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$$ -S_G = \frac{1}{4\pi K} \int d\tau dx \left[ i \partial_\tau \phi_R \partial_x \phi_R - i \partial_\tau \phi_L \partial_x \phi_L - ... \right]$$  It looks just like the perfect metal action, so I conclude that:  $$ \rho_R(x) = \frac{1}{2\pi} \frac{-1}{2\pi}  \partial_x \phi_R $$ $$ \rho_L(x) = \frac{1}{2\pi} \frac{-1}{2\pi}  \partial_x \phi_R $$