DFT EXERCISES. FELIPE CERVANTES SODI January 2006

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1 DFT EXERCISES FELIPE CERVANTES SODI January Dr. Gábor Csányi 1 Hydrogen atom Place a single H atom in the middle of a largish unit cell (start with a cube of 1 nm on each side), and compute the total energy. Plot the resultant wavefunction (or charge density) along a radial line and compare to the exact answer. How do the total energy and the wavefunction converge with increasing kinetic energy cutoff? Converge your answers with the size of the unit cell. Try a norm conserving pseudopotential. Now turn on spin polarization. How do your answers change? A comparison involving total energy for different cutoff energy values with norm conserving (NC) and ultra soft (US) pseudopotential is drawn in fig [1] below. The energy differences of total energy minus the total energy of the calculation with larger cutoff (1400eV) are plotted in continuous lines. Dotted lines represent the difference of total energy minus eV (Kittel), i.e. the theoretical value of the total energy for a hydrogen atom (the small number in the figure label refers to the length of the cubic unit cell side). Energy convergence Difference of energy (log scale) US 1nm NC 1nm NC 0.6nm US 1nm NC 1nm NC 0.6nm Fig Energy cutoff (ev) US = Ultra soft, NC = Norm conserving. Dotted line > Ei 13.6eV. Continuous line > Ei Ef 1

2 Absolute convergence with an ultra soft pseudopotential is much better than with a norm conserving one. Even the point with higher energy cutoff for a calculation done with norm conserving pseudopotential has a larger difference of total energy than the first point in a calculation with an ultra soft pseudopotential. The convergence with cell size is illustrated in fig. [2]. It shows that the convergence is reached using a unit cell of 1 nm length each side. The blue curve represents the difference of energy of each calculation minus the energy of the calculation with the biggest cell size while the difference of the total energy minus -13.6eV is represented in red Convergence with cell size. H. E = E E f E = E 13.6 ev E (Log scale) Fig Cell length (nm) A comparison between the theoretical charge density, i e ((1/π a 0 3 ) 1/2 exp(-r/a 0 )) 2, and the calculated charge density using 6 different values of energy cutoff with ultra soft pseudopotential in a cubic unit cell of 1 nm per side is shown in the next figure. Faraway from the core, all the calculations performs reasonable good predictions, but near the core all the calculations fails. Charge density obtained with a cutoff energy of 1400eV in the norm conserving pseudopotential has also been included. It presents the best performance in distance regions from the core and better performance than all the US calculations. Difference of charge density(log scale) Difference between calculated charge density (CD) and theoretical CD for different enegy cutoff eV US 1400eV NC eV US 1000eV US eV US 600eV US 400eV US Distance from the core (nm) Fig. 3 US = ultra soft, NC = norm conserving 2

3 2 Hydrogen Molecule Try a H 2 molecule, and a H 2 radical. Find the optimal bond length. Compare results of LDA with a gradient corrected functional (GGA). What is the ionization energy? Converge your answer with all parameters? The optimal bond length using a GGA functional is 0.75Å (Resnick 0.7Å). 5 0 Total Energy vs Distance nm Total Energy (ev) Fig Distance (nm) A comparison of the optimal distance and cutoff energy between GGA and LDA, shows that both energies converge at the same rate. Optimal distance for a GGA calculation is 0.75Å and for LDA is 0.77Å. Fig [5] shows the convergence with different values of energy cutoff using a 1.2nm cell (in the direction of the separation between atoms) for GGA and LDA. 3

4 Difference of energy (log scale) 10 0 Energy convergence cell = 1.2nm, d = 0.075nm, GGA " " " ", LDA Fig Energy cutoff (ev) According to fig [6], a cubic unit cell of 1 nm side length could be considered converged. Fig. 6 Difference of energy (log scale) 10 2 Energy convergence d = 0.075nm, E cutoff = 800eV Size of the lattice length paralel to a line drawn from atom 1 to atom 2. (nm) + H 2 radical. Total energy vs. inter nuclear distance of a H + 2 ion is drawn in fig [7]. The general behavior of the curve is similar to the behavior of the H 2 molecule, but the equilibrium distance and the minimum energy change from -31.7eV and 0.75 Å to -18.3eV and 1.1 Å respectively. From these values, the obtained ionization energy can be calculated as the difference of E H2 - E H2+, i.e. 13.4eV. 4

