Two different forms of Bloch theorem

Question

In condensed matter physics, Bloch theorem is very helpful in describing the band-structure of periodic systems as it breaks down the wavefunction into a plane-wave part and a periodic potential part. However, I noticed that there are two different ways to write down Bloch theorem:

1. $\Psi_k (\vec{r}) = \exp(i\vec{k}\cdot\vec{r})u_k (\vec{r})$
2. $\Psi_k (\vec{r}) = \sum_j \exp(i\vec{k}\cdot\vec R_j)u_k (\vec{r}-\vec{R}_j)$

What is the difference between these two forms? I know it has got to do with spatial position of the unit cells, the position of the origin etc. But I quite can't figure it out.

Answer 1

I believe in your equations $u_k (\vec{r})$ has different meanings in each equation:

• In equation 1, $u_k (\vec{r})$ refers to a periodic function in the Bravais lattice.
• In equation 2, it refers to the wavefunction $\phi(\vec{r})$ around each lattice site (it is written as $\phi(\vec{r}-\vec{R}_j)$ becasue $\phi(\vec{r})$ is centered at the origin).

Now, let's try to see if both forms are equivalent:

$$ \tag{1} \exp(i\vec{k}\cdot\vec{r})u_k (\vec{r}) \equiv^? \sum_j \exp(i\vec{k}\cdot\vec R_j)\phi (\vec{r}-\vec{R}_j)$$

Take the RHS:

$$ \tag{2} \sum_j \exp(i\vec{k}\cdot\vec R_j)\phi (\vec{r}-\vec{R}_j)= e^{i\vec{k}\cdot \vec{r}} \left (e^{-i\vec{k}\cdot \vec{r}} \sum_j e^{i\vec{k}\cdot\vec R_j}\phi (\vec{r}-\vec{R}_j) \right ) $$

Note that I only multiplied by $1=e^{i\vec{k}\cdot \vec{r}} e^{-i\vec{k}\cdot \vec{r}}$, then we must ask: are the terms in parenthesis a periodic function in the Bravais lattice?

Let's use the substitution $\vec{r}\rightarrow\vec{r}+\vec{R}_k$, where $\vec{R}_k$ is an arbitrary lattice vector. Then we have:

$$ e^{-i\vec{k}\cdot(\vec{r}+\vec{R}_k)} \sum_j e^{i\vec{k}\cdot\vec R_j}\phi (\vec{r}+\vec{R}_k-\vec{R}_j) =e^{-i\vec{k} \cdot \vec{r}}\sum_j e^{i\vec{k}\cdot(\vec{R}_j-\vec{R}_k)} \phi(\vec{r}+\vec{R}_k-\vec{R}_j) $$

Now, we can write our question as:

$$ \sum_j e^{i\vec{k}\cdot\vec R_j}\phi (\vec{r}-\vec{R}_j) \equiv^? \sum_j e^{i\vec{k}\cdot(\vec{R}_j-\vec{R}_k)} \phi(\vec{r}+\vec{R}_k-\vec{R}_j) $$

If we have an infinite lattice, then these equations are equivalent, so we know the part in parenthesis in equation (2) is periodic. If instead, we have periodic boundary conditions (for example, a 2D lattice wrapped into a torus), there are some terms in the LHS that don't appear in the RHS and vice-versa. In this case, however, they would cancel out, since these terms would be separated by a distance equivalent to the boundary conditions.

This means your equation 2. can be written as:

$$e^{i\vec{k}\cdot \vec{r}} \left (e^{-i\vec{k}\cdot \vec{r}} \sum_j e^{i\vec{k}\cdot\vec R_j}\phi (\vec{r}-\vec{R}_j) \right ) = e^{i\vec{k}\cdot \vec{r}} u_\vec{k}(\vec{r}) $$

where $u_\vec{k}(\vec{r}) = e^{-i\vec{k}\cdot \vec{r}} \sum_j e^{i\vec{k}\cdot\vec R_j}\phi (\vec{r}-\vec{R}_j)$ is a periodic function in the Bravais lattice.

A great reference is: https://web.science.uu.nl/ITF/teaching/2014/2014vanMiert.pdf