Are Electronic States in a 1D Atomic Chain Eigenstates of the Hamiltonian?

[squareroot]

Homework Statement

1D atomic chain with one atom in the primitive cell and the lattice constant a. The system in described within the tight binding model and contains N-->∞ primitive cells indexed by the integer n. The electronic Hamiltonian is $$H_{0} = \sum_{n} (|n \rangle E_{at} \langle n | -|n+1 \rangle \beta \langle n| - |n \rangle \beta \langle n+1 | )$$

with E_at being the energy on one electron in the state $|n \rangle$at site n and$\beta >0$ represents the energy overlap integral responsable for the interaction between first neighbors. We assume that the atomic orbitals | n \rangle are orthonormalized and neglect the overlap of atomic orbitals on different sites, thus $\langle n|n' \rangle = \delta_{nn'}$

First off, show that the electronic states described by :

$$ | k \rangle = \frac{1}{\sqrt{N!}}\sum_{n} e^{ikna} |n \rangle $$

are eigenstates of H₀ and calculate the corresponding eigenvalues E₀(k) in the first Brillouin zone

Homework Equations

above

The Attempt at a Solution

[/B]
We start by plugging H₀ into the equation $$ H_{0}|k \rangle = E_{0} | k \rangle $$ and thus obtaining

$$(\sum_{n} (|n \rangle E_{at} \langle n | -|n+1 \rangle \beta \langle n| - |n \rangle \beta \langle n+1 | ))| k \rangle$$

and now replacing the form of $|k \rangle"$ one gets

$$(\sum_{n} (|n \rangle E_{at} \langle n | -|n+1 \rangle \beta \langle n| - |n \rangle \beta \langle n+1 | ))(\frac{1}{\sqrt{N!}}\sum_{n} e^{ikna} |n \rangle)$$

moving on with

$$ H_{0}|k \rangle = \frac{1}{\sqrt(N!)}\sum_{n}|n \rangle E_{at} \langle n |e^{ikna}|n \rangle - |n+1 \rangle \beta \langle n|e^{ikna}|n \rangle - |n \rangle \beta \langle n+1 | e^{ikna}|n \rangle $$

and here is where I get stuck. I don t know how to evaluate $\langle n |e^{ikna}|n \rangle$ and $|n \rangle \beta \langle n+1 | e^{ikna}|n \rangle$

From my intuition I think that after solving the LHS of the Schrodinger equation like I started to do above, at one point I should get that the expression above is of form $Y|k \rangle$ with Y being a number and thus showing that the ket k is a eigenstate of H₀.Thank you

 

[Chandra Prayaga]

You should start by using a different summation index in the Hamiltonian and the state. You used n as the index of sites in the Hamiltonian. Use some other index, say m, when you write |k>. Then use the orthogonality of states at different sites. Also notice that a matrix element like <n| e^ikna |n> = e^ikna.