While reading lecture 9, I got confused about the basis for the conventional unit cell of the graphene lattice. The lattice vectors of said cell are given as \bf{a_1} = [\sqrt{3},0]a and \bf{a_2} = [0,3]a, which both make sense; Now the unit cell contains four carbon atoms which need to be specified in the basis. The basis is then given as C(0,0) for the lower left corner atom, C(1/2,1/2) for the middle atom, C(1/2,5/6) for the top-middle atom and, lastly C(1/3,0). However, I can’t figure out to which atom this last basis corresponds. Could anyone clarify this?
Have a look at panel D of the graphene figure in the lecture. Can you identify lattice vector a1? Then you should be able to determine the location of last atom from the basis coordinates.
Well I can see how it could be C(0,1/3), as there is an atom at \frac{1}{3}\bf{a_2}. However I don’t see an atom at \frac{1}{3}\bf{a_1}(?)
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I agree, I also think it should be C(0, 1/3)