Why are the K points in the Brillouin zone of graphene inequivalent?

[Newstein]

The Brillouin Zone (BZ) of both graphene and the (111) surface of metals like Ag(111) eihibit a hexagon, but I wonder why the BZ of graphene has two inequivalent ponits K and K', while the K point of BZ of the Ag(111) is equivalent.
Thank you in advance!

 

[DrDu]

In case of Graphite, a primitive elementary cell will contain 2 carbon atoms, while the Ag layer will contain only one. If I remember correctly, the reciprocal and direct lattice are identical in these two cases, so the Brillouin zone also has two K points in case of C and only one in case of Ag.

 

[Newstein]

Thank you, but I remember the BZ of both graphene and Ag(111) is a hexagon, identical to the honeycomb lattice of graphene in real space, while the lattice of Ag(111) in real space is the close packed structure. I want to know is the correlation of lattice in real space and the BZ, or does the inequivalence of K and K' of graphen originate from the two inequivalent sublattices?

 

[DrDu]

Yes, the BZ have the same shape. However, the symmetry groups acting on it are different as Graphene and Ag have different space groups, the group of Graphene being a sub group of the one of Ag. This is due to the lower symmetry of the elementary cell of Graphene as compared to Ag.
I can only make a guess on how to proceed:
Find the little group of the k vector belonging to the K point. Probably it is identical for Graphene and Ag. Then by the orbit stabilizer theorem, the set of equivalent K vectors has to be smaller for Graphene than for Ag.

 

[DrDu]

The Brillouin Zone (BZ) of both graphene and the (111) surface of metals like Ag(111) eihibit a hexagon, but I wonder why the BZ of graphene has two inequivalent ponits K and K', while the K point of BZ of the Ag(111) is equivalent.
Thank you in advance!

I pondered about this problem further and realized that Graphene and Ag(111) have the same (wallpaper) symmetry group. Hence, group theoretically I would expect also the K points to transform alike. While the adjacent K points cannot be mapped upon using only translations by lattice vectors, they can be mapped upon using mirror symmetry of the lattice, so on symmetry grounds, I would consider all K points equivalent, whether in Graphene or Ag.
So in what sense do you expect do you expect the points K to be (in)equivalent in Graphene/Ag?

 

[Newstein]

I pondered about this problem further and realized that Graphene and Ag(111) have the same (wallpaper) symmetry group. Hence, group theoretically I would expect also the K points to transform alike. While the adjacent K points cannot be mapped upon using only translations by lattice vectors, they can be mapped upon using mirror symmetry of the lattice, so on symmetry grounds, I would consider all K points equivalent, whether in Graphene or Ag.
So in what sense do you expect do you expect the points K to be (in)equivalent in Graphene/Ag?

Thank you! I think that what you said is right. Considering only the group theory, all K points are equivalent, whether in graphene or Ag(111). The inequivalence of K and K' in graphene might have its electronic origin. The problem I encountered is a √3 honeycomb reconstruction (structural reconstruction) of graphene, and I want to know how many inequivalent points in the BZ. So I consider first the origin of the inequivalence of the K points of graphene.