Parallels between Tight-binding wavefunctions and Bloch states

Question

I was reading Robert Knox's Theory of Excitons and I came across a certain statement that I have trouble reconciling, both mathematically and conceptually. The context concerns the electronic structure description by a Hamiltonian that includes terms akin to the Kohn-Sham equation. I'm familiar with this but then the author discusses the approximations to the wavefunction itself - apart from the well-known Hartree-Fock solutions. The two approximations are:

a) A tight-binding-esque wavefunction wherein one writes it down as an atomic function $ \psi_{lr} $ where l is a set of atomic quantum numbers and R is one of the lattice sites where $ \psi_{lr} $ is centered

b) A bloch state $ \psi_{nk} $ where n is a band index and k runs over the entire Brillouin zone

I am familiar with both models - With the tight-binding model, you have localized orbitals whereas with the bloch state, it is distributed uniformly over the crystal. The author then says that

In a crystal consisting of atoms in closed shells, the ground state is identical in the two schemes. This holds in the formal sense: if from a set of localized states, one constructs Bloch states, $$ \psi_{lk} = N^{-1/2} \sum_{R}e^{ik.R} \psi_{lR}(r) $$

The author goes on to say that the determinantal functions (basically a many-particle wavefunction written as a Hartree product) of the two are identical:

$ \psi_{lk_{1}\alpha}(r_1)\psi_{lk_{1}\beta}(r_2)..\psi_{lk_{N}\beta}(r_2N) $

$ \psi_{lR_{1}\alpha}(r_1)\psi_{lR_{1}\beta}(r_2)..\psi_{lR_{N}\beta}(r_2N) $

Here $\alpha$ and $\beta$ (I assume) run over spins -1/2 and +1/2, and the system has 2N valence electrons.

The justification , directly quoted from the author is: Once these two equations are written as determinants, and noting the elements form matrices related by simple multiplication by a unitary matrix ($U_{pq} = N^{-1/2}e^{ik_{p}.R_{q}}$) whose determinant is unity.

I have trouble understanding how both these schemes are the same in this context and would be grateful if anyone could break it down for me.

Answer 1

Bloch functions are the eigenstates of an electron in a spatially periodic potential, such as that of a crystal lattice: $$ \psi(\mathbf{r}) = e^{i\mathbf{k}\mathbf{r}}u(\mathbf{r}). $$ That the eigenstates in a crystal have the form of the Bloch states can be shown by quite general symmetry arguments. However, thsi leaves undefined the periodic part of the Bloch states - $u(\mathbf{r})$. There exist approximate methods for calculating these function, one of which is the tight-binding approximation.

In the tight-binding approximation one starts with localized orbitals (or Wannier states), which are coupled via hopping. Solving the tight-binding Hamiltonian results in the Bloch states, expressed in terms of these orbitals and the hopping matrix elements.

Wannier states are a special choice of localized orbitals which are orthogonal among themselves (for different atoms), that is the tight-binding hopping matrix elements are zero. Thus, Wannier states are just linear combinations of Bloch states and vice versa (however, Wannier states are not eigenstates of the Hamiltonian).

The many-particle states are the Slater determinants of one-particle states, which can be written in either basis. Knox's choice is thus to use for developing exciton tehory a basis, which is not the eigenbasis of non-interacting Hamiltonian (i.e., without Coulomb interaction).