I wanted to know how the electron density in metals behaves with temperature. I couldn't really find an answer online so I searched through my x-years old scripts and, surprisingly, I found a sentence in a lecture:

Metals: Electron density will be influenced only slightly with temperature because the fermi distribution will be shifted only with kT at the fermi level, increase of electron density (negative contribution of TCR).

TCR is the temperature coefficient of a resistance

$ \alpha = \frac{1}{R} \frac{dR}{dT} $ and $ R $ is $ R = \frac{1}{enµ} \frac{L}{A} $

But I think the cited explanation should be sufficient, yet, I don't get it. How does the Fermi distribution play a role here? I know, the higher the temperature, the more likely electrons will occupy higher energy states (right?). Shouldn't the electron density rise then? Or maybe rather "Why is it influenced only slightly by temperature?".

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    $\begingroup$ Because the conduction band already has a huge number of electrons in it. Promoting electrons from a valence band just isn’t going to do much. $\endgroup$
    – Jon Custer
    May 24 '20 at 15:32

Let me try to suggest an idea (maybe not very precise, but hopefully still useful).

First of all, the total electron density is fixed (number of electrons per unit volume), what changes with temperature is the energy profile of electronic density, which is described by the Fermi function $$ f(\varepsilon) = \frac{1}{e^{\beta(\varepsilon-\mu)}+1}. $$ where $\beta=1/k_BT$ is the inverse temperature. Now the chemical potential equals the Fermi energy up to quadratic corrections in $T/T_F$: $\mu = E_F + \mathcal{O}(T/T_F)^2$. Since typical Fermi temperatures are of order $10^4$K, while lab. temperatures are of order $10^2$K, the above assumption is fine.

Now the important part: in this temperature range the electric resistance is due to electronic scattering from phonons, so when you increase the number of electrons that may be involved in a scattering process, you increase electric resistence. However, electrons deep in the Fermi sea (with energies $\varepsilon \ll E_F$) can't scatter because they don't have available final states, due to Pauli exclusion principle. So roughly we can say that only electrons close to the Fermi level can be involved in a scattering process, in particular only electrons with energies in the range $E_F - k_BT < \varepsilon < E_F + k_BT$. In other words, thermal fluctuation can induce scattering processes of those electrons that in the process gain/lose energy of order $k_BT$ and still find an available final state, so around the Fermi level. This is mathematically translated in the following approximation: $-\frac{\partial f}{\partial \varepsilon} \approx \delta(\varepsilon - E_F)$.

Now if you look at Aschcroft-Mermin book (eq. 13.25) you get the following expression for the thermal conductivity $\sigma$ of a given electron band: $$ \sigma = \frac{e^2}{3} \int d\varepsilon \rho(\varepsilon) \tau(\varepsilon) v^2(\varepsilon) \left( - \frac{\partial f}{\partial \varepsilon} \right), $$ where $\tau(\varepsilon)$ is the average time between two scattering processes, and $v(\varepsilon)$ is the electron velocity when the electron's energy is $\varepsilon$. Notice that I have simplified a little bit, because the book is very general. Now the suppression of scattering of electrons in the Fermi sea means that we can use the Delta function and obtain $$ \sigma = \frac{e^2v_F^2}{3} \rho(E_F) \tau(E_F). $$

Now the only object which is temperature dependent is $\tau(E_F)$ (let me call it $\tau$ in the following). $\tau$ depends only on the scattering process, which in our case is a scattering with acoustic phonons of the lattice, and it can be estimated as $\tau \sim T^{-1}$, take a look at sec. 2.3.1 here: https://link.springer.com/chapter/10.1007/978-3-319-48933-9_2#Equ31 Finally the resistvity due to phonon scattering only is $1/\sigma \propto T$, as is well known.

Finally, what if you want to be a little more precise? you have to evaluate the full integral, but you should get a correction of order $(T/T_F)^2$ to your previous result. In conclusion: the fact that only the electrons at the Fermi level scatter with phonons is very important to solve your integral, but in some sense the thermal occupation of the conduction band (solution of the full integral) gives you only a correction to your main result (with the Delta function), which is most affected by the temeprature dependence of $\tau$.

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    $\begingroup$ Thanks a lot! You probably explained it but I have to admit I don't understand. I think I struggle with "Electrons deep in the Fermi sea can't scatter (...) only electrons close to the Fermi level can be involved in scattering processes". Why is that? In metals, the valence and conduction band overlaps, so all electrons at Fermi energy participate in conducting/resistance - but also many more, especially, all above the Fermi energy, or? And when I increase the temperature more and more electrons should possess $ E > E_F $? $\endgroup$
    – Ben
    May 25 '20 at 4:21
  • $\begingroup$ Well, I was reading a book about this and realized the story is quite complicated. However, consider an electron with very small energy $E_i\ll\mu$, then in a scattering process it can gain/loose an energy $k_BT$ right? This is a rough estimate, of course, but it's reasonable because phonons, impurities etc. have energy proportional to temperature. Well, now the electron in the final state should have an energy $E_f=E_i+k_BT$, again very rougly, right? But if $k_BT \ll \mu$, then $E_f\ll\mu$, which means that the final state is occupied by another electron, and the Pauli exclusion forbids it $\endgroup$
    – Matteo
    May 27 '20 at 16:12
  • $\begingroup$ Moreover, one thing that might be misleading is the following: in metals, the absence of a gap implies that resistance is in general small (compared to gapped insulators), but it is FALSE that the more electrons are excited, the more the material is conducting. Actually in conductors it is the opposite! The more electrons are excited, the more scattering with phonons are likely to happen (as argued above), so despite resistance keeps being small, it increases with temperature! This mechanism is completely different in insulators though! $\endgroup$
    – Matteo
    May 27 '20 at 16:37
  • $\begingroup$ I have edited my answer! The thing is not straightforward at all, as you can see, but now I think my answer is more convincing and well supported than before $\endgroup$
    – Matteo
    May 27 '20 at 18:58

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