Thursday, July 21, 2016

Bad metals and the unitary limit


In clean elemental metals the mean free path is much larger than the lattice constant and the Fermi wavelength. This means that an electron (or quasi-particle) has a well defined wavelength and momentum (wave vector) between collisions which change its momentum.
Thus, quasi-particles are a well defined entity.
However, consider that limit where the scattering becomes so strong that the mean-free path becomes comparable to the Fermi wavelength (or lattice constant).
Then clearly the idea of a quasi-particle with a well define wave vector and a mean free path does not make sense.

The resistivity (in a Boltzmann-Bloch) picture is inversely proportional to kF l.
Waving ones hands one can argue that in a metal there is a maximum value for the resistivity.
This is known as the Mott-Ioffe-Regel (MIR) limit.
Waving one hands  some more, one might argue that as the temperature increases (and inelastic scattering increases) towards the MIR limit the resistivity might saturate or even decrease because the material becomes an insulator.
Some people also debate whether the minimum value of kF l is 1, pi or 2 pi.

In reality, it is not clear whether the MIR limit exists in any known material.
Bad metals, by definition, violate it.

So what about the above argument? Obviously, there is significant hand waving.
The argument basically extrapolates results for weak scattering to the strong scattering limit.
Is there any way one can make some of the argument more rigorous?

One case where one can do better is for the specific case of elastic scattering due to impurities. Large scattering corresponds to what is known as the unitary limit.
[I often find the terminology obscure. I think it may relate to the optical theorem and the scattering S- matrix having the maximum possible value (unit = -1). I welcome clarification].

This post was stimulated by discussions with my colleagues who wrote the following paper, which has a brief discussion of some of the issues in Section II.

Breakdown of the universality of the Kadowaki-Woods Ratio in multi-band metals
D C Cavanagh, A C Jacko, and B J Powell

Here is my version of the argument, drawing on results found in Hewson’s wonderful book on The Kondo Problem.

Consider an electron scattering off a single impurity potential.
In the weak scattering limit the scattering cross section and mean-free path can be calculated in the Born approximation. However, the strong scattering limit can also be solved using a T-matrix which
sums all of the relevant Feynman diagrams.
Results can be expressed in terms of the scattering phase shifts.
In the limit of an infinite s-wave potential, the relevant phase shift becomes pi/2.
The scattering cross section is proportional to 1/k^2 which is must on dimensional grounds since there is no other well-defined length scale when the scattering length associated with the potential becomes infinite.

The scattering rate for the electrons is written in terms of phase shifts eta_l


If the only non-zero phase shift is the s-wave one and it equals pi/2, one sees that the scattering rate scales as 1/kF.
Note: this is what determines the resistivity in the Kondo problem at very low temperatures.
This (unitary limit) is similar to what one would get from a hand waving argument that takes the weak coupling result and sets kF l = 1.

So how does this relate to bad metals?
Well it should be stressed that the above argument is for elastic scattering and so does not necessarily carry over to inelastic scattering, which is actually what is relevant to bad metals.

I welcome clarifications on any of this.

5 comments:

  1. Fig. 4(a) of this paper, DOI: 10.1103/PhysRevB.84.085128 , shows that for the Hubbard model, where the scattering is inelastic at finite temperature, the resistivity saturates in the weak to intermediate correlation regime (U=6t here). According to DMFT, this does not happen when U is larger than the bandwidth, namely the resistivity continues to increase with temperature instead of saturating. See for example DOI: 10.1103/PhysRevLett.110.086401

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  2. Does it mean Dear Prof. that in the unitary limit there are ill defined quasiparticles?

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  3. In heavy fermion superconductos such as UPt3, there were some strong evidences that the compound is in the unitary regime.

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  4. So, how behaves the conductivity of UPt3 in its normal state?

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