[Wien] AFM calculations

[swati chaudhury]

Hello,
your Usb2.inst file will be:

> U
> Rn 3
> 5, 3,0.0  N      <----
> 5, 3,3.0  N      <----
> 6, 2,1.0  N
> 6, 2,0.0  N
> 7,-1,1.0  N
> 7,-1,1.0  N
> U
> Rn 3
> 5, 3,3.0  N      <----
> 5, 3,0.0  N      <----
> 6, 2,0.0  N
> 6, 2,1.0  N
> 7,-1,1.0  N
> 7,-1,1.0  N
> U
> Rn 3
> 5, 3,0.0  N      <----
> 5, 3,3.0  N      <----
> 6, 2,1.0  N
> 6, 2,0.0  N
> 7,-1,1.0  N
> 7,-1,1.0  N
> U
> Rn 3
> 5, 3,3.0  N     <----
> 5, 3,0.0  N     <----
> 6, 2,0.0  N
> 6, 2,1.0  N
> 7,-1,1.0  N
> 7,-1,1.0  N
total spin-up and spin down moment will be same.
best of luck.
swati
--- On Sun, 22/2/09, Jian-Xin Zhu <jxzhu at lanl.gov> wrote:

> From: Jian-Xin Zhu <jxzhu at lanl.gov>
> Subject: [Wien] AFM calculations
> To: "A Mailing list for WIEN2k users" <wien at zeus.theochem.tuwien.ac.at>
> Date: Sunday, 22 February, 2009, 3:03 AM
> Dear Wien2k users and Prof. Blaha,
> 
> I am doing a AFM calculation with s.o. in USb2.
> Several questions for your help:
> 
> 1. The nonmagnetic USb2 has the tetragonal layered
> structure with P4/nmm
> symmetry.
> 
> The cell has 6 atoms with lattice constant a, b=a, and c. 
> In the AFM
> state, the magnetic cell is doubled with respect to the
> chemical cell.
> Specifically, the lattice constant along the c direction is
> doubled.
> 
> When I prepare the USb2.struct_supergroup file of the
> nonmagnetic state,
> should I keep the lattice parameters as the original (i.e.,
> a, b=a, c), or
> take  them to match the magnetic cell (i.e., a, b=a, 2c)?
> 
> If I follow the first way, x afminput does not create the
> input file for
> the program CLMCOPY. If I follow the second way by
> specifying the symmetry
> type as P in w2web,
> after running sgroup, it came with a warning of reducing
> the lattice
> constant from 2c to c, essentially going back to the first
> way.
> 
> If I insisted to use the primitive type P and specify 12
> inequivalent atoms in the cell, the complaints with
> "incorrect symmetry"
> came out.
> 
> So what's the correct way to construct
> USb2.struct_supergroup?
> 
> 2. I assume the magnetism comes from the 5f state on U by
> flipping the
> corresponding occupation on U in the file of USb2.inst.
> 
> U
> Rn 3
> 5, 3,0.0  N      <----
> 5, 3,3.0  N      <----
> 6, 2,1.0  N
> 6, 2,0.0  N
> 7,-1,1.0  N
> 7,-1,1.0  N
> U
> Rn 3
> 5, 3,3.0  N      <----
> 5, 3,0.0  N      <----
> 6, 2,1.0  N
> 6, 2,0.0  N
> 7,-1,1.0  N
> 7,-1,1.0  N
> U
> Rn 3
> 5, 3,3.0  N      <----
> 5, 3,0.0  N      <----
> 6, 2,1.0  N
> 6, 2,0.0  N
> 7,-1,1.0  N
> 7,-1,1.0  N
> U
> Rn 3
> 5, 3,0.0  N     <----
> 5, 3,3.0  N     <----
> 6, 2,1.0  N
> 6, 2,0.0  N
> 7,-1,1.0  N
> 7,-1,1.0  N
> 
> 
> However, the UG also reminds that we should set a zero
> moment (identical
> spin up and spin down occupations) for all
> "non-magnetic" atoms.
> In my case, the non-magnetic atoms are Sb. However, there
> is
> one spin up  electron on n=5,kappa=1,
> no spin down electron on n=5, kappa=1,
> two spin up electrons on n=5, kappa=-2,
> no spin down electron on n=5, kappa=-2,
> as shown below.
> 
> I can split 2 electron on n=5 and kappa=-2 into one with
> spin up and the
> other with spin dn.
> Then what should do with the only one electron on n=5 and
> kappa=1? Can I
> put it as
> 0.5 and 0.5? I don't find a discussion on this
> situation.
> 
> Sb
> Kr 5
> 4, 2,2.0  N
> 4, 2,2.0  N
> 4,-3,3.0  N
> 4,-3,3.0  N
> 5,-1,1.0  N
> 5,-1,1.0  N
> 5, 1,1.0  N
> 5, 1,0.0  N
> 5,-2,2.0  N
> 5,-2,0.0  N
> 
> ============>
> 
> Sb
> Kr 5
> 4, 2,2.0  N
> 4, 2,2.0  N
> 4,-3,3.0  N
> 4,-3,3.0  N
> 5,-1,1.0  N
> 5,-1,1.0  N
> 5, 1,0.5  N      <----?
> 5, 1,0.5  N      <----?
> 5,-2,1.0  N    <----
> 5,-2,1.0  N    <----
> 
> 
> Thanks you in advance for help.
> 
> Jian-Xin Zhu
> 
> 
> 
> 
> 
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