# Help, first Brillouin zone and K points

1. Aug 5, 2007

### nicola_gao

As it's said, the number of k point in a first Brillouin zone is determined by the number of lattice sites. For exmaple, a 2-d n by m square lattice, its 1st BZ contains m by n k values and I assume these k values are equally separated.

My question is that how the layout of k point in the 1st BZ is determined? I mean, it's easy to think of a square lattice or a cubic structure. What about other shaped lattice, i.e. a triangular lattice?

2. Aug 5, 2007

### meopemuk

The shape of the 1st BZ is determined by the lattice periodicity. Positions of k points in the Brillouin zone are determined by the size and shape of the crystal. Usually, periodic boundary conditions are applied on the crystal boundary, so the crystal as a whole is assumed to form a periodic unit cell. The size of this unit cell is huge, so the distance between k-points is very small. For all practical purposes one can assume that the crystal is infinite, and that actual positions of the k-points have no physical significance, and summations over k-points can be replaced by integrations in the k-space.

Eugene.

3. Aug 6, 2007

### Gokul43201

Staff Emeritus
You construct the 1st BZ in the same way that you construct a Wigner-Seitz primitive cell. For a 2D triangular lattice (in k-space) of spacing c, this will give you a 1st BZ that is a regular hexagon of side $c/\sqrt{3}$

4. Aug 6, 2007

### nicola_gao

I am now constructing a crystal of theoretically finite size. Can I understand as that, for a triangular lattice, the positions of k points exactly form a "triangular lattice" too in the 1st BZ? Thanks a lot!

5. Aug 6, 2007

### Gokul43201

Staff Emeritus
I don't know. I've never thought about finite sized lattices before. I'd have to start from scratch and see what happens.

Last edited: Aug 6, 2007
6. Aug 6, 2007

### meopemuk

In the 2-dimensional case, a crystal cannot have the "triangular lattice". You probably meant a "parallelogram lattice". Each 2D crystal lattice has two basis vectors $\mathbf{e}_1$ and $\mathbf{e}_2$. Arbitrary lattice sites are linear combinations of these vectors with integer coefficients $\mathbf{e} = n\mathbf{e}_1 + m\mathbf{e}_2$. So, these sites form a "parallelogram" or a "distorted square" lattice.

The definition of the "reciprocal lattice" formed by $\mathbf{k}$-vectors is such that

$$\exp(i \mathbf{ke}) = 1$$............(1)

You can find in any solid state theory textbook that vectors $\mathbf{k}$ also form a "parallelogram lattice" whose basis vectors can be easily found by solving eq. (1).

In the case of a crystal model with periodic boundary conditions, basis translation vectors $\mathbf{e}_1$ and $\mathbf{e}_2$ are very large (presumably infinite), which means that basis vectors of the reciprocal lattice $\mathbf{k}_1$ and $\mathbf{k}_2$ are very small, so the distribution of $\mathbf{k}$-points is very dense (presumably continuous).

Eugene.

7. Aug 7, 2007

### Cthugha

The term triangular lattice is quite usual in solid state physics, especially in 2D spin models. There is nothing wrong about using that common term.