How to model surfaces: The slab model
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How to model surfaces: The slab model
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In nature, crystals are not infinite but finite 3-D objects terminated by surfaces. Many phenomena and processes occur at the interface between a condensed phase and the environment. Modeling surfaces is then of great theoretical and practical interest. A surface can be created by cutting a crystal, which we simulate as an infinite object, through a crystalline plane. Two semi-infinite crystals are then generated containing an infinite number of atoms in the direction orthogonal to the surface, where periodicity is lost. We then need further approximations to be able to treat this problem. We will focus here on the two-dimensional (2-D) slab model.
The Slab Model
The slab model consists of a film formed by a few atomic layers parallel
to the (hkl) crystalline plane of interest. The film, of finite thickness, is limited
by two surface planes, possibly related by symmetry. For sufficiently thick
slabs, this kind of model can provide a faithful description of the ideal surface.
In the following, we will use MgO as a case study.
We take MgO (rock salt structure) as a simple example.| CRYSTAL | Select a 3D periodic structure |
| 0 0 0 | Default options for space group |
| 225 | The space group |
| 4.21 | The lattice parameter in Angstrom |
| 2 | 2 symmetry inequivalent atoms |
| 12 0.0 0.0 0.0 | The Mg atom at the origin |
| 8 0.5 0.5 0.5 | The O atom at 1/2 1/2 1/2 (fractional coordinates) |
In the CRYSTAL output a description of the primitive cell contents and details on the direct space cell in the adopted Cartesian frame are printed, as follows:
CRYSTAL CALCULATION (INPUT ACCORDING TO THE INTERNATIONAL TABLES FOR X-RAY CRYSTALLOGRAPHY) CRYSTAL FAMILY : CUBIC CRYSTAL CLASS (GROTH - 1921) : CUBIC HEXAKISOCTAHEDRAL SPACE GROUP (CENTROSYMMETRIC) : F M 3 M LATTICE PARAMETERS (ANGSTROMS AND DEGREES) - CONVENTIONAL CELL A B C ALPHA BETA GAMMA 4.21000 4.21000 4.21000 90.00000 90.00000 90.00000 NUMBER OF IRREDUCIBLE ATOMS IN THE CONVENTIONAL CELL: 2 INPUT COORDINATES ATOM AT. N. COORDINATES 1 12 0.000000000000E+00 0.000000000000E+00 0.000000000000E+00 2 8 5.000000000000E-01 5.000000000000E-01 5.000000000000E-01 ******************************************************************************* << INFORMATION >>: FROM NOW ON, ALL COORDINATES REFER TO THE PRIMITIVE CELL ******************************************************************************* LATTICE PARAMETERS (ANGSTROMS AND DEGREES) - PRIMITIVE CELL A B C ALPHA BETA GAMMA VOLUME 2.97692 2.97692 2.97692 60.0000 60.0000 60.0000 18.65462 COORDINATES OF THE EQUIVALENT ATOMS (FRACTIONARY UNITS) N. ATOM EQUIV AT. N. X Y Z 1 1 1 12 MG 0.00000000000E+00 0.00000000000E+00 0.00000000000E+00 2 2 1 8 O -5.00000000000E-01 -5.00000000000E-01 -5.00000000000E-01 NUMBER OF SYMMETRY OPERATORS : 48 ****************************************************************************** |
This part of output reports the transformation of the crystallographic cell given in input (Figure 1) into a more convenient primitive unit cell (Figure 2).
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| Figure 1 The Crystallographic cell of MgO | Figure 2 The Primitive Cell of MgO |
Not all crystalline surfaces are physically stable or worthy of investigation. This is specially true for ionic or semi-ionic crystals. For example, in a polar surface consisting of charged layers alternating there is a net dipole moment per unit area, normal to the surface; such a surface is then unstable and can only be prepared with substantial reconstruction or with the adsorption of charged species.
Exercise: See, for instance, the MgO (111) surface.
| SLABCUT | Transform cell to have two lattice vectors in the chosen surface plane and create 2D translational symmetry |
| 1 0 0 | Miller indices of the surface plane |
| 1 3 | Select Layer 1 as the terminating layer and request a slab of 3 layers |
SLABCUT cuts a 2D-periodic slab of the required thickness. If requested, wave function and energy of 2D system are computed. The geometry of the surface layers and the surface unit cell for a slab with a single layer are displayed in the figure below.
