site symmetry in crystals : basis of the theory and ... · pdf filerobert a. evarestov st....
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Site Symmetry in crystals :Site Symmetry in crystals : basis of the theory and basis of the theory and
applications for applications for the electron and phonon states.the electron and phonon states.
Robert A. Evarestov Robert A. Evarestov
St. Petersburg State University St. Petersburg State University
RUSSIA RUSSIA
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Site symmetry approach establishes symmetry relations between the localized states ( atomic electron states and atomic displacements) and extended molecular or crystalline states( symmetry of molecular or crystalline orbitals and phonons ). The localized states transform according to the irreducible representations (irreps ) of the site symmetry point groups (in crystals- point symmetry groups of Wyckoff positions occupied by atoms ). The extended states transform according to the irreps of the space group of the crystal (point group of the molecule). The extended states are induced from the localized states.
KMnO4 Pnma (62-D2h16) Z=4
4c .m. x 0.25 z 8d 1 x y z K, Mn –4c O-4c,4c,8d( MnO4 )- Mn-Td O-C3v
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G=Td =HMn HO =C3v
Subduced irreps of Td Induced irreps of Td
(correlation table ) Td C3v C3v Td
a1 a1 a1 (z) a1 t2
a2 a2 a2 a2 t1
e e e (x,y) e t1 t2
t1 a2e
t2 (x,y,z) a1e Mechanical Representation a1 + e +t1 +3t2
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Perovskite CaTiOPerovskite CaTiO3 3 Anatase TiO Anatase TiO22Ca 1b (0.5 ,0.5, 0.50.5 ,0.5, 0.5 )
Ti 1a (0 ,0, 0)Ti 1a (0 ,0, 0)O 3d (0.5, 0, 0; 0,0.5,0; 0, 0, 0.5)O 3d (0.5, 0, 0; 0,0.5,0; 0, 0, 0.5)
(abd)= (bac)(abd)= (bac)
Z=1, Pm-3mZ=1, Pm-3m
(221-O(221-Ohh11)) Z=2, I41/amd
141-D4h19
two Ti atoms 2a(0 ,0 ,0; 0, 1/2, 1/4),four oxygen atoms 4e (0, 0, u; 1/2, 0, u −+ 1/2; 0, 1/2, u +1/4;1/2, 1/2, u− +1/2). The anatase structure is defined by three parameters: a, c, u. (a e) =(b e)
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Brillouin ZonesBrillouin Zones
• Simple cubic lattice Tetragonal bc latticeSimple cubic lattice Tetragonal bc lattice
Г(000) R(0.5 0.5 0.5)3X(0.5 0 0 ) 3M(0.5 0.5 0)
Г (000) M (0.5 0.5 0.5)2P(0.25 0.25 0.25) 2X (0 0 0.5 ) 4N (0 0.5 0)
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G(1) a1g 1+ ;G(2) a1u 1-; G(3) a2g 2+; G(4) a2u 2-; G(5) eu 3-; G(6) eg 3+; G(7) t2u 5-; G(8) t2g 5+; G(9) t1u 4-; G(10) t2u 4+
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R(1) a1g 1+ ;R(2) a1u 1-; R(3) a2g 2+; R(4) a2u 2-;
R(5) eu 3-; R(6) eg 3+; R(7) t2u 5-; R(8) t2g 5+;
R(9) t1u 4-; R(10) t2u 4+
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M(1) 1+ ; M(2) 1- ; M(3) 3+ ; M(4) 3- ; M(5) 2+; M(6) 2-
M(7) 4+ ; M(8) 4- ; M(9) 5- ; M(10) 5+
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M(1) 3; M(2) 4; M(3) 1 ; M(4) 2
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N(1) 1+; N(2) 1-; N(3) 2+ ; N(4) 2-
N(1) 1+; N(2) 1-; N(3) 2+ ; N(4) 2-
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N(1) 1+; N(2) 1-; N(3) 2+ ; N(4) 2-
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THANKS FOR ATTENTION!THANKS FOR ATTENTION!