# Perfect Metal (Transport) Conventions

I want $$K = K_R \oplus - K_L$$. And $$K_R = K_L = 1$$.

(opposite velocities to PM paper) $\mathcal{L}_f = \psi^\dagger_R i(\partial_t + v_F \partial_x) \psi_R + \psi^\dagger_L i(\partial_t - v_F \partial_x) \psi_L$

I agree with your equation 1.7. $\psi = e^{i k_F x} \psi_R + e^{-i k_F x} \psi_L$

Just like in the perfect metal paper: \begin{aligned} \psi_R & \propto & e^{i \phi_I} \\ \psi_L & \propto & e^{-i \phi_{N+I}}\end{aligned}

This implies the variations are $\delta \psi_{R,I} = \epsilon(i k_{F,I} +\partial_x \psi_{R,I} )$ $\delta \psi_{L,I} = \epsilon(-i k_{F,I} +\partial_x \psi_{L,I} )$ (in particular, we are assuming the Fermi momenta of the right and left movers in channel I coincide)

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 $P = - \sum_I k_{F,I} ( \psi^\dagger_{R,I} \psi_{R,I} - \psi^\dagger_{L,I} \psi_{L,I} ) + \sum_I (\psi^\dagger_{R,I} (i\partial_x) \psi_{R,I} + \psi^\dagger_{L,I} (i\partial_x) \psi_{L,I} )$

If I use the non-chiral action (Mike’s eqn 1.23 except with an overall minus sign in the definition of $$\Theta_I$$, this is correct given $$K = K_R \oplus K_L$$ and $$K_R = 1$$) \begin{aligned} \Theta_I &=& \phi_I - \phi_{N+I} \\ \Phi_I &=& \phi_I + \phi_{N+I} \\ \mathcal{L}_\text{nc} &=& \frac{1}{8\pi} \left[ 2 \partial_x \Theta_I \partial_t \Phi_I + ... + 4e A_t^{(I)} \partial_x \Phi_I + ...\right] \\ &=& \frac{1}{4\pi} \left[ \partial_t \phi_I \partial_x \phi_I - \partial_t \phi_{N+I} \partial_x \phi_{N+I} + ...+ 2 e A_t^{(I)} \partial_x \phi_I +... \right]\end{aligned}

... and if (crucially) I find the Noether charge of the non-chiral action, I get $P = \frac{1}{2\pi} \sum_I k_{F,I} \left(\partial_x \phi_I - \partial_x \phi_{N+I} \right) + \frac{1}{4\pi} \sum_I \left( (\partial_x \phi_I)^2 - (\partial_x \phi_{N+I})^2 \right)$

Fermion density. By starting at the coupling to the gauge field we can already see the densities should be $\rho_R = e \frac{1}{2\pi} \partial_x \phi_R$ $\rho_L = e \frac{1}{2\pi} \partial_x \phi_L$

To connect with another notation: Giamarchi claims (eq 2.55), for a non-chiral boson, $$\rho_R(x) + \rho_L(x) = \frac{-1}{\pi}\partial_x \phi$$. In addition $$\phi_R^{(G))} = K\theta - \phi, \phi_L^{(G))} = K \theta + \phi$$. 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 chiral action from above so we should be able to compare, but he has (eq 2.22) $$\psi_R \propto e^{i k_F x} e^{i \phi_R}, \psi_L \propto e^{-i k_F x} e^{i \phi_L}$$ whereas we have $$\psi = e^{i k_F x} \psi_R + e^{-i k_F x} \psi_L$$ where $$\psi_R \propto e^{i \phi_I}, \psi_L \propto e^{- i \phi_{N+I}}$$. That means our chiral fields are off by a minus sign: $\phi_I \equiv \phi^R_I = \phi_R^{(G))}$ $\phi_{N+I} \equiv \phi^L_I = -\phi_L^{(G))}$

so finally I conclude that: $\rho_R(x) = \frac{1}{2\pi} \partial_x \phi_R^{(G))} = \frac{1}{2\pi} \partial_x \phi_R$ $\rho_L(x) = \frac{-1}{2\pi} \partial_x \phi_L^{(G))} = \frac{1}{2\pi} \partial_x \phi_L$

