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This implies that total momentum of fermions is  $$ P = - \sum_I k_{F,I} ( \psi^\dagger_{R,I} \psi_{R,I} - \psi^\dagger_{L,I} \psi_{L,I} ) + \sum_I (\psi^\dagger_{R,I} (i\partial_x) \psi_{R,I} + \psi^\dagger_{L,I} (i\partial_x) \psi_{L,I} )$$  If I use the non-chiral action (Mike's eqn 1.23 except with an overall minus sign in the definition of $\Theta_I$)  \begin{align}  \Theta_I &=& \phi_I - \phi_{N+I} \\  \Phi_I &=& \phi_I + \phi_{N+I} \\  \mathcal{L}_\text{nc} &=& \frac{1}{8\pi} \left[ \partial_x \Theta_I \partial_t \Phi_I + ... + 4e A_t^{(I)} \partial_x \Phi_I + ...\right]  \end{align}  Bosons (we have a factor of 2 difference in the $k_F$ term Mike, eq 1.22)  $$ P = \frac{1}{4\pi} \sum_I k_{F,I} \left(\partial_x \phi_I - \partial_x \phi_{N+I} \right) + \frac{1}{4\pi} \sum_I \left( (\partial_x \phi_I)^2 - (\partial_x \phi_{N+I})^2 \right) $$