Yang Mills SU(N) charge couplings

by michael879
Tags: charge, couplings, mills, yang
 P: 585 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 g2/2 and all couplings to the SU(3) field are g (including self-coupling 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 higher-level 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 g2/2 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.
 P: 1,287 In a non-Abelian 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 Clebsch-Gordan coefficients. It gets complicated.
P: 796
 Quote by michael879 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 g2/2 and all couplings to the SU(3) field are g (including self-coupling 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).
The unified way of looking at this is that each particle has to be in a definite irreducible representation of each gauge group. The group U(1) has an infinite number of representations, each labeled by a number we call the charge. Nonabelian groups like SU(2) and SU(3) also have an infinite number of representations, but they are more varied and complicated. The representations of SU(2) are familiar because we have to work them out when discussing spin. All fermions in the standard model are in either the trivial representation of SU(2) or in the doublet ("spin-1/2") representation of SU(2). But in principle any representation is allowed; tomorrow we could discover a new fermion in the "spin-37" representation. SU(3) also has an infinite number of representations; all the quarks are in the "triplet" representation but there are more complicated ones. For example people sometimes consider the possibility of particles in the "adjoint" representation of SU(3), which is the same representation that the gluons are in.

 Quote by michael879 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 g2/2 I dont see how gauge invariance can be recovered..
I think we should be a little careful because in the SM the SU(2) gauge symmetry is spontaneously broken and so the actual physical particle states do not have definite charges under (that is, do not live in definite representations of) SU(2).

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

Mentor
P: 15,625

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.
 P: 379 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 $U(1)$? Of course $SU(2)_{Left} × U(1)_{Hypercharge}$ symmetry is SB in $U(1)_{Charge}$ Though the electric charge takes contributions from both the initial $SU(2)$ and $U(1)$ symmetry... For example the photon field $A_{μ}$ is given (through Weinberg's angle) by both the abelian $U(1)$ gauge boson field $B_{μ}$ as well as by the 3rd component $W^{3}_{μ}$ of the non-abelian $SU(2)$ symmetry. So you have the couplings { $\bar{f} A_{μ} γ^{μ} f$} with coupling constant $eQ$ $Q= T_{3} + \frac{Y}{2}$ (it has both generators of $SU(2)$ and $U(1)$) $e=\frac{gg'}{\sqrt{g'^{2}+g{2}}}$ So I don't think what you say is correct (the initial coupling constants appearing in $SU(2)$ and $U(1)$ - $g$ and $g'$ respectively- unbroken symmetry give you the electric charge)
P: 1,287
 Quote by ChrisVer 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 U(1)? Of course SU(2)LeftxU(1)Hypercharge symmetry is SB in U(1)Charge Though the electric charge takes contributions from both the initial SU(2) and U(1) symmetry... For example the photon field A is given (through Weinberg's angle) by both the abelian U(1) gauge boson field B as well as by the 3rd component W3 of the non-abelian SU(2) symmetry. So you have the couplings { (Fermion) A(slash) (Fermion) } with coupling constant eQ Q= T + Y/2 (it has both generators of SU(2) and U(1)V) and e= gg'/sqrt[gg+g'g'] So I don't think what you say is correct (the initial coupling constants appearing in SU(2) and U(1) - g and g' respectively- unbroken symmetry give you the electric charge)
The duck was talking about charge in general (which in this case turns out to be the hypercharge, not the electric charge)
P: 379
 Quote by dauto The duck was talking about charge in general (which in this case turns out to be the hypercharge, not the electric charge)
My question was to the poster, not to Duck's post :/
But I think I misinterpreted the question
P: 585
 Quote by dauto In a non-Abelian 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 Clebsh-Gordon coefficients. It gets complicated.
thanks for the response, but I still have two questions:

1) WHY are non-abelian 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
P: 585
 Quote by The_Duck The unified way of looking at this is that each particle has to be in a definite irreducible representation of each gauge group. The group U(1) has an infinite number of representations, each labeled by a number we call the charge. Nonabelian groups like SU(2) and SU(3) also have an infinite number of representations, but they are more varied and complicated. The representations of SU(2) are familiar because we have to work them out when discussing spin. All fermions in the standard model are in either the trivial representation of SU(2) or in the doublet ("spin-1/2") representation of SU(2). But in principle any representation is allowed; tomorrow we could discover a new fermion in the "spin-37" representation. SU(3) also has an infinite number of representations; all the quarks are in the "triplet" representation but there are more complicated ones. For example people sometimes consider the possibility of particles in the "adjoint" representation of SU(3), which is the same representation that the gluons are in.
I get what you're trying to do, but I still don't see the connection.. hypercharge can take on any real value, so its not part of any U(1) representation... SU(N) charges on the other hand are "N-component" but are necessarily normalized to the coupling constant g which is fixed!

 Quote by The_Duck I think we should be a little careful because in the SM the SU(2) gauge symmetry is spontaneously broken and so the actual physical particle states do not have definite charges under (that is, do not live in definite representations of) SU(2). In general, though, as dauto says you "add" representations of nonabelian groups in the same general way as you "add" spins, by working out Clebsch-Gordan coefficients.
Yea forget the higgs field, I'm talking about Yang-Mills theories. The SM is a good example, but the Higgs field is just a complication that isn't really important to the question
P: 585
 Quote by Vanadium 50 You get one coupling constant, but you can still have different charges: as a physical example, the quarks and gluons carry different color charges.
yes but each color charge has the same "value". You can't have fractional color charges, you cant even have something with 2x color charges (strictly talking about fermions here)!
 P: 1,287 The non-abelian 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 non-abelian). That is one of the most important evidences that GUT theory might turn out to be correct.
P: 313
 Quote by dauto 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 non-abelian). That is one of the most important evidences that GUT theory might turn out to be correct.
I did not realise this before. That is super cool.
P: 585
 Quote by dauto The non-abelian 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 non-abelian). That is one of the most important evidences that GUT theory might turn out to be correct.
This argument doesn't hold up classically though, and there is nothing intrinsic to the SM Lagrangian that makes it quantized. You could treat it as a classical field theory, and although it would behave somewhat differently, this strange disconnect between U(1) and SU(N) fields would still exist
 P: 379 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"
 P: 585 There is nothing wrong with a non-Abelian 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
 P: 1,287 It doesn't matter if the fields are not quantized. Even if the fields commute, the generators don't. It's the non-commutation of the generators which determines what the possible representations look like.

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