Figure 5: Crystal structure of GaN 6 Full file at I.4. Hartree-Fock approximation.
Soft Condensed Matter Solutions Download Report ViewCohen University of California, Berkeley Download Report View 491.LouieUniversity of California, Berkeley c Cambridge University Press 2016 1 Full file at AcknowledgementThe authors thank Mr.Meng Wu, together with Mr.Soft Condensed Matter Solutions Full File AtFelipe H. da Jornada, Mr. Fangzhou Zhao and Mr. Ting Cao, for their invaluable help in preparingthis problem solutions manual. Full file at Sec. II.1. Crystal structure of MgB2. The unit cell and Wigner-Seitz cell are displayed in Fig. Mg B1 B2 B1 B1 B2 B2 Top view Wigner-Seitz cell Unit cell Unitcell Wigner-Seitzcell Side view Mg Mg Mg B B B Figure 1: Unit cell and Wigner-Seitz cell of MgB2 If a is the B-B distance, then a1 a2 a 3. We will choose the following convention forthe lattice vectors, a1 ( 32a, 3 2 a, 0) a2 ( 32a, 32 a, 0 )a3 (0, 0, c), (1) in Cartesian coordinates.(b) The unit cell volume is prim (a1 a2) a3 3 3 2 a2c, and the Brillouin zone volume is BZ (2)3 prim. The reciprocal lattice vectors are given by bi 2prim aj ak ijk, whereijk is the Levi-Civita symbol, or b1 2 prim (3 2 ac, 32ac, 0 ) b2 2 prim (3 2 ac,32ac, 0 ) b3 2 prim (0, 0, 3 3 2 a2), (2) in Cartesian coordinates. Note that other reciprocal lattice vectors can be obtained dependingon the orientation of the real-space lattice vectors.The reciprocal lattice vectors are display in Fig. Using the convention from our unit cell,the atoms are at Mg a32, B1 a13 a23, B2 2a13 2a23. Full file at kx ky kz b1 b3 b2 b2 Brilluoinzone Brilluoin zone Reciprocalcell Reciprocalcell Top view Side view Figure 2: Brillouin zone (c) There are 24 point symmetry elements in MgB2 that leave the crystal structure invariant.There are 12 in-plane symmetry elements: 6 rotation operations, including the identity oper-ation: C6, 2C6,, 6C6 I. There are also 6 in-plane mirror symmetries that go along the Batoms or between the B atoms, including, for instance, the mirror symmetry along the x 0and y 0 planes, denoted by x and y, respectively. One can also compose any of these 12in-plane symmetry elements with a mirror operation along the z 0 plane, z. Note that this system has inversion symmetry, which isequivalent to zC2. The irreducible part of the Brillouin zone is sketched below. It can be easily verified that, ifwe apply all the 24 symmetry elements, we will recover the full Brillouin zone. Irr. BZ Irr BZ Top view Side view kx ky kz Figure 3: Irreducible BZ I.2. The GaN crystal. This problem is similar to Problem I.1, except that a here is the lengthof the in-plane lattice vectors. The details of the computation will be omitted. We will choose a different orientation for the unit cell vectors with respect to problem I.1. The 4 Full file at lattice vectors are given by a1 ( 12a, 3 2 a, 0) a2 ( 12a, 32 a, 0 )a3 (0, 0, c), (3) in Cartesian coordinates. Note that, in real space, the Wigner-Seitz cell is a hexagon rotated90 relative to that in problem I.1.The unit cell volume is (a1 a2) a3 3 2 a2c, and the reciprocal lattice vectors are, b1 2a (1, 1 3, 0) b2 2a (1, 1 3, 0) b3 2c (0, 0, 1), (4) in Cartesian coordinates. The Brillouin zone is sketched in Fig. Brilluoinzone Brilluoin zone Top view Side view Figure 4: BZ (b) If we inspect the unit cell of GaN. So, there are 4 atomsin the unit cell. A top view of the structure is displayed in Fig. There are 6 point symmetry elements that do not involve glide or screw axes: 3 mirror planesalong the atoms that form the hexagons, and 3 C3 rotation operations (including identity),performed around the center of the hexagons. Ga has 3 valence electrons, and N has 5 valence electrons. So, its not surprisingthat this material has a bandgap. ![]() Think about the simplest crudemodel that captures the physics of this problem. Imagine that we have two rigid ions with massM 5 Full file at and effective charge Q each, separated by a distance x. Consider that you put a single effectivepoint charge of mass me and charge 2Q between the two ions, which corresponds to the effectivecharge created by the bond. At rest, we know that x a.The electronic contribution to the energy can be written as a sum over kinetic energy and potentialenergy, and we assume that the kinetic energy comes essentially from quantum confinement viathe uncertainty principle px, E(x) T (x) U(x) (5) T (x) p2 2me 2 2mex2(6) U(x) Q2 x 2Q(2Q) x2 Q2 x(7) E(x) 2 mex2 Q 2 x(8) Our model has a free parameter, Q. ![]() This gives rise to a harmonic potential,where the vibrational modes are quantized, Evib. Finally, the rotational energy can be calculated if we assume that the rotational motion is quan-tized. This gives Erot 22 Ma2.So, the ratio of the electronic to vibrational to rotational energy is given by Eel: Evib: Erot 1: meM: meM. The consequence of this finding is that, since M me, we can separate the three degrees of freedom, and calculate the contribution to the vibrational and rotational degrees offreedom as a perturbation to the electronic one. This validates the use of the Born-Oppenheimerapproximation, which treats the ions as fixed particles when solving for the electronic contribution. Figure 5: Crystal structure of GaN 6 Full file at I.4. Hartree-Fock approximation.
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