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| Figure 3 Four primitive cells of the MgO(100) surface |
CRYSTAL chooses to place the origin of the surface unit cell at the centre of the slab and has rotated the lattice vectors by 45° relative to the bulk crystallographic cell. This produces the minimum number of atoms and maximum symmetry description of the 2D slab - both factors are exploited to reduce the computer resources required.
The calculation of the wave function (if complete input is supplied) will take a few minutes to run. Using the slab directive above and the computational parameters in the file mgo100_3 (extension d12) the energy of the slab will be -823.9327343 Hartree. Note that the tolerances on the integrals (set by the TOLINTEG keyword) have been raised from the default values ( 6 6 6 6 12 ) - a rather accurate calculation is required to converge the surface energy.
Exercise: Construct a MgO (110) 5 layers slab. Compare your result with:
ATOMS IN THE ASYMMETRIC UNIT 6 - ATOMS IN THE UNIT CELL: 10
ATOM X/A Y/B Z(ANGSTROM)
1 T 12 MG -1.110223024625E-16 0.000000000000E+00 2.976919548795E+00
2 T 8 O -1.110223024625E-16 -5.000000000000E-01 2.976919548795E+00
3 T 12 MG -5.000000000000E-01 -5.000000000000E-01 1.488459774398E+00
4 T 8 O -5.000000000000E-01 0.000000000000E+00 1.488459774398E+00
5 T 12 MG 0.000000000000E+00 0.000000000000E+00 0.000000000000E+00
6 T 8 O 0.000000000000E+00 -5.000000000000E-01 0.000000000000E+00
7 F 12 MG -5.000000000000E-01 -5.000000000000E-01 -1.488459774398E+00
8 F 8 O -5.000000000000E-01 1.054843728860E-16 -1.488459774398E+00
9 F 12 MG 1.110223024625E-16 0.000000000000E+00 -2.976919548795E+00
10 F 8 O 1.110223024625E-16 -5.000000000000E-01 -2.976919548795E+00
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Note: by default, the coordinates of atoms in 2-D systems are fractionary in the directions with translational symmetry, and Å in the non periodic direction.
Choosing the surface terminationMgO (100) surface is very simple case, there is 1 distinct layer only. When the structure is more complex, there are different possible terminations of the slab. The keyword SLABINFO, followed (next record) by the crystallographic (Miller) indices of the surface, defines a new 3D cell, with two lattice vectors perpendicular to the \(\left[hkl\right]\mid\) direction. The indices refer to the Bravais lattice of the crystal; the hexagonal lattice is used for the rhombohedral systems, the cubic lattice for cubic systems (non primitive).
The atoms in the new reference cell are re-ordered according to their \(z\mid\) coordinate, in order to recognize the layered structure, parallel to the \(\left(hkl\right)\mid\) plane. The layers of atoms are numbered.
