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Fermion density. Giamarchi claims (eq 2.55), for a non-chiral boson, $ \rho_R(x) + \rho_L(x) = \frac{-1}{\pi}\partial_x \theta$. In addition $\phi_R = K\theta - \phi, \phi_L = K \theta + \phi$. His final action for the chiral bosons looks like (C.12 upon above field redefinition)  $$ -S_G = \frac{1}{4\pi K} \int d\tau dx \left[ i \partial_\tau \phi_R \partial_x \phi_R - i \partial_\tau \phi_L \partial_x \phi_L - ... \right]$$  It looks just like the perfect metal action, so I conclude that: that (dropping the factor of K):  $$ \rho_R(x) = \frac{-1}{2\pi} \partial_x \phi_R $$  $$ \rho_L(x) = \frac{-1}{2\pi} \partial_x \phi_R $$