}[/itex] symmetry is SB in [itex]U(1)_{Charge}[/itex] Though the electric charge takes contributions from both the initial [itex]SU(2)[/itex] and [itex]U(1)[/itex] symmetry... For example the photon field [itex]A_{μ}[/itex] is given (through Weinberg's angle) by both the abelian [itex]U(1)[/itex] gauge boson field [itex]B_{μ}[/itex] as well as by the 3rd component [itex]W^{3}_{μ}[/itex] of the nonabelian [itex]SU(2)[/itex] symmetry. So you have the couplings { [itex] \bar{f} A_{μ} γ^{μ} f[/itex]} with coupling constant [itex]eQ[/itex] [itex]Q= T_{3} + \frac{Y}{2} [/itex] (it has both generators of [itex]SU(2)[/itex] and [itex]U(1)[/itex]) [itex]e=\frac{gg'}{\sqrt{g'^{2}+g{2}}}[/itex] So I don't think what you say is correct (the initial coupling constants appearing in [itex]SU(2)[/itex] and [itex]U(1)[/itex]  [itex]g[/itex] and [itex]g'[/itex] respectively unbroken symmetry give you the electric charge) 


#6
Dec313, 12:50 PM

Thanks
P: 1,948




#7
Dec313, 12:53 PM

P: 1,059

But I think I misinterpreted the question 


#8
Dec313, 02:11 PM

P: 632

1) WHY are nonabelian theories so different to abelian theories in this respect? In U(1) there is no unique coupling constant and particles can take on any charge 2) Is there some way to view this nonabelian behavior as a generalization of the U(1) situation? Because I'm not seeing it... isospin and hypercharge are two vastly different things... hypercharge is not REQUIRED to be quantized (at least not in the SM), but isospin is! *sorry I got interrupted in the middle of posting this and didnt see any of the other responses when I got back 


#9
Dec313, 02:16 PM

P: 632




#10
Dec313, 02:17 PM

P: 632




#11
Dec413, 12:46 PM

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P: 1,948

The nonabelian charges are quantized for the same reason that spin is quantized. The different group generators do not commute amongst themselves and that limits the possible representations. So, for instance, a particle with spin 1/2 will form a doublet with possible spins {1/2, 1/2} but nothing in between. A different representation for the group might be a particle with spin 1 which form a triplet with spins {1, 0, 1}, but nothing in between. The spin is quantized. Similarly the isospin is also quantized. Other groups such as SU(3) will have its own set of allowed representations, but always with quantized values for their charges. No such restrictions apply to the U(1) group since it has only one generator which obviously commutes with itself. And charge is allowed. And yet... the electric charge and hypercharge are quantized! There is no justification for that within the Standard Model (beyond the requirement of triangular anomaly cancellation). But within GUT theories that quantization is necessary because the U(1) generator is just a linear combination of some generators of the GUT group (which is nonabelian). That is one of the most important evidences that GUT theory might turn out to be correct.



#12
Dec413, 07:08 PM

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#13
Dec513, 01:37 PM

P: 632




#14
Dec513, 02:41 PM

P: 1,059

commutation relations brings quantization in your theory...
As for example, when you have a classical field theory and you want to do canonical quantization, you impose the [p,x] commutor... For the SU(n) that happens for the generators anyway, because it's not abelian. So I don't think that you should look at an SU(2)xU(1) lagrangian as "classical" 


#15
Dec513, 03:20 PM

P: 632

There is nothing wrong with a nonAbelian classical field theory. The generators don't commute but that doesn't make the theory quantized...
On a side note, I think I realize where my confusion is coming from. U(N) would be the true generalization of a U(1) theory. Since U(N)~U(1)xSU(N), it's pretty clear why SU(N) theories are "missing" that U(1) quality I was wondering about 


#16
Dec513, 09:43 PM

Thanks
P: 1,948

It doesn't matter if the fields are not quantized. Even if the fields commute, the generators don't. It's the noncommutation of the generators which determines what the possible representations look like.



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