It is commonly written in textbooks that in metals the electrons responsible for an electric current are the ones that have an energy about $E_F$ and a few $k_BT$ around that energy. See for example Datta's textbook "Electronic transport in mesoscopic systems" page 37 (book available as PDF from a Google search):

It is easy to see why the current flows entirely within a few $k_BT$ of the quasi-Fermi energy.

But then, no proof nor anything involving $k_BT$ is shown anymore. What's more, it is then showed that the number of electrons involved in electrical conduction is proportional to the applied electric field's magnitude, which makes entirely sense to me. More precisely, he shows that the difference in energy between the most energetic electrons that create a current and the least energetic ones that also carry a current is worth $2eEL_m$ where $L_m$ is the mean free path, which is worth about $10$ nm. In other words, the width of energy around $E_F$ that electrons that create the current have, has nothing to do with $k_BT$.

I can understand that regarding the calculations for the specific heat, it is indeed true that only electrons having an energy about $k_BT$ around the Fermi energy (of the order of $1$ eV for metals) can absorb thermal energy, which is itself of the order of $k_BT$ (so about $10^{-5}eV$ to $10^{-3}eV$). It is easy to realize when one uses the fact that electrons are fermions and that at room temperature a metal is similar to a cold Fermi gas. Thus the electrons are roughly forming a sphere (let's take alkali metals to make things simple) in k-space and all states below the surface are occupied. The surface of the sphere is blurred due to finite temperature, in an energy amount around $k_BT$. So that the electrons that are below the surface by more than $k_BT$ cannot absorb thermal energy because the states above them are all occupied. It is only in that $k_BT$ window-range that electrons can absorb thermal energy.

But when I apply the same logic to an electric current, i.e. we apply an electric field to the metal instead of a temperature, I do not get anything related to $k_BT$ anymore. By considering that we apply $1$ V on a $1$ cm sample, the electric's field magnitude is about $100$ V/m which translate as an energy of about $10^{-6}$ eV. In other words, the electric field is a very tiny perturbation to the system, it is about 40 times smaller than rising the temperature of a metal by 1 K. I would expect then that only electrons having an energy around the Fermi energy $E_F$ with a margin equal to that extremely small $10^{-6}$ eV amount would be able to react to the field and produce a current. This has absolutely nothing to do with $k_BT$ and is in fact proportional to $|\vec E|$, as it intuitively (to me at least) should. I.e. I get something linear in the strength of the perturbation, just like with the case of the thermal energy with its thermal perturbation.

So, I do not see, for the life of me, how to reach to the conclusion that only electrons that have an energy within $k_BT$ of $E_F$ are able to produce a current.

I am well aware of the Fermi-Dirac distribution and how its derivative with respect to energy is non zero only around $E_F$, also of the density of states and how temperature affects it, etc. But I fail to see how it is relevant to answer my question.

Edit regarding Jon Custer's comment:

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.

I had checked that chapter, and seen that the conductivity can be written as an integral with a term that contains $\partial f/\partial \varepsilon$ which involves $k_BT$ (as I wrote in the previous paragraph), but I fail to see how this implies that the electrons responsible for the current at the ones that are within $k_BT$ of $E_F$. But that is indeed probably the way to go. But still, I would have to see where I go wrong in my reasoning that I exposed above.

  • $\begingroup$ @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. $\endgroup$ May 30, 2019 at 12:40
  • $\begingroup$ @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. $\endgroup$ May 30, 2019 at 12:49
  • $\begingroup$ @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. $\endgroup$ May 30, 2019 at 12:51
  • $\begingroup$ @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?! $\endgroup$ May 30, 2019 at 13:01
  • $\begingroup$ 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. $\endgroup$
    – Jon Custer
    May 30, 2019 at 13:14

3 Answers 3


I finally figured out. The statement that only electrons within a few $k_BT$ around $E_F$ contributes to a current when an electric field is applied to a metal is not universally true. This roughly holds when $k_BT >> e|\vec E|L_m$ where $L_m$ is the mean free path. For a reasonable current, the statement holds for almost all temperatures, i.e. above $1$ K.

The reason can be understood by considering 2 cases.

