Shubnikov–de Haas effect

An

Physical process

At sufficiently low temperatures and high magnetic fields, the free electrons in the conduction band of a

Fig 1 shows the

The Landauer–Büttiker approach is used to describe transport of electrons in this particular sample. The Landauer–Büttiker approach allows calculation of net currents I_m flowing between a number of contacts 1 ≤ m ≤ n. In its simplified form, the net current I_m of contact m with chemical potential µ_m reads

${\displaystyle I_{m}=2{\frac {e\cdot i}{h}}\left(\mu _{m}-\sum _{l\neq m}T_{ml}\mu _{l}\right),\,}$

(1)

where e denotes the

Fig 2 shows a sample with four contacts. To drive a current through the sample, a voltage is applied between the contacts 1 and 4. A voltage is measured between the contacts 2 and 3. Suppose electrons leave the 1st contact, then are transmitted from contact 1 to contact 2, then from contact 2 to contact 3, then from contact 3 to contact 4, and finally from contact 4 back to contact 1. A negative charge (i.e. an electron) transmitted from contact 1 to contact 2 will result in a current from contact 2 to contact 1. An electron transmitted from contact 2 to contact 3 will result in a current from contact 3 to contact 2 etc. Suppose also that no electrons are transmitted along any further paths. The probabilities of transmission of ideal contacts then read

${\displaystyle T_{21}=T_{32}=T_{43}=T_{14}=1,}$

and

${\displaystyle T_{ml}=0}$

otherwise. With these probabilities, the currents I₁ ... I₄ through the four contacts, and with their chemical potentials µ₁ ... µ₄, equation (1) can be re-written

${\displaystyle \left({\begin{matrix}I_{1}\\I_{2}\\I_{3}\\I_{4}\end{matrix}}\right)={\frac {2e\cdot i}{h}}\left({\begin{matrix}1&0&0&-1\\-1&1&0&0\\0&-1&1&0\\0&0&-1&1\end{matrix}}\right)\left({\begin{matrix}\mu _{1}\\\mu _{2}\\\mu _{3}\\\mu _{4}\end{matrix}}\right).}$

A voltage is measured between contacts 2 and 3. The voltage measurement should ideally not involve a flow of current through the meter, so I₂ = I₃ = 0. It follows that

${\displaystyle I_{3}=0={\frac {2e\cdot i}{h}}\left(-\mu _{2}+\mu _{3}\right),}$

${\displaystyle \mu _{2}=\mu _{3}.}$

In other words, the chemical potentials µ₂ and µ₃ and their respective voltages µ₂/e and µ₃/e are the same. As a consequence of no drop of voltage between the contacts 2 and 3, the current I₁ experiences zero resistivity R_SdH in between contacts 2 and 3

${\displaystyle R_{\mathrm {SdH} }={\frac {\mu _{2}-\mu _{3}}{e\cdot I_{1}}}=0.}$

The result of zero resistivity between the contacts 2 and 3 is a consequence of the electrons being mobile only in the edge channels of the sample. The situation would be different if a

For a given sample, all factors including the electron density n on the right hand side of relationship (