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When I see the derivation of Bloch functions $\psi_{n\mathbf{k}}(\mathbf{r})=\mathrm{e}^{\mathrm{i}\mathbf{k}\mathbf{r}}u_{n\mathbf{k}}(\mathbf{r})$, the eigenfunctions of electrons in a periodic lattice and their properties, I always see a relation for the orthogonality of Bloch factors $u_{n\mathbf{k}}(\mathbf{r})$ with respect to the band index $n$:

$$\int u_{n\mathbf{k}}(\mathbf{r}) u_{n'\mathbf{k}}(\mathbf{r}) \ \mathrm{d^3}r=\delta_{nn'}$$

This relation always appears with respect to the same $\mathbf{k}$ for both factors. I wonder how this relation would look for different $\mathbf{k}$ values, $\mathbf{k}$ and $\mathbf{k}'$, while the band indices are also different. I.e. how would the general case of the scalar product of two Bloch factors look like?

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For a fixed $\textbf{k}$, a Bloch factor $u_{n\textbf{k}}$ satisfies the following eigenvalue equation: \begin{equation} \frac{(\hat{\textbf{p}} + \hbar \textbf{k})^2}{2m}u_{n\textbf{k}}(\textbf{r}) = E_{n\textbf{k}} u_{n\textbf{k}}(\textbf{r}) \end{equation} subject to a periodic boundary condition over a unit cell. Note that Bloch factors corresponding to different Bloch wave vectors ($\textbf{k}$'s) constitute separate eigensystems. Therefore, it is meaningless to consider an orthonormality relation between Bloch factors unless they have the same $\textbf{k}$. All we can say is that for each $\textbf{k}$ separately, the relation \begin{equation} \int_{\Omega} d^3 \textbf{r}\, u_{n\textbf{k}}^*(\textbf{r})u_{n'\textbf{k}}(\textbf{r}) = \delta_{nn'} \end{equation} holds. ($\Omega$ denotes a unit cell.)

Nevertheless, an orthonormality relation between Bloch wave functions $\psi_{n\textbf{k}}(\textbf{r}) = e^{i\textbf{k}\cdot\textbf{r}}u_{n\textbf{k}}(\textbf{r})$ still exists, and it is defined as an integral over the entire space: \begin{equation} \begin{split} \int d^3\text{r}\, \psi_{n\textbf{k}}^*(\textbf{r})\, \psi_{n'\textbf{k}'}(\textbf{r}) &= \int d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}'}(\textbf{r})\, e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{r}} \\ &= \sum_{\textbf{R}} e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{R}}\int_{\Omega} d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}'}(\textbf{r})\, e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{r}}\\ &=\frac{(2\pi)^3}{V_{\Omega}}\,\delta^{(3)}(\textbf{k}-\textbf{k}') \int_{\Omega} d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}'}(\textbf{r})\, e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{r}}\\ &= \frac{(2\pi)^3}{V_{\Omega}}\,\delta^{(3)}(\textbf{k}-\textbf{k}') \int_{\Omega} d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}}(\textbf{r})\\ &=\frac{(2\pi)^3}{V_{\Omega}}\,\delta_{nn'}\,\delta^{(3)}(\textbf{k}-\textbf{k}'). \end{split} \end{equation} Here, $\textbf{R}$ denotes a lattice vector of the crystal, and $V_{\Omega}$ the volume of each unit cell. Also, the property $u_{n\textbf{k}}(\textbf{r}) = u_{n\textbf{k}}(\textbf{r} + \textbf{R})$ was used to obtain the second line.

