Simulations for the topography and FFT of 2D

Chapter 5: Graphene and interlayer interaction

5.2 Single-step Growth of Graphene by Plasma-enhanced Chemical

5.2.3 Simulations for the topography and FFT of 2D

To simulate the FFT of the topography image, we need to first simulate the topography image pixel by pixel in order to find the corresponding functions that can generate the lattice structure that we would like to have.

Triangular lattice:

To generate lattice structures for one dimension is straightforward. We can use a periodic function like sinusoidal function like cos(𝑘𝑘𝑘𝑘). For 2D, the sinusoidal function becomes a plane wave. To create a 2D lattice, we need two plane waves (Figure 5.10 (a)), so that intuitively we would choose the function:

𝑇𝑇𝐶𝐶𝑖𝑖(𝑘𝑘,𝑦𝑦) = cos�𝑲𝑲�����⃗ ∙ 𝒓𝒓�⃗�𝟏𝟏 cos�𝑲𝑲�����⃗ ∙ 𝒓𝒓�⃗�𝟐𝟐 . (5.25)

𝑲𝑲_𝟏𝟏

�����⃗ and 𝑲𝑲�����⃗ are reciprocal lattice constants. _𝟐𝟐

Figure 5.10: (a) Attempt to use two plane waves to create a triangular lattice leads to the structure shown in (b) that lost the C3 symmetry. (c) Using three plane waves to create a triangular lattice shown in (d) leads to a structure satisfying the C3 symmetric. (e) Relationship between the lattice constant 𝑎𝑎 and wavelength of plane wave 𝜆𝜆. (f) the orientation of lattice can be controlled by the orientation of the wave vector.

Unfortunately, the lattice thus created did not have the C3 symmetry as it should be for a triangular lattice. As it turns out, we need three plane waves (Figure 5.10 (c)).

𝑇𝑇𝐶𝐶𝑖𝑖(𝑘𝑘,𝑦𝑦) =𝑇𝑇𝐶𝐶𝑖𝑖(𝒓𝒓�⃗) = cos²�𝑲𝑲���⃗_𝟏𝟏∙ 𝒓𝒓�⃗�cos²�𝑲𝑲���⃗_𝟐𝟐∙ 𝒓𝒓�⃗�cos²��𝑲𝑲���⃗_𝟏𝟏− 𝑲𝑲���⃗_𝟐𝟐� ∙ 𝒓𝒓�⃗�. (5.26)

The lattice generated from Eq. (5.26) indeed revealed the C3 symmetry (Figure 5.10 (d)).

Here the cosine square makes sure that the function value is between 0 and 1. The wavelength will become half of the original value. This will make controlling the size of atoms accessible by taking different order of root. The relationship between the wavelength of plane waves and the lattice constant a is

𝜆𝜆

2 = 𝜆𝜆^′= √3

2 𝑎𝑎 = 2𝜋𝜋

2𝐾𝐾 ⇒ 𝐾𝐾 = 2𝜋𝜋

√3𝑎𝑎. (5.27)

Therefore, to control the rotation of the lattice, we can simply assign 𝑲𝑲���⃗ as

⎩⎨

⎧𝑲𝑲���⃗_𝟏𝟏(𝜃𝜃) = 2𝜋𝜋

√3𝑎𝑎(cos𝜃𝜃, sin𝜃𝜃)

���⃗𝑲𝑲_𝟐𝟐(𝜃𝜃) = 2𝜋𝜋

√3𝑎𝑎(cos(𝜃𝜃+ 60°) , sin(𝜃𝜃+ 60°)). (5.28)

The location of the lattice can simply be controlled by shifting the 𝒓𝒓�⃗ by ��⃗ 𝑹𝑹

𝑇𝑇𝐶𝐶𝑖𝑖 �𝒓𝒓�⃗ − 𝑹𝑹��⃗,𝑲𝑲���⃗_𝟏𝟏(𝜃𝜃),𝑲𝑲���⃗_𝟐𝟐(𝜃𝜃)�. (5.29)

This is the general formula for triangular lattice for any rotation 𝜃𝜃, displacement 𝑹𝑹��⃗, and lattice constant 𝑎𝑎.

