[Wien] f orbitals

2012-09-19 Thread Gavin Abo
In Wien2k 12.1, $WIENROOT/SRC_qtl/ltext.f contains the following line:

txt(3,2)=' f,x3-3xy2,y3-3yx2,z(x2-y2),xyz,xz2,yz2,z3, real basis '

Is this the general set for f-orbitals, it looks like it?

ltext.f seems to be unused code.  Instead, $WIENROOT/SRC_qtl/qtltext.f 
is used, which contains:

txt(3,2)='f,A2,x(T1),y(T1),z(T1),ksi(T2),eta(T2),zeta(T2), real basis '
...
txf(1)=' A2=xyz  x(T1)=x(x2-3r2/5)  y(T1)=y(y2-3r2/5) z(T1)=z(z2-3r2/5) '
txf(2)=' ksi(T2)=x(y2-z2)   eta(T2)=y(z2-y2)   zeta(T2)=z(x2-y2)'

This should be used for l=3 and qsplit=2.  In 
$WIENROOT/SRC_qtl/QTL-tehnical-report.pdf,
it mentions octahedral potential.  Would it be proper terminology to 
call this the octahedral set
for f-orbitals that the program outputs?

Does this mean that the Wien2k code currently does not output the cubic 
set?

The following site has equations (cubic  general set) for the 5f 
orbitals that might be of interest:

http://winter.group.shef.ac.uk/orbitron/AOs/5f/equations.html

Sorry for giving more questions than answers.  The topic is currently 
beyond by current understanding,
but hopefully it will provide some insight.

On 9/18/2012 11:44 AM, Viktor Zano wrote:
 Hi
 As I said, I used the program QTL (and not lapw2 -qtl)
 The automatic  ISPLIT was -2.
 Sorry, I read the manual and I couldn't find it. I spent few weeks and 
 still don't have a clue!
 So again I ask your help

 2012/9/17 Peter Blaha pblaha at theochem.tuwien.ac.at 
 mailto:pblaha at theochem.tuwien.ac.at

 Don't play with ISPLIT. Leave it as set during initialization.

 You should use the program QTL (and not lapw2 -qtl) and its input
 file case.inq

 Read the UG.

 Am 16.09.2012 13:34, schrieb Viktor Zano:

 Dear Wien2k users
 I'm trying to find the DOS of the 5f orbitals for cubic set
 (whole 7 of them: 5fy^3, 5fz^3, 5fx^3, 5fx(z^2-y^2),
 5fy(z^2-x^2), 5fz(x^2-y^2), 5fxyz).
 Attached the struc file (UAl3_new4.struc).
 The QTL calculates special partial charge, and through it a
 proper input file (*.int).
 I couldn't find how to do it using qtl. Both Wien2k manual and
 other users didn't help.
 I used different QSPLIT, which didn't help.
 qsplit=-2
 ATOM  U: 1  tot,s,p,d,d-eg,d-t2g,f,A2,T1,T2,
 ATOM  Al: 2  tot,s,p,pxy,pz,

 qsplit=-1
 ATOM  U: 1
  tot,s,p,p1/2(-1/2),p1/2(1/2),p3/2(-3/2),,,p3/2(3/2),
 
 d,d3/2(-3/2),,,d3/2(3/2),(d5/2)(-5/2),d5/2(5/2),f,f5/2(-5/2),f5/2(5/2),f7/2(-7/2),,,f7/2(7/2),
 ATOM  Al: 2  tot,s,p,p1/2(-1/2),p1/2(1/2),p3/2(-3/2),,,p3/2(3/2),

 qsplit=0
 ATOM  U: 1  tot,s,p,p1/2,p3/2,d,d3/2,d5/2,f,f5/2,f7/2,
 ATOM  Al: 2  tot,s,p,p1/2,p3/2

 qsplit=1
 ATOM  U: 1
  
 tot,s,p,(1;-1),(1;0),(1;1),d,(2;-2),(2;-1),(2;0),(2;1),(2;2),f,(3;-3),(3;-2),(3;-1),(3;0),(3;1),(3;2),(3;3),
 ATOM  Al: 2  tot,s,p,(1;-1),(1;0),(1;1),

