In the nearly-free electron model a particle of mass $m$ moves in one dimension in a periodic potential of the form $V(x)=\lambda U(x)$, where $\lambda \ll 1$ is a dimensionless coupling and $U(x)$ has a Fourier series

$$U(x)=\sum_{l=-\infty}^{+\infty} U_{l} \exp \left(\frac{2 \pi i}{a} l x\right)$$

with coefficients obeying $U_{-l}=U_{l}^{*}$ for all $l$.

Ignoring any degeneracies in the spectrum, the exact energy $E(k)$ of a Bloch state with wavenumber $k$ can be expanded in powers of $\lambda$ as

$$E(k)=E_{0}(k)+\lambda\langle k|U| k\rangle+\lambda^{2} \sum_{k^{\prime} \neq k} \frac{\left\langle k|U| k^{\prime}\right\rangle\left\langle k^{\prime}|U| k\right\rangle}{E_{0}(k)-E_{0}\left(k^{\prime}\right)}+O\left(\lambda^{3}\right)$$

where $|k\rangle$ is a normalised eigenstate of the free Hamiltonian $\hat{H}_{0}=\hat{p}^{2} / 2 m$ with momentum $p=\hbar k$ and energy $E_{0}(k)=\hbar^{2} k^{2} / 2 m$.

Working on a finite interval of length $L=N a$, where $N$ is a positive integer, we impose periodic boundary conditions on the wavefunction:

$$\psi(x+N a)=\psi(x)$$

What are the allowed values of the wavenumbers $k$ and $k^{\prime}$ which appear in (1)? For these values evaluate the matrix element $\left\langle k|U| k^{\prime}\right\rangle$.

For what values of $k$ and $k^{\prime}$ does (1) cease to be a good approximation? Explain your answer. Quoting any results you need from degenerate perturbation theory, calculate to $O(\lambda)$ the location and width of the gaps between allowed energy bands for the periodic potential $V(x)$, in terms of the Fourier coefficients $U_{l}$.

Hence work out the allowed energy bands for the following potentials:

$$\begin{aligned} &\text { (i) } \quad V(x)=2 \lambda \cos \left(\frac{2 \pi x}{a}\right), \\ &\text { (ii) } \quad V(x)=\lambda a \sum_{n=-\infty}^{+\infty} \delta(x-n a) . \end{aligned}$$