Covariant deriv of matrix valued field(srednicki)

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The discussion centers on the covariant derivative of a matrix-valued field \(\Phi\) in the context of the gauge group SU(N), as presented in chapter 84 of Srednicki's work. The correct formulation of the covariant derivative is given by \(D_{\mu}\Phi=\partial_{\mu}\Phi-igA^a_{\mu}\left[T^a,\Phi\right]\), where the commutator arises due to the non-commutative nature of the adjoint representation. This formulation is essential for accurately capturing the dynamics of the gauge field interacting with the scalar field in the adjoint representation.

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LAHLH
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Hi

In ch84, Srednicki is considering the gauge group SU(N) with a real scalar field \Phi^a in the adjoint rep. He then says it will prove more convienient to work with the matrix valued field \Phi=\Phi^a T^a and says the covariant derivative of this is D_{\mu}\Phi=\partial_{\mu}\Phi-igA^a_{\mu}\left[T^a,\Phi\right]

Why is this covariant derivative not just D_{\mu}\Phi=\partial_{\mu}\Phi-igA^a_{\mu}T^a\Phi ?

I understand \Phi is a matrix and it does not commute with the generators, but I don't understand how this commutator is arising here in the second term of the covariant derivative? is it something to do with the adjoint rep?
 
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The covariant derivative as written by Srednicki indeed contains the commutator because it is the adjoint representation of the SU(N) group. The adjoint representation has a matrix structure, and for a given generator T^a, the operation [T^a, \Phi] is the matrix multiplication of the generator with the scalar field, which gives a matrix result. Thus, the second term of the covariant derivative contains this matrix multiplication.
 

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