Is one-body tight-binding Hamiltonian always diagonal in Fourier space? A generic tight-binding Hamiltonian for one-body operators is (assume one state per site)
$$
H=\sum_{ij}C_i^\dagger t_{ij}  C_j \quad; \quad t_{ij}=\int d\mathbf{r} \phi_i^*(\mathbf{r}) t(\mathbf{r})\phi_j(\mathbf{r})
$$
here $\phi_i(\mathbf{r})$ is the only atomic orbit at site $i$, and $t(\mathbf{r})$ is operator density. In general, $t(\mathbf{r})$ can be any one-body operator, i.e., for kinetic energy $t(\mathbf{r})=-\frac{\hbar^2\nabla^2}{2m}$ and for external potential or impurity potential $t(\mathbf{r})=V(\mathbf{r})$.
In Fourier space $H$ becomes:
$$
H=\frac{1}{N}\sum_{\mathbf{k,k'}}\sum_{ij} C_\mathbf{k}^\dagger t_{ij} e^{-i(\mathbf{k\cdot r_i-k'\cdot r_j})}  C_\mathbf{k'}
$$
If we open the summation over $j$, we always get diagonal Hamiltonian as I argue in the following:
$$
H=\frac{1}{N} \sum_{\mathbf{k,k'}}\sum_{i}C_\mathbf{k}^\dagger 
\left(
t_{ii}e^{-i(\mathbf{k\cdot r_i-k'\cdot r_i})}
+\sum_{\mathbf{\delta}} t_{i,i+\mathbf{\delta}}e^{-i(\mathbf{k\cdot r_i-k'\cdot (r_i+\mathbf{\delta})})}
+\sum_{\mathbf{\delta}'} t_{i,i+\mathbf{\delta}'}e^{-i(\mathbf{k\cdot r_i-k'\cdot (r_i+\mathbf{\delta}')})}+\cdots
\right)
C_\mathbf{k'}
$$
here $\delta$ and $\delta'$ run over all the nearest and next-nearest neighbors. Assume translation periodicity too, $t_{i,i}=t_{i+x,i+x}=t_0$ and $t_{i,i+\delta}=t_{i+x,i+x+\delta}=t_\delta$ and so on.
$$
H=\frac{1}{N} \sum_{\mathbf{k,k'}}\sum_{i}C_\mathbf{k}^\dagger 
\left(
t_{0}e^{-i(\mathbf{k-k'})\cdot \mathbf{r}_i}
+\sum_{\mathbf{\delta}} t_{\mathbf{\delta}} e^{-i(\mathbf{k - k'})\cdot \mathbf{r}_i} e^{i\mathbf{k'} \cdot \mathbf{\delta}}
+\sum_{\mathbf{\delta}'} t_{\mathbf{\delta}'}e^{-i(\mathbf{k - k'})\cdot \mathbf{r}_i} e^{i\mathbf{k'} \cdot \mathbf{\delta'}}
+\cdots
\right)
C_\mathbf{k'}\\
H=\sum_{\mathbf{k}}C_\mathbf{k}^\dagger 
\left(
t_{0}
+\sum_{\mathbf{\delta}} t_{\mathbf{\delta}}  e^{i\mathbf{k} \cdot \mathbf{\delta}}
+\sum_{\mathbf{\delta}'} t_{\mathbf{\delta}'} e^{i\mathbf{k} \cdot \mathbf{\delta'}}
+\cdots
\right)
C_\mathbf{k}
$$
It is diagonal. So no matter what kind of one-body operator we have, it will always be diagonal in Fourier space. Even the external potential $V(\mathbf{r})$ will be diagonal in tight-binding approximation. However, the field operator Hamiltonian for external potential $\int d\mathbf{r} \Psi^\dagger (\mathbf{r}) V(\mathbf{r}) \Psi (\mathbf{r})$ is not diagonal in Fourier space. It is $\sum_{\mathbf{kq}} V(\mathbf{q}) C_\mathbf{k}^\dagger C_\mathbf{k+q}$. Why is it so?
 A: This question can be answered without second quantization. The statement you've noticed is that the Hamiltonian only mixes states with the same momentum, i.e. $\langle \mathbf{k} | H | \mathbf{k}'\rangle \sim \delta_{\mathbf{k},\mathbf{k}'}$ in tight-binding models. This statement follows from the translation symmetry of the system. In the case of tight-binding models, the Hamiltonian contains the information that your system is spatially periodic. Intuitively, the gist is since the Hamiltonian is invariant under translations by lattice vectors, states with different "symmetries" don't mix.
To be precise but technical, by "symmetry", I am referring to the irreducible representation (irrep) that a state transforms under. In the case of translational symmetry, the irrep that a wavefunction transforms under is labeled by the momentum $\mathbf{k}$. One can then show from group/representation theory considerations that translational symmetry of the system ($[H,T(\mathbf{R})] = 0$ for all translations by lattice vectors $\mathbf{R}$) implies that different irreps of the translational group don't mix. This is intimately related to Bloch's theorem.
For a nice overview of representation theory, you can look into "Group theory and Chemistry" by Bishop. Dresselhaus' "Group Theory: Application to the Physics of Condensed Matter" goes into detail for specifically lattice systems, but is more dense.
As for your question of the impurity potential $V(\mathbf{r})$, note that an impurity breaks the translational symmetry of the system. As such, the discussion of the previous paragraph doesn't apply, and we generically expect the Hamiltonian to mix states with different momenta.
