# Why the electrons below the Fermi level do not conduct electricity?

Physically, why is it that the electrons need to excited above the Fermi level to conduct electricity? In other words, why is the current zero when the electrons lie below the Fermi level? Does Pauli exclusion principle play any role here?

## Electrons in Solids

Electrons in solids are treated by the so-called Band Theory, a quantum mechanical model for fermions interacting with a fixed crystal lattice. The solution to the Schrodinger's equation shows that atomic orbitals overlaps and the energy levels of the electrons form energy bands of a few electron volts each. If the lattice has $N$ atoms, then each band has $N$ k-states (momentum states) which are distributed symmetrically between positive and negative momenta. Since the number of atoms in solids is typically large, the energy levels in each band is so close that it looks like a continuum. The bands are separated by a finite energy, i.e. regions empty of states called bandgaps (or simply gaps). It is the filling of these bands and the size of the gaps that characterize electrical conductivity (as well as optical properties) of solids.

Let us assume that we are at zero (absolute) temperature. The electrons fill the bands in a way that energy is minimized but since they obey Pauli Exclusion Principle, there can be only two electrons (spin up and spin down) in each k-state. Thus the electrons occupy the bands from lowest to highest energy levels. At $0\, \mathrm K$ all the levels are completely filled from the bottom. The last band with electrons is called valence band and the energy of the highest occupied level is then said to be the Fermi energy.

## Electrical Conductivity from Band Theory

Note that all the electrons below the Fermi level cannot decay, all the states bellow are occupied and the electrons must satisfy the Pauli Principle. By the same reason, the only way they can be excited is by going above the Fermi level, if that is possible. Sometimes people refer to these electrons as being frozen.

When an electric field is applied to the solid, the electrons interact with the field and therefore can change their state. However an electron can only gain energy from the field if there is an available state for it to jumping on. Suppose we start with an arbitrarily weak field, the electrons deep bellow the Fermi level cannot be excited, the energy possibly gained from the field is not enough to bring it to an empty state. On the other hand, the electron near the Fermi level may or may not be excited and that is what distinguish between conductors and insulators.

## Electric Current

Let us say that our solid has one valence electron, i.e. the $N$ atoms have one electron in the last shell each. As we saw, each band has $2N$ states so that the valence band in only half filled. By the symmetric distribution of the k-states, the average momentum of the electrons is zero. This means they are not moving in a preferred direction, there is no current flowing.

If an electrical field is applied along some direction, say the $x$ axis, the electrons can accelerate since there are plenty of empty states to go. There will be a repopulation of electronic states. This means that electrons with negative k-states migrate to positive ones, which are above the Fermi level. Note that the average momentum is now non zero, there is a preferred direction to move which gives raise to a non zero current. This solid is a conductor.

On the other hand, if the valence band is completely filled, there can be no repopulation because there is a gap above the Fermi level. The average momentum is still zero and this solid does not conduct.

As we can see from this two examples above, when we say that only electrons above the Fermi level conduct what we really mean is that in order to have electrical conductivity there must be a repopulation of k-states and this is only possible if the electrons can jump to available states, the states above the Fermi level.

The Pauli exclusion principle is the key thing here. Focusing on momentum space and looking at isotropic space (not possible in a crystal, of course) at first. At cold temperatures, most electrons will try to get into the state of minimal energy. There is only one state with zero momentum. The next electron has to choose a state with non-zero momentum. The third electron will then choose the state on the other side of the momentum state such that their momenta add up to zero. Adding more and more electrons will result in a Fermi sphere, look at the image on this page for a representation. It is just a sphere in momentum space.

There are electrons moving the whole time, but always the same number into the opposite direction as well. So although they move, there is no net current. In a crystal, the Fermi surface will not look that simple, but more like the one shown on that page for copper. The Fermi surface cannot be compressed due to the Pauli exclusion principle.

When you apply a tiny external electric field, the states moving along the electric field will be favored. When all electrons start to drift a bit, the whole Fermi sphere in momentum space will be shifted to the side. Then it becomes asymmetric (in momentum space) and there will be a net current.

Looking at the differences between the Fermi surfaces with and without the external field, you will see that the bulk has not changed (electrons are indistinguishable) but that the surface has moved. One one side they are above the original Fermi surface, on the other the occupied states might no longer reach the Fermi surface.

At $T = 0$, all states below the Fermi surface are occupied, none with higher energy are occupied. With $T > 0$, the distribution is as $\exp(- E/T)$. Higher states can now be occupied while there are free states at lower momentum within the sphere. Therefore the Fermi surface will start to blur out.

I think the question itself is a bit misleading, or incorrectly stated. Strictly speaking, if there is current flowing in something, it does not have a Fermi level at all since the Fermi-Dirac distribution is a creature of equilibrium statistical mechanics. Current flow implies there is not equilibrium. So what makes the current flow?

