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)

  • $\begingroup$ 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. $\endgroup$ Aug 16, 2022 at 6:14
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    $\begingroup$ Did you see the answer to this previous question physics.stackexchange.com/questions/326127/… ? $\endgroup$ Aug 16, 2022 at 6:16
  • $\begingroup$ @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. $\endgroup$ Aug 16, 2022 at 6:29
  • $\begingroup$ @GiorgioP It's like proving 2y=4x from y=2x, but it doesn't say why x=1. $\endgroup$ Aug 16, 2022 at 6:31
  • $\begingroup$ 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. $\endgroup$ Aug 16, 2022 at 6:33

2 Answers 2


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 $H_\mathbf k u_{n\mathbf k} = E_{n\mathbf k}u_{n\mathbf k}$.

From some comments,

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 $e^{i\mathbf k\cdot \mathbf x}u_{n\mathbf k}(\mathbf x)$.

  • $\begingroup$ Can you show $H_K$ is self-adjoint? It contians a term of $ik$, appears not Hermitian. $\endgroup$
    – ytlu
    Aug 17, 2022 at 6:39
  • $\begingroup$ 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. $\endgroup$ Aug 17, 2022 at 6:41
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    $\begingroup$ @ytlu Note that $-\hbar^2(\nabla +i\mathbf k)^2 = (\hat P+\hbar\mathbf k)^2$. $\endgroup$
    – J. Murray
    Aug 17, 2022 at 12:03
  • $\begingroup$ @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. $\endgroup$
    – J. Murray
    Aug 17, 2022 at 12:09
  • $\begingroup$ @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. $\endgroup$
    – J. Murray
    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 $$ \int d^3\mathbf{r}\psi_{n\mathbf{k}}^*(\mathbf{r})\psi_{m\mathbf{k}}(\mathbf{r})= \int d^3\mathbf{r}u_{n\mathbf{k}}(\mathbf{r})u_{m\mathbf{k}}(\mathbf{r})=\delta_{n,m} $$


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