Skip to content
Snippets Groups Projects
Commit 34da6d6c authored by Lars kleyn Winkel's avatar Lars kleyn Winkel
Browse files

Update 7_tight_binding_model_solutions.md

parent 3680041a
No related branches found
No related tags found
No related merge requests found
Pipeline #29958 passed
......@@ -29,7 +29,7 @@ Hint: What kind of particles obey Bose-Einstein statistics? What kind of 'partic
### Subquestion 2
Group velocity is given as $v=\hbar^{-1}\frac{\partial E}{\partial k}$ with $E=\hbar\omega$ and $g(\omega) = \frac{dN}{d\omega} = \frac{dN}{dk}\frac{dk}{d\omega}$. So we find: $$ v(k) = \frac{a}{2}\sqrt{\frac{2\kappa}{m}}\cos(\frac{ka}{2})\sign(k)$$ $$ g(\omega) = \frac{L}{\pi}\frac{d}{d\omega} \bigg [\frac{2}{a}\sin^{-1}\bigg(\sqrt{\frac{m}{\kappa}}\frac{\omega}{2} \bigg) \bigg ] \\ = \frac{L}{\pi a} \frac{1}{\sqrt{\frac{4\kappa}{m}-\omega^2}}$$
Group velocity is given as $v=\hbar^{-1}\frac{\partial E}{\partial k}$ with $E=\hbar\omega$ and $g(\omega) = \frac{dN}{d\omega} = \frac{dN}{dk}\frac{dk}{d\omega}$. So we find: $$ v(k) = \frac{a}{2}\sqrt{\frac{2\kappa}{m}}\cos(\frac{ka}{2})\sgn(k)$$ $$ g(\omega) = \frac{L}{\pi}\frac{d}{d\omega} \bigg [\frac{2}{a}\sin^{-1}\bigg(\sqrt{\frac{m}{\kappa}}\frac{\omega}{2} \bigg) \bigg ] \\ = \frac{L}{\pi a} \frac{1}{\sqrt{\frac{4\kappa}{m}-\omega^2}}$$
### Subquestion 3
......
0% Loading or .
You are about to add 0 people to the discussion. Proceed with caution.
Please register or to comment