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]

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?

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 c/\sqrt{3}

 

[nicola_gao]

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.

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]

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!

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.

 

[Cthugha]

In the 2-dimensional case, a crystal cannot have the "triangular lattice". You probably meant a "parallelogram lattice".

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.