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 $c_{\mathbf{k}+\mathbf{G}}$ ?
Best Answer
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.