cancel the remaining hbar in g(E)

docs/7_tight_binding.md

@@ -337,7 +337,7 @@ Notice that the effective mass can be negative, which implies the electrons acce
 The DOS is the number of states per unit energy. In 1D we have
 
 $$
-g(E) = \frac{L}{2\pi}\sum |dk/dE| = \frac{\hbar L}{2\pi}\sum |v|^{-1}
+g(E) = \frac{L}{2\pi}\sum |dk/dE| = \frac{L}{2\pi \hbar}\sum |v|^{-1}
 $$
 
 The sum goes over all possible values of $k$ and spin which have the same energy $E$.
@@ -365,7 +365,7 @@ You can get to this result immediately if you remember the derivative of arccosi
 We now add together the contributions of the positive and the negative momenta as well both spin orientations, and arrive to the density of states
 
 $$
-g(E) = \frac{\hbar L}{2\pi}\frac{4}{a}\frac{1}{\sqrt{4t^2 - (E - E_0)^2}}.
+g(E) = \frac{L}{2\pi}\frac{4}{a}\frac{1}{\sqrt{4t^2 - (E - E_0)^2}}.
 $$
 
 A quick check: when the energy is close to the bottom of the band, $E = E_0 - 2t + \delta E$, we get $g(E) \propto \delta E^{-1/2}$, as we expect in 1D.