Number of bands in 1D tight-binding model

Question

I was reading about the one-dimensional tight-binding Hamiltonian (TBH) with one quantum state per atom $$H=E_0\sum\limits_{n}|n\rangle\langle n|-t\sum\limits_{n}\Big(|n\rangle\langle n+1|+|n+1\rangle\langle n|\Big)\tag{1}$$ where $E_0$ and $t$ denote the on-site energy and the hopping parameter, repectively. The Hamiltonian of Eq.(1) leads to the electron dispersion relation $$E(k)=E_0-2t\cos(ka)\tag{2}$$ where $a$ is the lattice spacing, and $k$ is the wavenumber.

$\bullet$ How does one draw the inference that this Hamiltonian leads to only one band and not more than one? Is it because of the energy $E(k)$ a single-valued function of $k$?

$\bullet$ What is(are) the simple possible modification(s) to the one-dimensional TBH of Eq.(1) so that more than one band is obtained? What is the corresponding physical situation?

Answer 1

The rule of thumb is that there are a number of bands equal to the "degrees of freedom" of the lattice. You can get additional degrees of freedom from having multiple species of atoms, multiple orbitals per atom, multiple coupling strengths, etc. These degrees of freedom all increase the dimension of your Hamiltonian matrix.

In this simplest 1-atom example, your Hamiltonian is 1x1, so you only have one Eigenvalue per wavevector. If you have two atoms, say an alternating line of atoms A and B, with different on-site energy, your Hamiltonian will be 2x2, yielding two Eigenvalues per wavevector.

A slightly more intuitive way to think about this is comparing a simple 1-d lattice with the alternating A-B lattice. In these lattices, you have the same number of total $k$-states (which equals the number of atoms in the crystal), but in the A-B case, the lattice vector is twice as large. This means that the Brillouin zone is half as large, and the band in the second Brillouin zone is folded into the first, leading to two bands.

Not actually sure if that last part will make sense to you, but hopefully this helps at least a little!