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.

  • $\begingroup$ The two wave functions are related by a unitary transformation. $\endgroup$
    – my2cts
    Commented Jan 12, 2021 at 10:27
  • $\begingroup$ So basically, rotating one basis set yields the other? @my2cts and hence they are equivalent, is the reasoning? $\endgroup$
    – user285482
    Commented Jan 12, 2021 at 18:29

1 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).

  • $\begingroup$ Thank you for the answer. I am still a bit iffy on the main premise listed in the question. I guess it might be because I'm reading between the lines. Knox states that these two basis sets are equivalent for the ground state in the case of atoms in closed shells. Can you explain a bit on the extra qualifier? $\endgroup$
    – user285482
    Commented Jan 12, 2021 at 18:31

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