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Is there a (somewhat) intuitive explanation for why a non-abelian gauge symmetry leads to its gauge bosons requiring boson-boson interactions (being charged)?

Any QFT text derives this result but in tens of pages of math and usually just end with concluding that for example yes, the weak interaction leds to 3 and 4 boson vertices.. I'm not allergic to the math per se but it feels as a quite important consequence and should have a condensed, intuitive explanation.

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    $\begingroup$ "Having interactions" and "being charged" are formally two different things - the first refers to the form of the Lagrangian while the second technically only means that the bosons transform in a non-trivial representation of the gauge group. $\endgroup$ – ACuriousMind Jul 2 '17 at 16:09
  • $\begingroup$ True about the definitions, but the latter implies the former right? (by the typical derivation) And the essence of the question is then how the (non-trivial) group transformations for the boson field intuitively leads to the higher-valued interaction terms, I guess.. $\endgroup$ – BjornW Jul 2 '17 at 16:42
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    $\begingroup$ Have you ever come across an intuitive explanation for anything involving nonabelian gauge theory? I haven't. $\endgroup$ – tparker Jul 10 '17 at 3:23
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In non-abelian gauge theories, two generators of infinitesimal gauge transformations don't commute: $$ [t^a,t^b]=f^{abc}t^c~,\tag{1} $$ where we don't distinguish upper and lower Lie algebra indices.

Now take a general definition of a field strength, $$ F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu - [A_\mu,A_\nu]~.\tag{2} $$ For an abelian gauge theory the last term (commutator) vanishes, because there is only one generator and it obviously commutes with itself. For a non-abelian gauge theory $$F_{\mu\nu}=F^a_{\mu\nu}~t^a~,~A_\mu=A^a_\mu~t^a~.$$ Plugging this into $(2)$ you can see that since generators don't commute $\rightarrow(1)$, the last term doesn't vanish and $~F_{\mu\nu}F^{\mu\nu}~$ leads to trilinear and quartic terms which are absent in abelian gauge theories.

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  • $\begingroup$ Right, so the tri/quartic terms arise from the kinematic part of the gauge field lagrangian when it's evaluated in the general sense.. I do know about the non-commutating structure, so it would be good if we could then also find an explanation for why the kinematic part behaves like this in a non-abelian gauge theory. I mean, it's usually defined from the commutator of the covariant derivative so I understand how the terms can appear, but I'd like to understand why somehow :) $\endgroup$ – BjornW Jul 2 '17 at 16:32
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The field strength tensor $\mathbf{F} \equiv F_{\mu\nu}^a T^a$ of the gauge field transforms in the adjoint representation. Under a gauge transformation $V$, the field strength tensor transforms explicitly as $\mathbf{F} \to V \mathbf{F} V^{-1}$. In the U(1) case, everything commutes, so $\mathbf{F}$ doesn't transform, and we say that the gauge boson is uncharged. However, in the general non-abelian case, the transformation is non-trivial, which is why we say that the gauge boson is charged.

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  • $\begingroup$ Yes, for U(1) the role of the gauge boson field as a connection is quite intuitive, as a measurement of the local phase convention, and furthermore the interaction itself of the "matter" field with the gauge field is understandable because the gauge field modifies the definition of kinetic energy for the matter field (covariant derivative etc.). But for the non-abelian case, it's quite more complex to make a "picture" of what happens when the gauge field interacts with itself.. $\endgroup$ – BjornW Jul 11 '17 at 10:41
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From the other way around it is perhaps easier to ask why boson-self interactions require a non-abelian gauge symmetry. If you write down the most general 3-particle amplitude for boson self-interaction you find that it violate bose-symmetry unless the coupling is anti-symmetric.

Taking it a step further you can glue together two 3-particle amplitudes to find a 4-particle amplitude. One then finds that the only way to preserve Unitarity and locality is if the coupling constant satisfy the Jacobi identity, i.e if it is a non-abelian theory. So in summary boson-self interactions leads to a non-abelian theory due to locality, unitarity and lorentz invariance. This is discussed in great detail in Schwartz & Elvang.

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  • $\begingroup$ How to understand "the coupling is anti-symmetric"? $\endgroup$ – JamieBondi Jul 3 '17 at 13:53
  • $\begingroup$ If you have say 3 bosons labeled a,b,c the amplitude will be some coupling constant $g_{abc}$ times a kinematic piece. Anti-symmetric in this context means $g_{abc}=- g_{bac}$. The anti-symmetric condition is there to compensate the anti-symmetric behaviour of the kinematic part of the amplitude. $\endgroup$ – Lunaron Jul 3 '17 at 14:17

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