[Pw_forum] question about projwfc.x

[Parwana HABIBI]

Hi,

what I mean is this :
   1    1Cr1    1    2    1
       1       1        0.0000239456
       1       2        0.0000330427
       1       3        0.0000002564
       1       4        0.0002903697
       1       5        0.0022361365
       1       6        0.0091035384
       1       7        0.0255469182
       1       8        0.0353335072
       1       9        0.0000000000
       1      10        0.0078272056
       1      11        0.0000000000
       1      12        0.0308578770
       1      13        0.0591287408
       1      14        0.0000000000
       1      15        0.0634927971
       1      16        0.0891558543
       1      17        0.0000000000
       1      18        0.0668713027
       1      19        0.0609816494
     ....

For each atom considered, I have in the output the k-point, then the 
band number and then the total weight of all the orbitals projected of 
the corresponding band. I would like to see the different contributions 
of each orbital listed exactly as this, but for each orbital.

THanks again,



NgocLinh Nguyen wrote:
> Parwana HABIBI wrote:
>   
>> Hi,
>>
>> I would like to study the character of a surface state in chromium (is 
>> it more dz2 or dxy or .. ?), and for that, I've done a projwfc calculation.
>> It gives me on each kpt, each band and the total weight of all the orbitals.
>>   
>>     
> Here, I am not clear about what you meant as kpt ???
>   
>> What I would like to get is the contribution for each orbital separate. 
>>   
>>     
> The fact that what you get from projwfc calculation are not only for all 
> orbitals of each atom but also Projected DOS on each valence orbital of 
> the each atom.
> Is it right ? If so, see more in 
> http://www.quantum-espresso.org/input-syntax/INPUT_PROJWFC.html
>
>
> Good luck,
> Linh
>
>   
>> Is that possible? Or should I change a piece of the code? and if so? how 
>>   
>>     
>
> p/s: Pls enclose your affiliation 's name in next discusses.
>   
>> do I know the right place?
>>
>> Thanks,
>>
>> Parwana
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>>   
>>     
>
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