Problem #\(1\): Describe how the classical Hall coefficient \(\rho^{-1}\) and explain why it’s “causally intuitive”.
Solution #\(1\): In the classical Hall effect, the “cause” is both an applied current density \(J\) together with an applied perpendicular magnetic field \(B\). The “effect” is an induced transverse electric field \(E\) whose magnitude and direction are such as to ensure a velocity selector steady state. So it seems reasonable to define the Hall coefficient by:
\[\rho^{-1}:=\frac{\text{effect}}{\text{cause}}=\frac{E}{JB}\]
where the notation \(\rho^{-1}\) is deliberately suggestive of being the reciprocal charge density which is also what the Hall effect is. Note that here \(E\) only represents the transverse component of the electric field, i.e. \(E=-\textbf E\cdot(\textbf J\times\textbf B)/JB\), as there may also be a longitudinal component e.g. to compensate for scattering and other resistances.
The simplest way to derive this is to just set the Lorentz force density \(\textbf f\) to zero:
\[\textbf f=\rho\textbf E+\textbf J\times\textbf B=\textbf 0\Rightarrow\rho^{-1}=-\textbf E\cdot (\textbf J\times\textbf B)/|\textbf J\times\textbf B|^2\]
Since \(J,B\) are applied by the experimentalist, they are readily known, and \(E\) can also be readily measured by measuring instead a suitable Hall voltage \(\Delta\phi_H=-\int d\textbf x\cdot\textbf E\) in the transverse direction (voltages are always experimentally accessible as well), so the classical Hall effect provides a simple way to directly measure the charge density \(\rho\) (via the Hall coefficient \(\rho^{-1}\)), and hence the number density of charge carriers \(n=\rho/{\pm e}\) (strictly this assumes a single charge carrier species; for semiconductors it would be a bit more complicated…so in fact strictly speaking maybe one shouldn’t denote it by \(\rho^{-1}\) but just by \(R_H\)).
Problem #\(2\): Delve into the quantum Hall effect.