If you have something described by the Schroedinger equation, it could be a molecule or a block of semiconductor and it could either have a continuum of states or a discrete number. Such an object will obey Fermi-Diract statistics and will have a Fermi level. I will call this object the 'device' from now on for simplicity. At ${T>0}$ there will be some electrons in states above the Fermi level and some states below will be empty. This is what the Fermi-Dirac distribution tells us: it tells us how states are filled by Fermions in equilibrium. Notice even though we have some electrons above the Fermi level, current does not flow.

Now if we attach some contacts to this system and wait a while we will see that there is still no current. Finally, if we hook up the contacts to a battery (or other voltage source) we will see a current flow through the device we solved the Schroedinger equation for. Why? Because the battery has changed the chemical potentials (a.ka. Fermi levels) in the contacts. If the 'window' of energy defined by $\mu_1-\mu_2$ corresponds to energy states in the device, current will flow. Conceptually, the contact at higher chemical potential $\mu_1$ fills the states in the device with energy $E<\mu_1$ and the contact at lower chemical potential $\mu_2$ empties the states with energy $E>\mu_2$ If there are no states in the window $\mu_1-\mu_2$, no current will flow. This is why you will not get much current from a block of quartz no matter how much voltage you apply. Depending on how we move the energy window we can have more or less states corresponding to it. This explains how we can get results like Negative Differential Resistance, where we increase the voltage but the current goes down and devices like the Resonant Tunneling Diode.

Notes and caveats

At this point you might be screaming, "did you not say that you could not have a Fermi Dirac distribution and a current at the same time?" Strictly speaking that is true. However, it is typical to assume that the contacts represent such a large reservoir of electrons that the current flowing through them does not perturb them far from equilibrium and the Fermi-Dirac distribution still approximately holds. This works well for sufficiently large contacts and sufficiently low voltages. The overall picture still works without this assumption if you have some other method of defining $\mu_1$ and $\mu_2$, but that is tricky to do. If you are really concerned about the occupation of states in your contacts, it is probably better to treat them as other small devices coupled to your central device and assume the Fermi-Dirac distribution holds elsewhere. At some point, your device has to be coupled to the outside world in order to measure anything about it.

Canny observers will note there are others way to change the chemical potentials in the contacts besides just applying a voltage like temperature gradients. This is also true, but a bit beyond the scope of this question and the overall picture still holds up: electrons flow if there are states in the energy window.

The Pauli Exclusion Principle comes in here via the derivation of the Fermi-Dirac distribution function.

Where to learn more I do not have time at the moment to make figures illustrating the above so in lieu of that for the moment I will direct the reader to Professor Supriyo Datta's webpage who has a lot of excellent videos and instruction in this area.

"Physically, why is it that the electrons need to excited above the Fermi level to conduct electricity?"

They don't. This is a popular misconception; the premise of your question is mostly wrong, as are all 6 answers supplied to date (June 22, 2018)!

A simple proof that the other answers are wrong is to consider that the number of electrons above the Fermi surface is zero at absolute zero. Therefore, if they are correct, and I am incorrect, the resistance at absolute zero would be maximum. For most substances, the resistance is actually minimal at absolute zero, as can be seen experimentally in this graph:

In a normal and pure metal, the electrons below the Fermi level are nearly superconducting, and move without any resistance to speak of.

Conceptually, one can view the superconductive fraction of the electron cloud like hockey pucks in a game of air hockey, with the disturbances from ions being small compared to the sizes of the pucks, allowing a smeared or average force of zero to it:

However, electrons above the Fermi the level have a relatively high energy and an effective size small enough to be able to "see" (and therefore interact with) the periodic ionic field due to the screened, tightly-spaced ions.

In my air hockey example, these high-energy contributions toward the electron cloud are small enough to get pushed backwards a little before going over an air spout or sliding without an air layer. This interaction produces momentum exchange, which, because the electrons have to be thought of as a cloud distribution, not to mention that splitting of the cloud would cause a strong field, gives the entire electron cloud a net drag force measurable as resistance, thus slowing it down to the drift velocity and preventing it from net acceleration.

In a non-pure conductor, impurities cause warps in the lattice, which the electron cloud will view as drag.

In my air hockey example, this is like dirt in the surface which pucks rub against.

In a nonconductor, all of the electrons are trapped on the atoms, and there is no conduction except for special cases like high fields (e.g., lightning) or external radiation (e.g., the photoelectric effect).

In my air hockey example, the number of free hockey pucks is zero, as they are all stuck between the air holes.

