Eugeniu Plamadeala edited untitled.tex  about 9 years ago

Commit id: 0af517f2a83faceb074de4d8be931be98b676b71

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which makes sense given our choice $\psi_R \propto e^{i \phi_I}, \psi_L \propto e^{- i \phi_{N+I}}$.  {\bf Charge vector}  We know given our conventions that total charge density is $\rho_e = \sum_I \frac{\partial_x \phi_I}{2\pi}$. We also know from looking at the commutation relations that the momentum conjugate to $\phi_I$ is $\Pi_I = -K_{IJ} \frac{\partial_x \phi_J}{2\pi}$. From the commutation properties of the momentum, we can derive that   $$ [\Pi_I, e^{i m_J \phi_J} ] = \sum_J \delta_{IJ} m_J e^{i m_J \phi_J} \delta(x-x') $$  Which means that   $$ [\rho_e, e^{i m_J \phi_J}] = \sum_I [ \frac{\partial_x \phi_I}{2\pi}, e^{i m_J \phi_J}] = - \sum_{I,L} K_{IL} [ \Pi_L, e^{i m_J \phi_J} ] $$  {\bf Current operators.}