Example: surface \(\left(001\right)\mid\) of corundum (see test 24 of CRYSTAL test cases):
TEST24 - CORUNDUM 001 2 LAYERS (3D-->2D) - SEE TEST04 CRYSTAL 0 0 0 167 4.7602 12.9933 2 13 0. 0. 0.35216 8 0.30624 0. 0.25 SLABINFO 0 0 1 TESTGEOM END |
The above input defines a bulk corundum crystal and the analysis of the layered structure parallel to the surface of crystallographic indices is requested. The program stops after the geometry step (TESTGEOM) and no further input data are required. The following output is obtained:
CRYSTAL CALCULATION
(INPUT ACCORDING TO THE INTERNATIONAL TABLES FOR X-RAY CRYSTALLOGRAPHY)
CRYSTAL FAMILY : HEXAGONAL
CRYSTAL CLASS (GROTH - 1921) : DITRIGONAL SCALENOHEDRAL
SPACE GROUP (CENTROSYMMETRIC) : R -3 C
LATTICE PARAMETERS (ANGSTROMS AND DEGREES) - CONVENTIONAL CELL
A B C ALPHA BETA GAMMA
4.76020 4.76020 12.99330 90.00000 90.00000 120.00000
NUMBER OF IRREDUCIBLE ATOMS IN THE CONVENTIONAL CELL: 2
INPUT COORDINATES
ATOM AT. N. COORDINATES
1 13 0.000000000000E+00 0.000000000000E+00 3.521600000000E-01
2 8 3.062400000000E-01 0.000000000000E+00 2.500000000000E-01
*******************************************************************************
INFORMATION : FROM NOW ON, ALL COORDINATES REFER TO THE PRIMITIVE CELL
*******************************************************************************
LATTICE PARAMETERS (ANGSTROMS AND DEGREES) - PRIMITIVE CELL
A B C ALPHA BETA GAMMA VOLUME
5.12948 5.12948 5.12948 55.29155 55.29155 55.29155 84.992234
COORDINATES OF THE EQUIVALENT ATOMS (FRACTIONAL UNITS)
N. ATOM EQUIV AT. N. X Y Z
1 1 1 13 AL 3.52160000000E-01 3.52160000000E-01 3.52160000000E-01
2 1 2 13 AL 1.47840000000E-01 1.47840000000E-01 1.47840000000E-01
3 1 3 13 AL -3.52160000000E-01 -3.52160000000E-01 -3.52160000000E-01
4 1 4 13 AL -1.47840000000E-01 -1.47840000000E-01 -1.47840000000E-01
5 2 1 8 O -4.43760000000E-01 -5.62400000000E-02 2.50000000000E-01
6 2 2 8 O 2.50000000000E-01 -4.43760000000E-01 -5.62400000000E-02
7 2 3 8 O -5.62400000000E-02 2.50000000000E-01 -4.43760000000E-01
8 2 4 8 O 4.43760000000E-01 5.62400000000E-02 -2.50000000000E-01
9 2 5 8 O -2.50000000000E-01 4.43760000000E-01 5.62400000000E-02
10 2 6 8 O 5.62400000000E-02 -2.50000000000E-01 4.43760000000E-01
NUMBER OF SYMMETRY OPERATORS : 12
*******************************************************************************
* GEOMETRY EDITING - INPUT COORDINATES ARE GIVEN IN ANGSTROM
*******************************************************************************
GEOMETRY NOW FULLY CONSISTENT WITH THE GROUP
*******************************************************************************
* CELL ROTATION
*******************************************************************************
DIRECT-LATTICE FUNDAMENTAL VECTORS
X Y Z
A1 2.7483 0.0000 4.3311
A2 -1.3742 2.3801 4.3311
A3 -1.3742 -2.3801 4.3311
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There are 10 atoms in the primitive cell of bulk corundum. The shape of the cell is modified in order to have two lattice vectors perpendicular to the \(\left[001\right]\mid\) direction. The indices \(\left(001\right)\mid\) given in input refer to the crystallographic cell, they are transformed into the equivalent indices \(\left(111\right)\mid\) referred to the primitive cell:
**********************************************************************
* DEFINITION OF THE NEW LATTICE VECTORS *
* TWO OF WHICH BELONG TO THE SELECTED PLANE *
**********************************************************************
PLANE INDICES = 0 0 1
PRIMITIVE INDICES = 1 1 1
NEW FUNDAMENTAL DIRECT-LATTICE VECTORS B1,B2,B3
B1= 0 A1 -1 A2 1 A3
B2= -1 A1 1 A2 0 A3
B3= 0 A1 0 A2 -1 A3
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The coordinate of the atoms are computed in the new reference system:
UNIT CELL ATOM COORDINATES : LAB AT.NO. CARTESIAN (ANG) CRYSTALLOGRAPHIC 1 13 0.000 -2.748 4.086 0.295680 -0.352160 0.943520 2 13 -2.380 -1.374 2.410 -0.295680 -0.147840 0.556480 3 13 -2.380 1.374 0.245 -0.295680 0.352160 0.056480 4 13 0.000 0.000 1.921 0.295680 0.147840 0.443520 5 8 0.729 1.486 1.083 0.500000 0.443760 0.250000 6 8 0.922 -1.374 1.083 0.193760 -0.250000 0.250000 7 8 -1.651 -0.112 1.083 -0.193760 0.056240 0.250000 8 8 -3.109 -2.860 3.248 -0.500000 -0.443760 0.750000 9 8 -3.302 0.000 3.248 -0.193760 0.250000 0.750000 10 8 -0.729 -1.262 3.248 0.193760 -0.056240 0.750000 |
the sequence of the atoms is reordered according to the \(z\) coordinate. Each value of \(z\) corresponds to a "layer". The top layer contains an Aluminum:
ATOMS CLASSIFIED ACCORDING TO THE Z COORDINATE :
LAYER 1 Z= 4.0865; LABEL AT.NO. X,Y,Z (ANG.)
