How can localization of electrons in a solid be defined in a basis-independent way?

Question

In the tight binding model, it's said that in a certain limit, we can regard electrons in the solid as localized to individual atoms. This statement shows up in most introductory condensed matter textbooks, like Ashcroft and Mermin, Kittel, and Altland and Simons.

However, the statement is basis-dependent! If we expand in the Wannier basis, then it's true that our electron wavefunctions are each localized about a particular atom. However, we can also expand in the Bloch basis, in which case "the" wavefunction of each electron is delocalized. Moreover the Bloch basis feels more natural to me since it diagonalizes the Hamiltonian.

The equivalence of these two perspectives follows from second quantization, as you can write the second-quantized ground state of the system in terms of creation operators in either basis. Hence there's no basis-independent meaning of "the" state of a single electron. How can localization be defined in a basis-independent way?

Answer 1

The potential experienced by electrons in a crystal is exceedingly complex. It is, therefore, desirable to try to find simple approximations for this potential.

In the tight binding model we write down a set of trial wavefunctions (the Wannier basis) in the hope that the can write the action of the Hamiltonian on these states in a simple form. We have not said that the electron is in any of these states, we have only asked the question of what would happen if it was. The physical insight of the model is that if the potential well at each site in the lattice is fairly deep and narrow, then

1. The probability of an electron being between sites is (probably) small (in a sense the electrons are localised "on the atoms" as opposed to "between" them)
2. The eigenstates of an isolated deep, narrow well will be widely spaced, so the dynamics of an electron within a single well are probably negligible
3. If an electron were to be in a state localised to a single atom, the probability of tunnelling from one site to another would probably drop off quickly with distance, so only a small number of tunnelling terms would be needed.

This allows use to write down a good approximation for the matrix elements of the Hamiltonian in this particular basis. It does not, however, tell us what the electrons are actually doing. To do that for a system at finite temperature we need the density matrix, which, in thermal equilibrium, has the same eigenstates as the Hamiltonian. As you know, for any lattice system, these are the Bloch states and so these are the states which will determine the expectation values of any observables. Consequently, it is the non-localised Bloch states which are important to the physics of the system and not the Wannier states, which are simply an intermediate convenience.

Answer 2

Here is the basis-independent statement that the electrons are "spatially localized:" in certain limits (which physically correspond to a strong insulator), the many-body electron wavefunction is unentangled in the position basis - i.e. it is (approximately) a product state of single-particle Wannier wavefunctions, whose spatial wavefunctions have negligible overlap with each other. In this case we can think of an "individual electron" living on each lattice site, and ignore the subtleties that come up with quantum entanglement. However, for a conductor, there will be very high spatial entanglement between the different lattice sites, so there is no sense in which the many-body wavefunction is simply a product of spatially localized single-particle wavefunctions.

In both cases, if the Hamiltonian is translationally invariant and noninteracting, then the many-body wavefunction will be unentangled in the momentum basis. The difference lies in the position basis.