[Wien] AFM calculation

[蔡 孟秋]

Dear Peter Blaha , Stefaan Cottenier’s and wien user:
     I am doing a AFM calculation. In according to the UG and the Stefaan 
Cottenier’s suggestion, I construct the AFM case.struct. In my original 
case.struct, I split the positions of the Fe-atoms according to their spin. 
In the Fe position with MULT=2 has one atom with spin up and one with spin 
dn, I make one positions with MULT=1 out of it. Give these two different 
spin types a different label. Run now init_lapw, and accept all structure 
changes proposed by nn and sgroup. Moreover, I put the correct up and dn 
occupations for  Fe in the final case.inst file before proceeding with 
lstart. Then I begin always to use a regular 
runsp_lapw. The result is :  MAGNETIC MOMENTS OF MIXED CHARGE DENSITY
:MMINT: MAGNETIC MOMENT IN INTERSTITIAL =    0.00000
:MMI001: MAGNETIC MOMENT IN SPHERE   1    =   -0.00256
:MMI002: MAGNETIC MOMENT IN SPHERE   2    =    0.00255
:MMI003: MAGNETIC MOMENT IN SPHERE   3    =    3.66726
:MMI004: MAGNETIC MOMENT IN SPHERE   4    =   -3.66725
:MMI005: MAGNETIC MOMENT IN SPHERE   5    =    0.05963
:MMI006: MAGNETIC MOMENT IN SPHERE   6    =   -0.05963
:MMTOT: TOTAL MAGNETIC MOMENT IN CELL   =    0.00000

The original-case.inst and the flip-case.inst as follow:
Oringial-case.inst
Bi
Xe 7 5
4, 3,3.0  N
4, 3,3.0  N
4,-4,4.0  N
4,-4,4.0  N
5, 2,2.0  N
5, 2,2.0  N
5,-3,3.0  N
5,-3,3.0  N
6,-1,1.0  N
6,-1,1.0  N
6, 1,1.0  N
6, 1,1.0  N
6,-2,1.0  N
6,-2,0.0  N
Bi
Xe 7 5
4, 3,3.0  N
4, 3,3.0  N
4,-4,4.0  N
4,-4,4.0  N
5, 2,2.0  N
5, 2,2.0  N
5,-3,3.0  N
5,-3,3.0  N
6,-1,1.0  N
6,-1,1.0  N
6, 1,1.0  N
6, 1,1.0  N
6,-2,1.0  N
6,-2,0.0  N
Fe
Ar 3 5
3, 2,2.0  N
3, 2,2.0  N
3,-3,2.5  N
3,-3,0.0  N
4,-1,1.0  N
4,-1,0.5  N
Fe
Ar 3 5
3, 2,2.0  N
3, 2,2.0  N
3,-3,2.5  N
3,-3,0.0  N
4,-1,1.0  N
4,-1,0.5  N
O
He 3 5
2,-1,1.0  N
2,-1,1.0  N
2, 1,1.0  N
2, 1,1.0  N
2,-2,2.0  N
2,-2,0.0  N
O
He 3 5
2,-1,1.0  N
2,-1,1.0  N
2, 1,1.0  N
2, 1,1.0  N
2,-2,2.0  N
2,-2,0.0  N
****
****         END of input (instgen_lapw)

flip-case.inst
Bi
Xe 7 5
4, 3,3.0  N
4, 3,3.0  N
4,-4,4.0  N
4,-4,4.0  N
5, 2,2.0  N
5, 2,2.0  N
5,-3,3.0  N
5,-3,3.0  N
6,-1,1.0  N
6,-1,1.0  N
6, 1,1.0  N
6, 1,1.0  N
6,-2,0.5  N    <=== equal occup for up/dn
6,-2,0.5  N    <=== equal occup for up/dn
Bi
Xe 7 5
4, 3,3.0  N
4, 3,3.0  N
4,-4,4.0  N
4,-4,4.0  N
5, 2,2.0  N
5, 2,2.0  N
5,-3,3.0  N
5,-3,3.0  N
6,-1,1.0  N
6,-1,1.0  N
6, 1,1.0  N
6, 1,1.0  N
6,-2,0.5  N      <=== equal occup for up/dn
6,-2,0.5  N      <=== equal occup for up/dn
Fe
Ar 3 5
3, 2,2.0  N
3, 2,2.0  N
3,-3,2.5  N
3,-3,0.0  N
4,-1,1.0  N
4,-1,0.5  N
Fe
Ar 3 5
3, 2,2.0  N
3, 2,2.0  N
3,-3,0.0  N       <=== spin flipped
3,-3,2.5  N       <=== spin flipped
4,-1,0.5  N       <=== spin flipped
4,-1,1.0  N       <=== spin flipped

O
He 3 5
2,-1,1.0  N
2,-1,1.0  N
2, 1,1.0  N
2, 1,1.0  N
2,-2,1.0  N        <=== equal occup for up/dn
2,-2,1.0  N        <=== equal occup for up/dn
O
He 3 5
2,-1,1.0  N
2,-1,1.0  N
2, 1,1.0  N
2, 1,1.0  N
2,-2,1.0  N     <=== equal occup for up/dn
2,-2,1.0  N     <=== equal occup for up/dn
****
****         END of input (instgen_lapw)
I have two question:
1 Is it right for my claculational method?
2 I do no kmow why the MAGNETIC MOMENT of the Bi abd O atom is non-zero?
   Any help and suggestion is welcome!


Yours Meng-Qiu Cai

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