1
$\begingroup$

I'd like to study an inhomogeneous system, i.e., momentum is not a good quantum number therein. Therefore, I tried to calculate temperature Green functions like $\mathcal{G}(x_1,x_2;\tau)$, or its twofold Fourier transformation $\mathcal{G}(p_1,p_2;\tau)$.

But how can I get any transport property, e.g., conductivity, from these Green functions? I checked Mahan's overwhelming book, however, it only deals with the formalism of $\mathcal{G}(p;\tau)$ for homogeneous systems. Thanks in advance for any useful information.

$\endgroup$
1
  • $\begingroup$ If your system is inhomogeneous or has messy boundaries, then the conductivity is probably not given by just one number but rather would be anisotropic. $\endgroup$
    – Nanite
    Commented Feb 21, 2014 at 22:25

3 Answers 3

1
$\begingroup$

By inhomogeneous I assume you mean disordered, i.e., a system with a noisy/random potential landscape.

I'm not sure which Mahan book you are referring to, however I found Akkermans and Montambaux' Mesoscopic Physics of Electrons and Photons to give a good discussion of the problem of wave propagation in disordered media.

Essentially the problem is "solved" by not trying to calculate the Green function for an individual system, but instead calculate averages for an ensemble of systems sharing similar characteristics, i.e., all being generated from the same random process. As long as the correlations in the inhomogeneity are spatially invariant, you restore a sort of general uniformity to the system and you can talk about a real conductivity, at least, in terms of the average conductivity. Also you can talk about average Green function (this decays rapidly, mind you), and correlation functions of conductivity.

In the end these ensemble average calculations end up being even more useful than directly computing Green functions for individual systems, since in experiment the disorder pattern is usually not known and so it would be impossible to make a direct comparison anyway.

$\endgroup$
1
  • $\begingroup$ Thanks for the reference. It seems to be a good book that I missed. By ensemble average, do you mean sth. like averaging over the possible positions which the impurities may have in the solid? Thereafter, momentum conservation is recovered and we can calculate $\mathcal{G}(\vec{p};\mathrm{i} \omega_n)$, which is the "average Green function" you mentioned? Why does it decay rapidly? Thanks! $\endgroup$
    – xiaohuamao
    Commented Oct 23, 2014 at 9:11
0
$\begingroup$

Losing translation invariance is one of the big problem when studying disordered system (for example). In that case, one usually averages over the disorder to render the system translation invariant (on average), with its own technical difficulties. The applications for transport properties are explained in the Mahan or most good text book.

If the inhomogeneity comes in your case from another source, there might be other possibilities. You might also have to generalize the definition of the conductivity in your special case, but it is hard to tell without more details.

$\endgroup$
0
$\begingroup$

There are a few ways to extract transport properties from your single-particle temperature Green's functions. By analytically continuing it to real time $t$, one gets information about how a particle propagates in the medium. More exactly, you get the probability of the particle traveling a distance $x = |x_1-x_2|$ in the interval of $t$. From this, you could estimate the diffusion constant $D \sim x^2/t$ and then you relate it to conductivity $\sigma$ via Einstein's relation $\sigma \sim e^2 D N(E_F)$. In this picture, the particle moves like a Brownian dust. You may consult the book Conduction in noncrystalline Materials by N.F. Mott for further treatment.

Another way is to use Kubo formula. This formula expresses conductivity in terms of the electron-hole propagator, which can be written as a product of two single-particle Green's functions like yours. Thence you get the conductivity. This method is accurate but less physically transparent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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