In the post Bloch wave function orthonormality, it is proved that the Bloch wavefunctions $$\Psi_{n\vec{k}}\left(\vec{r}\right) = \frac{1}{\sqrt{V}}e^{i\vec k \vec r}u_{n\vec k}\left(\vec r\right),$$ are orthogonal. But the proof all depends on the factor $u_{n\vec{k}}(\vec{r})$ being orthogonal, which I cannot understand.

Since the Bloch factor $u_{n\vec{k}}(\vec{r})$ is found using the fact that $T_R$ and $H$ commute, hence admitting simultaneous eigenfunctions $u_{n\vec{k}}(\vec{r})$ . I argue that when $H$ is degenerate, even if its has orthogonal eigenstates $|E_n \rangle$, the simultaneous eigenstates $u_{n\vec{k}}(\vec{r})$ constructed out of $|E_n\rangle$ may not be orthogonal.

So, why is the factor $u_{n\vec{k}}(\vec{r})$ orthogonal? Are there other assumptions I did not see?

Note: What I mean by "constructed" can be seen in this proof, more specifically, in the "Proof that commuting observables possess a complete set of common eigenfunctions", second part: "When $A$ has degenerate eigenvalues".


1 Answer 1


Maybe you can think of it the other way around.

Let's impose that the Bloch wavefunctions are orthonormal, i.e. $$\int\Psi^*_{nk}(r)\Psi_{mk'}(r)d^3r=\delta_{n,m}\delta(k-k').$$ Then we can define $u_{nk}$, $$\Psi_{nk}=e^{ikr}u_{nk}(r).$$ Substituting this definition into the orthogonality relation we find \begin{align} \int\Psi^*_{nk}(r)\Psi_{mk'}(r)d^3r & = \int e^{i(k'-k)r}u_{nk}^*(r)u_{mk'}(r)d^3r\\&=\sum_Re^{i(k'-k)R}\int_{UC}u_{nk}^*(r)u_{mk'}(r)d^3r\\&=\frac{(2\pi)^3}{V_{UC}}\delta(k'-k)\int_{UC}u_{nk}^*(r)u_{mk'}(r)d^3r\\&=\frac{(2\pi)^3}{V_{UC}}\delta(k'-k)\int_{UC}u_{nk}^*(r)u_{mk}(r)d^3r \end{align} Which means that


i.e. the $u_{nk}(r)$ are also orthogonal.

  • $\begingroup$ You mean that Bloch waves can be non-orthogonal, it's just that $u(\mathbf{r})$ must be assumed to be orthogonal when we want to use Bloch waves as an orthonormal complete basis for a set of Schrodinger equations. Please do I understand right? $\endgroup$ Aug 15, 2022 at 8:45
  • $\begingroup$ I think I'm saying that we want our Bloch waves to be orthonormal. If we impose that condition, we find an orthogonality relation for the $u(r)$ as a consequence. $\endgroup$
    – aRockStr
    Aug 16, 2022 at 14:33
  • $\begingroup$ Yes, you have proved that the orthogonality of $u_{n\mathbf{k}}(\mathbf{r})$ is equivalent to the orthogonality of $\psi_{n\mathbf{k}}(\mathbf{r})$, but I don't understand how the orthogonality of is$u_{n\mathbf{k}}(\mathbf{r})$ derived. I asked a question physics.stackexchange.com/questions/723252/… , do you know the answer? $\endgroup$ Aug 17, 2022 at 2:09

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