help, first Brillouin zone and K points

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?

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.

Gokul43201

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 [itex]c/\sqrt{3}[/itex]

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!

Gokul43201

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

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 [itex]\mathbf{e}_1[/itex] and [itex]\mathbf{e}_2[/itex]. Arbitrary lattice sites are linear combinations of these vectors with integer coefficients [itex]\mathbf{e} = n\mathbf{e}_1 + m\mathbf{e}_2[/itex]. So, these sites form a "parallelogram" or a "distorted square" lattice.

The definition of the "reciprocal lattice" formed by [itex]\mathbf{k}[/itex]-vectors is such that

[tex]\exp(i \mathbf{ke}) = 1 [/tex]............(1)

You can find in any solid state theory textbook that vectors [itex]\mathbf{k}[/itex] 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 [itex]\mathbf{e}_1[/itex] and [itex]\mathbf{e}_2[/itex] are very large (presumably infinite), which means that basis vectors of the reciprocal lattice [itex]\mathbf{k}_1[/itex] and [itex]\mathbf{k}_2[/itex] are very small, so the distribution of [itex]\mathbf{k}[/itex]-points is very dense (presumably continuous).

Eugene.

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.