The size of the triangular lattice (can be deemed as half-height width) can be controlled by taking different order of root of the triangular lattice functions

𝑇𝑇𝐶𝐶𝑖𝑖(𝑘𝑘,𝑦𝑦)^𝑛𝑛. (5.30)

Because the triangular lattice function is between 0 and 1, for 𝑛𝑛 > 1, atoms displayed in the lattice will become smaller or sharper with increasing n. On the other hand, for 0 <

𝑛𝑛 < 1, atoms displayed will become larger with decreasing n. However, as shown in Figure 5.11, larger order of root of function hugely amplifies the small value of the function and thus amplifies the small feature we do not want to see. In Figure 5.11, the only function with n = 1/2 shows a perfect triangular FFT. For n = 1/6, the FFT becomes hexagons mixed with triangles. For n = 1/16, the lattice points along the three straight lines are amplified.

Figure 5.11: Triangular lattices and the corresponding FFTs generated from the functions shown in Eqs. (5.28) – (5.30). First row: triangular function with different exponent n. For n >

0, if n is smaller, the size of the lattice point is bigger. This helps us control the size of the atoms.

Second row: The corresponding FFTs for the lattice structures shown in the first row.

With the approach outlined above, no matter how big the atoms may be, they will never touch each other. However, to form bonding between atoms like atomically resolved images of STM, we need to make atoms overlap. To do this, we can combine different triangular lattices to form a denser triangular lattice. Given that the size of a real lattice is constant, a denser simulated lattice will correspond to an effectively larger size when we scale the denser lattice back to the real lattice constant, as shown in Figure 5.12. This construction is effectively like putting more atoms into the basis of a unit cell. For a doubly dense lattice, we have put 3 more lattices inside the original lattice. Assume the basis primitive vector 𝒅𝒅

𝒅𝒅_𝟏𝟏 =𝒂𝒂

𝟐𝟐(−sin𝜃𝜃, cos𝜃𝜃) 𝒅𝒅𝟐𝟐= 𝒂𝒂

𝟐𝟐(cos(𝜃𝜃+ 30°) , sin(𝜃𝜃+ 30°)). (5.31)

Figure 5.12: Illustration of the scheme to enlarge the size of atoms in a triangular lattice by making the effective lattice denser. First row from left to right: single lattice, doubly dense lattice, triply dense lattice. For doubly dense lattice, the lattice constant b = a/2; for triply dense lattice, the lattice constant c = a/3. Second row: Effective lattice images by scaling b and c back to a. The size of atoms is enlarged and begins to overlap to nearby atoms. Third row: The FFTs corresponding to the lattices shown in the second row. Even though doubly dense and triply dense lattice shows some large scale patterns, triangular lattice points remain dominant.

The doubly dense function becomes

𝑇𝑇𝐶𝐶𝑖𝑖2(𝜃𝜃) =𝑀𝑀𝑎𝑎𝑘𝑘 �𝑇𝑇𝐶𝐶𝑖𝑖(𝒓𝒓�⃗,𝜃𝜃),𝑇𝑇𝐶𝐶𝑖𝑖�𝒓𝒓�⃗ − 𝒅𝒅��⃗𝟏𝟏,𝜃𝜃�,𝑇𝑇𝐶𝐶𝑖𝑖�𝒓𝒓�⃗ − 𝒅𝒅��⃗𝟐𝟐,𝜃𝜃�,𝑇𝑇𝐶𝐶𝑖𝑖�𝒓𝒓�⃗ − 𝒅𝒅��⃗𝟏𝟏− 𝒅𝒅��⃗𝟐𝟐−,𝜃𝜃��. (5.32)

Notice that we choose the max of all four functions rather than adding them together. This helps us eliminate unwanted small value features and makes the FFT image look clean.

Square lattice:

A square lattice structure is easier to construct. We just need a product of two sinusoidal functions with two perpendicular wave vectors.

𝑆𝑆𝑆𝑆𝑆𝑆(𝑘𝑘,𝑦𝑦) =𝑆𝑆𝑆𝑆𝑆𝑆(𝒓𝒓�⃗) = cos⁴�𝑲𝑲���⃗_𝟏𝟏∙ 𝒓𝒓�⃗�cos⁴�𝑲𝑲���⃗_𝟐𝟐∙ 𝒓𝒓�⃗�. (5.33)

Here we use the fourth-order power to enhance the lattice points.

