Quantum Mechanics – Wannier Functions as Discrete Basis in Solid-State Physics

discretequantum mechanicssolid-state-physics

In solid state physics, using Bloch's theorem we know that the one-electron energy eigen-function can be written as $\psi_{\lambda,\vec{k}}(\vec{r})$ where $\lambda$ indexes eigenvalues of $\hat{H}$ and $\vec{k}$ indexes eigenvalues of $\hat{T}_\vec{R}$, the translation by lattice-vector-$\vec{R}$ operator.

Because of Bloch's theorem, we know that $\psi_{\lambda,\vec{k}}(\vec{r})$ is periodic in $\vec{k}$ by reciprocal-lattice-vectors $\vec{K}$ and so we may write a Fourier series of it as: $$\psi_{\lambda,\vec{k}}(\vec{r})=\sum_{\vec{R}}\tilde{\psi}_{\lambda,\vec{R}}(\vec{r})\exp\left({i\vec{R}\cdot\vec{k}}\right)$$

$\tilde{\psi}_{\lambda,\vec{R}}(\vec{r})$ are called Wannier functions. One can prove that $\tilde{\psi}_{\lambda,\vec{R}}(\vec{r})$ depends only on the difference $\vec{R}-\vec{r}$ as well as show that the Wannier functions are orthogonal between different values of $\lambda$ or $\vec{R}$.

As a result we may write an arbitrary wave-function $\psi(\vec{r})$ as $$ \psi(\vec{r}) = \sum_{\lambda,\,\vec{R}} \alpha_{\lambda,\,\vec{R}}\tilde{\psi}_{\lambda,\vec{R}}(\vec{r})$$

where $\alpha_{\lambda,\,\vec{R}}=\int_{\mathbb{R}^3}\psi(\vec{r})\tilde{\psi}_{\lambda,\vec{R}}(\vec{r})d^3\vec{r}$ are the expansion coefficients in the Wannier basis.

My question is:

1) Expressed in the Wannier basis $\left\{\tilde{\psi}_{\lambda,\vec{R}}(\vec{r})\right\}_{\lambda,\,\vec{R}}$, why is it true that the potential energy of the electron is proportional to a delta-function?

$$ V_{\lambda,\lambda',\vec{R},\vec{R'}} \propto \delta_{\vec{R},\vec{R'}} $$

It is clear intuitively because $$ V_{\lambda,\lambda',\vec{R},\vec{R'}} \equiv \int_{\mathbb{R}^3}\tilde{\psi}_{\lambda,\vec{R}}(\vec{r})^{\ast}V(\vec{r})\tilde{\psi}_{\lambda',\vec{R'}}(\vec{r})d^3\vec{r}$$

and coming from the tight-binding model (or alternatively using some dubious claim that the Wannier functions can always be chosen to be maximally-localized using the gauge-freedom in the definition of the wave functions) we know that $\tilde{\psi}_{\lambda,\vec{R}}(\vec{r})$ is "sort of" around one particular lattice site $\vec{R}$ and so the integrand is non-zero only if the two lattice sites match.

But I would like a more rigorous (or hopefully much more direct) derivation of this. I am assuming that there is a simple easier reason I don't see.

2) My second question is how to express the kinetic energy, $\frac{-\hbar^2}{2m}\vec{\nabla}^2$, in the Wannier basis? I can't think of how to proceed from $$ T_{\lambda,\lambda',\vec{R},\vec{R'}} = \frac{-\hbar^2}{2m} \int_{\mathbb{R}^3}\tilde{\psi}_{\lambda,\vec{R}}(\vec{r})^{\ast}\vec{\nabla}^2\tilde{\psi}_{\lambda',\vec{R'}}(\vec{r})d^3\vec{r}$$

Best Answer

So it turns out that Adam was correct. Even when using the maximally localized Wannier functions, the answer to question 1) is that it is merely an approximation, and the answer to the second question is that there is nothing further to do with that expression as it stands. Thanks Adam.

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