1
$\begingroup$

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

  1. $\quad$ $H_{ext}^{(1)} = \int d^3r \rho(r)\phi(r) \tag{1}$
  2. $\quad$ $H_{ext}^{(2)} = \int d^3r J(r)\cdot A(r) \tag{2}$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

I think there is a baseline of linear response plus there are the external effects to be considered separately.

First consider the linear response without extra effects: in this case you have two scenarios one is that you have a probe electric field e.g. your perturbation $H^1_{ext}$ and then you measure the current, and second scenario is that you have a probe electric current $H^2_{ext}$ and you see the outcoming Electric field, in total you are still computing the conductivity $\sigma_{ij}$ or the inverse matrix i.e. the resistivity $\rho_{ij}$. You have a derivation of formula (3) from perturbation (1), But If you see Tong QHE pdf he derives your formula (3) from perturbation (2) and the two of them are the two faces of the same medal.

In the simple scenario you do this linear response without a background field, but you can also do it in a fixed given external background, that you can imagine as the magnetic or electric field that you enforce from the outside of the sample or the "lab".

Then on top of these two fields the paper is adding a thermoelectric effect and therefore an additional current term. In your case I expect you want to add some paramagnetic or diamagnetic effect and that one you can also add it again as a separate term (i.e. separate extra magnetic fields or extra electric fields that comes back in the perturbation). The magnetization currents are actually an extra background that is generated inside the sample (and that one may have it's own modeling).

$$ A_{tot}=A_{probe} + A_{bg} + A_{dia} + A_{para}\\ \phi_{tot} = \phi_{probe} + \phi_{bg} + \phi_{th}$$

And each of these if the total response is still linear, i.e. if essentially each contribution is small, will lead to a separate contribution to the conductivity.

PS: The 2 kubo papers are actually a good place to learn linear response.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.