⎩⎪

⎨

⎪⎧𝑲𝑲���⃗_𝟏𝟏(𝜃𝜃) =𝜋𝜋

𝑎𝑎(cos𝜃𝜃, sin𝜃𝜃) 𝑲𝑲���⃗_𝟐𝟐(𝜃𝜃) =𝜋𝜋

𝑎𝑎(cos(𝜃𝜃+ 90°) , sin(𝜃𝜃+ 90°))

=𝜋𝜋

𝑎𝑎(−sin𝜃𝜃, cos𝜃𝜃).

(5.34)

The size of atoms is also controlled by 𝑆𝑆𝑆𝑆𝑆𝑆(𝑘𝑘,𝑦𝑦)^𝑛𝑛. Again, even though we can make lattice points bigger, its size never exceeds the lattice constant. Therefore, to create overlaps between neighboring atoms, we have to add two square lattices together to make it denser.

The simplest choice is to put the lattice point into the center of the original lattice by choose basis primitive vector 𝒅𝒅_𝟏𝟏

𝒅𝒅𝟏𝟏 =𝒂𝒂

𝟐𝟐(cos𝜃𝜃 −sin𝜃𝜃, sin𝜃𝜃+ cos𝜃𝜃). (5.35) This will generate a doubly dense square lattice or simply an FCC lattice. This choice is convenient because we can simply use the Cu FCC lattice constant directly without changing the formula (Figure 5.13). The doubly dense square lattice function becomes

𝑆𝑆𝑆𝑆𝑆𝑆₂(𝜃𝜃) =𝑀𝑀𝑎𝑎𝑘𝑘 �𝑆𝑆𝑆𝑆𝑆𝑆(𝒓𝒓�⃗,𝜃𝜃),𝑆𝑆𝑆𝑆𝑆𝑆�𝒓𝒓�⃗ − 𝒅𝒅��⃗_𝟏𝟏,𝜃𝜃��. (5.36)

Figure 5.13: Illustration of the scheme to enlarge the size of atoms in a square lattice by making the effective lattice denser. First row from left to right: single square lattice, doubly dense lattice, quadruple dense lattice. For doubly dense lattice, the lattice constant 𝑆𝑆=𝑎𝑎/√2; for quadruple dense lattice, the lattice constant c = a/2. Second row: Effective lattice images by scaling b and c back to a. Third row: The FFTs corresponding to the lattices shown in the second row. The reciprocal lattice size is bigger when the spatial lattice is denser.

From simulations of the triangular lattices and square lattices, we found that lattice points only contribute to the FFT when the size of lattice points is small. If the lattice points overlap to each other too much, the minimum between lattice points will contribute more to the FFT signals. That is, the lattice points themselves become a background.

Hexagonal lattice:

Figure 5.14: Illustration of different approaches to construct a hexagonal lattice. First row: a schematic explanation for the method used to generate hexagon #1, hexagon #2, and hexagon

#3. Hexagon #1 combines two triangular lattices. Hexagon #2 uses a constant background with the deduction of a triangular lattice. Hexagon #3 combines hexagons #1 and #2, so it has both bond-like and atom-like hexagonal structures. Second row: Lattice images based on the methods of construction shown in the first row. Third row: FFTs of the lattice images shown in the second row.

There are two ways to construct a hexagonal lattice. The first is to combine two triangular lattices together, which gives rise to the atom-like hexagonal lattice (hexagon #1 in Figure 5.14). The second is to use a constant background minus triangular lattice, which results in a bond-like hexagonal lattice (hexagon #2 in Figure 5.14). Alternatively, we can simply combine both methods to produce hexagons shown as hexagon #3 in Figure 5.14. Although using a triangular lattice minus another triangular lattice with greater lattice constant can

also generate a hexagonal lattice, this approach will make lattice points more dilute and deviate from the real situation. In the end, hexagon #3 is closer to the real STM topography of graphene, so we will use it for our simulations.