1
$\begingroup$

In the integral quantum Hall effect, one has that in regions where $R_{xy}$ (the Hall resistance) is a constant, $R_{xx}$ surprisingly goes to zero. Why does that happen? Do impurities in the material play a role in this?

$\endgroup$
1
$\begingroup$

$R_{xx}$ is the longitudinal resistance, i.e. the resistance along the direction of current flow -- to be clear, this is the direction where the voltage difference from the battery or power source is. So a vanishing $R_{xx}$ means that there is dissipationless flow, and is kinda the "big thing" about the IQHE (Integer Quantum Hall Effect) and why it relates to topological insulator.

The fact that $R_{xx}=0$ is not directly caused by impurities. Impurities guarantee that there are plaeatux in $R_{xy}$ and hence large regions where $R_{xx}=0$.

The reason why $R_{xx} = 0$ is that at that precise value for magnetic field strength $B$ and carrier number $N$, the Landau levels are completely filled. So there are literally no other states an electron could scatter into. If you had Ohmic (i.e. dissipationful) conduction, you'd have an electron being scattered around and hence moving at a constant velocity (Drude model). This would require it to change motional state at each scattering event. But if there are no states available, then it cannot scatter and it just keeps going along. Hence the carrier flow is dissipationless (and topologically protected, in that it does not matter whether you put a dent on the edge of the material, electrons will go around it and still move along the edge).

The presence of impurities just broadens the otherwise fixed-energy Landau levels, thereby guaranteeing that the dissipationless flow occurs for a range of $B$ (or $N$) and not just at the specific combination filling the Landau level.

Just to be clear, the plot we are talking about is ($R_{xy} =R_{\mathrm{H}}$ and $R_{xx} = R_x$): enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So if I understand correctly, the Landau levels in this system are the various possible wave functions whose centers are at a fixed separation from each other. Also, a_xy=(ne^2/h), where a is the Hall conductance and n is the number of Landau levels that are filled. The characteristic step graph of Hall resistance "is" because Landau levels are getting filled one by one and not that all are filled from the onset. Then according to you, Rxx should be zero only when all the levels are filled, and not throughout the whole graph. Also, how would you explain the spikes in Rxx then? $\endgroup$ – Ayush Raj May 29 at 6:46
  • $\begingroup$ The Landau level is just the energy level. The collection of the individual closed orbits (the ones you are talking about) is the just the degeneracy of the Landau level, i;e. how many particles there are per energy level. The spikes in $R_{xx}$ are when you are filling the next broadened Landau level. Now there are states that electrons can scatter into, so they do not have to go only forward. This causes usual Ohmic dissipation. $\endgroup$ – SuperCiocia May 29 at 6:57
  • $\begingroup$ So when you say that Landau level is filled, you mean that all the possible degenerate states associated with that Landau level is filled? $\endgroup$ – Ayush Raj May 29 at 7:00
  • $\begingroup$ Yes. All the states in a given energy level are filled. $\endgroup$ – SuperCiocia May 29 at 7:16
  • $\begingroup$ Oh! Got it. Thanks a lot :) $\endgroup$ – Ayush Raj May 29 at 7:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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