is due to the repetition of the unit cells $N$ times.

The band Hamiltonian
--------------------

![Dispersion relation obtained from bulk Hamiltonian plotted for different parameter cases[]{data-label="sshdispersion"}](fig-dispersion.pdf)

The two-site unit cell adds another band to the band structure we explored previously. To see this, we need to solve for its dispersion relation by only focusing on the bulk part of the chain, that is to say, assuming the chain forms a loop (periodic boundary). Because the system is translationally invariant, it allows us to perform Fourier transform in the external Hilbert space similar to what we did earlier. $$\begin{aligned}
\ket{n}\otimes\ket{A(\text{or }B)} = \frac{1}{\sqrt{N}} \sum_k e^{ikn} \ket{k}\otimes \ket{A(\text{or }B)}\end{aligned}$$ Plugging this back into the Hamiltonian (with periodic boundaries), we get $$\begin{aligned}
H&=&\sum_k (v+e^{-ik}w)\ket{k,B}\bra{k,A}\nonumber \\
&& + (v+e^{ik}w)\ket{k,A}\bra{k,B}\end{aligned}$$ The external and internal parts can be separated and this can further be reduced to $$\begin{aligned}
H&=&\ket{k}\bra{k}\otimes\big[ (v+e^{ik}w) \ket{A}\bra{B}\nonumber \\
&&+ (v+e^{-ik}w) \ket{B}\bra{A} \big]\end{aligned}$$ In a compact form, $H=\sum_k \ket{k}H(k)\bra{k}$.

$$\begin{aligned}
H(k)=
\begin{pmatrix}
0 & v+e^{ik}w \\
v+e^{-ik}w  & 0
\end{pmatrix}\end{aligned}$$

$H(k)$ is the band Hamiltonian that acts on the internal Hilbert space [@rshankar2]. With the help of Fourier transform we have block diagonalised the full Hamiltonian into $N$ such $2\times 2$ band Hamiltonians for 