[Wien] d_z2 DOS from TETRA and QTL

[Stefaan Cottenier]

I asked Doru Torumba for advice, and this is the result:

Yes, you have indeed a 10x10 matrix with complex entries. It consists of 
4 5x5 blocks: up-up, up-dn, dn-up, dn-dn. The first 5 rows are 
m=-2,-1,0,1,2. If you put a '*' then you sum in the output all lines 
above the star without star, and the line with the star itself. 
Therefore, if you put a star at the 3th line, you obtain the sum of -2, 
-1 and 0. Not the m=0 itself as you want. To obtain that one, just 
remove your star, and you will get every m separately.

Stefaan

>Dear Stefaan,
>  The case.cf_d_nonrel from SRC_templates looks like the following
>
>1. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.
>0. 0.  1. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.
>0. 0.  0. 0.  1. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.
>0. 0.  0. 0.  0. 0.  1. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.
>0. 0.  0. 0.  0. 0.  0. 0.  1. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.
>0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  1. 0.  0. 0.  0. 0.  0. 0.  0. 0.
>0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  1. 0.  0. 0.  0. 0.  0. 0.
>0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  1. 0.  0. 0.  0. 0.
>0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  1. 0.  0. 0.
>0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  0. 0.  1. 0.
>
>It is a 10x10 matrix with complex entries. Do you have a different
>case.cf?
>
>Thanks,
>Chiung-Yuan
>
>On 12/5/05, Stefaan Cottenier <Stefaan.Cottenier at fys.kuleuven.be> wrote:
>  
>
>>I did not check, but as far as I remember the matrix you are refering to
>>is a 5x5 _complex_ matrix, and not a 10x10. You have to read it in pairs
>>of numbers (real and imaginary part). If true, then your input was not
>>correct for m=0.
>>
>>Stefaan
>>
>>    
>>
>>>Dear all,
>>>  I just get started using QTL to calculate the partial DOS of 3d for
>>>different m's (m=-2,-1,...,2). In order to check whether I am doing it
>>>right, I compare the spin-up DOS plot of 3d(m=0) by QTL with the one
>>>of 3dz^2 by TETRA. I expect that they have exactly the same curve but
>>>they are kind of different. Their main peaks have different values
>>>(QTL is 1.7 times larger than TETRA) are off each other by 1 eV, and
>>>the satellite peaks around are quite different.
>>>  Here is the way I did for 3d(m=0) by QTL: I use case.cf_d_nonrel
>>>      
>>>
>>>from SRC_templates. If I understand it right, the 1st~5th diagonal
>>    
>>
>>>entries of this 10x10 identity matrix represent 3d(m=-2,-1,...,2) for
>>>spin up and 6th~10th for spin down. To obtain 3d(m=0) spin up, I put
>>>only one star in front of the 3rd row, rename it to case.cf1, use the
>>>following case inq
>>>
>>>SUMA
>>>DOSYM       /NOSYM
>>>-.5 1.4
>>>0.17601
>>>1                      ( number of atoms for which the DOS are calculated)
>>>1 2                    ( jatom  latom)
>>>
>>>and then run x qtl -up. This generates case.qtlup1. Then I change
>>>"case.qtlup" to "case.qtlup1" in uptetra.def, and run tetra
>>>uptetra.def.
>>> My questions are: did I do the right thing for 3d(m=0) by QTL? Did I
>>>compare QTL 3d(m=0) and TETRA 3dz^2 in a correct way? Does such a
>>>comparison make sense?
>>>
>>>Thank you,
>>>Chiung-Yuan
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>>>Wien at zeus.theochem.tuwien.ac.at
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>>>
>>>
>>>
>>>      
>>>
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