[Pw_forum] Wrong ELF values

[Stefano de Gironcoli]

I see...
how large is your calculation? if it is not too big, would you mind send 
me (at my email) one set of input and output to have a look ?
stefano

On 02/15/2011 04:42 PM, Claudia Loyola wrote:
> Hello!
>
> yes, you are right if I define the vectors in function of "a" and I define
> alat=a , but let me show you how I define the vectors in the scf input file:
>
> &SYSTEM
>    ibrav = 0 ,
>    celldm(1) =1.0 ,
> .
> .
> .
> CELL_PARAMETERS (alat=1.0)
>    19.109476287  -0.014917030   0.000000953
>    -9.542147268  16.554932043   0.000000170
>     0.000000706   0.000000344  14.136501037
>
> so in this case the vector basis are the real large in bohr, and the alat is
> 1.0 like celldm(1). Indeed, the value of the vectors for pp.x are in
> function of alat, but my alat is 1.0, so I can write directly in bohr the
> values of the planes that I take, in fact I take "two replicates" in each
> direction. I think this is correct too.
>
> Thank you very much for your comments :)
>
> regards,
> Claudia Loyola
>
> On Tue, Feb 15, 2011 at 2:01 AM, Stefano de Gironcoli<degironc at sissa.it>wrote:
>
>> Dear Claudia,
>>     I may not remember correctly what the format adopted by pp.x to
>> define planes is... but i think it is Cartesian coordinates in unit of
>> alat (aka celldm(1)).
>>
>>     so when you write
>>       e1(1) =38.218, e1(2)=0.0, e1(3) = 0.0,
>>      e2(1) =0.0, e2(2)=33.108, e2(3) = 0.0,
>> this means presumably 38 x 33 UNIT CELLS, sampled with 100x100 points
>> (that is about 2-3 points in each unit cell). No surprise the FFT
>> interpolation used gets confused.
>>
>> I think you should use more sensible values for e1(1) and e2(2)... what
>> is your value of celldm(1)?
>>
>> At least this is my guess.
>>
>> stefano
>>
>>
>> On 02/14/2011 06:57 PM, Claudia Loyola wrote:
>>> Well, this is one of the input file:
>>>
>>> &inputpp
>>>    &plot
>>>       nfile = 1 ,
>>>       filepp(1) = 'Ba38_33_10.charge' ,
>>>       weight(1) = 1.0 ,
>>>       iflag = 2 ,
>>>       output_format = 3 ,
>>>       fileout = 'ba.rho38_33_10.xsf' ,
>>>       x0(1) =0.0, x0(2)=0.0, x0(3) = 10.602,
>>>       e1(1) =38.218, e1(2)=0.0, e1(3) = 0.0,
>>>       e2(1) =0.0, e2(2)=33.108, e2(3) = 0.0,
>>>       nx=100, ny=100
>>>    /
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>
>
>
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