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And similarly, for the bosons, but with no minus sign:  $$ \delta \phi_I = \epsilon( k_{F,I} + \partial_x \phi_I ) $$  $$ \delta \phi_{N+I} = \epsilon( k_{F,I} + \partial_x \phi_{N+I}) $$ This implies that total momentum of fermions is  $$ P = - \sum_I k_{F,I} ( \psi^\dagger_{R,I} \psi_{R,I} - \psi^\dagger_{L,I} \psi_{L,I} + \sum_I \psi^\dagger_{R,I} (i\partial_x) \psi_{R,I} + \psi^\dagger_{L,I} (i\partial_x) \psi_{L,I} $$