# Reduce integration over crystal to integration over unit cell

I am wondering when I can reduce integrals over a periodic crystal to a an integral over the unit cell. Especially I consider the following two-electron integral $$I=\langle \varphi_i \varphi_j | V | \varphi_a \varphi_b \rangle = \int \text d^3 r_1 \int \text d^3 r_2 \frac{\varphi_i^*(\vec r_1)\varphi_j^*(\vec r_2)\varphi_a(\vec r_1)\varphi_b(\vec r_2)}{|\vec r_1-\vec r_2|}$$ where the integrals go over the whole periodic crystal. Here the $\varphi$'s are one-electron orbitals (e.g. solutions of the Hartree-Fock or Kohn-Sham equations). In Computational Physics the integrals are often evaluated only in one unit cell instead of the whole crystal. But I do not really understand why?

My attempt to explain goes like that: Due to the Bloch's theorem the one-electron orbitals are Bloch waves such that $$\varphi_i(\vec r+\vec R) = \text e^{\text i \vec k_i \vec R}\varphi_i(\vec r)$$ where $\vec R$ is a vector in the Bravais lattice. Hence I can write my integral $I$ as $$I= \sum_{lm} \int_{\Omega_l} \text d^3 r_1 \int_{\Omega_m} \text d^3 r_2 \frac{\varphi_i^*(\vec r_1)\varphi_j^*(\vec r_2)\varphi_a(\vec r_1)\varphi_b(\vec r_2)}{|\vec r_1-\vec r_2|}\\ = \sum_{lm} \int_{\Omega_0} \text d^3 r_1 \int_{\Omega_0} \text d^3 r_2 \frac{\varphi_i^*(\vec r_1+\vec R_l)\varphi_j^*(\vec r_2+\vec R_m)\varphi_a(\vec r_1+\vec R_l)\varphi_b(\vec r_2+\vec R_m)}{|\vec r_1+\vec R_l-\vec r_2+\vec R_m|}\\ = \sum_{lm} \text e^{-\text i (\vec k_i - \vec k _a)\vec R_l} \text e^{-\text i (\vec k_j - \vec k _b)\vec R_m} \int_{\Omega_0} \text d^3 r_1 \int_{\Omega_0} \text d^3 r_2 \frac{\varphi_i^*(\vec r_1)\varphi_j^*(\vec r_2)\varphi_a(\vec r_1)\varphi_b(\vec r_2)}{|\vec r_1+\vec R_l-\vec r_2+\vec R_m|}$$ Here $\Omega_m$ is the $m$th unit cell and $\vec R_m$ is the lattice vector that connects $\Omega_m$ with some fixed chosen unit cell $\Omega_0$. So I almost achieved the integration which goes only over a single unit cell. Yet the denominator is not in the right form. I tried to solve that by using the Fourier transformation of the Coulomb kernel $$\frac{1}{|\vec r|} = \int \frac{\text d^3 G}{(2\pi)^3} \frac{\text e^{\text i \vec G \vec r}}{\vec G ^2}$$ to obtain $$I=\sum_{lm} \text e^{-\text i (\vec k_i - \vec k _a)\vec R_l-\text i (\vec k_j - \vec k _b)\vec R_m}\int \frac{\text d^3 G}{(2\pi)^3} \text e^{\text i(\vec R_l - \vec R_m)\vec G}\\\cdot \int_{\Omega_0} \text d^3 r_1 \int_{\Omega_0} \text d^3 r_2 \varphi_i^*(\vec r_1)\varphi_j^*(\vec r_2)\varphi_a(\vec r_1)\varphi_b(\vec r_2)\frac{\text e^{\text i \vec G (\vec r_1-\vec r_2)}}{\vec G^2}$$ Still the $\text {exp}{(\text i(\vec R_l - \vec R_m)\vec G)}$ term prevents from reaching the goal. If I could argue that $\vec G$ are reciprocal lattice vectors the term is always $1$. In this case I would obtain $$I= \left(\sum_{lm} \text e^{-\text i (\vec k_i - \vec k _a)\vec R_l-\text i (\vec k_j - \vec k _b)\vec R_m}\right)\cdot \int_{\Omega_0} \text d^3 r_1 \int_{\Omega_0} \text d^3 r_2 \frac{\varphi_i^*(\vec r_1)\varphi_j^*(\vec r_2)\varphi_a(\vec r_1)\varphi_b(\vec r_2)}{|\vec r_1-\vec r_2|}$$ which is exactly what I want. But I can't since the integration over $\vec G$ is a continous integration.

Who can help me?

• Very well posed question. – boyfarrell Oct 16 '15 at 16:52
• @thyme: Please participate in the discussion about your question. There exist posted answers; see whether they are appropriately answering your question or further clarification is needed. – AlQuemist Dec 5 '15 at 15:39

I think the problem is not in the calculational steps above, but in the very first expression, the ‘two-electron integral’, which seemingly aims to calculate the expectation value of the Coulomb interaction in terms of some (non-interacting) Bloch wave-functions, $\phi_{n{\bf k}} ({\bf r})$. However, the notation seems to be incorrect, since the wave-functions used are denoted by $\varphi_i(\vec r)$ which does not match with Bloch wave-functions that need a band-index $n$ and a momentum ${\bf k}$, but with the Wannier functions which are some linear superpositions of the Bloch wave-functions. Therefore, the question above is notationally ambiguous, if not incorrect, and should be clarified.
If the original aim has been to calculate the expectation value of the Coulomb interaction in terms of Bloch wave-functions, one should note that the Bloch functions have the following normalization due to their periodicity (up to a phase factor) with respect to a translation by a Bravais lattice vector $\mathbf{R}$: $$\langle \phi_{n{\bf k}} | \phi_{m {\bf q}} \rangle = \frac{1}{v_{c}} \int_{v_{c}} \mathrm{d}\mathbf{r} \, \phi_{n{\bf k}}^\ast ({\bf r}) \, \phi_{m{\bf q}} ({\bf r}) = \delta_{n,m} \, \delta_{{\bf k + G}, {\bf q}} ~,$$ where $n,m$ are band indices and $\mathbf{G}$ is any vector in the reciprocal lattice. Notice that the domain of integration is $v_c$, the volume of the primitive cell, not the whole volume of the lattice. Hence, to obtain the expectation value of some operator (e.g., the Coulomb interaction) in position representation, one should restrict the domain of spatial integrals to the volume of the primitive cell, $v_c$; for example, $$\langle \phi_{n{\bf k}} , \phi_{m{\bf q}} | \hat{V}_{Coulomb} | \phi_{n'{\bf k'}}, \phi_{m'{\bf q'}} \rangle = \\ \frac{1}{v_{c}} \int_{v_{c}} \mathrm{d}\mathbf{r}_1 \; \frac{1}{v_{c}} \int_{v_{c}} \mathrm{d}\mathbf{r}_2 \, \frac{\phi_{n{\bf k}}^\ast ({\bf r}_1) \, \phi_{m{\bf q}}^\ast ({\bf r}_1) \, \phi_{n'{\bf k'}} ({\bf r}_2) \, \phi_{m'{\bf q'}} ({\bf r}_2) }{|\mathbf{r}_1 - \mathbf{r}_2|} ~,$$ in which a constant appearing in the actual form of the Coulomb potential is dropped for brevity.
Note Added: Notice that appearance of the $\frac{1}{v_c}$ factors in front of volume integrals depends on the normalization of the Bloch waves used; that is, one can absorb them in the definition of the Bloch waves.