Using a Drude model of charge carriers with a charge $q$ and a mass $m$ (which I allow to take either sign at this stage) in a sample with an applied electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$ we can find that $$E=\frac{\mathbf{p}}{q\tau}-\mathbf{v}\times\mathbf{B}$$ where $\tau$ represents the mean scattering time of the electrons. Using $\mathbf{p}=m\mathbf{v}$ and $\mathbf{j}=nq\mathbf{v}$ this can be written as $$E=-\frac{1}{nq}\mathbf{j}\times\mathbf{B}+\frac{m}{nq^2\tau}\mathbf{j}$$ and assuming we have a current density $\mathbf{j}=j\hat{\mathbf{x}}$ and an applied magnetic field $\mathbf{B}=B\hat{\mathbf{z}}$ this reduces to $$\mathbf{E}=\frac{jB}{nq}\hat{\mathbf{y}}+\frac{mj}{nq^2\tau}\hat{\mathbf{x}}$$ The Hall coefficient is defined as $R_{H}=\frac{E_{y}}{jB}$ and so here is given by $R_{H}=\frac{1}{nq}$.
I now consider three cases
- Electrons in the bottom of the conduction band. Here $q=-e$ and the Hall coefficient is negative.
- Holes at the top of the valence band. Here $q=e$ and the Hall coefficient is positive.
- Electrons at the top of the valence band. Here as in the first case the Hall coefficient is negative. These electrons have a negative effective mass but this does not affect the Hall coefficient as defined above.
However the second and third cases should be physically equivalent - if I had a conductor with a nearly full valence band the Hall coefficient would be positive, and I should be able to calculate this by considering either holes or the electrons. There appears to be a contradiction. The only way I can think of resolving this is to note that in the third case because the electrons have negative mass the second term in the final expression for $E$ is actually reversed - does this somehow change the measured Hall coefficient (the precise definition of the Hall coefficient in terms of which directions to consider is clearly relevant in this case)?
