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In the derivation of Bloch Wave, I encountered a problem. First of all this is the definition of Bloch Wave: $$ \psi_{n\mathbf{k}} (\mathbf{r} ) = e^{i\mathbf{k} \cdot \mathbf{r} } u_{n\mathbf{k}} (\mathbf{r} ) $$ Then we prove its orthogonality: $$ \begin{align} \langle \psi_{m\mathbf{k}'}| \psi_{n\mathbf{k}}\rangle =& \int_{\mathrm{all \\space } } e^{-i\mathbf{k}' \cdot \mathbf{r} } u_{m\mathbf{k}}^* (\mathbf{r} ) e^{i\mathbf{k} \cdot \mathbf{r} } u_{n\mathbf{k}} (\mathbf{r} ) \\ =& \dfrac{(2\pi)^3}{V} \delta(\mathbf{k} - \mathbf{k}') \delta_{nm} \end{align} $$ Where $\delta (i-j) $ is the Dirac function: $$ \int \delta(x) ~ dx = 1 $$ and $\delta_{ij}$ is Kronecker delta: $$ \delta_{ij} = \begin{cases} 1 & i = j \\ 0 &i \ne j \end{cases} $$ May I ask what is the difference between this and the orthogonality of the sdve form, and why there are both the orthogonality of the Dirac delta function and the orthogonality of the Kronecker delta function。


The above question is a simplification of this except that I didn't add the coefficients

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This formula is taken from this article Maximally localized Wannier functions

Zhao Dazhuang
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Where did you get this? It's kind of similar in that, if $i\neq j$, then the inner product is $0$. At the very least they are not orthonormal, because if $i=j$, the inner product isn't $1$.

agaminon
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