# Why are bloch factor orthogonal?

The Bloch wave can be expressed as: $$\psi_{n\mathbf{k}}(\mathbf{r}) = u_{n\mathbf{k}}(\mathbf{r})\,e^{i\mathbf{k}\cdot \mathbf{r}} \tag{A1}$$ In this problem Bloch wave they say that $$u_{n\mathbf{k}}(r)$$ is orthogonal. I would like to ask whether $$u_{n\mathbf{k}}(r)$$ itself can be non-orthogonal, but if the Bloch wave is a set of orthonormal basis, the premise is that $$u_{n\mathbf{k}}(r)$$ must be orthogonal, so we mandate: $$\int_{\mathrm{unit \,cell}} u_{n\mathbf{k}}(\mathbf{r})\,u_{m\mathbf{k}}(\mathbf{r})d\mathbf{r} = \delta_{nm} \tag{A2}$$ $$\delta$$ is the Dirac Function.

Thanks to a commenter for the reminder that in both answers one and two they give the origin of the $$u_{n\mathbf{k}}(\mathbf{r})$$ quadrature and state that this is derived from such an equation: $$\left[\dfrac{(i\hbar\nabla + \hbar\mathbf{k})^2}{2m} + V(\mathbf{r})\right] u_{n\mathbf{k}}(\mathbf{r}) = E_{n\mathbf{k}}u_{n\mathbf{k}}(\mathbf{r}) \tag{B1}$$ I know where this wave equation came from, First use the momentum operator $$\hat{p}=-i\hbar \nabla$$:
\begin{align} \hat{p}\psi_{n\mathbf{k}}(\mathbf{r}) =& e^{i\mathbf{k}\cdot \mathbf{r}}(\hat{p} + \hbar \mathbf{k})u_{n\mathbf{k}}(\mathbf{r}) \\ \hat{p}^2\psi_{n\mathbf{k}}(\mathbf{r}) =& e^{i\mathbf{k}\cdot \mathbf{r}}(\hat{p} + \hbar \mathbf{k})^2u_{n\mathbf{k}}(\mathbf{r}) \end{align}

Substituting this into the Schrodinger equation gives eq(B1), but I don't know how to derive eq(A2) from eq(B1)

• Orthogonality is a relation between two or more functions. You should modify your first formula by adding at least one band index to the function(s) $u$. Moreover, it would also be useful to explicitly state the integration domain in the last formula. Aug 16, 2022 at 6:14
• Did you see the answer to this previous question physics.stackexchange.com/questions/326127/… ? Aug 16, 2022 at 6:16
• @GiorgioP Thanks, I just saw that answer and came to ask. That answer is to deduce the orthogonality of $u_{n\mathbf{r}}(\mathbf{r})$ from the orthogonality of the Bloch wave, to show that the orthogonality of $u_{n\mathbf{r}}(\mathbf{r})$ and the orthogonality of the bloch wave are equivalent, but it does not state that the orthogonality of u(r) is how to get. Aug 16, 2022 at 6:29
• @GiorgioP It's like proving 2y=4x from y=2x, but it doesn't say why x=1. Aug 16, 2022 at 6:31
• They are eigenfunctions of a self-adjoint operator (and as such can be chosen orthonormal), often denoted by $H(k)$. I can't give a reference right now, but this should be discussed in any text book in solid state physics. Aug 16, 2022 at 6:33

Bloch's theorem tells us that the energy eigenvectors of a Hamiltonian with a periodic potential can be written $$\psi_{n\mathbf k}(\mathbf x) = e^{i\mathbf k \cdot \mathbf x} u_{n\mathbf k}(\mathbf x)$$ where $$n\in \mathbb Z$$, $$\mathbf k\in \mathrm{BZ}$$ (the first Brillouin zone), and $$u_{n\mathbf k}(\mathbf x)$$ is periodic with the same periodicity as the lattice. Applying the Hamiltonian operator yields $$\big(H \psi_{n\mathbf k}\big)(\mathbf x)= \left[-\frac{\hbar^2}{2m} \nabla^2 + V(\mathbf x)\right]e^{i\mathbf k \cdot \mathbf x}u_{n\mathbf k}(\mathbf x)$$ $$= e^{i\mathbf k\cdot \mathbf x}\left[-\frac{\hbar^2}{2m}(\nabla + i\mathbf k)^2 + V(\mathbf x) \right]u_{n\mathbf k}(\mathbf x) = E_{n\mathbf k} e^{i\mathbf k\cdot \mathbf x} u_{n\mathbf k}(\mathbf x) \tag{\star}$$ Cancelling the factor $$e^{i\mathbf k \cdot \mathbf x}$$ from both terms in $$(\star)$$ yields that $$u_{n\mathbf k}$$ is a solution of the equation $$H_\mathbf k u_{n\mathbf k} = E_{n\mathbf k} u_{n\mathbf k}$$, where $$H_{\mathbf k} \equiv -\frac{\hbar^2}{2m} (\nabla +i\mathbf k)^2 + V(\mathbf x)$$, defined on the unit cell with periodic boundary conditions.

