Often when conductivity is explained through band theory, the term "free" tends to crop up. As an example, I often come across descriptions of the valence band as a highest filled set of states occupied by electrons bound to their specific atoms; raising them to the conduction band supposedly "frees" them so that they can move freely in the metal, thereby enabling them to contribute to a current when an electric field is applied. In fact, in my introductory solid-state physics book, the additional electron contributed by a donor in a doped semiconductor is referred to as "loosely bound" to the donor ion, requiring a push into the conduction band to break free and become a charge carrier.

At the same time, I was also given to understand that the electrons of a valence band do not contribute to a current when there is an electric field, because their respective velocities balance each other out perfectly; there is no net velocity and hence no net movement. Raising an electron into the conduction band essentially means creating a hole in the valence band so that the electrons can now redistribute (in k-space) and thereby achieve a non-zero net velocity.

But according to this latter statement, the electrons in the valence band should contribute to a current across the metal when an electric field is applied.

A) How then can the valence band electrons be bound to a specific atom, as the former statement claims, if they are simultaneously capable of acting as charge carriers? Also, how then can the donor electron - which occupies an energy state above the valence band - be "loosely bound" to the donor atom, when the electrons below aren't?

B) Assume that we raise the temperature enough (without the metal somehow disintegrating) so that some electrons from even the lowest band leave for higher energy bands. Will the holes that are left behind in this lowest band also mean that the remaining electrons in this band can carry charge, similar to how the electrons in the valence band with holes were able to carry charge?

I'll be grateful for anything that can help me clear up this mess!


1 Answer 1


A free electron is, roughly speaking an electron whose wavefunction is delocalised in the lattice. Which means that there is a non negligible probability of finding the electron at any lattice point (atom) in the crystal (metal). In contrast, the wavefunction of an electron in the core are more or less localised to the atom, meaning that the probability of finding them at a neighbouring atom is very negligible. The delocalisation of an electron occurs due to the fact that there is a continuum of possible energy levels for the electron to exist in and if the energy level has higher energy than the binding energy of the electron to the atom it is “free”. This is the case in metals. The temperature provides energy for the electrons to transition through the energy levels and once it gains energy greater than the binding energy, its “free”. It is to be noted that as long as the electron is bound to the atom, there is no conduction. This is because it’s localised to the atom. Thus external electrons can’t enter the system either. But once the electron is free, the external electrons can enter the system.

Addendum: In the k-space, initially the states are in a ground state. Meaning the lowest energy state they’d be respecting the exclusion principle. Think of it as you have N electrons and many possible states. Now you must fill them such that the total energy is the lowest with the constrain that no single state has more than 2 electrons (one for up spin and one for down). These allowed states are discrete. Meaning they have finite space between them in the k-space. But at higher energies, the distance between the subsequent states decreases and can be approximated by a continuum. Also, in a crystal, the continuum limit is reached at comparatively lower energies (overlap of neighbouring electronic orbitals). As long as the electron isn’t promoted to one of these states, it doesn’t conduct. Also, an atom accepting a hole is equivalent to an electron from the same atom jumping to the site where the hole previously was, which is what is happening physically. This can only occurs if the attraction by the hole (absence of electron) attracting the neighbouring electron stronger than what it’s nucleus is attracting with. Think of it as each of the electrons being given a site in k space and then rearranging the electrons. The number of electrons remains the same and they always try to minimise the total energy. Now without any external energy, they’ll all be in the lowest energy state and no free electrons exist. Now of external energy is provided, the electrons have a tendency to use that energy to go to higher states. And only when they successfully go to another state, there’s a vacancy created in its previous site which now others can jump to. This vacancy didn’t exist before. Thus no conduction.

  • $\begingroup$ Thanks for your reply! How do I reconcile this with the notion that the electrons in the valence band with holes can conduct current? In a filled valence band, I was given to understand that the electrons were unable to rearrange themselves in k-space so as to achieve a non-zero net velocity (drift velocity), hence no current contribution. According to this sense of "free", this should mean that the valence band electrons are bound to an atom. Yet without gaining energy, just by accepting a hole, these electrons may now conduct current and hence have broken free. Do I misunderstand something? $\endgroup$
    – Stinkfly
    Nov 30, 2018 at 15:57
  • $\begingroup$ Added a section in my answer. See if it helps. $\endgroup$ Nov 30, 2018 at 16:36
  • $\begingroup$ Again, thank you so much for your elaborate response! As I understand it, the flow of holes is actually a flow of electrons (hole = a vacant state in k-space, as you too suggest). Now, in a full v-band there is only one possible state configuration for the electrons in k-space, no? The influence of an electric field doesn't change that, assuming it's weak enough not to promote electrons to the c-band. But introducing a hole (= a vacancy), the electrons can now reconfigure so as to yield a non-zero net k and hence a current - this without any promotion to the c-band. Is this incorrect? $\endgroup$
    – Stinkfly
    Dec 1, 2018 at 17:54
  • $\begingroup$ Yes. Locally no promotion. But there is an electron in the c-band somewhere in the system. Total number of holes = total number of electrons in c-band (unless ionised). $\endgroup$ Dec 2, 2018 at 9:20
  • $\begingroup$ I see. You seem to be suggesting that the current in this case is really only made up by the flow of electrons in the c-band, and that the flow of holes in the v-band is an alternate but equivalent way in which to portray the current, is this correct? I was under the impression that they were two independent flows which contributed to a current together. My book always talks about the current as a sum of both parts, I = J_{e} + J_{h}. Also, in a doped semiconductor you could have an empty c-band and holes in the v-band and have a current. How would you account for this? $\endgroup$
    – Stinkfly
    Dec 2, 2018 at 14:23

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