0
$\begingroup$

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 $|v\rangle$.

$\endgroup$
  • 2
    $\begingroup$ Small typographical note: In your equations, please use \langle and \rangle instead of < and >, respectively. $\endgroup$ – Danu Oct 20 '14 at 17:38
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.