Quantum Hall Effect resistivity

In summary, the conversation discusses the behavior of resistivity in relation to the density of states for the quantum hall effect. The diagrams (A) and (B) show different scenarios where the resistivity remains constant or increases due to scattering. The XX and XY plots also play a role in determining the behavior of resistivity. The system becomes a metal when the gap closes and the quantization of the hall conductance is only for sigma xy, not sigma xx. Impurities do not cause backscattering, as the edge states are chiral.
  • #1
rwooduk
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I'm having trouble understanding why the resistivity behaves as it does in comparison to the density of states for the quantum hall effect. Take the following two diagrams:
(A)
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(B)
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I understand that there can be no scattering in (A) because all states are full (i.e. no elastic) and the gap is too big to scatter inelastically. But why should the resistivity remain constant?

Also on (B) why does the resistivity increase when there are free states to scatter into? Is it saying scattering hinders conduction?

Having real trouble relating the DOS to resistivity. Any suggestions more than welcome.
 
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  • #2
Ok, I'll update this. Since there are more free states in (b) the scattering increases and so does the resistivity.

For (A) when there are no states available it drops to zero.

^^ This is for the XY component. I'm still a little unsure of the difference (aside from direction) of the XX and XY plots. So any advice would be appreciated.
 
  • #3
There is no back scattering because the edge states are chiral, the direction is determined by the B field. The diagonal components of the conductivity matrix (when you have the chemical potential between bands) are zero because there is no current along the direction of the electric field except when the gap closes. When this happens the system becomes a metal, both the conductivity and resistivity are then nonzero. The resistivity is the inverse of conductivity, there can be no resistivity for this reason, the inverse of conductivity is not diagonal it is off diagonal. When the gap closes there is a phase transition between an insulator with hall conductivity ne^2/h and (n+1)e^2/h. When there are impurities there is still no backscattering of the edge states, they will just go around the impurity. There can't be backscattering, the states are chiral.

The quantization of the hall conductance is not for the sigma xx it is only for sigma xy. If there is a gap, the system is an insulator, there can't be a current in the bulk (xx direction).
 
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  • #4
radium said:
There is no back scattering because the edge states are chiral, the direction is determined by the B field. The diagonal components of the conductivity matrix (when you have the chemical potential between bands) are zero because there is no current along the direction of the electric field except when the gap closes. When this happens the system becomes a metal, both the conductivity and resistivity are then nonzero. The resistivity is the inverse of conductivity, there can be no resistivity for this reason, the inverse of conductivity is not diagonal it is off diagonal. When the gap closes there is a phase transition between an insulator with hall conductivity ne^2/h and (n+1)e^2/h. When there are impurities there is still no backscattering of the edge states, they will just go around the impurity. There can't be backscattering, the states are chiral.

The quantization of the hall conductance is not for the sigma xx it is only for sigma xy. If there is a gap, the system is an insulator, there can't be a current in the bulk (xx direction).

Thats very helpful! Thank you!
 

FAQ: Quantum Hall Effect resistivity

1. What is the Quantum Hall Effect resistivity?

The Quantum Hall Effect resistivity is a phenomenon observed in two-dimensional electron systems at low temperatures and high magnetic fields. It is characterized by the quantization of the Hall resistance, which is the ratio of the transverse voltage to the longitudinal current in the material.

2. How is the resistivity quantized in the Quantum Hall Effect?

The resistivity is quantized in the Quantum Hall Effect because the energy levels of the electrons in a two-dimensional material become discrete in the presence of a magnetic field. This leads to the formation of Landau levels, which are energy bands that are separated by a fixed amount known as the cyclotron energy. As a result, the Hall resistance takes on discrete values that are determined by fundamental physical constants.

3. What causes the Quantum Hall Effect resistivity to be so precise?

The Quantum Hall Effect resistivity is so precise because it is a topological effect, meaning it is determined by the geometry and topology of the material rather than its microscopic details. This makes it robust against impurities and imperfections in the material, allowing for extremely accurate measurements of the Hall resistance.

4. How is the Quantum Hall Effect resistivity used in scientific research?

The Quantum Hall Effect resistivity is used in scientific research to study the behavior of electrons in two-dimensional materials and to investigate fundamental physical phenomena such as topological phases and quantum entanglement. It also has practical applications in the development of precise electrical standards and in the creation of electronic devices with unique properties.

5. Can the Quantum Hall Effect resistivity be observed in any material?

The Quantum Hall Effect resistivity can only be observed in materials that have a two-dimensional electron gas, such as semiconductor heterostructures or graphene. These materials must also be cooled to very low temperatures and subjected to high magnetic fields to observe the effect. However, researchers are constantly exploring new materials and conditions in which the Quantum Hall Effect resistivity can be observed.

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