Update 11_nearly_free_electron_model_solutions.md - typos

src/11_nearly_free_electron_model_solutions.md

@@ -134,24 +134,24 @@ 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 $H=H_n+H_{\hat{n}}$, where $H_n$ describes a particle in a single delta-function potential well, and $H_\hat{n}$ is the perturbation by the other delta functions:
+We split the Hamiltonian into two parts $H=H_n+H_{\overline{n}}$, where $H_n$ describes a particle in a single delta-function potential well, and $H_\hat{n}$ is the perturbation by the other delta functions:
 \begin{align}
-H_n = & \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} - V_0\delta(x-na) \\
-H_\hat{n} = & - V_0 \sum_{m\neq n}\delta(x-ma)
+H_n = & \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} - \lambda\delta(x-na) \\
+H_\overline{n} = & - \lambda \sum_{m\neq n}\delta(x-ma)
 \end{align}
-such that $H_n|n\rangle = \epsilon_0|n\rangle = -\hbar^2\kappa^2/2m |n\rangle$ with $\kappa=mV_0/\hbar^2$. We can now calculate 
+such that $H_n|n\rangle = \epsilon_0|n\rangle = -\hbar^2\kappa^2/2m |n\rangle$ with $\kappa=m\lambda/\hbar^2$. We can now calculate 
 $$
-\langle n | H |n \rangle = \epsilon_0 + \langle n |H_\hat{n}|n\rangle
+\langle n | H |n \rangle = \epsilon_0 + \langle n |H_\overline{n}|n\rangle
 $$
 Note that the last term represents the change in energy of the wavefunction $|n\rangle$ that is centered at the $n$-th delta function caused by the presence of the other delta functions. This term yields
 $$
-\langle n |H_\hat{n}|n\rangle = \kappa \sum_{m \neq 0 }\int e^{-2\kappa|x|}\delta(x-ma)  = \kappa \sum_{m \neq 0 } e^{-2\kappa|ma|} = 2\kappa(\frac{1}{1-e^{-2\kappa a}}-1)
+\langle n |H_\overline{n}|n\rangle = -\kappa \lambda \sum_{m \neq 0 }\int e^{-2\kappa|x|}\delta(x-ma)  = -\kappa \lambda \sum_{m \neq 0 } e^{-2\kappa|ma|} = -2\kappa\lambda(\frac{1}{1-e^{-2\kappa a}}-1)
 $$
 Note that the result should not depend on $n$, so we chose $n=0$ for convenience.
 
 Similarly, we can calculate 
 $$
-\langle n-1 | H |n \rangle = \epsilon_0\langle n-1  |n \rangle + \langle n-1 |H_\hat{n}|n\rangle
+\langle n-1 | H |n \rangle = \epsilon_0\langle n-1  |n \rangle + \langle n-1 |H_\overline{n}|n\rangle
 $$
 where $\langle n-1|n\rangle$ is the overlap between two neighbouring wavefunctions:
 $$
@@ -159,8 +159,8 @@ $$
 $$
 and
 \begin{align}
-\langle n-1|H_\hat{n}|n\rangle = & \kappa \sum_{m \neq 0 }\int e^{-\kappa|x-a|} \delta(x-ma)  e^{-\kappa|x|} \\
-=& \kappa \sum_{m \neq 0 } e^{-\kappa a|m-1|} e^{-\kappa a |m|} =\kappa(e^{ka}+e^{-ka}) \sum_{m=1}^{m=\infty} e^{-2\kappa a m}
+\langle n-1|H_\overline{n}|n\rangle = & -\kappa \lambda \sum_{m \neq 0 }\int e^{-\kappa|x-a|} \delta(x-ma)  e^{-\kappa|x|} \\
+=& -\kappa \lambda \sum_{m \neq 0 } e^{-\kappa a|m-1|} e^{-\kappa a |m|} =-\kappa \lambda(e^{ka}+e^{-ka}) \sum_{m=1}^{m=\infty} e^{-2\kappa a m}
 \end{align}
 
 ### Subquestion 4