Semiconductors and localization of the electrons

When looking at the band diagram of a semi-conductor, direct conclusion of the invariance under discrete translations, for a filled state with an electron, one does know precisely it's momentum, so my question is: does it imply, by Heisenberg, uncertainty that this electron is spread over the whole lattice? Is it the case of all the other electrons?

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As you might have guessed, these two representations of an electron in a periodic potential are related by Fourier transforms (or series). Say you pick a state with crystal momentum $\mathbf{k}$. Then the Bloch wave function has the form $$\psi_{\mathbf{k}}(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}u_{\mathbf{k}}(\mathbf{r})$$where $u_{\mathbf{k}}(\mathbf{r}) = u_{\mathbf{k}}(\mathbf{r}+\mathbf{R})$ with $\mathbf{R}$ being the lattice vector. Therefore, the wave function would look like an oscillating function with period $\mathbf{R}$ with an envelope function with period $2\pi/|\mathbf{k}|$. In other words, it will (as you correctly pointed out) be spread out in space. You can see a picture of this in Figure 1 of the article: Maximally localized Wannier functions: Theory and applications.
Now, if you define the origin of your spatial axes as the lattice site at $\mathbf{R}_{0}$ then the probability amplitude (or wave function) of finding the electron relative to that site is $$\psi_{\mathbf{R}_{0}}(\mathbf{r})=\sum_{\mathbf{k}}e^{-i\mathbf{k}\cdot\mathbf{R}_{0}}\psi_{\mathbf{k}}(\mathbf{r})$$This is the definition of a Wannier function. For the sake of this argument you can ignore subtleties like the gauge degree of freedom (or non-uniqueness) of this mapping. Now, looking at the above equation, and from our intuition of Fourier transforms (series), the weighted sum of many functions that are spread out in space can give you a function localized in space. The fact that we chose the weighting function $e^{-i\mathbf{k}\cdot\mathbf{R}_{0}}$ means we will find an electron state localized about $\mathbf{R}_{0}$.