# QM approach to the static dielectric constant for non conductors

I'm into a procedure for obtaining a formula for the static dielectric constant of a non conducting material, from a QM point of view.

I start with an umperturbed hamiltonian $H_{0}$ proper of the system, which I perturb with a potential of the form $$\delta u(r)= \hat u(q)e^{i\vec q \cdot \vec r}+\hat u^*(q)e^{-i\vec q \cdot \vec r}$$ Now, the eigenvectors of $H_0$ are the Bloch states $$\phi^{(0)}_{\vec k (r)}=\langle\vec r|\vec k , a\rangle_0 = e^{i\vec k \cdot \vec r}w(\vec r,a)=e^{i\vec k \cdot \vec r}\sum_{\vec g} e^{i\vec g\cdot\vec r}\hat w_{\vec k} (\vec g,a)$$ where the last equality is the Fourier expansion over all the vectors $\vec g$ of the reciprocal space, the script $0$ indicates the unperturbed items, and $a$ is the band index. It's valid the following orthonormality relation:

$${}_0\langle \vec{k'} , a'|\vec k , a\rangle_0 = \delta_{a,a'} \delta_{\vec{k},\vec{k'}}$$

This means that Bloch's states in different bands are orthogonal even if they have the same $\vec k$, meaning that the orthogonality lies inside the periodic part of the wavefunction.

I want to find the perturbed eigenvalues and eigenvectors of $H= H_0 +\delta u(r)$ , using the stationary non-degenerate perturbative theory at the first order in $\delta u (r)$. So first of all, I know that $$E_a(\vec k) = E^{(0)}_a(\vec k) + {}_0\langle \vec{k},a|\delta u(\vec r)|\vec{k},a\rangle_0$$ This is the point I don't handle comfortably. Giving that I discard the umklapp processes, I have to show that the above equation reduces to $E_a(\vec k) = E^{(0)}_a(\vec k)$; how can I calculate the action of $\delta u (\vec r)$ over the state $|\vec{k},a\rangle_0$ ?

For example, in the procedure for the dielectric constant of a metal, I can focus only in one band, and I can cast the Bloch eigenstates in the form of plane waves (using a jellium-like model) $$\langle\vec r|\vec k , a\rangle_0 = e^{i\vec k \cdot \vec r} w(\vec r,a)= e^{i\vec k \cdot \vec r} \frac{1}{\sqrt{V}}$$

This means that

$$\delta u(\vec r)|\vec{k},a\rangle_0 = \hat u(\vec q)e^{i\vec q \cdot \vec r}e^{i\vec k \cdot \vec r} \frac{1}{\sqrt{V}} + \hat u^*(\vec q)e^{-i\vec q \cdot \vec r}e^{i\vec k \cdot \vec r} \frac{1}{\sqrt{V}}=$$ $$= u(\vec q)|\vec{k}+\vec{q},a\rangle_0 + u^*(\vec q)|\vec{k}-\vec{q},a\rangle_0$$ so I obtain two eigenstates orthogonal with $\langle \vec{k},a |$.

But in the case of an insulator, the equation runs as

$$\delta u(\vec r)|\vec{k},a\rangle_0 = \hat u(\vec q)e^{i\vec q \cdot \vec r}e^{i\vec k \cdot \vec r}\sum_{\vec g} e^{i\vec g\cdot\vec r}\hat w_{\vec k} (\vec g,a) + \hat u^*(\vec q)e^{-i\vec q \cdot \vec r}e^{i\vec k \cdot \vec r}\sum_{\vec g} e^{i\vec g\cdot\vec r}\hat w_{\vec k} (\vec g,a)$$ Basically, I think it becomes (considering only the first term) $$= \hat u(\vec q)e^{i\vec q + \vec k \cdot \vec r}\sum_{\vec g} e^{i\vec g\cdot\vec r}\hat w_{\vec k} (\vec g,a)$$ but it seems it should go as $$= \hat u(\vec q)e^{i\vec q + \vec k \cdot \vec r}\sum_{\vec g} e^{i\vec g\cdot\vec r}\hat w_{\vec k +\vec q} (\vec g,a)$$. Can someone help?

P.S.: I know I have not wrote down the expressions for $\delta q(\vec r) |\vec k , a \rangle$ in a proper manner in the coordinate representation (expansion over the $|\vec r \rangle$ basis); forgive me, I wanted to write as less as i could :)

So, I calculate in the coordinate representation $${}_0\langle \vec{k},a|\delta u(\vec r)|\vec{k},a\rangle_0 = \int_\Omega d\vec r \Big(\sum_{\vec{g'} }e^{-i \vec{g'}\cdot \vec r} \hat w_{\vec{k}}^*(\vec{g'},a) \Big) e^{-i \vec{k}\cdot \vec r}e^{i \vec{k}\cdot \vec r} \Big(\sum_{\vec{g} }e^{-i \vec{g}\cdot \vec r} \hat w_{\vec{k}}^*(\vec{g},a) \Big)* *\Big( e^{i \vec{q}\cdot \vec r} \hat u(\vec{q}) + e^{-i \vec{q}\cdot \vec r} \hat u^*(\vec{q})\Big)$$ I'll calculate only one integral; the one for the complex coniugate is analogous. The complex exponantials with $\vec k$ cancel out, leaving $$\int_\Omega d\vec r \Big(\sum_{\vec{g'} }e^{-i \vec{g'}\cdot \vec r} \hat w_{\vec{k}}^*(\vec{g'},a) \Big) e^{i \vec{q}\cdot \vec r} \hat u(\vec{q}) \Big(\sum_{\vec{g} }e^{-i \vec{g}\cdot \vec r} \hat w_{\vec{k}}^*(\vec{g},a) \Big)$$ I assume that the integral and the summations converge, so it's legitimate to invert the order between integration and summation, that has a justification, obtaining
$$\sum_{\vec{g'},\vec g } \,\hat w_{\vec{k}}^*(\vec{g'},a) \, \hat w_{\vec{k}}^*(\vec{g},a) \, \hat u(\vec{q})\Big[\int_\Omega d\vec r e^{-i \vec{g'}\cdot \vec r} e^{i \vec{q}\cdot \vec r} e^{-i \vec{g}\cdot \vec r}\Big]$$ The quantity between the brackets, $$\int_\Omega d\vec r e^{-i (\vec{g}-\vec{g'}+\vec{q})\cdot \vec r}$$ is different from 0 only when $\vec{g}-\vec{g'}+\vec{q} = \vec{g''} + \vec q= 0$, but this can happen only when $\vec q$ is a vector of the reciprocal lattice.
For this reason, if we discard the umklapp processes from the calculation, we deny the possibility for $\vec q$ to be a reciprocal vector, and so the above integral vanishes.