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By applying a voltage and a magnetic field on a (let's say metallic to keep things as simple as possible) sample, one is able to create the Hall effect and to obtain the Hall coefficient $R_H \sim 1/n$ where $n$ is the charge carrier density. But what is $n$, really? Is it the $n$ that appears in the conductivity formula $\vec J = en\vec v$? If so, I face a huge problem.

Indeed, one can find in numerous sources that either Drude's model or a quantum mechanics treatment lead to the same formula for $R_H$. This implies that whatever model is used to explain a metal, from experiment one finds that $n$ is a sort of universal value that does not depend on the model of the solid. But it is well known (e.g. Ziman "Physics of solids") that a QM treatment of a solid where electrons satisfy the Pauli exclusion principle and that they obey Fermi-Dirac statistics, very few of the free electrons actually participate in electrical conduction. When one looks at the Fermi sphere with and without an applied $\vec E$ field, the application of the $\vec E$ field has the same end result as a displacement of the Fermi sphere in an opposite direction than the $\vec E$ field. Thus only electrons at the Fermi surface that were moving in the $\vec E$ field direction and had a momentum near $-\vec p_F$ get their momentum changed to near $\vec p_F$ (actually slightly higher than that, and with a constant that depends linearly on the $\vec E$ field). Those are very few electrons compared to the total number of free electrons (those that constitute the Fermi sphere), and they move extremely fast compared to the drift velocity that arises from the Drude's model.

Indeed, I would expect that from a QM treatment, since so few electrons actually participate in electrical conduction and that they move about 2 orders of magnitude slower than the speed of light in vacuum, if we want to describe a particular $\vec J$, then $\vec J \approx en'\vec v_F$ where $n'$ would be a tiny fraction of $n$ that appears in Drude's model. But it turns out that $n'=n$, so I do not see any way to explain consistently a current density from Drude's mode and a QM treatment.

So I do not understand exactly what is "n", the charge carrier density. How can it be the same regardless of the model used to describe a solid, while the models give very different values for the number of electrons participating in electrical conduction and very different values in "drift velocity"?

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  • $\begingroup$ Do you want to propose a new theory of conductivity and the Hall effect? $\endgroup$ – Alex Trounev Dec 24 '18 at 22:23
  • $\begingroup$ @AlexTrounev Not in my plans. I just want to understand what's going on. $\endgroup$ – thermomagnetic condensed boson Dec 24 '18 at 22:30
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Actually, Drude model is consistent with free electron model, because when you apply electric field, Fermi sphere is shifted by $k_{d}= \frac{eE\tau}{\hbar}$enter image description here where $\tau$- scaterring time. Every electron from shadow areay have counterpropagating partner, so they don't contribute to the current. The number of electrons from white area inside circle is $\delta n=\frac{v_{drift}}{v_{F}}n$, so the current is: $$J=e\delta n v_{F}=env_{drift}$$ (Here we assume that this white area is very small, so every electron moves with Fermi velocity). If you want to be more precise you need to use band theory for electrons in solid state.

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  • $\begingroup$ Thanks! So $n$ would be related to the total number of electrons, not the ones that participate in electrical conduction. Is the formula for $\delta n$ you provide an approximation? Because the area of a crescent is slightly more complicated than that. I have read in several papers though, that from the Hall experiment they can get information on pockets on the Fermi surface and other strange things (such as condensation of the electrons into another phase that does not contribute to current). It seems like they measure $\delta n$ somehow, not $n$. Could you elaborate on that? $\endgroup$ – thermomagnetic condensed boson Dec 27 '18 at 19:44

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