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I'm refering to the Paper: PHYSICAL REVIEW B 80, 195310 (2009) "Möbius graphene strip as a topological insulator" Z. L. Guo, Z. R. Gong, H. Dong, and C. P. Sun.

The paper is also available as a preprint version via: http://arxiv.org/abs/0906.1634

When I'm refering to an equation, I'll also use the arxiv-reference (since it's freely available).

I'm refering to Section II: Edge States in Möbius Graphene Strip, Equations (12), (13).

I can show that these linear combinations do indeed satisfy the periodic boundary conditions. What I do not see is why you have to use y>0 and y<0 in the proof? At which stage of the periodicity proof does one need to distinguish between y>0 and y<0?

It's true that is seems to be a natural distinction, because one can cut the moebius strip in the middle and obtain two cylinders as shown in Fig 3.

I'd be more than happy on some advice.

Best regards.

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When the Möbius strip is cut down the middle you don't get two cylinders. See here and here for example.

Fig. 3(b) should be interpreted as two cylinders, each with an extra (and different, thus two cases, $y<0$ and $y>0$) on-site potential that accounts for the twist. After the transformation the field operators obey periodic boundary conditions so we have access to Fourier transformation in the $x$ direction.

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But to be very explicit if you show the periodicity of say Eq (12) step by step. Can you show me at which point you explicitly use y>0 in the calculation? – user9361 May 23 '12 at 8:24
Another way to put the question: Why does one use Eq(12) as the unitary transformation for y>0 and Eq(13) for y<0 and not the other way round? – user9361 May 23 '12 at 9:00
@user9361: I believe you could do this, but it would flip the sign of the on-site potential. Without that flip the result wouldn't be consistent with Eq. (15). – oen May 23 '12 at 18:44
Okay. Yet another two question: 1) Is there some intuitive way on how to come up with the Eq (12) and (13)? 2) The fact that one distinguishes between y>0 and y<0 is not limited to the case of a hexagonal lattice, is it? For say a square lattice it would be the same I assume? – user9361 May 23 '12 at 20:54
By the way I'm always interested in finding people to have discussions on theoretical solid state physics in. I don't know your overall expertise (I'm a grad student). But I'd be more than happy if you would be interested. – user9361 May 23 '12 at 20:59
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