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In the classical Drude model, we find that the electrical conductivity of a metal is given by $$\sigma = \frac{ne^2\tau}{m} $$ where $\tau$ is the relaxation time, $m$ is the electron mass and $n$ is the conduction electron density, taken to be equal to the number of valence electrons per atom multiplied by the atomic density.

We can easily show that the same formula holds in the free electron model (by considering the displacement of the Fermi sphere) if we again assume the presence of some scattering mechanism. Again, it makes sense here to take $n$ to be the number of valence electrons per atom multiplied by the atomic density.

Even in the band structure picture we can keep the formula by replacing the mass $m$ by the effective mass $m^*$ (assuming 1D for simplicity) to get $$\sigma = \frac{ne^2\tau}{m^*} $$

However, it is now to so clear to me how we should understand $n$ and its relation to the number of valence electrons per atom. From my understanding of band structure theory, $n$ should be the density of electrons in the unfilled band. For example, Al has 3 valence electrons and crystallizes (in an fcc structure) with 1 atom per unit cell, meaning that one band is completely filled and one is half empty. We would then expect $n$ to be given simply by the atomic density.

However, if the conductivity of the metal is also well described by the free electron model (which I thought was often the case), it seems that this would give a value too small for the conductivity $\sigma$, provided $m^*$ does not differ too much from the actual electron mass $m$.

Any suggestions?

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    $\begingroup$ Aluminium is not a simple metal. The Hall effect can be positive, with about one hole per atom. $\endgroup$ – Pieter Feb 25 '18 at 7:57
  • $\begingroup$ Thanks! Ok, but I think we could still use the formula above if we consider the occupied states in the highest (partially) filled band for Al. Should we then use the value $n=1$, seeing as the band is half-filled? What about my questions for other metals? In general, for atoms with more than one valence electron, the number of electrons per atom in the partially filled band will then be lower than the number of valence electrons (depending on the crystal structure also, of course). $\endgroup$ – Étienne Bézout Feb 25 '18 at 12:26
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    $\begingroup$ One can fit the optical conductivity $\sigma(\omega)$ with the Drude model, estimate a value for $\tau$, assume an electron density, adjust discrepancies with an effective mass. But when that effective mass does not apply in other kinds of experiments (for example with a band dispersion as measured in angle-resolved photoemission or with low-T $c_v$), such parameters are not so meaningful. $\endgroup$ – Pieter Feb 25 '18 at 13:29

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