normal case, where the lattice spacing is $a$, the electronic density is uniform. But if we consider a case where every atom distorts itself such as the resulting lattice has two sublattice sites in a unit cell and the inter-unit cell distance is $2a$. The electronic density will become periodically modulated. This is known as a **charge density wave** [@thorne]. The final energy of the distorted lattice atoms alone will, of course, be higher than the original undistorted case. In fact, estimating this energy per unit length in the elastic approximation gives $$\begin{aligned}
\Delta E_{\text{lattice}} = \frac{1}{4}K\delta^2\end{aligned}$$ which is quadratic in the distortion $\delta$. $K$ is the elastic constant determined by the material. However, if we calculate the electronic energy per unit length of the new charge configuration due to distortion, in the low distortion limit we get, $$\begin{aligned}
\Delta E_{\text{electronic}} \approx \frac{A^2\delta^2}{\pi} \ln{|\delta|}\end{aligned}$$ the total energy per unit length, $\Delta E_{\text{lattice}} + \Delta E_{\text{electronic}}$ is extremized as, $$\begin{aligned}
\frac{d}{d\delta} [ \Delta E_{\text{total}} ] = \frac{1}{2} K \delta + \frac{2A^2}{\pi}\delta \ln{|\delta|} + \frac{A^2}{\pi}\delta = 0\end{aligned}$$ The solution $\delta = 0$ corresponds to a maxima which can be checked explicitly by calculating the second derivative. The minima corresponds to $$\begin{aligned}
\delta_{\text{min}} 