More concretely, let $$\mathscr u$$ denote the unit cell. Consider the Hilbert space $$L^2(\mathscr u)$$ of square-integrable functions on the unit cell equipped with the standard inner product $$\langle \psi,\phi\rangle := \int_{\mathscr u} \mathrm d^n x \ \overline{\psi(\mathbf x)} \phi(\mathbf x)$$

Further define the Bloch Hamiltonian $$H_\mathbf k$$ to act on the twice-weakly differentiable elements of $$L^2(\mathscr u)$$ with periodic boundary conditions. One can show that $$H_{\mathbf k}$$ is self-adjoint with discrete spectrum, and therefore that one can construct an orthonormal basis $$\{u_{n\mathbf k}\}$$ of solutions to the eigenvalue equation $$H_\mathbf k u_{n\mathbf k} = E_{n\mathbf k}u_{n\mathbf k}$$.

Why do we go and change the relevant Hilbert space here? Put differently, I think that during this procedure, in eq. (2), $$H_k$$ is an operator of $$L^2(V)$$ (sq. int. function on $$V$$, the sample volume, respecting the BvK b.c.) and moreover $$u_{nk}\in L^2(V)$$. Can't we say directly that $$H_k$$ is self-adjoint on this Hilbert space (but ofc. defined only on a dense subspace) and that then we can choose the $$u_{nk}$$ as orthonormal (or orthogonal and normalize it on the unit cell)? Why do you (and all textbooks I've seen) go first to $$L^2(u)$$ for this?

[...]

Why can't we leave the BvK boundary conditions? Through the calculation you've done, we simply find that $$u_{nk}$$ are eigenfunctions of $$H_k$$, which are lattice periodic due to the Bloch theorem anyway and hence also fulfill the BvK boundary conditions. [From another point of view: One can see $$V$$ as a single unit cell of a crystal with volume $$NV$$ and the math would be the same as you did, namely treating $$H_k$$ as an operator over the unit cell $$V$$ of this new crystal.] I don't see a reason why we would need to change boundary conditions to be lattice periodic.

It's worth taking a moment to think about what Bloch's theorem actually says. We begin with the Hamiltonian $$H = -\frac{\hbar^2}{2m} \nabla^2 + V$$ where $$V$$ is a periodic function. We take the Hilbert space to be $$L^2(V)$$, where $$V$$ is the sample volume, and impose periodic boundary conditions on $$V$$. These are the so-called Born-von Karman boundary conditions.

The fact that $$V$$ is periodic means that $$H$$ commutes with a set of translation operators $$T_i$$. As per the usual recipe, since the $$T_i$$'s commute among themselves, we seek to simplify the problem by finding a simultaneous eigenbasis of $$H$$ and the $$T_i$$'s.

One can show without excessive difficulty that if a function $$\psi$$ is an eigenfunction of the $$T_i$$'s, then it must be of the form $$\psi(x) = e^{i\mathbf k \cdot \mathbf x}u(\mathbf x)$$ where $$u$$ is periodic with the same periodicity as the lattice. This identification is not unique, however. If $$\mathbf b$$ is any arbitrary inverse lattice vector, then the function $$\exp[i\mathbf b \cdot \mathbf x]$$ is also periodic on the lattice, and so $$e^{i\mathbf k\cdot \mathbf x}u(\mathbf x) = e^{i\mathbf k \cdot \mathbf x} e^{i \mathbf b \cdot \mathbf x} e^{-i\mathbf b \cdot \mathbf x} u(\mathbf x) = e^{i\tilde{\mathbf k} \cdot \mathbf x} \tilde u(\mathbf x)$$ where $$\tilde{\mathbf k}= \mathbf k + \mathbf b$$. In order to resolve this ambiguity, we restrict $$\mathbf k$$ to the first Brillouin zone, which makes the decomposition of $$\psi$$ into $$e^{i\mathbf k \cdot \mathbf x}u(\mathbf x)$$ unique.