First case: T= absolute zero. At that temperature, the Fermi surface is perfectly sharp and if the statement would be true, then only the electrons exactly at the Fermi surface would contribute to a current, but this is wrong as can be seen from the countless displaced Fermi sphere pictures found in textbooks (and showed here in the answer by Pieter). Even at $0$ K, as Datta mathematically shows, the electrons that have an energy above $E_F - e|\vec E|L_m$ all do contribute to the current. In that case the energy window around $E_F$ is indeed of width $2e|\vec E|L_m$. In Pieter's figure of the Fermi sphere, only the crescent between the displaced and non displaced spheres contributes to the current. The maximum energy of these electrons is proportional to the applied $\vec E$ field strength ($v_d$ is proportional to it).

Second case: Finite temperature. In that case before applying the electric field, the Fermi surface isn't sharp, it is blurred. This means that there unoccupied states below $E_F$ and occupied states above $E_F$, all within a few $k_BT$ (because of the Pauli exclusion principle, as you've already pointed out). Nevertheless, it is very important to realize that there are unoccupied states within a few $k_BT$ around $E_F$. So that when another perturbation, such as an electric field, is applied, then all these electrons around $E_F$ by a few $k_BT$ can interact with the $\vec E$ field and get their energy increased (because they have unoccupied states above them). Here it is assumed that the electric field is a smaller perturbation than $k_BT$. For if the electric's field magnitude was gigantic, then even electrons with a much lower energy than $E_F-k_BT$ would be able to interact with the field and contribute to the current. You can picture this in the usual Fermi sphere figure as a huge displacement compared to the radius of the sphere, rather than a very tiny displacement (for ordinary current the real "displacement" is so tiny that it wouldn't be distinguishable to the naked eye on these figures).


In the free-electron model, I prefer to say that all valence electrons contribute to the current. This gives the correct value of the drift velocity as measured by the Hall effect. It also makes it clear that the drift velocity does not depend on temperature.

The image to visualize this is the displacement of the Fermi sphere by the drift velocity in velocity-space: enter image description here

The statement that only states within $kT$ of the Fermi level contribute to the current may lead students to conclude that resistivity of metals should go up at low temperature. I am not arguing that the statement is wrong, but I cannot find it very helpful to explain the phenomena. It needs so much more explanation than to say that the number of conduction electrons is constant.

  • $\begingroup$ The statements that only electrons above E_F contribute and that more electrons are free as T increases are not wrong. However, electron mobility goes down faster than number density goes up. Both factors are needed. I'd wonder otherwise why not just give the right explanation rather than hand-washing the integrity of the system. $\endgroup$ May 30, 2019 at 22:40
  • $\begingroup$ These statements are wrong @JeffreJWeimer. Please have a look at figure 1.7.2 of Datta's reference. $\endgroup$ May 31, 2019 at 11:23
  • $\begingroup$ Thanks for the picture Pieter, but this does not answer the question. In your picture when you calculate F+-F- where F+ is the energy associated to electrons with wavevector kf+kd (most energetic electrons creating the current) and F- is the energy associated to the least energetic electrons that create current (wavevector is kf-kd), you get an energy window around E_F of magnitude proportional to the electric field strenghth and kBT doesn't show up. Please see the Datta's reference... $\endgroup$ May 31, 2019 at 11:29
  • $\begingroup$ @thermomagneticcondensedboson I was aware that the figure is not directly pertinent to the question. This would be a figure for $T=0$. At higher temperature there was a simulation by Silsbee. I had made a movie of that (text in Swedish), and at the frames linked it shows conduction at finit temperature: youtu.be/Y9Q2vgch490?t=158 $\endgroup$
    – user137289
    May 31, 2019 at 15:10
  • 1
    $\begingroup$ Thanks for the video Pieter, I'll take a look. Meanwhile I have finally figured out I think! I posted an answer... I am so happy!!! And, your picture was actually very helpful! I considered it in my answer. $\endgroup$ May 31, 2019 at 20:17

The Fermi sphere is the $E(k)$ boundary between the occupied (bonding) and unoccupied (non-bonding) states in a metal at zero Kelvin 1. In a metal, conduction is due primarily if not exclusively to the motion of free electrons. Free electrons are those that are not in occupied (bonding) states. At 0 K with no electric field, all electrons are in occupied states. Therefore, the metal carries no electrical current.

Let's perturb the metal in one of two ways.

  • Put an electric field on the metal. This can distort the Fermi surface. Such a distortion is NOT the cause for the conduction of current. The distortion is analogous to how the shape of the Fermi surface is different along different crystallographic orientations. All that is being changed is the position of the Fermi energy. Nothing is said about the motion of the free electrons.

