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Wikipedia says, in its page on hexagonal and honeycomb lattices:

The honeycomb lattice is a special case of the hexagonal lattice with a two-atom basis. The centers of the hexagons of a honeycomb form a hexagonal lattice, and the honeycomb lattice can be seen as the union of two offset triangular lattices. In nature, carbon atoms of the two-dimensional material graphene are arranged in a honeycomb lattice.

But graphene does not have a two-atom basis, so why is it getting called a honeycomb lattice?

I'd think it's more accurate to refer to it as a hexagonal lattice.

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Graphene does have a two-atom basis. The point is made graphically very clearly in the Wikipedia page you quote, which includes this figure:

The honeycomb lattice includes two different site types, which are labelled in red and green in the image. In graphene, both of these sites are populated with identical carbon atoms, but that is irrelevant to the periodicity of the lattice structure. In essence, there simply is no way to create a proper unit cell (where "proper" here is equivalent to "shaped as a parallelogram") which only includes one green atom and no red atoms (or vice versa) and which tiles the plane as a Bravais lattice.

You can, of course, cut up the unit cell of the Bravais lattice into two equivalent triangular sectors, and tile the plane with those. But the resulting tiling is not a Bravais lattice, which requires every copy of the unit cell to be obtained from the primal unit cell by a displacement of the form $\mathbf d_{n_1,n_2} = n_1 \mathbf a_1 + n_2 \mathbf a_2$ (and, therefore, without any rotations or reflections or additional symmetry operations).

That said, of course, when you say

I'd think it's more accurate to refer to it as a hexagonal lattice.

all honeycomb lattices are also hexagonal lattices, as you should already be aware of (since it is in the quote in your question). But graphene is in the special case, not in the general hexagonal-lattice category.

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  • $\begingroup$ Hi. Thank you for your answer. So basically two atoms can be "inequivalent" even if they are identical. I have a follow-up question. I have read in some places (here on PhysicsStackExchange also: physics.stackexchange.com/questions/142553/…) that it is technically wrong to say that "The Bravais lattice of graphene is clearly not honeycomb". What are your thoughts on this? $\endgroup$
    – Lost
    Commented Apr 4, 2022 at 15:41
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    $\begingroup$ @Lost Because the honeycomb lattice is not a Bravais lattice. The point is that the two atoms in the basis have different environments: the one on the left has two nearest neighbors above and below to the left, whereas the basis atom on the right has two nearest neighbors and below to the right. $\endgroup$
    – march
    Commented Apr 4, 2022 at 15:55
  • $\begingroup$ @march "In essence, there simply is no way to create a proper unit cell (where "proper" here is equivalent to "shaped as a parallelogram") which only includes one green atom and no red atoms (or vice versa) and which tiles the plane as a Bravais lattice." This statement implies that if we do include red and green atoms we will be able to tile the whole lattice as a Bravais lattice. $\endgroup$
    – Lost
    Commented Apr 4, 2022 at 16:02
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    $\begingroup$ @Lost. Right, and when you do that, you get a hexagonal Bravais lattice. The question is, ignoring the fact that there are atoms there, do the points at which the atoms sit form a Bravais lattice? And the answer is no, they don't. You have to remove one of those points from each of the unit cells shown in this answer, and what's left is a hexagonal lattice. The problem is partly one of terminology. We aren't careful when we say "lattice". Perhaps we should say a honeycomb crystal structure is a hexagonal lattice with a two-atom basis. $\endgroup$
    – march
    Commented Apr 4, 2022 at 16:05
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    $\begingroup$ @Lost The definition is here. It's a set of points that is closed under addition (addition being the vector addition of the position vectors representing those points). Alternatively, it can be defined as the span of a linearly independent set of vectors, but with integer coefficients. In either case, you can show that the honeycomb "lattice" doesn't satisfy these conditions. I think we should stop spamming Emilio Pisanty's inbox now though. $\endgroup$
    – march
    Commented Apr 4, 2022 at 16:18

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