I am looking at the Bose-Hubbard model. Specifically the solution in the limit of weak interactions $U = 0$. I understand the proof to show the Hamiltonian is diagonal in momentum space, however I would like clarify the origin of the transformation
\begin{equation} \hat{b}_{i}^{\dagger} = \frac{1}{\sqrt{M}}\sum_{\mathbf{k}}\hat{c}_{\mathbf{k}}^{\dagger}e^{-i\mathbf{k}\cdot\mathbf{r}_{i}} \end{equation}
where $\hat{b}_{i}^{\dagger}$ is the creation operator for site $i$. $\mathbf{k}$ is the momentum vector, $\mathbf{r}_{i}$ is the position vector of the $i$th lattice site. $\hat{c}_{\mathbf{k}}^{\dagger}$ is the momentum creation operator. $M$ is the number of lattice sites.
Several questions. Firstly I understand the sum $\sum_{\mathbf{k}}$ is a sum over momentum space which is discrete because the position space is discrete, why does a discretised position space imply a discretised momentum space?
Secondly I understand this equation is a sort of fourier transform between position and momentum space operators. But it can also be determined by changing the basis of the creation operator, could somebody show me how this is explicitly done?
I guess I am struggling to understand this dual space picture (between position and momentum).
Lastly \begin{equation} \sum_{i = 1}^{i = M}e^{i(\mathbf{k}^{'} - \mathbf{k})\cdot\mathbf{r}_{i}} = M\delta_{\mathbf{k},\mathbf{k^{'}}} \end{equation} where $\mathbf{k}^{'}$ is another momentum vector. Is the orthonormality of this sum because $e^{i\mathbf{k}\cdot\mathbf{r}_{i}}$ is eigenfunction of the momentum operator and so it is orthogonal to any other eigenfunction $e^{i\mathbf{k^{'}}\cdot\mathbf{r}_{i}}$. Does this plane wave form some sort of of basis for both position and momentum space or is that wrong and it is just representative of the overlap between the spaces?