5 Total energy (ev) Fig. 7 Total energy vs distance, H Distance (nm) Fig [9] is a comparison between charge densities of the hydrogen molecule and of the hydrogen radical in their optimal configuration. The charge density is plotted in the direction of a line that links the respective nuclei. Difference of energy (log scale) Fig Energy convergence (H + 2 ) cell = 1.2nm, d = 0.115nm, GGA Energy cutoff (ev) 600 Charge density vs position (H 2 +, H2 ) Charge Density (e/a 3 ) d = nm H 2 + d = nm H 2 Fig r (0.01 nm) 5

6 3 Chlorine Molecule and radicals Compare the energy levels of the Cl atom with the experimental spectrum. Now consider Cl2, is the convergence with energy cutoff better than for H 2? Try Cl-, and plot its energy as a function of unit cell size. Can you observe something strange? The equilibrium distance between two chlorine atoms is nm Total energy vs distance, Cl Total energy (ev) Fig Distance (nm) The convergence of total energy with energy cutoff is faster in the chlorine molecule than in the hydrogen one (Figs [11] & [12]). Difference of energy (log scale) 10 0 Energy convergence cell = 1.2nm, d = 0.075nm, GGA " " " ", LDA Fig. 11 Energy cutoff (ev) Fig. 12 H 2 Difference of energy (log scale) 10 0 Energy convergence. (Cl 2 ) Energy cutoff (ev) d atm = nm 6

7 When total energy is plotted against the size of unit cell for a Cl + radical, an energy minimum is reached when the length of each side of the cubic unit cell is near to 0.7nm. If the size is smaller the energy presents an abrupt growth, and if the size of the cell is larger than 0.7nm, the energy approaches an asymptote of eV. The form of this curve is similar to the curve of total energy vs. inter-atomic distance for the system of one hydrogen molecule in a largish unit cell Energy convergence. E cutoff = 500eV Total Energy (ev)) Fig size of cell (nm) 7

8 4 Gold Plot the valence orbitals (wave functions) of an isolated gold atom, and identify the orbitals with different symmetry. Are the true orbitals spherically symmetric? compare the results with the relativistic treatment and also spin-orbit coupling. The form of the valence orbitals that arises from the calculations using either a relativistic or non relativistic pseudopotential, could be one of three different symmetries. In fig [14], the first two images on the left are d-orbitals, the third is the s-orbital and the last images correspond to the first conduction p-orbital. There are 5 d-orbitals, they are labeled d x 2 -y2, d xy, d xz, and d yz, all with the same form and distinct orientation and the d z 2 with a different configuration. The s orbital is only one and the p-orbitals are 3 with the same form and different orientation, i.e. 4p z, 4p x, and 4p y. d x 2 -y 2 d z 2 s Fig. 14 Some valence orbitals (top) and first pconduction orbital (bottom) for gold. Test with three different pseudopotentials: Spin-orbit coupling: Ultrasoft potential generated using the setting suggested by Prof. Lee group Relativistic treatment: Qc-tuned Optimized Au Pseudopotential updated by Victor Milman. Non-relativistic treatment: no extra data. 8

9 The resulting isosurfaces of the relativistic and non-relativistic treatments are identical; however, there is a notable difference in the spatial distribution of isosurfaces of same energetic level between the calculations done with the ultrasoft pseudopotential. Near the nucleus the difference is more evident. Fig. 15 Comparison of the orbitals d x 2 -y 2 (bottom) d z 2 (top) for a norm conserving pseudopotential (left) and a ultrasoft pseudopotential (right). 9

10 5 Oxygen Consider the oxygen molecule O 2. Try different spin configurations. Which is the ground state? Think about what singlet and triplet states really mean in a single particle context. The configuration of an oxygen molecule with 12 e -, and total z component of spin = 2 is the ground state of the system. Total z component of # spin # spin Energy electronic spin. up down Ground state: 5 and 7. The singlet and triplet states are the restrictions in the energy eigenfunctions that arise from the exchange force between electrons. In a single particle system there are no interactions between particles because we are dealing with a single particle. The energy levels that would be found in a multi-particle system due to Exclusion Principle and Exchange forces do not exist. The energy states of the particle are restricted only by the potential. 10

11 6 Water Molecule Investigate the H 2 O molecule. Optimize bond lengths and the angle. What are the vibrational frequencies? How well does this match with experiment? Angle and length optimization using a 12x12x12 cubic cell and energy cutoff of 600eV with a norm conserving pseudopotential converges to o. ( o ) 7 Hydrogen Bond Using two water molecules, figure out the binding energy of the hydrogen bond (the binding that occurs between an H atom of the one water molecule and the O atom on the other water molecule). The energy in the hydrogen bond is equal to: H bond = E2 H2O-2E H2O = (2)( ) = E 0.16 ev (Kittel, H bond 0.1eV) Where: E2 H2O = Energy of two bonded water molecules. E H2O = Energy of 1 water molecule. Fig

8 Benzene Compute the ground state of the C 6 H 6 ring molecule, by making a series of SCF calculations, each time allowing one more SCF iteration than before.

