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I am a high school student, trying to better understand quantum point contacts. Would appreciate a simple explanation if possible.

Source: https://arxiv.org/abs/cond-mat/0512609

Since the conductance quantum $e^2/h$ contains only constants of nature, the conductance quantization might be expected to occur in metals as well as in semiconductors.A quantum point contact in a semiconductor is a mesoscopic object, on a scale intermediate between the macroscopic world of classical mechanics and the microscopic world of atoms and molecules.This separation of length scales exists because of the large Fermi wave length in a semiconductor.In a metal, on the contrary, the Fermi wave length is of the same order of magnitude as the atomic separation.A quantum point contact in a metal is therefore necessarily of atomic dimensions.

Anonymous
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  • Can you provide a source for this claim? I am not sure that the Fermi wavelength in a semiconductor is large than in a metal in general – By Symmetry Nov 15 '21 at 14:13
  • Have included the source. – Anonymous Nov 15 '21 at 14:16
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    Note we strongly discourage posting images of text - it cannot be searched by the site engine. Also it is preferred that links to papers are to the abstract page and not the document itself. This reduces the chances of link rot if e.g. a site reorganizes it's structure. Giving the title and authors also helps avoid issues with link rot – StephenG - Help Ukraine Nov 15 '21 at 14:59

3 Answers3

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Fermi wave length is $$\lambda_F=\frac{2\pi}{k_F}, \text{ where } k_F=\frac{1}{\hbar}\sqrt{2m^*\epsilon_F},$$ where the Fermi energy, $\epsilon_F$, is measured in respect to the bottom of the conduction band.

The Fermi energy of a (n-doped) semiconductor is typically rather close to the band edge. In fact, by controlling the doping, it can be made it as close to the band edge as we want. On the other hand, in metal, where the conduction band is partially filled with electrons, the Fermi energy is necessarily quite big.

Remarks:

  • Note the difference between the Fermi level and Fermi energy, see here, which are often confounded.
  • We are talking here about an n-dope semiconductor, which is essentially a metal.
Roger V.
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  • Do you mean that the Fermi_level is close to the band edge? – my2cts Mar 01 '22 at 10:12
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    @my2cts You mean that technically it is within the gap, even in an n-doped semiconductor? Btw, this is not necessarily the case, if we deal with nanostructures, where the gate potential may shift the Fermi level. – Roger V. Mar 01 '22 at 10:14
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The Fermi wavelength of a free electron gas is $$\lambda_F = \left( \frac{8\pi V}{3N} \right)^{\frac{1}{3}} \,,$$ so it is proportional to the separation of the electrons. In a semiconductor this distance is much larger than in a metal, for which the electron density is much higher.

my2cts
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The presence of a band gap and a lower Fermi energy in semiconductors leads to a larger Fermi wavelength compared to metals.

Priyanka Garai
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