Rep:Mod:HB2515MgORND
Results and discussion:
Structures used:
For the Mgo crystal structure, two different types of unit cells exist: the primitive unit cell which is a rhombohedron with an internal angle of 60 degrees and contains 1 Mg atom and 1 O atom, or a conventional unit cell, which is 4 times the volume of the primitive unit cell, containing 4 Mg atoms and 4 O atoms. The conventional unit cell is face centered cubic (fcc) and better represents the crystal structure, which is why it is more widely used when for describing the lattice of MgO. These two cells both have the same density, which means that inter-converting between them is just a question of multiplying by 4. This is interesting when performing computational calculations, as it is possible to perform the calculations on the primitive unit cell (smaller therefore easier and faster to process) before converting the results to the conventional unit cell.
SCREENSHOT OF A PRIMITIVE AND CONVENTIONAL UNIT CELL OF MGO
Lattice dynamics:
Phonon Dispersion and Density of States
The phonon dispersion plot give a relationship between the frequency and the wavevector k. While computing the phonon dispersion curve, 50 different wavevectors were calcuated giving a plot as follows:
GRAPH OF DISPERSION PLOT.
It is then possible to generate a density of states diagram, by summer over all the different states for a particular frequency, but only one wavevector. This is represented by the following plot with is for a 1x1x1 cubic cell.
DOS STATES PLOT
This gives sharp peaks of the vibrations are at precise frequencies. We should be expecting 6 different normal modes, as mentioned previously. This is what we observe in the previous plot as the peaks come in a 2:2:1:1 ratio. Therefore we can identify that this corresponds to the set of wavevectors at the L point in the dispersion plot by comparing with the number of branches at that point in the dispertsion plot. This can be checked by examining the log file of the output file, it confirms the ratio that was suspected, labelling the accoustic modes give peaks at 286 and 351 cm-1 and 0.333 and the optical modes give peaks at 676 and 806 cm-1 as 0.167. The two accoustic peaks are both doubly degenerate. The log file also reinforces this by examining the frequencies at point L matching the dentsity of states plot at (0.5, 0.5, 0.5).
This is not however a good representation of reality because a crystal if formed of a infinite number of cells next to each other arranged periodically. To increase the grid size, large shrinking factors are selected, which will spead the frequency bands over a larger number of states in the DOS plot whilst increasing the number of wavevectors at which the frequency is calculeted and displayed in the plot. Many different grid sizes were tested, and 40x40x40 was selected giving a smooth cruve, whilst maintaining a realistic time per calculation.
Grid size and free energy
As mentionned before, by summing over all the states in the DOS plot, it is possible to obtain the free energy of the crystal. As the grid size increases, the accuracy of the ernegy also increases. As seen on the graph below, the energy converges to 6dp quite rapidly, and by a shrinking factor of 20 the differences in the values are too small to be reccorded.
| Grid size | Free energy/eV | Accuracy |
|---|---|---|
| 1x1x1 | -40.9303 | 0.003818 |
| 2x2x2 | -40.9266 | 0.000126 |
| 3x3x3 | -40.9264 | -5.1E-05 |
| 4x4x4 | -40.9265 | -3.3E-05 |
| 10x10x10 | -40.9265 | -3E06 |
| 15x15x15 | -40.9265 | -1E06 |
| 20x20x20 | -40.9265 | 0 |
| 30x30x30 | -40.9265 | 0 |
| 40x40x40 | -40.9265 | 0 |
| 50x50x50 | -40.9265 | 0 |
In this exercise the binding energy of the crystal lattice of one unit cells was calculated and found to be -40.9265 eV. Increasing the grid size beyond 40x40x40 only changes the DOS plot a little, and doesnt change the value of the free energy. Therefore the computaional time would be much longer, and with no real benefit. The optismal shrinking factor was therefore selected as 40.
The grid size is dependent on the the size of the reciproqual space, therefore the brouillon zone. As mentionned previously, as a and a* are inversly proportional, if the lattice constant (a) is larger, the reciproqual space constant (a*) will be smaller, and a saller grid size will be enough to acieve the same accuracy. The opposite is also true.
So when looking at a crystal of CaO , which has a lattice parameter of a=4.811 A [1], compared to a=4.21 A [2] for MgO and has the same structure (fcc, high in symmetry), the same grid size would be applicable to CaO. The symmetry is key also, as the progrmme will recognise this and perform less calculation, by just applying the same results by symmetry.
In the case of the Zeolite, the lattice constant a is much larger as the molecule is a large complex framwork. It is found in litterature tu be anywhere between a=24 A and a=26 A [3]. With the same logic as previously applied, a much smaller grid may be use for the zeolite.
Finally in the case of a metal such as Li, a=3.5092 A [4], a larger grid would be required. This smaller lattice constant is the result of mettallic bonding and small Li ions, with very stong and hard charges.
Free energy and lattice contant:
PLOT OF FREE ENERGY
Above in graph A, the Helmholtz free energy is plotted in funtion of the temperature, varying from 0 to 2900 K. The equation relating the free energy and temperature is: A=U-TS. The graph is decreasing as T increases due to the entropy term becoming more dominant as the temperature increases (because -TS). In the graph, at high temperature the line becomes more straight. This would mean that at low temperature, the internal energy U and entropy S are still weighting on the free energy term. But when T is very large, the change in U and S is not important enough compared to the change in temperature, meaning that the graph takes a y=mx+p shape with a negative gradient.
PLOT OF LATTICE CONSTANT
The lattice volume was also reccorded for the same temperature interval, and is plotted above (Graph B). To find this value, simple calculation were performed, converting the volume of the primitive unit cell in the output file to the volume of the conventional unit cell (mulitplying by 4) to give the result. As expected, as the temperature increases, the average volume of the unit cell increases. This is because the atoms are vibrating the a greater extent. As observed above, the unit cell continues increasing in volume, and this is where the methods does not provide a good approximation anymore. At thos high temperature the bonds should breaks, however this is not taken into account in the QHA, which does not take into account anharmonincity.