# Number of bands in 1D tight-binding model

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?

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.