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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

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    It's not really clear to me what the question here is, but if you're concerned about the $\delta(i-j)$ not actually being a function that could take the values 1 or 0, then see e.g. https://physics.stackexchange.com/q/208596/50583, https://physics.stackexchange.com/q/273423/50583 and their linked questions for various issues with the idea of continuous bases. – ACuriousMind Aug 30 '22 at 08:48
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    Please include the details of the question in the body of the question. You can edit your question. – Mauricio Aug 30 '22 at 08:50
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    Related: https://physics.stackexchange.com/q/89958/2451 and links therein. – Qmechanic Aug 30 '22 at 10:00
  • "If $\langle \psi_i| \psi_j \rangle = \delta (i-j)$ not $\delta_{ij} $, are these two wave functions orthogonal?" YES, they are orthogonal. You have the Dirac delta instead of the Kronecker one when the states are "non-normalizable", see https://physics.stackexchange.com/q/273423/226902 and links therein – Quillo Aug 30 '22 at 14:21

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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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