12 8 Benzene Compute the ground state of the C 6 H 6 ring molecule, by making a series of SCF calculations, each time allowing one more SCF iteration than before. Plot the individuals orbitals, and observe how they converge with iteration number. Do all orbitals converge uniformly? Find the delocalized orbital that gives rise to the special properties of such aromatic compounds. In the next three pages fig [19,20,21] there are plotted the resultants orbitals of three calculations with different allowance number of SCF iterations. The first block, fig [19], corresponds to a calculation with 1 SCF iteration, the second block, fig [20], arises from a calculation with 2 SCF iterations, and in the last block, fig [21], the converged orbitals with 33 iterations are depicted. From these three blocks, it is clear that orbitals corresponding to lower energy levels converge faster than the outermost ones. For example, the first 3 orbitals have almost the same shape for all the calculations and the 6 th orbitals with only 3 SCF cycles reaches a form very similar to the converged orbital (fig [17]) 1 SCF (6 th orbital) 3 SCF (6 th orbital) 33 SCF (6 th orbital) Fig. 17 The delocalized orbital of the benzene, the ring orbital is the 11 th. Fig

13 a) b) c) d) e) f) g) h) i) j) k) l) m) n) o) Fig

14 a) b) c) d) e) f) g) h) i) j) k) l) m) n) o) Fig

15 a) b) c) d) e) f) g) h) i) j) k) l) m) n) o) Fig

16 9 Silicon Find the equilibrium lattice parameter and bulk modulus of crystalline Si, using an 8 atom cubic unit cell. Converge your answer with the number of k points that you sample the Brillouin zone with. Try shifting the k point mesh so that it symmetrically straddles the origin (but does not include it). Plot the band structure, and compare with experiment. How large is the band gap? Difference of Total Energy (log scale) 10 0 Energy Convergence. Si k point mesh includes (0,0,0) k point mesh does not include (0,0,0) Equilibrium lattice parameters and Bulk modulus: a = Å. B = GPa. Reported value in literature is 5.43 Å and 98.8 GPa respectively (Kittel) Number of k points Fig. 22 In this calculation, using 20 k-points and a cutoff energy of 310eV, crystalline silicon presents a band gap of 0.68 ev (fig down-right). The reported value in literature is 1.1 ev for experimental measurements and Brust et. al. has reported 0.9 ev in their paper from which the next fig [23] was taken. Band Structure. Silicon. 2 2 Energy(eV) Γ Fig. 24 X W L Γ K K Point Fig. 23 (Source Brust et al. PRL 9, 389 (1962)) 16

17 10 Sodium Plot the density of states and the band structure. Compare the dispersion of the semi core 2s state with the tight binding cosine band. 8 Band Structure. Na. 6 Energy(eV) 4 2 Ef 2 4 Γ Fig. 25 H P Γ N k point Fig. 26 (Source: Takao Kotani, et. al. PRB 52, (1995)) From the complete band Band Structure. Na. structure fig [27] it is worthy to 20 remark on the dispersion process 10 depending on each the energy levels of each band. On the one hand, the Ef lower green and blue lines are 10 similar to the lower energy levels of an isolated atom (the blue line has 6-20 fold degeneracy and the green line 30 has 2-fold degeneracy) On the other hand, outermost levels, fig [25], 40 present a wider dispersion and lower 50 agreement with the levels of an isolated atom. These two facts are in 60 Γ H P Γ N agreement with the tight binding Fig. 27 k point method where the Hamiltonian is approximated by the Hamiltonian of an atom plus a corrective function. The corrective function vanishes near the nucleus and grows at long distances from it, while the atomic Hamiltonian behaves in the other way round. Energy(eV) 17

18 Na Crystal Density of States 4 DOS (e/ev) Energy (ev) 0 50 Fig. 28 Energy (ev) total 18

11 Graphite Use variable occupancies to compute the band structure of graphite. Why is it called a semimetal? Find the equilibrium lattice constant along the c axis.

