I'm slightly confused with the following situation:

Suppose you have an electron in a tight-binding model, and let's say we are in one dimension with $N$ lattice sites.

Add to this a single impurity at a specific site $i$.

The pure system is translationally invariant and has as eigenfunctions the plain waves. As a consequence, the probability to find the (free) electron on the impurity site should scale as $1/N$ and I'd imagine that the full forward-scattering Green function for the problem has some form

$$G(k,\omega) = G_0(k,\omega) + \frac{1}{N} T(k,\omega) G_0(k,\omega)^2$$

where $T(k,\omega)$ would be some function independent of $N$.

As a consequence, in the limit of an infinite system $N \rightarrow \infty$ I would expect to not see the impurity effect at all as a consequence of that $1/N$ factor.

However, if I imagine that the impurity is a temperature effect (an example would be a flipped local spin in a ferromagnetic ground state, or an excited phonon) then the impurity could occur on any of the $N$ sites of the system. If I confine myself to very low temperatures where I assume that the entire system contains at most one impurity, the temperature averaged propagator would then look like

$$G_T(k,\omega) = 1/Z \times \left[ \left\langle \phi_0 | \hat G(k,\omega) |\phi_0 \right\rangle + \sum_i e^{-\beta E_I} \left\langle i | \hat G(k,\omega + E_I) | i \right\rangle \right]$$

where $|\phi_0\rangle$ is the free system's ground state and $| i \rangle$ is the state with an impurity at site $i$ and $E_I$ is the energy cost of the impurity.

$Z$ is the partition function and in our simple example just gives $1 + Ne^{-\beta E_I}$.

If that's all correct, then the finite temperature Green's function should just be $$G(T,k,\omega) = \frac{G_0(k,\omega) + Ne^{-\beta E_I} \left( G_0 + G_0^2 1/N T_{kk}(\omega)\right)}{1 + Ne^{-\beta E_I}}$$ which can be simplified to $$G_0(k,\omega) + \frac{e^{-\beta E_I} G_0^2 T_{kk}}{1 + Ne^{-\beta E_I}}$$

What confuses me here is that if we let $N$ go to infinity, the impurity term will again vanish at small temperature and thus seems to play no role at all in determining the Green function or the spectral function, but clearly that cannot be correct, since we know that impurities can, for example, lead to bound states.

Am I missing something really obvious here?

EDIT: With your answers so far, let me just add what I think I could do right now.

At low temperature, I'd assume a low density of thermally excited impurities, so I will do an expansion to first order in that density. The partition function is $$Z = (1 + e^{-\beta E_I})^N = \sum_{n}^N \begin{pmatrix}N\\n\end{pmatrix} e^{-\beta n E_I}.$$

The numerator for $\langle G \rangle$ becomes approximately, only accounting for interactions with one impurity at a time, justified by the low density: $$\sum_n e^{-\beta n E_I} \begin{pmatrix}N \\ n \end{pmatrix} \left[ G_0(\omega) + n T(\omega) G_0^2(\omega)\right]$$ Now the $G_0$ don't depend on anything we're summing over, so we can pull them out of the sum and obtain

$$\langle G \rangle = G_0 + \langle n \rangle T G_0^2$$ where I omit the momentum and frequency arguments because they're the same everywhere anyway.

I like this answer, it makes intuitive sense: The extra term due to impurity scattering is, to first order, linear in the expected number of thermally excited impurities. I understand that this gets the physics of disordered systems wrong since it can't account for interference due to multi-impurity-scattering and thus for example won't ever see Anderson localization. But just to get started on this, I think it's at least the correct way to start thinking about things.

  • $\begingroup$ Why would you expect to see any effects of the impurity in your calculation? You are sending the density of impurities to zero - why would you see anything unless you had a finite density in the thermodynamic limit? $\endgroup$ Commented Jun 25, 2013 at 0:24
  • $\begingroup$ "Am I missing something really obvious here?" -> Why, of course I am. In some sort of Born approximation, could I then assume that scattering involves only one of my impurities so that all I have to do in my formulas above is multiply the scattering part with $N_I$, the number of impurities, and then have $N_I/N = n = const$ as $N$ goes to infinity? $\endgroup$
    – Lagerbaer
    Commented Jun 25, 2013 at 1:13
  • $\begingroup$ @Lagerbear: Basically, but you should be a little careful. See the answer I've written. Also as you probably realize, you are doing thermal perturbation theory wrong - it need to be set up so you get a finite correction to the free energy density. In the diagrammatic notation this is the re-summation of disconnected diagrams. $\endgroup$ Commented Jun 26, 2013 at 16:03

1 Answer 1


You are working a finite number of impurities and expecting an answer in the thermodynamic limit. This should be clear if you do your calculation at a fixed density of impurities, and resum the infinite series, taking the leading term at each power of the impurity density. This will give you a correction to the self-energy.

Alternatively, as you suggest, you could do the Born approximation for a single impurity. Note that to get anything non-trivial you need to work to second order in the scattering potential. You would then need to multiply this by the density of impurities, (or maybe something like the mean free time between scatterings, don't know off the top of my head) to get the appropriate correction to the self energy.

Also note that the second Born approximation knows about only a fragment of the behavior we expect from an electron in a disordered system. It know only that the momentum becomes randomized by repeated scattering. It does not know anything about even simple things like diffusion. To get that you need to look at other diagrams in the perturbation theory.


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