# Two different forms of Bloch theorem

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.

• What are the R_j ? Commented Jun 1, 2020 at 9:36
• R_j locates an atomic site in the crystal lattice. This is the same notation as in the tight binding formulation- Check the mathematical formulation section in the wikipedia article for Tight binding. Commented Jun 1, 2020 at 18:10
• are you sure about the second equation? Does u have a k-dependence? Commented Jun 3, 2020 at 20:49
• This might help : en.wikipedia.org/wiki/Wannier_function. But then u_k in equation 1) differs from the one in 2) Commented Jun 11, 2020 at 12:26
• Are you sure that $u_k(\vec{r})$ is the same in 1. and 2. ?
– Ivan
Commented Aug 14, 2020 at 7:10

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