 qsplit=2
 ATOM  U: 1
  
 tot,s,p,px,py,pz,d,dz2,d(x2-y2),dxy,dxz,dyz,f,A2,x(T1),y(T1),z(T1),ksi(T2),eta(T2),zeta(T2),
 ATOM  Al: 2  tot,s,p,px,py,pz,

 qsplit=3
 ATOM  U: 1
  
 tot,s,p,pxy,pz,d,dz2,d(x2-y2),d(yz+xz),dxy,f,A2,[x(T1)+y(T1)],z(T1),[ksi(T2)+eta(T2)],zeta(T2),
 ATOM  Al: 2  tot,s,p,pxy,pz,

 qsplit=4
 ATOM  U: 1
  
 tot,s,p,pxy,pz,d,dz2,d[(x2-y2)+xy],d[yz+xz],f,A2+zeta(T2),x(T1)+ksi(T2),y(T1)+eta(T2),z(T1),
 ATOM  Al: 2  tot,s,p,pxy,pz,

 qsplit=5
 ATOM  U: 1  tot,s,p,d,d-eg,d-t2g,f,A2,T1,T2,
 ATOM  Al: 2  tot,s,p,

 qsplit=88
 ATOM  U: 1  tot,s,p,d,f,xdos(i,i),i=1,lxdos2)
 ATOM  Al: 2  tot,s,p,d,f,xdos(i,i),i=1,lxdos2)

 qsplit=99
 ATOM  U: 1  tot,s,p,d,f,xdos(i,j),j=1,i),i=1,lxdos2)
 ATOM  Al: 2  tot,s,p,d,f,xdos(i,j),j=1,i),i=1,lxdos2)

 Please help, Victor



 -- 
 -
 Peter Blaha
 Inst. Materials Chemistry, TU Vienna
 Getreidemarkt 9, A-1060 Vienna, Austria
 Tel: +43-1-5880115671
 Fax: +43-1-5880115698
 email: pblaha at theochem.tuwien.ac.at
 mailto:pblaha at theochem.tuwien.ac.at
 - 


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[Wien] case.inhf

2012-09-19 Thread ali ghafari
Dear Prof. Blaha

For hybrid functionals case.inhf is necessary as discussed in the UG on pages 
49 and 97. But the value of nband' at case.inhf is confusing. on the page 97 
of UG has mentioned ?  nband should be at least equal to the number of 
(partially) occupied bands plus one. While at the line 5 of case.in1 the value 
of nband is automatically determined ( UG page 102) by ?? nband = ne ? 2.0 + 5 
 which is number of eigenvalues. Can we consider inband of case.inhf is 
equals: (eigenvalues -5)/2+1 which is extracted from case.in1?
?? 

Best Regards
Ali
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[Wien] case.inhf

2012-09-19 Thread t...@theochem.tuwien.ac.at
Why not, but I think that it is not really more simple than just looking
at the number n_occ of (partially) occupied bands in case.scf and choosing
nband = n_occ + a few more bands. But remember that nband in case.inhf is
a parameter (similar to R*K_max) which has to be tested.

One last thing: the purpose of nband in case.in1 is not at all the same
as nband in case.inhf.

F. Tran

On Wed, 19 Sep 2012, ali ghafari wrote:

 Dear Prof. Blaha
 
 For hybrid functionals case.inhf is necessary as discussed in the UG on pages 
 49 and 97. But the value of nband' at case.inhf is confusing. on the page 97 
 of UG has mentioned ?  nband should be at least equal to the number of 
 (partially) occupied bands plus one. While at the line 5 of case.in1 the 
 value of nband is automatically determined ( UG page 102) by ?? nband = ne ? 
 2.0 + 5  which is number of eigenvalues. Can we consider inband of case.inhf 
 is equals: (eigenvalues -5)/2+1 which is extracted from case.in1?
 ?? 
 
 Best Regards
 Ali