Timeline for How to show that the electrons responsible for a current have an energy within $k_BT$ of the Fermi energy?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 31, 2019 at 20:16 | comment | added | untreated_paramediensis_karnik | @JonCuster I think I figured it out!!! I posted an answer... man I am so happy, I think I have finally understood what's going on! | |
| May 31, 2019 at 20:15 | answer | added | untreated_paramediensis_karnik | timeline score: 2 | |
| May 30, 2019 at 21:47 | answer | added | user137289 | timeline score: 0 | |
| May 30, 2019 at 19:32 | history | edited | untreated_paramediensis_karnik | CC BY-SA 4.0 |
added 256 characters in body
|
| May 30, 2019 at 14:10 | comment | added | untreated_paramediensis_karnik | @JeffreyJWeimer I am afraid your statement is wrong. The Fermi sphere describes the k-states of the free electrons, there is no single bound (to the nuclei) electron pictured. | |
| May 30, 2019 at 14:02 | comment | added | untreated_paramediensis_karnik | @JonCuster I see, this is confusing (to me). I think it does not matter. For example a thermal perturbation of $k_BT$ does make electrons in k-space gain an energy of about $k_BT$, too. For the electrical current if I apply the same logic, then I apply a small perturbation and electrons gain an energy proportional to $|\vec E|$, no $k_BT$ involved. I understand that in k-space the FS is just an equi-energy surface where, in the free electron model, E goes like $k^2$. | |
| May 30, 2019 at 13:51 | comment | added | Jon Custer | No, the energy 'window' is the regions in $k$ space where the electron density can evolve under the external influence. The fact that the energy gained from the field during such evolution is much smaller than $kT$ only serves to show that treating the applied field as a perturbation is justified. | |
| May 30, 2019 at 13:42 | comment | added | untreated_paramediensis_karnik | @JonCuster I get that part, and thanks for the details I hadn't thought about. But as I wrote in my post, to me the energy window around $E_F$ should be of the order of $e|E|l$, which is in many cases, much smaller than $k_BT$. If you look in Datta's textbook, it is clear that the most energetic electrons have an energy of $E_F+e|E|l$ while the least energetic (that still produce a current) have an energy of $E_F-e|E|l$. There is no $k_BT$ involved. | |
| May 30, 2019 at 13:30 | comment | added | Jon Custer | I guess they way I internalized it was that for an electron/volume of electrons far from the Fermi energy, as it evolves under the field, its surroundings look basically like a full band. That is, all nearby levels are full, and the volume moves in $k$, hits the edge of the zone, and ends up back over on the other side, having done no net conduction. | |
| May 30, 2019 at 13:24 | comment | added | untreated_paramediensis_karnik | @JonCuster Thanks for the reference. I had read that chapter already (several times! And several others from that book, too). I still don't see it... See my edit of my post. | |
| May 30, 2019 at 13:23 | history | edited | untreated_paramediensis_karnik | CC BY-SA 4.0 |
added 835 characters in body
|
| May 30, 2019 at 13:14 | comment | added | Jon Custer | Ashcroft and Mermin discuss this in their Chapter 13, the Semiclassical Theory of Conduction in Metals. Following a volume of electrons as they move through phase space, their ends up being a factor of the derivative of the Fermi function with energy which is non-zero only within a few kT of the Fermi energy. | |
| May 30, 2019 at 13:02 | answer | added | Jeffrey J Weimer | timeline score: -1 | |
| May 30, 2019 at 13:01 | comment | added | untreated_paramediensis_karnik | @JeffreyJWeimer I do not see how it is possible, due to Pauli's exclusion principle. When the energy of interaction is not high enough for the low energetic electrons to go to higher non occupied states, they cannot interact. That's how superconductivity work, in a way. Here an applied field is a very small perturbation which cannot excite almost any of the free electrons. And there aren't 10 billions of them. I said that the E field can change the momentum of about 1 per 10 billion of them, that's quite different... but anyway how does this help answering the question?! | |
| May 30, 2019 at 12:53 | history | edited | untreated_paramediensis_karnik | CC BY-SA 4.0 |
added 7 characters in body
|
| May 30, 2019 at 12:51 | comment | added | untreated_paramediensis_karnik | @JeffreyJWeimer it isn't affecting the shape of the sphere. Again, that's an incredible small perturbation, the shift is insanely small (look at the numbers I wrote in my post). So I am not sure where you're leading me at. On a sketch it looks like as if the whole Fermi sphere had shifted, but physically this is not what happens. I do not see how this helps answering the question. | |
| May 30, 2019 at 12:49 | comment | added | untreated_paramediensis_karnik | @JeffreyJWeimer I would, as I wrote in my post already, look at $E_F$ plus and minus an energy range that's proportional to $|\vec E|$. The math is done in Datta's textbook (available with google as a browsable PDF), page 39. | |
| May 30, 2019 at 12:45 | history | edited | untreated_paramediensis_karnik | CC BY-SA 4.0 |
added 59 characters in body
|
| May 30, 2019 at 12:40 | comment | added | untreated_paramediensis_karnik | @JeffreyJWeimer It looks like the whole Fermi sphere is shifted by a tiny bit when we apply an E field, but it isn't so (as Ziman puts it, it is misleading). Due to Pauli exclusion principle, the E field only affects the electrons going in the E field's direction and change their momentum's direction against the E field. About less than 1 in ten billions free electrons are able to do so. I can give you several references if you want. | |
| May 30, 2019 at 12:29 | history | asked | untreated_paramediensis_karnik | CC BY-SA 4.0 |