1 13 0.00000 -2.74830 4.08648
LAYER 2 Z= 3.2483; LABEL AT.NO. X,Y,Z (ANG.)
8 8 1.65122 -2.85999 3.24832
9 8 -3.30244 0.00000 3.24832
10 8 -0.72888 -1.26246 3.24832
LAYER 3 Z= 2.4102; LABEL AT.NO. X,Y,Z (ANG.)
2 13 -2.38010 -1.37415 2.41017
LAYER 4 Z= 1.9209; LABEL AT.NO. X,Y,Z (ANG.)
4 13 0.00000 0.00000 1.92093
LAYER 5 Z= 1.0828; LABEL AT.NO. X,Y,Z (ANG.)
5 8 0.72888 1.48584 1.08278
6 8 0.92234 -1.37415 1.08278
7 8 -1.65122 -0.11169 1.08278
LAYER 6 Z= 0.2446; LABEL AT.NO. X,Y,Z (ANG.)
3 13 -2.38010 1.37415 0.24462
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The repetitive unit contains 5 atoms, Al - 3 Oxygens - Al, in 3 layers.
A model of the surface (001) of corundum could be a slab of 6 layers (2 repetitive units), with Aluminum in the outer layer is::
TEST24 - CORUNDUM 001 2 LAYERS (3D-->2D) - SEE TEST04 CRYSTAL 0 0 0 167 4.7602 12.9933 2 13 0. 0. 0.35216 8 0.30624 0. 0.25 SLABCUT 0 0 1 1 6 the top layer is the layer # 1; the slab is 6 layers thick. END ........ basis set, .... scf input follows |
The figure shows top view and lateral view of 12 layers \(\left(001\right)\mid\) slab of α-alumina parallel to the (001) plane.
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| Alumina slab (001) - lateral view | Alumina slab (001) - top view |
The input of a slab can be also defined starting directly form a 2D structure (see CRYSTAL User's Manual).
The 2D input for the 6 layer slab defined
above (starting from the 3D structure) has the following form:
TEST04 - CORUNDUM 111 (0001) 2 LAYERS- 2D SLAB 66 4.7602 3 13 0. 0. 1.9209 8 0.333333333 -0.027093 1.0828 13 -0.333333333 0.333333333 0.2446 TESTGEOM END |
The gometry analysis part of the output shows the following informations:
TEST04 - CORUNDUM 111 (0001) 2 LAYERS- 2D
SLAB CALCULATION
SYSTEM AND LATTICE : HEXAGONAL
TWO-SIDED PLANE GROUP N. 66 : P -3
CORRESPONDING SPACE GROUP N. 147
LATTICE PARAMETERS (ANGSTROMS AND DEGREES) - CONVENTIONAL CELL
A B GAMMA
4.76020 4.76020 120.00000
NUMBER OF IRREDUCIBLE ATOMS IN THE CONVENTIONAL CELL: 3
INPUT COORDINATES (X AND Y IN FRACTIONARY UNITS, Z IN ANGSTROMS)
ATOM AT. N. COORDINATES
1 13 0.000000000000E+00 0.000000000000E+00 1.920900000000E+00
2 8 3.333333330000E-01 -2.709300000000E-02 1.082800000000E+00
3 13 -3.333333330000E-01 3.333333330000E-01 2.446000000000E-01
LATTICE PARAMETERS (ANGSTROMS AND DEGREES) - PRIMITIVE CELL
A B C ALPHA BETA GAMMA VOLUME
4.76020 4.76020 500.00000 90.0000 90.0000 120.0000 19.62371
COORDINATES OF THE EQUIVALENT ATOMS (X AND Y IN FRACT. UNITS, Z IN ANGSTROMS)
N. ATOM EQUIV AT. N. X Y Z
