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$$ [\rho_L, e^{i \phi_{N+I}} ] = [ \frac{1}{2\pi} \partial_x \phi^L_I, e^{i \phi^L_I} ] = \delta(x-y) e^{i\phi^L_I(y)} $$  which makes sense given our choice $\psi_R \propto e^{i \phi_I}, \psi_L \propto e^{- i \phi_{N+I}}$. {\bf Current operators.}  We agree that