[Pw_forum] to extract k-point from bilbao crystallographic server

[Huiqun Zhou]

They are just written down there in your mail, aren't they? For Fd-3m, they are
K, L, U, W, X.

You can click on the "Brillouin zone" to see the graph for further understanding what 
they present.


zhou huiqun
@earth sciences, nanjing university, china

  ----- Original Message ----- 
  From: Abolore Musari 
  To: PWSCF Forum 
  Sent: Friday, June 10, 2011 2:26 PM
  Subject: [Pw_forum] to extract k-point from bilbao crystallographic server





  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 [Fd-3m]
  (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


        k-vector description Wyckoff Position ITA description 
        CDML* Conventional-ITA ITA Coordinates 
        Label Primitive 
        GM 0,0,0 0,0,0 a 2 m-3m 0,0,0  
        X 1/2,0,1/2 0,1,0 b 6 4/mm.m 0,1/2,0  
        L 1/2,1/2,1/2 1/2,1/2,1/2 c 8 .-3m 1/4,1/4,1/4  
        W 1/2,1/4,3/4 1/2,1,0 d 12 -4m.2 1/4,1/2,0  
        DT u,0,u 0,2u,0 e 12 4m.m 0,y,0 : 0 < y < 1/2 
        LD u,u,u u,u,u f 16 .3m x,x,x : 0 < x < 1/4 
        V 1/2,u,1/2+u 2u,1,0 g 24 mm2.. x,1/2,0 : 0 < x < 1/4 
        SM u,u,2u ex 2u,2u,0 h 24 m.m2 x,x,0 : 0 < x <= 3/8 
        S 1/2+u,2u,1/2+u ex 2u,1,2u h 24 m.m2 x,1/2,x : 0 < x < 1/8 
        S~SM1=[K M] h 24 m.m2 x,x,0 : 3/8 < x < 1/2 
        SM SM1=[GM M] h 24 m.m2 x,x,0 : 0 < x < 1/2 
        Q 1/2,1/4+u,3/4-u 1/2,1-2u,2u i 48 ..2 1/4,1/2-y,y : 0 < y < 1/4 
        A u,-u+v,v ex -2u+2v,2u,0 j 48 m.. x,y,0 : 0 < x < y <= 3/8 U
        U x,y,0 : 0 < x < 3/4-y < y < 1/2  
        B 1/2+u,u+v,1/2+v ex 2v,1,2u j 48 m.. x,1/2,z : 0 < z < x <= 1/4-z 
        B~B1=[K M W] j 48 m.. x,y,0 : 3/4-y <= x < y < 1/2 
        A B1=[GM M X] j 48 m.. x,y,0 : 0 < x < y < 1/2 
        C u,u,v ex v,v,-v+2u k 48 ..m x,x,z : 0 < z < x <= 3/8-z/2 
        J u,v,u[GMXUL] ex v,-v+2u,v k 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] k 48 ..m x,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] k 48 ..m x,x,z : 0 < z < 1/2 -x < 1/2, x!= z 
        GP u,v,w -u+w+v,u+w-v,u-w+v l 96 1 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    
       
          






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