To recap, a simultaneous eigenfunction of all of the $$T_i$$'s can always be uniquely expressed in the form $$e^{i\mathbf k \cdot \mathbf x} u(\mathbf x)$$, where $$\mathbf k$$ lies in the first Brillouin zone and $$u$$ is periodic with the same periodicity as the lattice. Until this point, we've said nothing of the Hamiltonian, but we can plug such a function in to the eigenvalue equation to obtain $$H \psi = E \psi$$ $$\underbrace{\left[-\frac{\hbar^2}{2m}\left(\nabla + i\mathbf k\right)^2 + V(\mathbf x)\right]}_{\equiv H_\mathbf k} u(\mathbf x) = E u(\mathbf x)$$ where we have done some algebra to cancel out the $$e^{i\mathbf k \cdot \mathbf x}$$ from each side.

As a result, we find that a simultaneous eigenvector of the $$T_i$$'s and the Hamiltonian takes the form $$e^{i\mathbf k \cdot \mathbf x} u(\mathbf x)$$ where $$\mathbf k$$ lies in the first Brillouin zone, $$u$$ is periodic with the same periodicity as the lattice, and $$u$$ is an eigenfunction of $$H_\mathbf k$$ as defined above. This is the Bloch theorem.

The point of all this is that $$H$$ has a complicated, continuous spectrum, which is difficult to deal with. Bloch's theorem allows us to choose a $$\mathbf k$$ from the first Brillouin zone and then seek the lattice-periodic eigenfunctions $$u_{n\mathbf k}$$ of $$H_\mathbf k$$, which form a nice discrete set; from there, the corresponding eigenfunction of $$H$$ is $$e^{i\mathbf k\cdot \mathbf x}u_{n\mathbf k}(\mathbf x)$$.

• Can you show $H_K$ is self-adjoint? It contians a term of $ik$, appears not Hermitian.
– ytlu
Aug 17, 2022 at 6:39
• Thank you for your answer, I am not familiar with Hilbert space, I will look at the textbook to help understand your answer. In addition, I would like to ask a question, if $u_{n\mathbf{k}}(\mathbf{r})$ satisfies eq{B1}, it must be orthogonal, without any other assumptions or requirements. Aug 17, 2022 at 6:41
• @ytlu Note that $-\hbar^2(\nabla +i\mathbf k)^2 = (\hat P+\hbar\mathbf k)^2$. Aug 17, 2022 at 12:03
• @ZhaoDazhuang It doesn't mean anything for an individual wavefunction $u_{n\mathbf k}$ to be orthogonal; that's like asking whether some vector in 3D space is perpendicular. What you may ask is whether $u_{n\mathbf k}$ and $u_{m\mathbf k}$ are orthogonal if $n\neq m$. The answer is that if $E_{n\mathbf k}\neq E_{m\mathbf k}$, then $u_{n\mathbf k}$ and $u_{m\mathbf k}$ are guaranteed to be orthogonal. If the energies are the same, then they are not guaranteed to be orthogonal - however, in this case we are free to choose them to be orthogonal if we wish. Aug 17, 2022 at 12:09
• @ZhaoDazhuang To make the latter case more clear, it's like picking a basis for $\mathbb R^2$. The basis can be any two vectors which aren't linearly dependent, so they don't need to be orthogonal, but you can certainly choose an orthogonal basis if you wish. Aug 17, 2022 at 12:11

$$\psi_{n\mathbf{k}}(\mathbf{r})$$ are eigenfunctions of the Hamiltonian and therefore orthogonal, that is $$\int d^3\mathbf{r}\psi_{n\mathbf{k}}^*(\mathbf{r})\psi_{m\mathbf{q}}(\mathbf{r})=\delta_{n,m}\delta_{\mathbf{k},\mathbf{q}}$$ (assuming for simplicity periodic boundary conditions, so that the wave vectors are discrete.) Considering now the case $$\mathbf{q}=\mathbf{k}$$ we have $$\int d^3\mathbf{r}\psi_{n\mathbf{k}}^*(\mathbf{r})\psi_{m\mathbf{k}}(\mathbf{r})= \int d^3\mathbf{r}u_{n\mathbf{k}}(\mathbf{r})u_{m\mathbf{k}}(\mathbf{r})=\delta_{n,m}$$