It is well known that the electric field can be written in terms of scalar $\phi(r)$ and vector $A(r)$ potential as $E=-\nabla\phi - \partial_tA$. Then the Hamiltonian $H_{ext}$ for the electromagnetic field perturbation can be written in two ways
- $\quad$ $H_{ext}^{(1)} = \int d^3r \rho(r)\phi(r) \tag{1}$
- $\quad$ $H_{ext}^{(2)} = \int d^3r J(r)\cdot A(r) \tag{2}$
here $\rho(r)$ and $J(r)$ are particle density and particle current density, respectively. The electrical conductivity $\sigma_{xy}$ is defined by equation $J_x=\sigma_{xy} E_y$.
After a long derivation of the Kubo formula, the DC conductivity for perturbation $H_{ext}^{(1)}$ become $$ \sigma_{xy}^{(1)} = \lim_{\omega\to0}\frac{1}{i\omega} \left[\Pi(\omega) - \Pi (\omega)\right] \quad ; \quad \Pi(\omega) = \int dt e^{-i\omega t} \langle[J_x(t),J_y(0)]\rangle \tag{3} $$ However, when we use $H_{ext}^{(2)}$ as perturbation, we get two parts of $J(r)$, namely paramagnetic $J^{(p)}$ and diamagnetic $J^{(d)}$. The DC response is given as calculated from the sum of both para and diamagnetic currents. For example $$ \sigma_{xy}^{(2)} = \lim_{\omega\to0} \left[\sigma_{xy}^{(p)}+\sigma_{xy}^{(d)}\right] \tag{4} $$
It is argued that for correct conductivity, one should take care of the diamagnetic part, which is connected with the "magnetization currents" - currents that are present in the system even in the absence of external perturbation. In non-magnetic materials, Eq $(3)$ gives correct results as there are no magnetization currents. However, in magnetic materials, we must add a term of magnetization current to get the correct conductivity, for example in this article.
Question:
I know that when we use $H_{ext}^{(1)}$ as a perturbation, we must add an extra term of magnetization current. I want to know if we need to add this magnetization current term when we use $H_{ext}^{(2)}$. Does not diamagnetic term take care of magnetization current?