How to prove Bloch function is periodic in reciprocal lattice?

Question

How to prove Bloch function is periodic in reciprocal lattice?

I saw in some textbooks this formula: $$ \Psi_{\mathbf{k}} (\mathbf{r}) = \sum_{\mathbf{G}} c_{\mathbf{k}+\mathbf{G}}e^{i(\mathbf{k}+\mathbf{G})\cdot \mathbf{r}} $$ which makes the statement of this question obvious. ($\mathbf{G}$ is reciprocal lattice vectors)

But I don't understand this formula. I know $$ \Psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}}u_{\mathbf{k}}(\mathbf{r}) $$ and $u_{\mathbf{k}}(\mathbf{r})$ is periodic function of lattice, therefore can be written in Fourier series: $$ u_{\mathbf{k}}(\mathbf{r}) = \sum_{\mathbf{G}} c_{\mathbf{k},\mathbf{G}}e^{i\mathbf{G}\cdot\mathbf{r}} $$ Now I don't understand why $c_{\mathbf{k},\mathbf{G}}$ can be written as $c_{\mathbf{k}+\mathbf{G}}$ ?

Answer 1

Bloch functions are not necessarily periodic in reciprocal space. By the translation symmetry of the lattice, the wave function $\psi_{nk}(r)$ must satisfy the Bloch condition:

$$ \psi_{nk}(r-R) = e^{-ik\cdot R}\psi_{nk}(r) $$ where $R$ is a lattice vector. Now this is generically satisfied by a function of the form $$\psi_{nk}(r) = e^{ik\cdot r}u_{nk}(r) $$ where $u_{nk}(r-R)=u_{nk}(r)$. But the choice of $u_{nk}(r)$ is not unique. There is a gauge freedom meaning that we can take $u_{nk}(r)\mapsto e^{-iG\cdot r}u_{nk}(r)$ and the new wavefunction will still satisfy the Bloch condition. So does it matter which one we choose?

Well the convention is to choose the so-called periodic gauge condition, i.e. we choose to have the wavefunction $\psi_{nk}$ be periodic in reciprocal space: $\psi_{n,k+G}(r)=\psi_{nk}(r)$. For this to be true, we must choose a $u_{nk}(r)$ which satisfies

$$ u_{n,k+G}(r) = e^{-iG\cdot r} u_{nk}(r) $$

So this is what makes $\psi_{nk}(r)$ periodic in reciprocal space. We do not have to satisfy this condition, but it is conventional and convenient.

Answer 2

because the index of summation only relates to G,you can forget about "k",and also k=G+k(that shows the transnational symmetry). and look here.

Answer 3

Because the reciprocal lattice $G$-periodic, the state with a wave vector $k+G$ describes the same state as that of wave vector $k$. You can therefore reduce your study to the first Brillouin Zone ($-\pi<k\leq\pi$). This means that the coefficient in your Fourier expansion will only depend on where you are within this zone. You can add or subtract as many times $G$ as you like from your $k$ vector, and the result will stay the same. At least in simple descriptions where no further corrections make the Bloch theorem only an approximation.

Answer 4

I'm also not happy with the exposition found in most (solid state physics) textbooks and think one cannot rigorously prove this without group theory. The argument would be the following in a setting with periodic boundary conditions (Born-von Karman) where $\Psi (x + Na) = \Psi (x)$ (for simplicity in 1d):

Using $$ [H, T] = 0~, $$ where the translation operator is defined as $$ T f(x) = f(x + a)~, $$ $k$ labels the $N$ unique solutions $\Psi_k$ that can be distinguished by $T$ and yield $$ T \Psi_k (x) = {\rm e}^{{\rm i} k a} \Psi_k (x) \quad \text{with}\quad k \in \left\{ \frac{2 \pi n}{N a} : n \in \mathbb N^{[0, N)}\right\}~. $$ Now, for any $\Psi_{k'}$ with $k' = k + G$, where $G = 2\pi / a \cdot m$ is an integer multiple of the reciprocal lattice vector $b = 2\pi / a$, we would find $$ T \Psi_{k+G}(x) = {\rm e}^{{\rm i} k a} \Psi_{k+G} (x)~, $$ i.e. $\Psi_{k'}$ yields the same eigenvalues of $T$ as $\Psi_k$ and is therefore not distinguishable from $\Psi_k$ as it belongs to the same irreducible representation. We can therefore define $$ \Psi_{k + G} (x) \equiv \Psi_k (x) \quad\text{for any}\quad G=2\pi/a \cdot m~. $$ All the properties of the Fourier representation of $\Psi_k$ are a consequence of this and not the other way round.

Literature:

• Dresselhaus, Group Theory, Chp. 10.2
• Zee, Group Theory in a Nutshell, Chp. III.1