Solving Confusing Indices in Operator Calculus

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SUMMARY

The discussion focuses on the operator calculus involving the covariant derivative defined as D_\mu \phi^a = \partial_\mu \phi^a - iA^{r}_{\mu}{(T_r)^a}_b\phi^b. The key question is how to compute D_\mu D_\nu \phi^a, which is resolved by recognizing that D_\mu D_\nu \phi^a can be expressed as D_\mu D_\nu \phi^a = (\partial_\mu{\delta^a}_b - iA^{r}_{\mu}{(T_r)^a}_b)(\partial_\nu{\delta^b}_c - iA^{s}_{\nu}{(T_s)^b}_c)\phi^c. The discussion clarifies that the differential operator has two indices, leading to the formulation D_\mu D_\nu \phi^a = (D_\mu)^a_{\phantom{a}{b}} (D_\nu)^b_{\phantom{b}{c}}\phi^c.

PREREQUISITES
  • Understanding of operator calculus in gauge theories
  • Familiarity with covariant derivatives and their notation
  • Knowledge of gauge fields and their representations
  • Basic concepts of tensor indices and their manipulation
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  • Study the properties of covariant derivatives in gauge theories
  • Learn about the role of gauge fields in quantum field theory
  • Explore the implications of tensor index notation in physics
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The discussion is beneficial for theoretical physicists, graduate students in physics, and anyone studying gauge theories and operator calculus in quantum field theory.

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Sorry if this is a stupid question but I really want to understand this.

If we have an operator that's defined in the following way:

D_\mu \phi^a = \partial_\mu \phi^a - iA^{r}_{\mu}{(T_r)^a}_b\phi^b

How would we go about working out:

D_\mu D_\nu \phi^a

What's confusing me is that the operator D acts on one component of the scalar (the gauge field) on the LHS, but acts on two components of the scalar on the RHS.

FYI: the "answer" is that D_\mu D_\nu\phi^a=(\partial_\mu{\delta^a}_b - iA^{r}_{\mu}{(T_r)^a}_b)(\partial_\nu{\delta^b}_c - iA^{s}_{\nu}{(T_s)^b}_c)\phi^c

How does the above expression come about?
 
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The first thing to realize is that the your differential operators has two indices:

D_\mu \equiv (D_\mu)^a_{\phantom{a}{b}}.​

Therefore, the combination D_\mu\phi^a is really

D_\mu\phi^a\equiv (D_\mu)^a_{\phantom{a}{b}}\phi^b.​

With this in mind, it should be pretty clear that

D_\mu D_\nu \phi^a\equiv (D_\mu)^a_{\phantom{a}{b}} (D_\nu)^b_{\phantom{b}{c}}\phi^c.​

Does that help?
 
thanks that was a good explanation:)
 

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