Last Modified 17 December 1999
In gold the close packed planes stack in the <111> direction. In a cubic system this is the long diagonal from, say, the lower left front corner of the cube to the top right back corner. In order to study the stacking fault energy, then, we will stack 12 layers of atoms with regular fcc ordering and adjust the primitive vectors of the unit cell so that we can move the interface between these layers and the next twelve layers from fcc to hcp stacking.
In our calculation we'll use the <11(-1)> direction for stacking. Because of the cubic symmetry of the fcc lattice this is entirely equivalent to <111>. Then the primitive lattice vectors of the supercell are:
a1 = ½ a Y + ½ a Z
a2 = ½ a X + ½ a Z
a3 = (4 + q/3) a X + (4 + q/3) a
Y - (4 - 2 q/3) a Z
Note that a1 and a2 are both
perpendicular to the <11(-1)> direction. When
q = 0 then a3 points along the
<11(-1)> direction, and the unit cell is a supercell of the
fcc lattice (ABCABCABCABCABCABCABCABC). When
q = ½ we have the full stacking fault
(ABCABCABCABCBCABCABCABCA). If we continue on to
q = 1 then we have the stacking ABCABCABCABCCABCABCABCAB, i.e., two "C" atoms are stacked
right on top of one another, mimicking a simple hexagonal lattice.
This is obviously an unstable situation.
(Note: we choose 12 layers so that the periodically repeated
stacking faults will be well separated and non-interacting. This
allows us to calculate the "isolated" stacking fault energy.)
Now where do the atoms go? In this calculation we will ignore
relaxation around the surface. Then the atoms in the "primitive"
unit cell will look just as they looked in the pure fcc
calculation, while the q parameter will create the stacking
fault. In that case, we can choose one atom from each <11(-1)>
layer in the primitive cell as part of our basis. One such choice
is
Note that these basis vectors do not move with q (no
relaxation). The periodic boundary conditions, and hence the
stacking fault, are imposed by the primitive lattice vectors of the
supercell. That is, every atom Bi has an infinite
number of images
When q><0 the stacking fault is periodically
repeated along the <11(-1)> direction every time we go from atom
B12(N1,N2,N3,q)
to
B1(N1,N2,N3+1,q).
One more thing: to generate the
proper k-points for this calculation we need the proper space
group file. The cases q = (0,½,1) have
higher symmetries, but for general values of q we need this (Cartesian) space group file. We will also
need a k-point mesh. This can be generated by static, but it is
best to use a pre-generated set which takes advantage of the fact
that the underlying lattice is fcc. We have prepared two
sets:
We are now ready to set up the SKIN file
for each of these calculations.
Previous: Installation
Next: Setting up the SKIN file
Look at other examples.
Get other parameters from the Tight-binding periodic
table. Return to the static Reference
Manual.
B1 = 0 B2 = ½ a X + ½ a Y B3 = a X + a Y B4 = a X + a Y - a Z B5 = 1½ a X + 1½a Y - a
Z B6 = 2 a X + 2 a Y - a Z B7 = 2 a X + 2 a Y - 2 a Z B8 = 2½ a X + 2½ a Y - 2 a
Z B9 = 3 a X + 3 a Y - 2 a Z B10 = 3 a X + 3 a Y - 3 a Z B11 = 3½ a X + 3½ a Y - 3 a
Z B12 = 4 a X + 4 a Y - 3 a Z