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1. Are there two definitions of TIs, one for usual TIs, and another (wider) including HgX? No, there is only one definition a 3D band topological insulator (TI): the $\mathbb{Z}_{2}$ classification scheme proposed by Fu, Kane, and Mele: Liang Fu, Charles L. Kane, and Eugene J. Mele. “Topological insulators in three dimensions.” Physical Review ...

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This was shown by Konstantin Iakoubovskii and Guy J Adriaenssens 2001 J. Phys.: Condens. Matter 13 6015 doi:10.1088/0953-8984/13/26/316. Their optical absorption experiments show that single substitutional nitrogen centers trap vacancies about eight times more efficiently than the substitutional nitrogen pairs. In my own reasoning, I would say that have to ...

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I think Volker (@vbraun) nailed it in his answer. Continuing where I left off: $$\cdots = {N_\mathrm{cell}\over\Omega_\mathrm{BZ}} \sum_{\mathbf{G}} e^{+i\mathbf{G} \cdot \mathbf{x}} \int_{\Omega_\mathrm{BZ}} \tilde f(\mathbf{G}+\boldsymbol\omega) e^{+i\boldsymbol\omega \cdot \mathbf{x}}\,d^3 \omega =$$ ...

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The laboratory made diamonds are as good as the naturally found ones. It is the same crystal structure. They are not used much as gemstones, (2% of the market) because of the objections of the diamond industry which relies on mined diamonds and dominates the markets. Gem-quality diamonds grown in a lab can be chemically, physically and optically ...

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The Fourier transform of a periodic function has discrete support, so your $\tilde{f}(G+\omega)$ is zero unless $\omega=0$ in your fundamental domain. The regulator needs some care, the crystal volume and the (related) number of cells are infinite. Its probably easier to think of the combination $\tilde{f}(G+\omega) \cdot N/\Omega_{BZ} = \tilde{f}(G)\cdot ... 0 The simplest derivation is probably to take the first equation and substitute into the second: $$F^{-1}[\tilde f(\mathbf{G})] = f(\mathbf{x}) = \sum_{\mathbf{G}} \tilde f(\mathbf{G}) e^{+i\mathbf{G} \cdot \mathbf{x}} =$$$$= \sum_{\mathbf{G}} \left({1\over\Omega_\mathrm{cell}} \int_{\Omega_\mathrm{cell}} f(\mathbf{x'}) ... 1 Let$I \sim \sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R} \int_{V_{UC}} d^3r \Psi^*_{n\vec{k}}\left(\vec r\right) \Psi_{n'\vec{k'}}\left(\vec r\right)$The term$\sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R}$gives you a$\sim \delta(\vec{k} - \vec{k'})$term. Now, you have :$\Psi^*_{n\vec{k}}\left(\vec r\right) ...

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It's because of the integral: $\int d^3r e^{i\vec{r}\cdot\vec{k}} = \delta^{(3)}(\vec{k})$ When you combine the two exponential factors you get (for the first case of $\vec{R}=0$): $\int d^3r e^{i\vec{r}\cdot(\vec{k}-\vec{k}')}$ Which using the above result is just $\delta^{(3)}(\vec{k}-\vec{k}')$. In your last line where you have a Kronecker delta with ...

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In a real-life misshapen blob of metal, strictly speaking the cyclic boundary conditions cannot be applied, since the blob only has a trivial group of spatial symmetries. However, the blob is approximately invariant under lattice translations (with the only mismatch occurring with the extremely small number of atoms at the surface) so it is tempting to ...

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