  • Put an electric field on the metal. This promotes electrons from occupied to non-bonding (initially unoccupied) band states. This action is independent of the above change in the shape of the surface. This promotion is NOT the root cause for the conduction of current. It is however a step toward that result.

  • Put an electric field on the material. This applies a force to the free electrons (those in non-bonding states). The free electrons move (accelerate). This is electrical current.

  • Put the material at a temperature above 0 K. This promotes electrons from occupied to non-bonding (initially unoccupied) band states. Those free electrons are just as free to move as are the electrons that were promoted by the electric field.

In conclusion, the initial shape of the Fermi surface has nothing by itself to say about electrical conduction. The perturbation that occurs in the shape by the electrical field has nothing to say by itself about electrical conduction. Finally, the promotion of electrons above the Fermi energy, whether by an applied field or by thermal means, is only a first (and required) step to determine electrical conduction.

The most important concern we have to determine conductivity is not any of these steps by themselves. It is the combination of how many electrons are free to carry current (due to promotion by the field and temperature) and how fast they are moving. In short, to determine the electrical conductivity of a metal, we must determine the number density of free electrons and the velocity of the free electrons under the applied electric field. In a metal, the number density of states depends on $\sqrt{E}$. At 0 K, we fill this with the appropriate number density of bonding electrons following Pauli exclusion principles. Then, we promote electrons using Fermi-Dirac statistics because electrons are fermions. This happens regardless of whether a field is applied or not. Using a convolution integral, we obtain a picture of electron density as a function of energy and temperature $\rho_E(E,T)$ as shown here. Those electrons above the Fermi energy are free to carry current.

The thermal energy that is applied to promote electrons from occupied states is on the order of $k_BT$ (this is the dashed line in the picture). When the energy of an electric field that is applied in a metal is below $k_BT$, more electrons are free due to thermal promotion than are free due to promotion by the electric field.

Alternative insights can also be obtained using a basic muffin-tin potential system. The electrons that are bound are localized in the muffin potentials. The electrons that are free are delocalized throughout the entire set of potentials above the Fermi energy. Electrons are free primarily because they are thermally promoted above the Fermi energy.

In summary, the Fermi surface is the boundary between occupied and free electrons. The number density of free electrons is obtained by an integral over the convolution of the density of states and the Fermi-Dirac function. The electric field perturbs the free electrons (those above the Fermi energy, whether by thermal or electrical means) by causing them to move throughout the lattice. This perturbation happens independently of whether the free electron has only an infinitesimal energy above the Fermi energy or is at the limit of $k_B T$ or above the Fermi energy.

  • $\begingroup$ "ALL free electrons can move, regardless of their energy above the Fermi energy" Do you mean interact instead of move? Or "change their energy"? Because of course, all but 2 electrons are constantly moving if we consider a Fermi gas. (yes, there are 2 electrons with exactly zero energy, though they are still spread across the whole sample). $\endgroup$ May 30, 2019 at 13:05
  • $\begingroup$ Essentially your answer is that the premise of my question is wrong. Then I would like to know where Datta (and many, many others) went wrong. $\endgroup$ May 30, 2019 at 13:07
  • $\begingroup$ Also I'm not sure why you complicate things by mentioning the bound electrons. My question would still stand if I replaced "metal" by "free electron gas". It would still be entirely valid, and your point regarding the lattice potential(s), useless... I just saw your edit that now says that free (unbound) electrons do not satisfy the Pauli exclusion principle. This is wrong. A free electron gas for example satisfies the PEP, and alkali metals are close to that ideal model. The density of electrons in metals is so high compared to an ordinary gas at atmosphere pressure that the PEP has to be $\endgroup$ May 30, 2019 at 13:14
  • $\begingroup$ taken into account. This is only from lowly doped semiconductors that the PEP starts to become less relevant. But in metals, it is of utmost importance to understand any basic property, like the electrical and thermal conductivities. $\endgroup$ May 30, 2019 at 13:18
  • $\begingroup$ I feel did not post any insult to your content. Feel free to report my posts if you think so, and moderators will take over. The free electron model (en.wikipedia.org/wiki/Free_electron_model) is an idealized model of a metal that takes into account PEP. It works well for most alkali metals where the effects of the lattice potential is not very strong. It has its limitations, but it is used to understand how to compute the conductivities and other basic properties of metals. It is much better than Drude's original model (which is still taught). $\endgroup$ May 30, 2019 at 13:35

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