Wannier Hamiltonian in Momentum Space

Question

In connection to a previous question,

We can write the one-particle Hamiltonian in the Wannier basis working on a general vector $v$ as : $$ \langle\vec{R},\,\lambda|\hat{H}|v\rangle = \sum_{\lambda',\,\vec{R}'}\sum_{\vec{k}}\langle\psi_{\lambda',\,\vec{k}}|v\rangle\exp\left(i\vec{k}\cdot\vec{R}'\right)\langle\vec{R},\,\lambda|\hat{H}|\vec{R}',\,\lambda'\rangle \tag{1}$$

Where $\left\{|\vec{R,\lambda}\rangle\right\}_{\vec{R},\lambda}$ is the basis of the Wannier functions ($\lambda$ indexes the band) and $|\psi_{\vec{k},\lambda}\rangle$ are the Bloch wave-functions (eigen-functions of both the Hamiltonian (with eigenvalues indexed by $\lambda$ and the translation-by-lattice-vector operator (with eigenvalues indexed by $\vec{k}$)).

In this sense we can think of $\left(\sum_{\vec{k}}\langle\psi_{\lambda',\,\vec{k}}|v\rangle\exp\left(i\vec{k}\cdot\vec{R}'\right)\right)=:v_{\lambda',\vec{R}'}$ as expansion coefficients of $|v\rangle$ in the Wannier basis $\left\{|\vec{R,\lambda}\rangle\right\}_{\vec{R},\lambda}$ and the first equation gets written in a natural matrix way as: $$ (\hat{H} |v\rangle)_{\lambda,\vec{R}} = \sum_{\lambda',\,\vec{R}'} H_{\lambda,\vec{R},\,\lambda',\vec{R}'} v_{\lambda',\vec{R}'} \tag{2} $$

My question is, how to apply Bloch-decomposition to (2).

By Bloch decomposition, I mean that we write $\psi_{\vec{k},\lambda}(\vec{r})=\exp\left(i\vec{k}\cdot\vec{r}\right)u_{\lambda,\vec{k}}\left(\vec{r}\right)$ where $u_{\lambda,\vec{k}}\left(\vec{r}\right)$ is periodic in $\vec{r}$, plug this into the Schroedinger equation to get an eigenvalue equation for $u_{\lambda,\vec{k}}\left(\vec{r}\right)$ alone: $$ \left[-\frac{\hbar^2}{2m}(\vec{\nabla}+i\vec{k})^2+V\left(\vec{r}\right)\right]u_{\lambda,\vec{k}}\left(\vec{r}\right) = E_{\lambda}\left(\vec{k}\right) u_{\lambda,\vec{k}}\left(\vec{r}\right) \tag{3}$$

So the question is how to obtain the Bloch basis of $H_{\lambda,\vec{R},\,\lambda',\vec{R}'}$, so as to be able to write (2) only for the periodic "$u$" part of the $|v\rangle$ coefficients in the Wannier basis. In other words, how to write $$ \underbrace{\left[-\frac{\hbar^2}{2m}(\vec{\nabla}+i\vec{k})^2+V\left(\vec{r}\right)\right]}_{H_{\vec{k}}(\vec{r})=\mbox{Bloch-decomposed Hamiltonian}}u_{\lambda,\vec{k}}\left(\vec{r}\right) \tag{4}$$ in the Wannier basis, but instead of with the "eigen-function" $u_{\lambda,\vec{k}}\left(\vec{r}\right)$ with a general superposition corresponding to the periodic part of $|v\rangle$.

Answer 1

So it turns out that the attempt is misguided, because the Wannier functions don't act on the $u$ functions: the $u$ functions live in a different vector space.

The correct way to proceed from equation (2) is to apply Bloch's theorem once more on it (noting that the matrix $H_{\lambda,\vec{R},\lambda',\vec{R}'}$ is invariant under $H_{\lambda,\vec{R},\lambda',\vec{R}'}\mapsto H_{\lambda,\vec{R}+\vec{R}'',\lambda',\vec{R}'+\vec{R}''}$ and so by Bloch's theorem we should expect that the eigenvectors to the matrix $H_{\lambda,\vec{R},\lambda',\vec{R}'}$ should be simultaneous eigenvectors of the operator that translates by $\vec{R}''$. As a result we could write a general eigenvector $\psi_{\lambda,\vec{R}}$ of $H_{\lambda,\vec{R},\lambda',\vec{R}'}$ as $$\psi_{\lambda,\vec{R},\vec{k}}=\exp\left(i\vec{k}\cdot\vec{R}\right)u_{\lambda,\vec{R},\vec{k}}$$ where $u_{\lambda,\vec{R},\vec{k}}$ obey the condition $$u_{\lambda,\vec{R}+\vec{R}',\vec{k}}=u_{\lambda,\vec{R},\vec{k}}$$ for any Bravais lattice vector $\vec{R}'$. However, this condition means that there is no need to carry on the index $\vec{R}$ for the $u$'s as they are all the same across the entire lattice.

Thus, a general eigenvector of $H_{\lambda,\vec{R},\lambda',\vec{R}'}$ can be written as: $$\psi_{\lambda,\vec{R},\vec{k}}=\exp\left(i\vec{k}\cdot\vec{R}\right)u_{\lambda,\vec{k}}$$

Plug this into equation (2), "cancel out" the $\exp\left(i\vec{k}\cdot\vec{R}\right)$ on both sides and get something similar to equation (3) in the Wannier basis.