1 1 1 13 AL 0.00000000000E+00 0.00000000000E+00 1.92090000000E+00
2 1 2 13 AL 0.00000000000E+00 0.00000000000E+00 -1.92090000000E+00
3 2 1 8 O 3.33333333000E-01 -2.70930000000E-02 1.08280000000E+00
4 2 2 8 O 2.70930000000E-02 3.60426333000E-01 1.08280000000E+00
5 2 3 8 O -3.60426333000E-01 -3.33333333000E-01 1.08280000000E+00
6 2 4 8 O -3.33333333000E-01 2.70930000000E-02 -1.08280000000E+00
7 2 5 8 O -2.70930000000E-02 -3.60426333000E-01 -1.08280000000E+00
8 2 6 8 O 3.60426333000E-01 3.33333333000E-01 -1.08280000000E+00
9 3 1 13 AL -3.33333333000E-01 3.33333333000E-01 2.44600000000E-01
10 3 2 13 AL 3.33333333000E-01 -3.33333333000E-01 -2.44600000000E-01
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To model the surface (\(\left(110\right)\mid\), analysis of the layered structure orthogonal to the \(\left[110\right]\mid\) direction:
TEST27 - CORUNDUM - SLAB (3D-->2D) - SEE TEST07 CRYSTAL 0 0 0 167 4.7602 12.9933 2 8 0.30624 0. 0.25 13 0. 0. 0.35216 SLABINFO 1 1 0 TESTGEOM END |
gives:
ATOMS CLASSIFIED ACCORDING TO THE Z COORDINATE :
LAYER 1 Z= 1.6512; LABEL AT.NO. X,Y,Z (ANG.)
1 8 -0.85440 0.67444 1.65122
2 8 -3.91909 -4.15525 1.65122
LAYER 2 Z= 1.4578; LABEL AT.NO. X,Y,Z (ANG.)
3 8 -2.38675 -1.74041 1.45776
LAYER 3 Z= 0.9223; LABEL AT.NO. X,Y,Z (ANG.)
6 8 0.17800 -1.74041 0.92234
LAYER 4 Z= 0.7289; LABEL AT.NO. X,Y,Z (ANG.)
4 8 -1.35435 -4.15525 0.72888
5 8 1.71034 0.67444 0.72888
LAYER 5 Z= 0.0000; LABEL AT.NO. X,Y,Z (ANG.)
7 13 -1.26595 -2.45160 0.00000
8 13 1.62194 -1.02921 0.00000
9 13 1.26595 2.45160 0.00000
10 13 -1.62194 1.02921 0.00000
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Exercise: Model the \(\left(110\right)\mid\) surface of corundum (solution1, solution2).
\[ E_{\text{ns}}=\frac{E(n)-n \cdot E_\text{bulk}}{2\cdot A} \mid \]
where \(A\mid\) is the area of the primitive surface unit cell (8.862 Å2) as reported in the
output) and \(E_\text{bulk}\mid\) is the energy of a single layers
worth of bulk material - in this case the energy of the bulk primitive unit cell which can
be obtained by running mgobulk (extension d12):
-274.6641410 Hartree.
Thus, \(E_{3\text{s}}\mid\) = 3.368 mHa/Å2 = 1.468 J/m-2
Note: 1 Ha/Å2 = 435.9748 J/m-2.
As more layers are used in the calculation \(E_{\text{ns}}\mid\) will converge to the surface
energy if numerical errors in the calculation of \(E(n)\mid\) and \(E_{\text{bulk}}\mid\) cancel exactly.
As CRYSTAL constructs the Hamiltonian matrix in direct space consistently in the SLAB and
CRYSTAL calculations the main source of numerical differences between \(E(n)\mid\) and \(E_{\text{bulk}}\mid\)
is due to k-space sampling.
