Electron localization in crystal

Question

I tried to use indetermination principle to explain why, electrons that are strongly bounded to the nucleus, are more localized. Unfortunately, the result that I've obtained says the exact opposite. What is wrong with the following reasoning?

The indetermination principle is $\Delta x \Delta p \approx h$. The dispersion relation is, approximately, $E=p^2/2m$, from which we get $\Delta E \approx (p/m) \Delta p$. Substituting the last one in the first: $\Delta x \Delta E \approx (p/m) h$

I interpret this as if, the larger the band, the more the electron is localized. Which is wrong, because large bands are those far from the nucleus, where electrons are not really localized.

Answer 1

The calculation in the OP (with some adjustments) is appropriate for estimating the binding energy of an electron in an atom. On the other hand, the bandwidth is determined by the probability of en electron jumping from one nucleus to another - a result that is easily obtained within the tight-binding approximation. This hopping integral can be roughly represented as $$ \gamma_{m,l}(\mathbf{R}_n)=-\int\varphi_m(\mathbf{r})\Delta U(\mathbf{r})\varphi_l(\mathbf{r}-\mathbf{R}_n)d^3\mathbf{r}, $$ where $\Delta U(\mathbf{r})$ is the interatomic potential and $\varphi_m(\mathbf{r})$ are the atomic orbitals. The deeper is the orbital -> the less its spatial extent -> less the product $\varphi_m(\mathbf{r})\varphi_l(\mathbf{r}-\mathbf{R}_n)$ -> less the hopping integral -> less the bandwidth.