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This is a follow up to my previous question. In general for a system with a given action, from Noether's theorem we have conserved currents for corresponding infinitesimal transformations leaving the action invariant. However $R$-charge as defined in the linked post doesn't generate symmetry transformations on fields constituting the action, but rather acts on the conserved charges.

My problem is, given the fact that $R-$charges don't act on the fields in the Lagrangian itself, what is the utility of introducing them in the first place? Do they serve any convenience upon quantization of the field theory, as classically they don't seem to serve much?

Michael Williams
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Supercharges transform in the fundamental representation of R-symmetry. The way to construct SUSY states is to start with a state which is annihilated by all $Q^i_{\alpha}$'s and build up states by acting with $Q^{i \dagger}_{\dot{\beta}}$'s. Since the $Q^{i \dagger}_{\dot{\beta}}$'s transform under the R-symmetry, the constructed states also transform under some representation of R-symmetry, and so the states can be labelled by which representation they transform under.

Also it is not correct to say that fields in the Lagrangian are invariant under R-symmetry transformations.

Contrary to what the question indicates, R-symmetry is not "introduced" for convenient manipulations, rather it is inbuilt into the SUSY algebra. It is only natural that states are labelled using the same.

Bruce Lee
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