The discussion in this section was inspired by a set of lecture notes\cite{oxford2013} dealing with magnetoresistance in crystals.
Currently, equation \eqref{eq:eom} assumed no external magnetic field. In the presence of a magnetic field \(\mathbf{B}\), it becomes: \[\frac{d\mathbf{p}}{dt} = - \frac{1}{\tau} \mathbf{p} + q \mathbf{E} + q \mathbf{v} \times \mathbf{B} \label{eq:eom-B}\] Equation \eqref{eq:eom-B} will produce one main effect: the resulting current will no longer be proportional to the applied electric field, as long as the electric and magnetic field are not parallel. Without loss of generality, let’s take the magnetic field parallel to the \(z\) axis, \[\mathbf{B} = \left(0, 0, B\right)\] Since any current parallel to the magnetic field will not be affected, let’s take both the electric field \(\mathbf{E}\) and current \(\mathbf{j}\) to lie in the \(x\)–\(y\) plane.1 Therefore, \[\mathbf{j} = \left(j_x, j_y, 0\right)\] \[\mathbf{E} = \left(E_x, E_y, 0\right)\] Substituting the components of \(\mathbf{B}\), \(\mathbf{E}\) and \(\mathbf{j} = n q \mathbf{p} / m^*\) into the equation of motion for the charge carrier in the presence of a magnetic field \eqref{eq:eom-B} gives: \[\tau \frac{d\mathbf{j}}{dt} = - \mathbf{j} + \sigma_0 \mathbf{E} + (\omega_\textrm{c} \tau) \mathbf{j} \times \left(\frac{\mathbf{B}}{B} \right) \label{eq:simplified-eomB}\] where \[\sigma_0 = \frac{n q^2 \tau}{m^*}\] is the electrical conductivity in the absence of a magnetic field and \[\omega_\textrm{c} = \frac{q B}{m^*}\] is the cyclotron frequency. Writing equation \eqref{eq:simplified-eomB} in components and solving for \(j_x\) and \(j_y\) gives: \[j_x = \frac{\sigma_0}{1 + (\omega_\textrm{c} \tau)^2} E_x + \frac{\sigma_0 \omega_\textrm{c} \tau}{1 + (\omega_\textrm{c} \tau)^2} E_y \label{eq:currentx-B}\] \[j_y = -\frac{\sigma_0 \omega_\textrm{c} \tau}{1 + (\omega_\textrm{c} \tau)^2} E_x + \frac{\sigma_0}{1 + (\omega_\textrm{c} \tau)^2} E_y \label{eq:currenty-B}\]
Let’s now look at equations \eqref{eq:currentx-B}, \eqref{eq:currenty-B} in the context of a though experiment. Assume we apply a magnetic field directed along the \(z\) axis and an electric field directed along the \(x\) axis to a conductive medium. We also allow the free flow of charge carriers along the x axis by attaching contacts at two ends of the sample.
Initially, \(E_y\) is zero. However, since the magnetic field drives a current in the \(y\) direction and there is no mechanism for charge carriers moving in the \(y\) direction to escape the sample, they will accumulate at the surface, creating an electric field counteracting \(j_y\). This will decrease the magnitude of \(j_y\), until, in equilibrium \(j_y = 0\). From equations \eqref{eq:currentx-B} and \eqref{eq:currenty-B} \[E_y = \omega_\textrm{c} \tau E_x\] \[j_x = \sigma_0 E_x\] This effect is just the classical Hall effect: the conductance in the \(x\) direction remains unchanged, with the magnetic field creating an electric field in the \(y\) direction due to redistribution of charge. Therefore, a conductor with a single type of charge carrier does not exhibit magnetoresistance – its conductivity is a constant with respect to the applied magnetic field.
However, this derivation hinges on a crucial assumption: all charge carriers have the same decay time and thus lead to the same contribution to the conductivity. We’ve seen that not to be true in the case of the Landau quasiparticles – their decay time depends on how far away their energy is to the Fermi energy.
Therefore, in a Fermi Liquid, there will be a population of decay timescales \(\tau_k\) and conductivities \(\sigma_0^{(k)}\). Without a magnetic field, these populations will just average out to result in an average timescale \(\overline{\tau}\) and conductivity \(\overline{\sigma_0}^{(k)}\). Thus, in the absence of a magnetic field, a distribution of decay timescales will not affect the conductivity. However, in the presence of a magnetic field, the picture changes. Let’s assume we have \(N\) different decay times. Let2 \[\gamma(1) := \frac{1}{N} \sum_{k=1}^{N} \frac{\sigma_0^{(k)}}{1 + (\omega_\textrm{c} \tau_k)^2}\] \[\gamma(\tau) := \frac{1}{N} \sum_{k=1}^{N} \frac{\sigma_0^{(k)}}{1 + (\omega_\textrm{c} \tau_k)^2} \tau_k\] Equations \eqref{eq:currentx-B}, \eqref{eq:currenty-B} will now read: \[j_x = \gamma(1) E_x + \omega_\textrm{c} \gamma(\tau) E_y\] \[j_y = -\omega_\textrm{c} \gamma(\tau) E_x + \gamma(1) E_y\] As previously, \(j_y = 0\) in equilibrium, giving \[E_y = \frac{\omega_\textrm{c} \gamma(\tau)}{\gamma(1)} E_x\] and resulting in a current density of \[j_x = \left(\gamma(1) + \frac{\omega_\textrm{c}^2 \gamma(\tau)^2}{\gamma(1)}\right) E_x\]
Therefore, in the presence of charge carriers with different lifetimes, the conductance in the \(x\) direction does depend on the magnetic field.
We can simply do so by setting \(E_z\) to zero. In the absence of a \(z\) component of the electric field, no current will flow in the \(z\) direction. The same is not true with the \(x\) and \(y\) direction, since the magnetic field will result in a nonzero component of the current in a direction perpendicular to the electric field.↩
With a slight abuse of notation, the way I see these equations is as a weighted average, with a weight of \(\sigma_0 / (1 + (\omega_\textrm{c} \tau_k )^2)\)↩