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    $\begingroup$ This is correct for $\psi$, but I don’t think it addressed the question, which asks about $u$. $\endgroup$ – Jahan Claes Feb 14 at 16:58
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    $\begingroup$ @JahanClaes My point is that it is generally not an interesting quantity to think about because $u_{n\textbf{k}}$'s for different $\textbf{k}$ constitute separate eigensystems. If one really wants to evaluate the overlap between Bloch factors with different $\textbf{k}$, it can be done, but one could get anything as the answer. $\endgroup$ – higgsss Feb 14 at 17:24
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    $\begingroup$ I disagree. The overlaps of the $u$s at different $k$ are a very interesting quantity. For example, they’ve involved in computing polarization, Wilson loops, and the $Z_2$ invariant. $\endgroup$ – Jahan Claes Feb 14 at 18:22
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    $\begingroup$ @JahanClaes Well, you mean $A_{nn'}(\textbf{k}) = \int d^3 r \,u_{n\textbf{k}}^*(\textbf{r})\nabla_{\textbf{k}}u_{n'\textbf{k}}(\textbf{r})$, or more precisely, gauge-invariant quantities built out of it. I agree that if a simple formula for the overlap between arbitrary Bloch factors existed, it would tell us everything about $A_{nn'}(\textbf{k})$. But I doubt such a formula can be derived because overlaps between Bloch factors are highly dependent on the details of the band structure. $\endgroup$ – higgsss Feb 14 at 18:40
  • $\begingroup$ @JahanClaes That sounds interesting. Can you point me to publications (and preferrably text books) that cover this? $\endgroup$ – HerpDerpington Feb 14 at 20:35
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I don't think there is any similar orthonormality condition on the $u_{n,k}$ for different $k$'s. In the limit when the potential $V(\mathrm{r}) \to 0$, the Bloch wavefunctions are essentially plane waves so that, in $1\mathrm{D}$ and up to a normalization factor: $$u_{n,k}(\mathrm{r}) = \exp \left(i \left( k + (-1)^n \mathrm{sign}(k) \mathrm{ceil} \left(\frac{n}{2} \right)\frac{2 \pi}{a} \right) r\right)$$ (please check the expression, I could easily have made an error as band mapping is a nightmare).

The important result is that, while we still have orthonormality condition at fixed $k$:

$$\int u_{n\mathbf{k}}^*(\mathbf{r}) u_{n'\mathbf{k}}(\mathbf{r}) \ \mathrm{d^3}r=\delta_{nn'},$$

We don't have a similar condition for different $k$. For instance, staying in $1\mathrm{D}$ and looking at the lowest band ($n=0$), we should have something like:

$$\int u_{0\mathbf{k}}^*(\mathbf{r}) u_{0\mathbf{k'}}(\mathbf{r}) \ \mathrm{d^3}r=\mathrm{sinc}\left((k-k')a \right).$$

(But you can also have a non-zero overlap between $u_{n,k}$ and $u_{n',k'}$ for $n \neq n'$ if $k \neq k'$.)

In the other limiting case, when $V(r)$ is infinitely strong, the bands are flat and $u_{n,k}(\mathrm{r}) = 1$ up to a phase and a normalization factor. In this case you will have something looking like:

$$\int u_{n\mathbf{k}}^*(\mathbf{r}) u_{n'\mathbf{k'}}(\mathbf{r}) \ \mathrm{d^3}r=\delta_{nn'}.$$

I hope this helps.

PS : what I've written when integrating two $u_{n,k}(\mathbf{r})$ is still a simplification, because it is possible to add a global phase depending on $\mathbf{k}$ for all $n$ i.e setting $u'_{n,\mathbf{k}}(\mathbf{r}) = \exp (i \theta(\mathbf{k})) u_{n,\mathbf{k}}(\mathbf{r})$ without any physically significant change. This would change the previous results by an extra phase factor of $\exp (i(\theta(\mathbf{k}) - \theta(\mathbf{k'})))$ but this does not change the main argument. If anything, it should convince you that comparing two $u_{n,k}$ with different values of $k$ is not so significant most of the time.

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  • $\begingroup$ Why do you need this complicated expression $k + (-1)^n \mathrm{sign}(k) \mathrm{ceil} \left(\frac{n}{2} \right)$? $\endgroup$ – HerpDerpington Feb 14 at 11:30
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    $\begingroup$ I've just added it for completion, but you don't need it to convince yourself that there are no general expression for the overlap with different $k$'s. You can also convince yourself that the "folding" of a free particle quadratic dispersion into the first Brillouin zone is non-trivial, as depending on the value of $n$ and the sign of $k$, folding occurs either "from the left" or "from the right" (in $1\mathrm{D}$). $\endgroup$ – QuantumApple Feb 14 at 11:45

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