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First of all, we set the position of the rotation axis. Let’s consider only a single graphene layer and say the rotation axis passes through a lattice point of the graphene, perpendicular to the layer. We can put this lattice point wherever we want on the graphene, but for simplicity, let’s say the lattice point is in the center of a hexagon at point O (Origin). The graphene plane, lattice vectors (~a1and ~a2), origin O where the rotation axis pass through and other directions and vectors to be explained later are shown in figure 6.2 below. Now, we define a vector ~R1

as in equation 6.3.

Figure 6.2: Unit cell generation figure; O is the origin (The point rotation axis pass through), ~a1 and ~a2 are lattice vectors for graphene, ~R1 and ~R2 are the vectors that are rotated onto p = 0 and p = q lines to generate the necessary unit cells.

R~1 = p1~a1+ q1~a2 (6.1)

where p1 and q1 are positive integers (p1 ≤ q¹); ~a1 and ~a2 are the lattice vectors of graphene. We can define a similar vector ~R^′₁ symmetric with respect to the p = 0 line shown in figure 6.2. Both of the vectors we defined are translation vectors since p₁ and q₁ are integers. Therefore, these vectors define another lattice point of graphene as well. Now, we consider two graphene layers with AA stacking (AA and AB stacking has no difference in the end), i.e. bilayer graphene. Let R~1 be a vector at the top layer and ~R^′₁ be a vector at the bottom layer. If we rotate the top layer (so the ~R1 vector) by some angle θ clockwise, such that R~1 coincides with the p = 0 line and rotate the bottom layer (as well as ~R^′₁ vector) by the same angle θ counter clockwise, the two vectors will just sit on top of each other, therefore, two lattice points will be one over another, just like the lattice points the rotation axis pass through. If p1 and q1 are prime among themselves, there is no other lattice points coinciding with each other on these vectors. Then, with such a rotation, we find a second lattice point of the rotated system. Moreover, these two lattice points are the closest possible ones. Since lattice points of graphene has six fold rotational symmetry, we have similar lines and vectors with p = 0 line (q = 0,p = −q), ~R1 ( ~R^′′₁, ~R^′′′₁) and ~R^′₁ vectors at ±60^o. Therefore, other lattice points are generated as well, which can be easily shown by rotating the original lattice point around the rotation axis ±60^o. After finding the three lattice points on the lines p = 0, p = −q and q = 0 we use these points with origin to make a parallelogram which is the unit cell. Since the two vectors are both rotated in opposite directions by angle θ, the top layer is rotated by 2θ with respect to the bottom layer. Therefore, the rotation angle is said to be 2θ which can be given by

θRotation= 2θ = 2 arctan

√3p1

2q₁ . (6.2)

For every rotation angle Moire pattern occur because in equation 6.2 p1 and q1

can take any value as long as p1 ≤ q¹ being positive integers. Therefore, if one takes a bilayer graphene and rotate the top layer with respect to bottom one, for every angle there will occur Moire pattern. However, geometrical and STM periodicity will differ if p₁ > 1, which will be explained later. In figure 6.3, atomic positions for some rotation angle are shown (the figures do not show the unit cell) to give an idea of how it looks like.

Figure 6.3: Rotated bilayer graphene structures with angle a) 2^o, b) 3^o, c) 4^o, d) 5^o.

Now, if we return to the generation of unit cell, we set the line that ~R1 vector will be rotated onto, as p = 0 line which is an armchair direction. However, instead of p = 0 line, we can choose p = q line which is in zigzag direction and ~R2 vector as the translation vector. And then apply the same procedure to create the unit cell. Will this make a difference? The answer to this question is affirmative. Actually, if the first nearest lattice point to the origin is found by rotating the vector to armchair direction, then, the second nearest lattice point is the one with a vector rotated to the zigzag direction and vice versa. Also, the second nearest lattice point is √

3 times further to the origin than the first one.

Therefore, for some p1 and q1 values, we can find a closer lattice point on p = q line. We made the necessary geometrical calculations and found the relations between these lattice points as follows; let ~R2 to be defined as

R~2 = p2~a1+ q2~a2, (6.3) then, if ~R1 defines the nearest lattice point (so ~R1 defines the second nearest

lattice point),

p1 = p₂− q2

3 , (6.4)

q1 = p₂+ 2q₂

3 ;

and if ~R2 defines the nearest lattice point (so ~R2 defines the second nearest lattice point);

p1 = p2− q² (6.5)

q1 = p2+ 2q2.

Now, lets say ~R₂ defines the nearest lattice point and lets subtract p₁ from q₁ using equation 6.6 to get

q1 − p¹ = 3q2. (6.6)

(6.7) Since q2 is an integer we can only have ~R2 defining the nearest lattice vector when q1−p¹ = 0 (mod3). Therefore, for ²₃ of possible cases ~R1 defines the nearest lattice vector. For the purpose, we are trying to have as small angle as possible (since STM images show those). Therefore, we set p1 = 1. Then, for the values where q₁− 1 = 0 (mod3), i.e. q1 = 4, 7, 10, 13, ... we have ~R₂ defining the nearest lattice point to the origin. For those special cases we will have half the unit cell we would for ~R1 defining the nearest lattice point case. And the angle of rotation will be 2θ = 2 arctan_2q^√3

1.

In this work, the smallest unit cell had 4 carbon atoms, whereas, the greatest had 628 atoms inside. We made single runs, i.e. geometric relaxation was off, only electronic relaxation is done. The distance between the graphene layers is set to be 3.35 ˚A. The system is periodic in plane, whereas, there is a vacuum more than 10 ˚A long perpendicular to the planes.