In a semiconductor, the number of free electrons in the conduction band increases with temperature, as electrons are forced out of the $0$K Fermi sphere in momentum space. In this case, raising the temperature can increase the conductivity if the conduction band was relatively empty.

In my air hockey analogy, the number of free pucks is small but increases as one increases the temperature, dislodging them from their ions. This happens for atoms with intermediate valances, like carbon, germanium, silicon, etc.

For this case, the premise of your question would seem to make sense, except that your question uses the word "need" without any restriction instead of something like "..need in semiconductors..." Therefore, it is wrong, along with all the other 6 answers to date for failing to notice this as well as for saying incorrect things, like that conducting electrons always need to jump to available states above the Fermi level (the current #1-voted answer), the momentum of electrons below the Fermi level fully cancel each other out (the current #2-voted answer), etc.

Incidentally, once a conduction band in a semiconductor becomes populated, further increases in temperature should decrease conductivity and increase resistance as phonons can interact more strongly with the free electron cloud.

In my air hockey example, the free, conducting pucks become small enough to have to slide without an air layer. Therefore, even in semiconductors more electrons being above the Fermi level does not always correlate to lower resistances or more current flow.

Turning now to your question about the relevance of the Pauli Exclusion Principle, it comes into play because electrons forming the cloud are trapped like the proverbial particle in a box of size determined by $Z$ and the ionic spacing. If these electrons are unpaired, they act like Fermions. This controls the number of electrons that can fill the lowest levels, which in turn controls the number of electrons that can fill the upper levels, like the conduction bands in semiconductors. Without this, there wouldn't be electrons up so high in these conduction-band energy levels, and nature would be much different.

• I was with you up to the "nearly superconducting" part. The best way to visualize conductivity in metals is the model of the Fermi sphere, slightly displaced from the origin: homepage.lnu.se/staff/pkumsi/1FY805/Fermi_sphere.png – Pieter Jun 23 '18 at 14:09
• You do realize that that sphere represents a $T=0$K, zero resistance, metal, don't you? It is consistent with what I wrote. – Jason Arthur Taylor Jun 23 '18 at 14:27
• No, resistance is not zero at $T=0$. – Pieter Jun 23 '18 at 14:49
• Look more closely. Resistance is not zero at $T=0$ unless there was a a superconducting transition at $T>0$. Copper does not do this. The mean free path may become long in very pure well-annealed copper, but the extrapolated resistance at $T=0$ remains finite. – Pieter Jun 23 '18 at 15:07
• You can weasel what you want but you do not understand what superconductivity is then. A superconducting ring can support a persistent current. Experimentally one can only establish a lower bound on the resistivity of a superconductor. But the resistivity of copper is always measurable and finite. – Pieter Jun 23 '18 at 15:49

Electron bands are symmetric about $k = 0$, so for every electron in a filled band, there exists another electron with opposite momentum which cancels out its current, resulting in zero net current flow. An infinitesimal applied electric field just tilts the bands by an infinitesimal amount, so if the whole band lies below the Fermi level, then it will remain filled if infinitesimal electric field is applied, and this current cancellation will remain robust.

• If it is below the Fermi levels of both the external contacts, it will remain filled regardless of the bands' symmetry. There are many materials that are not symmetric about k=0 and any applied strain will also break the symmetry. – Cogitator Feb 21 '17 at 20:08

I think of the TV commercial where the escalator stops and all the people are trapped. The steps are energy levels, and there are two people on each step (spin up and spin down). No third person can occupy a step (Pauli exclusion). Only the ones at the top (Fermi surface) are free to move (gain energy). When they do, the steps are free and more can move up.

Here is my way of thinking of it intuitively:

Each electron in an atom resides at a specific energy level (or state). Pauli's exclusion principle prevents any of them to occupy the same exact energy state.

When putting two equal atoms close together, they have identical electrons at identical energy states. Again Pauli's exclusion principle prevents two electrons to occupy the same energy state. So they must "fit" their energy states to be slightly lower or slightly higher, fitting in underneath and above one another to make it work.

When infinitely many equal atoms are packed together to form a solid material, infinitely many energy states must be crammed together and fitted underneath and above each other. This stack of many, many energy states is what we call a band, so closely packed that the band is as good as continuous.

All energy states are occupied, meaning that all spots in the band are taken. No electron can move anywhere. A current is impossible.

If an election would manage to reach the next band of tightly packed but this time empty energy states, then it would be alone. Any negligible amount of energy would make it move from spot to spot within this empty band. A current is now possible.

This is the idea. The Pauli exclusion principle is key. That is the reason why no electrons can move around in filled bands.

• "No electron can move anywhere" - no, they are all moving. The PEP forbids joint occupation of the same quantum state, not energy state. – Rob Jeffries Feb 20 '17 at 0:24