2
$\begingroup$

In the integer quantum Hall effect, with the applied magnetic field reduced, more and more LLs get filled and one can observe higher and higher plateaus in the Hall conductivity $\sigma_H(B)$. Superficially, $\sigma_H(B\rightarrow0)$ will simply diverge roughly as $1/B$ because the LL spacing $\propto B$. But this must not be the case in reality. How should $\sigma_H(B)$ behave as $B$ approaches zero?

One way I can think of is introducing the relaxation time $\tau$, but not so clear how it regularizes the behaviour.

$\endgroup$

2 Answers 2

0
$\begingroup$

Drude expressions for the diagonal $\sigma_{xx}$ and Hall $\sigma_{xy}$ conductivities:

$$\begin{align} \sigma_{xx}&=\frac{\sigma_0}{1+\omega_c^2\tau^2}\\ \sigma_{xy}&=\sigma_0\frac{\omega_c\tau}{1+\omega_c^2\tau^2} \end{align}$$

So, the divergence of $\sigma_{xy}$ at $B\to 0$ is the case only if you assume $\omega_c\tau>>1$. When $B$ tends to zero, this inequality violates and the Hall conductance tends to zero.

$\tau$ is the momentum relaxation time, so $\sigma_{xy}$ is finite if scattering is taken into account.

$\endgroup$
1
  • $\begingroup$ Thanks. Any reference for these formulae? $\endgroup$
    – xiaohuamao
    Commented Oct 16, 2023 at 1:29
-1
$\begingroup$

As a basis in Hilbert space, in the linear gauge, the quantum Hall states may be oscillators in the direction between the boundaries and free waves along the strip.

Width of the oscillator states is determined by $B$, the strength of the oscillator potential by $B^2$ in the linear gauge.

The mid point of the oscillator is given by the wave vector.

So for $B \to \infty$ all occupied states are ground states of the oscillators, lined up in parallels along the strip with midpoints given by the individual $k$-vector, playing the role of a coordinate, with a large energy gap to the first exited state.

On the other side, with the width of the strip small compared to the extension of a very weak oscillator, the spectrum has very small gaps, the magnetic oscillator potential is only a small disturbation for free states between the boundaries, forming a quasi-continuous conduction band.

The exact eigenstates are not of the form $e^{-x^2}H_n(x-x_0)$ but a combination of parabolic cylinder functions fitting the bouundary condtions of current zero, that become the free $\sin$ solutions of the square well in a constant electric field.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.