Abstract

This report is aimed at answering the following question: “why do real metals oppose the flow of current?” The simple answer is “because electrons can’t be accelerated indefinitely by an external field without loosing energy”. I aim at discussing this question in the framework of Fermi Liquids – however, a large part of the argument only depends on weakly interacting charge carriers, irrespective of them being Landau quasiparticles. The main elements that make Landau quasiparticles unique in the context of conductivity is the fact that they have an intrinsic decay time. As we’ll see, this phenomenon leads to both a \(T^2\) temperature dependence of the resistivity and a form of magnetoresistance.

The two key elements for conducting a current are mobile charged particles and an energy dissipation mechanism.^{1} Since a current is just the motion of charged particles, you need the former by definition. In order to prevent these particles accelerating indefinitely, you also need an energy dissipation mechanism – without it, you have no steady state current for nonzero electric field.

We’ll now derive Ohm’s law starting only from these assumptions, following the argument set by Drude (1900). Assume that we have a uniform density \(n\) of particles, carrying a charge \(q\). These carriers are accelerated by an electric field \(\mathbf{E}\) and scatter via a certain mechanism with a characteristic lifetime \(\tau\). Assuming they have a well defined momentum \(\mathbf{p}\), their equation of motion is:^{2} \[\frac{d\mathbf{p}}{dt} = - \frac{1}{\tau} \mathbf{p} + q \mathbf{E}
\label{eq:eom}\]

The force proportional to \(-\mathbf{p}\) serves to dissipate energy, while the electric field \(E\) serves to accelerate the particle. The steady state velocity \(\mathbf{v}\) is: \[\frac{d\mathbf{p}}{dt} = 0 \Leftrightarrow \mathbf{v} = \frac{q \tau}{m} \mathbf{E} \label{eq:v}\] The current density is given by multiplying the charge density of carriers with their velocity:(Halliday 2008) \[\mathbf{j} = n q \mathbf{v} \label{eq:j}\] Substituting the value of \(\mathbf{v}\) from equation \eqref{eq:v} gives \[\mathbf{j} = \frac{n q^2 \tau}{m} \mathbf{E}\]

This equation is essentially Ohm’s Law: the current density is linearly proportional to the applied field. We can calculate the conductivity tensor from its definition, \[j_\alpha \equiv \sigma_{\alpha\beta} E_\beta\] giving a diagonal conductivity tensor of \[\sigma_{\alpha\beta} = \frac{n q^2 \tau}{m} \delta_{\alpha\beta}\]

This result is intuitive: the conductivity increases with the carrier density and their charge, while it decreases with increasing mass. As long as the charge carriers are weakly interacting, all that is left to do is to calculate \(\tau\) for different scattering mechanisms. We’ll look at this problem in the next section.

The discussion in this chapter is partially inspired by Nozieres et al. (1999), Zocco (2006) and Hooley (2013).

Assuming that the interactions between electrons in a crystal can be described by Fermi Liquid theory, we are guaranteed weakly interacting particles in the form of Landau quasiparticles. Instead of *bare* electrons acting as the charge carriers, we have *dressed* electrons instead: electrons coupled to a “cloud” of electron–hole pairs. Furthermore, these quasiparticles have an intrinsic decay time, over which their energy is dispersed throughout the electron–hole sea. It is this intrinsic decay time that gives a Fermi Liquid its non-zero resistivity (or non-infinite conductivity).

In the next subsections, we’ll derive the dependence of this decay timescale on \(1/T^2\), where \(T\) is the temperature of the Fermi Liquid, assumed small on the scale of the Fermi temperature \(T_F\).

From the assumptions of the Fermi Liquid theory, we have a well defined Fermi surface in \(\mathbf{k}\)–space, with a crystal momentum \(\hbar \mathbf{k}_F\). Assume we now excite one of the Landau quasiparticles to an energy \(\epsilon\) above the Fermi surface and we’d like to know the characteristic time for it to lose energy.

The mechanism for this decay is by scattering with other quasiparticles. Assuming that the energy lost by our test particle is \(\omega\), an equal amount of energy must be gained by another quasiparticle due to conservation of energy. However, since the only particle above the Fermi surface is the test particle, the energy \(\phi\) of the scatterer must be at most \(\omega\) below the Fermi energy.^{1} Therefore, for each value of \(\omega\), there’s \(O(\omega)\) number of scattering processes that can result in the decay of the test particle.

Therefore, since \(0 < \omega < \epsilon\) and \(0 < \phi < \omega\), there are \(O(\epsilon^2)\) possible decay processes (see Figure \ref{fig:quasiparticle-lifetime} for an illustration). Assuming each decay process has a similar matrix element, the decay cross section will therefore be proportional to the number of decay processes, therefore proportional to \(\epsilon^2\). Thus, since lifetime is inversely proportional to decay cross section, the lifetime \(\tau\) of a Fermi Liquid quasiparticle is: \[\tau \sim \frac{1}{\epsilon^2}\]

Otherwise, there’s no other state for it to scatter in↩