Electron in an incommensurate potential: Consider the Hamiltonian
$$
\hat{\mathcal{H}}=\mathrm{t} \sum_l\left\{\frac{1}{2}(|l\rangle(l+1|+| l\rangle\langle l-1|)+\cos (2 \pi l \tau)|l\rangle\langle l|\right\} .
$$
The problem becomes particularly interesting when the potential $\cos (2 \pi l \tau)$ is incommensurate with the lattice sites $|l\rangle$, which happens when $\tau$ is irrational. For example, one might take $\tau$ to be the golden mean,
$$\tau=\frac{\sqrt{5}+1}{2} .$$
(a) Suppose that instead of taking $\tau$ to be exactly the golden mean, one replaces it by rational approximants to the golden mean:
$$\tau_n=\frac{F_{n+1}}{F_n},$$
where $F_n$ is the $n$th Fibonacci number, $F_n=F_{n-1}+F_{n-2}$,
$$F_0=1, F_1=1, F_2=2, F_3=3, F_4=5, F_5=8 \ldots$$
How many bands does (18.149) have when one uses $\tau_n$ for $\tau$ (compare with Problem 7 in Chapter 8).
(b) For $\tau$ now given by Eq. (18.150), the wave functions of the Hamiltonian are a curious intermediate between localized and extended. Demonstrate this fact by assuming that there is an eigenstate at $\mathcal{E}=0$, taking $\psi_0=\langle 0 \mid \psi\rangle=1$ and $\psi_1=\langle 1 \mid \psi\rangle=1$. Then use Eq. (18.149) to compute $\psi_m=\langle m \mid \psi\rangle$ for $m$ on the order of $10^4$ and observe how the magnitude of $\psi_m$ scales with $m$.
One way to do this is to plot the participation ratio
$$
P(m)=\frac{\sum_{l=1}^m \psi_l^4}{\left(\sum_{l=1}^m \psi_l^2\right)^2}
$$
In order to calculate $P(m)$, make use of quantities calculated in order to find $P(m-1)$; do not carry out a sum starting at 0 and going up to $m$ for each individual $P(m)$.
Compare the behavior of the participation ratio with what would be expected for localized states, and what would be expected for extended states.