Landauer-Büttiker formalism

The Landauer-Büttiker formalism , which can be traced back to the work of Rolf Landauer and Markus Büttiker , generally describes the current transport in the marginal channel model in systems of finite extent and is used in particular to describe the quantum Hall effect .

General

It is assumed that a system with any number j of contacts with different chemical potentials µ _k (k = 1,2,3, ..., j) is present. This forms i one-dimensional edge channels, which arise because the Fermi energy lies between the (i-1) -th and i-th Landau level .

If one assumes a transmission _probability T _{kk + 1} for the electrons scattered from contact k + 1 into contact k and R _kk denotes the reflection probability of an electron from contact k back into contact k, a rate equation for the net current I _k at the contact can be derived set up k:

${\displaystyle I_{k}=e\int _{0}^{\mu _{k}}{D(E)v(E)dE}-e\int _{0}^{\mu _{k}}{R_{kk}D(E)v(E)dE}-e\int _{0}^{\mu _{k+1}}{T_{kk+1}D(E)v(E)dE}}$

Here refers to the density of the one-dimensional edge of the channel and v (E) is the group velocity of the electrons. The integral is made up of the outgoing positive and incoming negative part of contact k. ${\displaystyle D(E)={\frac {1}{\pi \hbar v(E)}}}$

An evaluation of the integral allows conclusions to be drawn about the conductivity of an individual (spin-started) edge channel, which

${\displaystyle G=2{\frac {e^{2}}{h}}}$

amounts. The formalism is not restricted by the sample geometry or the number of edge channels and allows the calculation of the net current I _k at contact k for any geometries and edge channels .

Example of QHE with two occupied edge channels

For the following example we assume that there is only one edge canal with spin degeneration. The experimental setup corresponds to that of a normal Hall measurement with four contacts. A current is sent through two (contacts 1 and 3), the resulting voltage is tapped at the other two (contacts 2 and 4). The potentials of the contacts are denoted by μ ₁ to μ ₄ . Since only one voltage is tapped at contacts 2 and 4, their net current is zero, since the charge transported via the edge channels traverses the contact without loss.

After evaluating the integral given above, we can set up the following system of equations (it is assumed that no backscattering occurs, _i.e. R _kk = 0):

${\displaystyle I_{1}=I=2{\frac {e}{h}}(\mu _{1}-\mu _{4});\,T_{14}=1,\,T_{12}=T_{13}=0}$

${\displaystyle I_{2}=0=2{\frac {e}{h}}(\mu _{2}-\mu _{1});\,T_{21}=1,T_{23}=T_{24}=0}$

${\displaystyle I_{3}=-I=2{\frac {e}{h}}(\mu _{3}-\mu _{2});\,T_{32}=1,T_{31}=T_{34}=0}$

${\displaystyle I_{4}=0=2{\frac {e}{h}}(\mu _{4}-\mu _{3});\,T_{43}=1,T_{41}=T_{42}=0}$

From equation 2 it follows that μ ₁ = μ ₂ . Substituting this into equation 1, it follows

${\displaystyle I=2{\frac {e}{h}}(\mu _{2}-\mu _{4})=2{\frac {e^{2}}{h}}U_{24}}$.

This results in the conductivity of a spin-started edge channel

${\displaystyle G=2{\frac {e^{2}}{h}}}$.

In general, the Hall resistance can then be calculated for an edge channel:

${\displaystyle R_{Hall}={\frac {U_{24}}{I}}={\frac {h}{2e^{2}}}}$

The known relationship results analogously for i edge channels

${\displaystyle R_{Hall}={\frac {1}{2i}}{\frac {h}{e^{2}}}={\frac {1}{\nu }}{\frac {h}{e^{2}}}}$

where denotes the fill factor and also explains why at lower magnetic fields without spin splitting only plateaus with an even fill factor are observed. ${\displaystyle \nu }$

Individual evidence

1. ↑ M. Büttiker: Absence of backscattering in the quantum hall effect in multiprobe conductors, Phys. Rev. B 38, 9375 (1988)

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