Dear all QE users I am so sorry to ask this question I have been to crystallographic server to extract the coordinate for my k-poiint card for space group Fd-3m. the page is displayed below but I dont know how to extract my k-point from this page i would be grateful if U can explain how I can get the coordinates for my k point. Please I would be grateful for your assistance.
Abolore Musari Dept Of Physics University Of Agriculture, Nigeria. The k-vector types of space group 227 [*F**d*-3*m*] (Table for arithmetic crystal class m -3 mF)Fm-3m-Oh5 (225) to Fd-3c- Oh8(228)Reciprocal space group (Im-3m)*, No.229 Brillouin zone<http://www.cryst.ehu.es/cgi-bin/cryst/programs/nph-kv-list?gnum=227&fig=fm3qmf> k-vector descriptionWyckoff Position ITA description*CDML***Conventional-ITA * *ITA**Coordinates**Label**Primitive* GM0,0,00,0,0a2 m-3m 0,0,0 X1/2,0,1/2 0,1,0b 64/mm.m 0,1/2,0 L1/2,1/2,1/21/2,1/2,1/2c 8.-3m 1/4,1/4,1/4 W 1/2,1/4,3/41/2,1,0d 12-4m.2 1/4,1/2,0 DTu,0,u0,2u,0e 124m.m 0,y,0 : 0 < y < 1/2 LDu,u,uu,u,uf 16.3m x,x,x : 0 < x < 1/4 V1/2,u,1/2+u2u,1,0g 24mm2..x,1/2,0 : 0 < x < 1/4 SMu,u,2u ex2u,2u,0h 24m.m2 x,x,0 : 0 < x <= 3/8 S1/2+u,2u,1/2+u ex2u,1,2uh24 m.m2 x,1/2,x : 0 < x < 1/8 S~SM1=[K M]h24m.m2 x,x,0 : 3/8 < x < 1/2 SM SM1=[GM M]h24m.m2 x,x,0 : 0 < x < 1/2 Q1/2,1/4+u,3/4-u1/2,1-2u,2ui 48..2 1/4,1/2-y,y : 0 < y < 1/4 Au,-u+v,v ex-2u+2v,2u,0j 48m.. x,y,0 : 0 < x < y <= 3/8 U U x,y,0 : 0 < x < 3/4-y < y < 1/2 B1/2+u,u+v,1/2+v ex2v,1,2uj 48m.. x,1/2,z : 0 < z < x <= 1/4-z B~B1=[K M W]j48m.. x,y,0 : 3/4-y <= x < y < 1/2 A B1=[GM M X]j48m.. x,y,0 : 0 < x < y < 1/2 Cu,u,v exv,v,-v+2uk 48..m x,x,z : 0 < z < x <= 3/8-z/2 Ju,v,u[GMXUL] exv,-v+2u,vk 48..m x,y,x : 0 < x < y <= 1/2-x U U x,y,x : 1/4 < y < 1/2, 1/2-y < x < 3/8-y/2 J~J1=[GM L X3] + [L K M]k48 ..mx,x,z : 0 < x < z <= 1/2-x U U x,x,z : 0 < z < 1/4, 3/8-z/2 < x < 1/2-z C + J1=[GM M X3] \ [GM L]k48 ..mx,x,z : 0 < z < 1/2 -x < 1/2, x!= z GPu,v,w-u+w+v,u+w-v,u-w+vl 961 x,y,z : 0 < z < x < y < 1/2-x U U x,y,z : 0 < z < 1/2-y < x < y < 1/2 U U x,y,1/2-y : 1/4 < y < 1/2; 1/2-y < x < 1/4. * Cracknell, A. P., Davies, B.L., Miller, S. C., and Love, W. F. (1979). Kronecker Product Tables. Vol. 1. General Introduction and Tables of Irreducible Representations of Space Groups. New York: IFI/Plenum. The asymmetric unit of ITA is obtained from that used in these tables by reflectionthrough the plane x,x,z . The asymmetric unit is obtained from the representation domain of CDML by the equivalence [L K W M] ~[L U W X] through the two-fold rotation around the axis Q. Wing: [GM L X3] x,x,z: 0 < x < z < 1/2-x The transformation matrix that relates the primitive (CDML) base with the conventional-ITA is -*a*+*b*+*c*,* **a*-*b*+*c*,* **a*+*b*-*c* If you want to identify a *k*-vector you have to introduce: 1. The reciprocal bases: primitive (CDML) conventional dual (ITA) 2. The *k*-vector: kx ky kz -------------- next part -------------- An HTML attachment was scrubbed... URL: http://www.democritos.it/pipermail/pw_forum/attachments/20110610/3ffab9c4/attachment.htm