Exercise: Compute \(E(n)\mid\) and thus \(E_{\text{ns}}\mid\) for n=3, 4, 5, 6, 7... layers and confirm that for MgO the \(\left(100\right)\mid\) surface energy converges to 1.468 Jm-2.
MgO is a wide band gap insulator and the convergence of \(E_{\text{bulk}}\mid\) and \(E(n)\mid\) with
respect to k-space sampling is excellent.
\(E_{\text{ns}}\mid\) converges rapidly and the surface energy is well defined. In metals or
small band gap semiconductors this may not be the case. A more numerically stable
definition of the surface energy is given by:
\[
E_{\text{ns}}=\frac{E(n)-n\cdot\left[E(n) - E(n-1)\right]}{2\cdot A}\mid
\]
In this expression \(E_{\text{bulk}}\mid\) has been replaced by \(E(n) - E(n-1)\mid\). If you consider each
additional layer in the slab to be added as the central layer it is clear that this term should converge
to the energy of a single layer in the bulk crystal. However, in this case the bulk energy has effectively been computed
under exactly the same numerical conditions as the surface layers leading to a cancellation of any
systematic error.
Exercise: Plot \(E(n)-E(n-1)\mid\) as a function of \(n\) and confirm that it converges to
\(E_{\text{bulk}}\mid\)
# MgO(100), Tols 7 7 7 7 14, Basis 861/851 Eb=--2.746641058E+02, confac = 435.9748
# nlayer E(n) Es(n) Es(n) E(n)-E(n-1) Es(n)_method2
hartree Ha/A^2 J/m^2 hartree J/m^2
1 -274.60490128 0.00334234 1.45717757 -274.6049013 0.00000000
2 -549.26856072 0.00336951 1.46902251 -274.6636594 1.44533263
3 -823.93273428 0.00336767 1.46822105 -274.6641736 1.47062542
4 -1098.59687684 0.00336759 1.46818209 -274.6641426 1.46833792
5 -1373.26102035 0.00336744 1.46811979 -274.6641435 1.46843133
6 -1647.92516488 0.00336724 1.46803242 -274.6641445 1.46855660
7 -1922.58931106 0.00336695 1.46790457 -274.6641462 1.46879951
8 -2197.25345917 0.00336655 1.46772903 -274.6641481 1.46913331
9 -2471.91760952 0.00336601 1.46749869 -274.6641503 1.46957179
10 -2746.58176238 0.00336535 1.46720639 -274.6641529 1.47012934
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Many of the interesting properties of surfaces (optical, chemical, electronic, magnetic...) are related to the change in electronic structure which occurs in the near surface region. CRYSTAL provides a variety of tools for analyzing the electronic structure. These will be briefly reviewed here and discussed more extensively in the second tutorial on surfaces.
Charge distributions (Mulliken, charge density plots)
The Mulliken atomic charges are printed at each SCF cycle, and on top of properties
module output. To obtain full Mulliken population analysis insert the keyword PPAN
in SCF input block or in properties input.
Mulliken charges for the 6-layer slab (keyword PPAN):
MULLIKEN POPULATION ANALYSIS - NO. OF ELECTRONS 120.000000
ATOM Z CHARGE SHELL POPULATION
1 MG 12 10.038 2.000 7.986 0.052
2 O 8 9.957 2.007 4.597 3.353
3 MG 12 10.025 2.000 7.985 0.039
4 O 8 9.981 2.007 4.617 3.357
5 MG 12 10.021 2.000 7.986 0.035
6 O 8 9.979 2.007 4.617 3.355
7 MG 12 10.021 2.000 7.986 0.035
8 O 8 9.979 2.007 4.617 3.355
9 MG 12 10.025 2.000 7.985 0.039
10 O 8 9.981 2.007 4.617 3.357
11 MG 12 10.038 2.000 7.986 0.052
12 O 8 9.957 2.007 4.597 3.353
OVERLAP POPULATION CONDENSED TO ATOMS FOR FIRST 6 NEIGHBORS
ATOM A 1 MG ATOM B CELL R(AB)/AU R(AB)/ANG OVPOP(AB)
2 O ( 0 0 0) 3.978 2.105 -0.006
1 MG ( -1 0 0) 5.626 2.977 0.000
4 O ( 1 0 0) 6.890 3.646 0.000
1 MG ( -1 -1 0) 7.956 4.210 0.000
2 O ( -1 0 0) 8.895 4.707 0.000
3 MG ( -1 0 0) 9.744 5.156 0.000
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Measured in this way the bulk crystal is highly ionic, very nearly Mg2+O2-. At the surface the ionicity is very slightly reduced - by about 0.02|e|. The perturbation in the charges is rapidly screened and they converge rapidly reaching bulk values at the third layer.
