# Orthogonality of Bloch factors in $k$

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?

For a fixed $$\textbf{k}$$, a Bloch factor $$u_{n\textbf{k}}$$ satisfies the following eigenvalue equation: $$$$\frac{(\hat{\textbf{p}} + \hbar \textbf{k})^2}{2m}u_{n\textbf{k}}(\textbf{r}) = E_{n\textbf{k}} u_{n\textbf{k}}(\textbf{r})$$$$ subject to a periodic boundary condition over a unit cell. Note that Bloch factors corresponding to different Bloch wave vectors ($$\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}$$ separately, the relation $$$$\int_{\Omega} d^3 \textbf{r}\, u_{n\textbf{k}}^*(\textbf{r})u_{n'\textbf{k}}(\textbf{r}) = \delta_{nn'}$$$$ holds. ($$\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: $$$$\begin{split} \int d^3\text{r}\, \psi_{n\textbf{k}}^*(\textbf{r})\, \psi_{n'\textbf{k}'}(\textbf{r}) &= \int d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}'}(\textbf{r})\, e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{r}} \\ &= \sum_{\textbf{R}} e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{R}}\int_{\Omega} d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}'}(\textbf{r})\, e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{r}}\\ &=\frac{(2\pi)^3}{V_{\Omega}}\,\delta^{(3)}(\textbf{k}-\textbf{k}') \int_{\Omega} d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}'}(\textbf{r})\, e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{r}}\\ &= \frac{(2\pi)^3}{V_{\Omega}}\,\delta^{(3)}(\textbf{k}-\textbf{k}') \int_{\Omega} d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}}(\textbf{r})\\ &=\frac{(2\pi)^3}{V_{\Omega}}\,\delta_{nn'}\,\delta^{(3)}(\textbf{k}-\textbf{k}'). \end{split}$$$$ Here, $$\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.

• This is correct for $\psi$, but I don’t think it addressed the question, which asks about $u$. – Jahan Claes Feb 14 at 16:58
• @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. – higgsss Feb 14 at 17:24
• 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. – Jahan Claes Feb 14 at 18:22
• @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. – higgsss Feb 14 at 18:40
• @JahanClaes That sounds interesting. Can you point me to publications (and preferrably text books) that cover this? – HerpDerpington Feb 14 at 20:35

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.

• Why do you need this complicated expression $k + (-1)^n \mathrm{sign}(k) \mathrm{ceil} \left(\frac{n}{2} \right)$? – HerpDerpington Feb 14 at 11:30
• 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 $1\mathrm{D}$). – QuantumApple Feb 14 at 11:45