Hi, so I'm trying to derive the charge conservation law for a general SU(N) gauge field theory by using gauge invariance. For U(1) this is trivial, but for the more general SU(N) I seem to be stuck... So if anyone sees any flaws in my logic below please help!(adsbygoogle = window.adsbygoogle || []).push({});

Starting with the Lagrangian density:

[itex]L=-\dfrac{1}{2}Tr(F_{\mu\nu}F^{\mu\nu}) + 2Tr(J_\mu A^\mu)[/itex]

where:

[itex]F_{\mu\nu} = \dfrac{i}{g}[D_\mu, D_\nu][/itex]

[itex]D_\mu = \partial_\mu - igA_\mu[/itex]

The gauge symmetry transformation is:

[itex]A_\mu \rightarrow UA_\mu U^\dagger + \dfrac{i}{g}U\partial_\mu U^\dagger[/itex]

giving:

[itex]D_\mu \rightarrow UD_\mu U^\dagger[/itex] *note this is still a derivative operator

[itex]F_{\mu\nu} \rightarrow UF_{\mu\nu}U^\dagger[/itex]

so that trivially, the first term in the Lagrangian is unchanged. My problem is with the second term. I know from previous research I've done that the answer is:

[itex][D_\mu,J^\mu] = 0[/itex]

and that J transforms as:

[itex]J_\mu \rightarrow UJ_\mu U^\dagger[/itex]

HOWEVER, if I plug this gauge transformation into the Lagrangian I arrive at:

[itex]\delta{L} = 2\dfrac{i}{g}Tr(J^\mu\partial_\mu{U^\dagger}U)[/itex]

and I see absolutely no way to derive the continuity equation from this, considering it doesn't even depend on A!

The reason I know something is fishy (and why I've finally come here after hours of staring at this problem) is that if I make J transform like:

[itex]J_\mu \rightarrow J_\mu U^\dagger[/itex]

I get:

[itex]\delta{L} = 2\dfrac{i}{g}Tr([D_\mu,J^\mu]U)[/itex]

which is exactly what I was looking for! The only problem is, if J transforms like this then the continuity equation doesn't hold up under gauge transformations!! So there is a clear contradiction here, I just don't see where it is. Any help would be greatly appreciated!

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# Gauge invariance and conserved current in SU(N)

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