Note: the slab is symmetric - layer 6 is identical to layer 1. It is often more convenient to study systems with symmetric slabs - even when surface adsorption is being studied.
ECHG computes the charge density in a plane orthogonal to the surface. Run properties with the following input (3 layer slab):
ECHG keyword to compute charge density A----------D 0 order of the charge density derivative | | 65 number of points along the B-A segment | | ATOMS point A, B, C at atomic sites | | 1 0 0 0 point A: atom 1 (Mg top layer) in cell (0 0 0) B----------C 5 0 0 0 point B: atom 5 (Mg bottom layer) in cell (0 0 0) 6 0 0 0 point C: atom 6 (O bottom layer) in cell (0 0 0) Mg MARGINS | 4. 4. 6. 6. margins along AB, CD, AD, BC | END end of definition of the map | END end of properties input Mg---------O |
******************************************************************************* * ATOM X(ANGSTROM) Y(ANGSTROM) Z(ANGSTROM) ******************************************************************************* 1 12 MG -3.310539949135E-16 0.000000000000E+00 2.108500000000E+00 2 8 O 1.490934648132E+00 1.490934648132E+00 2.108500000000E+00 3 12 MG 1.490934648132E+00 1.490934648132E+00 0.000000000000E+00 4 8 O 1.441895329408E-16 1.026461361182E-16 0.000000000000E+00 5 12 MG 3.310069232598E-16 0.000000000000E+00 -2.108500000000E+00 6 8 O 1.490934648132E+00 1.490934648132E+00 -2.108500000000E+00 |
The script runprop executes the program to draw contour lines, and visualize the postscript file generated.
Electrostatic potential
POTM computes the electrostatic potential in a plane orthogonal to the surface with the following input:
POLI multipole calculation 4 0 0 maximum order of the poles POTM electrostatic potential 0 5 see CRYSTAL User's Manual 65 number of points along the B-A segment ATOMS point A, B, C at atomic sites 1 0 0 0 point A: atom 1 (Mg top layer) in cell (0 0 0) 5 0 0 0 point B: atom 5 (Mg bottom layer) in cell (0 0 0) 6 0 0 0 point C: atom 6 (O bottom layer) in cell (0 0 0) MARGINS 4. 4. 6. 6. margins along AB, CD, AD, BC END END end of properties input |
Electrostatic potential is positive on Mg ion and negative on Oxygen.
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| Charge density of 3-layer slab of MgO | Electrostatic potential of 3-layer slab of MgO |
Hint: The electrostatic potential can also be mapped on top of a charge density isosurface, see the "How to create colorcoded 3D maps" tutorial for further details.
If the primitive cell of the surface is defined by the lattice vectors a,
b then a reconstruction introducing
a new periodicity involving two cells in a and three cells in b is called a (2x3) reconstruction.
CRYSTAL provides a facility for creating supercells which can be used to describe any reconstruction which is
commensurate with the underlying lattice.
The new surface lattice vectors (a', b') are generated from the primitive ones by introducing a
SUPERCEL matrix,
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SUPERCEL |
Transform cell to have two lattice vectors in the chosen surface plane |
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2 0 |
S11 and S21 |
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0 3 |
S12 and S22 |
Exercise: Figure out the supercell which corresponds to the bulk crystallographic unit cell terminating at the surface - as shown in the figure below. How does the energy of a three layer slab for this cell compare to that of the primitive surface unit cell?
A supercell of the MgO(100) surface.
| SLABCUT | Transform cell to have two lattice vectors in the chosen surface plane |
| 5 0 1 | Miller indices of the surface plane |
| 1 10 | Select Layer 1 as the terminating layer and request a slab of 10 layers |
The primitive unit cell of this surface is 3.714 Angstrom thick and contains 20 atoms in 10 layers at distance of 0.413 Angstrom. The system has only 4 symmetry operators.
