Update 11_nearly_free_electron_model_solutions.md - fix solution (work in progress)

src/11_nearly_free_electron_model_solutions.md

@@ -134,6 +134,23 @@ E_-(k) = -\frac{\lambda}{a}+\frac{\hbar^2}{4m}\left[k^2+\left(k-\frac{2\pi}{a}\r
 See the lecture notes!
 
 ### Subquestion 3
+We split the Hamiltonian into two parts $H=H_0+H_1$, where$H_0$ describes a particle in one delta-function potential well, and $H_1$ is the perturbation by the other delta functions:
+\begin{align}
+H_0 = & \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} - \V_0\delta(x-na) \\
+H_1 = & - V_0 \sum_{m\neq n}\delta(x-ma)
+\end{align}
+such that $H_0|n\rangle = -\epsilon_0|n\rangle = -\hbar^2\kappa^2/2m |n\rangle$ with $\kappa=mV_0/\hbar^2$.
+
+It follows that 
+$$
+\langle n | H |n \rangle = \epsilon_0 + \langle n |H_1|n\rangle
+$$
+where
+$$
+\langle n |H_1|n\rangle = \kappa \sum_{m \neq 0 }\int e^{-2\kappa|x|}\delta(x-ma)  = \kappa \sum_{m \neq 0 } e^{-2\kappa|ma|}
+$$
+
+In progress of being updated....
 
 \begin{equation}
 \varepsilon_0=\braket{n|\hat{H}|n}=\braket{n|\hat{K}|n}+\braket{n|\hat{V}(x)|n}