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The Brillouin Zone (BZ) refers to a region of reciprocal space corresponding to the primitive cell. That is, a Brillouin Zone is a subset of the reciprocal space which contains all the information necessary to describe the crystal.

In the case of a crystal which only has a single type of atom, the procedure for determining the corresponding BZ is simple:

  1. Take the lattice in the real space and convert it to the recirpocal space

A lattice in real space

The same lattice in reciprocal space

  1. Choose any one lattice site as the origin and draw lines connecting it to all of its nearest neighbours Connecting the neighbouring atoms

  2. Draw the perpendicular bisectors of these connecting lines. The area bounded by the bisectors is the first Brillouin Zone Finding the Brillouin Zone

Wikipedia has a very clear illustration of the process.


It is further possible to find the Irreducible part of the Brillouin Zone (IBZ), which is the smallest possible subset of the BZ after reducing it along its symmetries in the same ways as the symmetries present in the point group associated with the lattice.

Irreducuble Brilllouin Zone


The Problem:

The above formulation is clear and well-known. However, I am not very sure about what to do if the crystal has more than one atom. There can be several possibilities:

  1. We could treat the question as purely concerning the lattice and treat all the atoms as if they were identical. Then follow the earlier process.
  2. We could treat the crystal as having a motif consisting of groups of the different atoms. Then, try to obtain a resulting lattice and carry out the standard process on it.
  3. Connect nearest neighbours using sets of either only identical or only dissimilar atoms and continue the earlier process.

Unfortunately, none of these options seems obviously correct. It feels like there should be a clear answer based on the underlying theory, but I have had no luck finding it so far.

What is the correct way to find the Brillouin Zone/ Irreducible Brillouin Zone for a crystal cconsisting of more than one type of atom? (such as the one shown below)

enter image description here

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2 Answers 2

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Lattice and crystal structure are two different things.

A crystal structure is a convolution of a (Bravais) lattice with a basis (an atom or a group of atom).

A lattice is a collection of geometrical points (not atoms) that have the same geometrical properties. Your picture represents a structure with two different kinds of atoms (blue and red), not a lattice. The orientation of the blue points around a red point is different that of red points around a blue point. In this case the lattice point are located at the centre of the hexagons and the basis is constituted of 1 blue atom plus 1 red atom (1/3 of each of the 3 blue atoms at vertexes + 1/3 of each of the 3 red atoms at vertexes).

So your reciprocal lattice is obatined by the direct lattice, independently on the associated basis.

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    $\begingroup$ The lattice points can be actually located anywhere with respect to the atoms, not necessarily at the center of the hexagons. $\endgroup$ Oct 16, 2022 at 18:13
  • $\begingroup$ That's true, but it's also half of the story. In practice it is better to choice the most symmetrical configuration. For example in the figure above the position of the blue and red atoms is defined by a 3-fold rotation axis at the centre of the hexagon. So the better choice for reference lattice is to select the position of this 3-fold axis at (0,0,z), rather than (x,y,z). At the end are the positions of the atoms that are defined by the reference lattice, not the opposite (see the space group table in the International Tables for Crystallography vol A). $\endgroup$
    – gryphys
    Oct 17, 2022 at 6:30
  • $\begingroup$ Indeed, I wasn't disputing that from a practical point of view, in performing calculations, certain choices are better than others. I just wanted to point out that from a theoretical point of view one can arbitrarily choose the location of the lattice. $\endgroup$ Oct 17, 2022 at 6:48
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Option 2 is essentially the correct approach. Just as in the your Bravais lattice example, you begin by writing down the lattice vectors $\mathbf{a}_i$ of the real-space lattice. The lattice vectors specify the distance you have to go for the motif (in the case of your honeycomb lattice it consists of two sites, one blue and one red) to repeat itself. Then determine the reciprocal space lattice vectors $\mathbf{b}_i$, which specify the periodicity of the lattice in $\mathbf{k}$-space and thus the size of the Brillouin zone. In the case of the honeycomb lattice, you'll find a hexagonal reciprocal lattice, for which you can then determine the first Brillouin zone following the approach you describe.

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    $\begingroup$ Yes, you separate the crystal structure from the atom basis. $\endgroup$
    – Jon Custer
    Oct 15, 2022 at 16:50
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    $\begingroup$ Thanks a lot to both of you! Unless I am misunderstanding you, we are supposed to choose any point and find the lattice vectors taking us to all corresponding points as per the symmetry of the crystal. So that the procedure should be something like this. Is this the correct? @JonCuster $\endgroup$
    – user0
    Oct 15, 2022 at 17:45
  • $\begingroup$ @user0 The first Brillouin zone looks good, but can't you reduce the irreducible part further by mirror symmetry about the $a^\star$ vector? $\endgroup$
    – Anyon
    Oct 16, 2022 at 17:08
  • $\begingroup$ Ah, true. What I have done only accounts for the rotational symmetry. Thanks for pointing it out. $\endgroup$
    – user0
    Oct 16, 2022 at 17:46

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