NEW FUNDAMENTAL DIRECT-LATTICE VECTORS B1,B2
X Y Z
B1 4.2100 0.0000 0.0000
B2 0.0000 10.7334 0.0000
AREA OF THE 2D CELL 45.188
COORDINATES OF THE ATOMS BELONGING TO THE SLAB
LAB AT.N. CARTESIAN (ANG) FRAC (2D) Z(ANG)
1 12 1.05250 -1.44489 1.85771 0.250 -0.135 1.85771 layer # 1
2 8 -1.05250 -1.44489 1.85771 -0.250 -0.135 1.85771
3 12 -1.05250 -3.50901 1.44489 -0.250 -0.327 1.44489 layer # 2
4 8 1.05250 -3.50901 1.44489 0.250 -0.327 1.44489
5 12 1.05250 5.16031 1.03206 0.250 0.481 1.03206 layer # 3
6 8 -1.05250 5.16031 1.03206 -0.250 0.481 1.03206
7 12 -1.05250 3.09618 0.61924 -0.250 0.288 0.61924 layer # 4
8 8 1.05250 3.09618 0.61924 0.250 0.288 0.61924
9 12 1.05250 1.03206 0.20641 0.250 0.096 0.20641 layer # 5
10 8 -1.05250 1.03206 0.20641 -0.250 0.096 0.20641
11 12 -1.05250 -1.03206 -0.20641 -0.250 -0.096 -0.20641 layer # 6
12 8 1.05250 -1.03206 -0.20641 0.250 -0.096 -0.20641
13 12 1.05250 -3.09618 -0.61924 0.250 -0.288 -0.61924 layer # 7
14 8 -1.05250 -3.09618 -0.61924 -0.250 -0.288 -0.61924
15 12 -1.05250 -5.16031 -1.03206 -0.250 -0.481 -1.03206 layer # 8
16 8 1.05250 -5.16031 -1.03206 0.250 -0.481 -1.03206
17 12 1.05250 3.50901 -1.44489 0.250 0.327 -1.44489 layer # 9
18 8 -1.05250 3.50901 -1.44489 -0.250 0.327 -1.44489
19 12 -1.05250 1.44489 -1.85771 -0.250 0.135 -1.85771 layer #10
20 8 1.05250 1.44489 -1.85771 0.250 0.135 -1.85771
CHARGE NORMALIZATION FACTOR 1.00000000; TOTAL ATOMIC CHARGES: (last SCF cycle)
10.0620844 9.9289767 10.0472148 9.9609506 10.0412394 9.9588169
10.0411268 9.9598695 10.0454670 9.9542538 10.0454670 9.9542538
10.0411268 9.9598695 10.0412394 9.9588169 10.0472148 9.9609506
10.0620844 9.9289767
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Exercise: Run the (501) 10-layer slab in CRYSTAL. How do the Mulliken charges on the low coordinate sites compare to those on the flat surface ? It should be clear from the charges that this slab is not thick enough to converge the electronic structure or the surface energy.
Exercise: If you have the patience try running a 20 layer version of this
slab. Check the disk space necessary to write mono- and bi- electron integrals
by inserting TESTRUN in the third input block.
If the space occupied by the integrals exceed 2Gb, check if extended files are
allowed by your installation. Otherwise split the integrals in several files
using the keyword BIESPLIT (third input block). If the space is not
available, run in SCF direct mode by inserting the keyword SCFDIR in the
third input block (remove TESTRUN). This is a reasonable model
of the [100] step and a starting point for considering the adsorption of small molecules on the stepped
surface.
Exercise:
Modify the number of layers of the slab MgO (001) and see how
the number of symmetry operators change.
Exercise:
Define a 3-layers slab parallel to the MgO (110) crystallographic plane.
Exercise:
Define a MgO surface with higher indices,
e.g. (103). Such planes can be used to simulate a step on the surface.
Note: play with the number of atomic layers in order to define a suitable
surface model.
We thank Professor N.M. Harrison, who prepared the first version of this tutorial for MSSC2000.
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