19 11 Graphite Use variable occupancies to compute the band structure of graphite. Why is it called a semimetal? Find the equilibrium lattice constant along the c axis. Compare the answers given by LDA and GGA. The energy in the conduction band edge is slightly lower than in the valence band edge. This small overlap causes a low carrier concentration in graphite, n e = n h = 3 x10 18 /cm 3 (Ashcroft) concentration that is several orders of magnitude lower than the /cm 3 typical of ordinary metals, but still is a good electric conductor. Because of this it is called a semimetal. The equilibrium lattice is constant along the c axis = Å, reported value = Å (Kittel) The band structure for graphite is plotted in fig [29]. There is no difference between the results using GGA or LDA Band Structure. Graphite. GGA LDA 10 Energy(eV) 5 Ef Fig Γ K M Γ M A k point Γ Fig. 30 (Source: Helmut Bross & Walter Alsheimer. Sektion Physik, University of Munich) 19

12 Bonds For each of the following systems, compute the ground state in the bulk phase, look at a slice of the charge density and observe the different bonding characters.

The zone in red and orange is the area with higher charge density where the electrons are clearly localized. C C C Fig. 31 b) Ionic. NaCl.

20 12 Bonds For each of the following systems, compute the ground state in the bulk phase, look at a slice of the charge density and observe the different bonding characters. Note: The small white letter in the density plots represent the position of the respective element. a) Covalent. Diamond. The zone in red and orange is the area with higher charge density where the electrons are clearly localized. C C C Fig. 31 b) Ionic. NaCl. The electronic distribution is high near the atomic cores, especially near the chlorine atom. There is a charge transfer since the electrons of the last occupied level of sodium atoms are more likely to fill the last unoccupied level of the chlorine atoms. Cl Na Fig. 32 c) Mixed Ionic-Covalent. GaAs. The strong localization of electrons near the non metallic element, characteristic of the Ionic bond, is not so clear. 20

because electrons of the outer bands are delocalized

Na Fig. 34 e) Hydrogen bond. Ice. O H O H Fig.

In the next diagram, the Van derwaals bond is between

21 Fig. 33 Ga As d) Metallic bond. Na. The localization of the inner bands appears diffused because electrons of the outer bands are delocalized and form a nearly uniform gas filling all the space. Na Fig. 34 e) Hydrogen bond. Ice. O H O H Fig. 35 H O f) Cubane. Van DerWaals. In the next diagram, the Van derwaals bond is between two H2 atoms of different cubane molecules. This bond is marked with red circles. C H H C Fig

22 13 Carbon Nanotube Compute the band structure of a simple carbon nanotube (e.g. (6,0)) and plot the density of states. Compare with that of graphite and diamond. The metallic behavior of the (6,0) CNT is supported by its band structure Band Structure. CNT (6,0). 6 4 Energy(eV) 2 Ef Γ k point X Fig. 37 Fig. 38 Band Structure & DOS of a (6,0) CNT ( Source: PRL (1994)) Carbon Nanotube CASTEP Density of States (6,0) Diamond DOS (e/ev) 0 18 Density of States (electrons/ev) Graphite Energy (ev) Energy (ev) Fig. 39 The carbon nanotube density of states presents its characteristic van Hove singularities and occupied levels at the Fermi energy. Diamond and graphite present respectively a large band gap and a semimetalic behavior with a very narrow ocuppancy at the Fermi Energy. beta DOS. Diamond (up), graphite (bottom). (Source: Phys. Rev. B 60, (1999)) Fig

23 14 Silicon Take a slab of silicon with an open (100) surface, and optimize the geometry of the atoms. What happens to the dangling bonds on the surface? Converge with the slab thickness and the vacuum size (distance between periodic images perpendicular to the surface) The outermost atomic layer of the structure has free bonds in the original slab. During the optimization those atoms relax and attract each other reaching a stable state in which all the superficial atoms bond in pairs form dimers. Fig [41] shows the differences between relaxed an non relaxed surface. Hydrogen atoms (white atoms) were used to atract the bonds of the bottom Si atoms. Side Front Top Fig. 41 Silicon surface. The optimized slab is indicated by the arrow and is compared with the original one in different views. 23

24 The energy convergence tests for these calculations are depicted in the last two figures. Diff of Energy/No layers (log scale) 10 0 Energy Convergence. Si surface Number of Si Layers Difference of Total Energy (log scale) 10 0 Energy Convergence. Si surface Vacuum gap (A) Fig. 42 Fig

Module 6 : PHYSICS OF SEMICONDUCTOR DEVICES Lecture 32 : Bonding in Solids

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