From the commutation relation of Giamarchi (2.25) $$[ \phi(x), \partial_y \theta (y) ] = i \pi \delta(x-y)$$ we can backtrack the commutation relations of the chiral fields after we substitute the field redefinition into the above commutator: $\phi = \frac{\phi_L^{(G)} - \phi_R^{(G)}}{2}$ $\theta = \frac{\phi_L^{(G)} + \phi_R^{(G)}}{2K}$ $\frac{1}{4K} \left[ \phi_L^{(G)}, \partial_x \phi_L^{(G)} \right] + \frac{1}{4K} \left[ -\phi_R^{(G)}, \partial_x \phi_R^{(G)} \right] = i \pi \delta(x-y)$

So, both for Giamarchi’s ($$\phi_{R/L}^{(G)}$$ and our chiral fields $$\phi^{R/L} = \phi_{I/N+I}$$: $\left[ \phi_R, \partial_x \phi_R \right] = -i 2\pi \delta(x-y)$ $\left[ \phi_L, \partial_x \phi_L \right] = i 2\pi \delta(x-y)$

This matches eqn 1.17 of Mike’s notes under the identification $$\phi_I \equiv \phi^R_I,\phi_{N+I} \equiv \phi^L_I$$. Moreover, by using the above we can convince ourselves that:

$\left[ \rho_R, e^{-i\phi_R^G} \right] = \left[ \frac{1}{2\pi}\partial_x \phi_R^G, e^{-i \phi_R^G} \right] = \delta(x-y) e^{- i \phi_R^G}$ $\left[ \rho_L, e^{-i\phi_L^G} \right] = \left[ \frac{-1}{2\pi}\partial_x \phi_L^G, e^{-i \phi_L^G} \right] = \delta(x-y) e^{- i \phi_L^G}$

and for our chiral fields $[\rho_{R,I}, e^{-i \phi_I}] = [ \frac{1}{2\pi} \partial_x \phi^R_I, e^{-i \phi^R_I} ] = \delta(x-y) e^{-i\phi^R_I(y)}$

$[\rho_L, e^{i \phi_{N+I}} ] = [ \frac{1}{2\pi} \partial_x \phi^L_I, e^{i \phi^L_I} ] = \delta(x-y) e^{i\phi^L_I(y)}$

which makes sense given our choice $$\psi_R \propto e^{i \phi_I}, \psi_L \propto e^{- i \phi_{N+I}}$$.

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 \begin{aligned} \left[ \rho_e, e^{i m_J \phi_J} \right] &=& \sum_I [ \frac{\partial_x \phi_I}{2\pi}, e^{i m_J \phi_J}] \\ &=& - \sum_{I,L} K^{-1}_{IL} [ \Pi_L, e^{i m_J \phi_J} ] \\ &=& -\sum_{I,L,J} K^{-1}_{IL} \delta_{L,J} m_J e^{i m_J \phi_J} \delta(x-x') \\ &=& -\sum_{I,J} K^{-1}_{IJ} m_J e^{i m_J \phi_J} \delta(x-x') \\ &=& \sum_{I,J} t_I K^{-1}_{IJ} m_J e^{i m_J \phi_J} \delta(x-x')\end{aligned}

which allows us to immediate conclude that the correct charge vector in our conventions is $t_I = -1$

Current operators.

We agree that $$\rho_e = \rho_R + \rho_L = e \sum_I \left( \psi^\dagger_{R,I} \psi_{R,I} + \psi^\dagger_{L,I} \psi_{L,I} \right) = \frac{e}{2\pi} \sum_{I=1}^N \left( \partial_x \phi_I + \partial_x \phi_{N+I} \right)$$.

Then the continuity equation implies $j_e(x) = 2 e \sum_{I=1}^N \sum_{J=1}^{2N} \left[ V_{I,J} - V_{N+I,J} \right] \partial_x \phi_J(x)$ which is off by a factor of $$4\pi$$ from Mike’s 1.22. (A $$2\pi$$ came from the commutator, and another 2 from the fact that one has to commute past 2 $$\partial_x \phi$$’s)

Thermal current operator. From Noether’s theorem: $T^{xt} = \frac{1}{4\pi} \left( \partial_t \phi_I K_{IJ} \partial_t \phi_J - 2 \partial_t \phi_I V_{IJ} \partial_x \phi_J \right)$

Basis of almost conserved charges. We will take the charge densities in each channel, and a piece of the total momentum: \begin{aligned} Q^R_I = \frac{e}{2\pi} \int dx \partial_x \phi_I \\ Q^L_I = \frac{e}{2\pi} \int dx \partial_x \phi_{N+I} \\ P_T = \frac{1}{4\pi} \sum_I \left( (\partial_x \phi_I)^2 - (\partial_x \phi_{N+I})^2 \right)\end{aligned}