[CP2K-user] [CP2K:15341] Ar lattice constant calculation
sumit agrawal
sumitag... at gmail.com
Wed May 12 11:42:09 UTC 2021
Dear Fabian,
Thanks for the quick response. Actually I prepared a lattice structure
(2x2x2) in VESTA software. Is it the correct tool to prepare lattice
structure, any idea?
I tried again to prepare a lattice structure. So 2x2x2 will contain 32
atoms. Then I performed cell_opt in cp2k. I am attaching my coordinate
file, input file and out file.
Where will I find the lattice constant in the output file?
One more query: will the lattice constant for 1x1x1 unit cell and nxnxn
unit cell be same or different?
On Wed, May 12, 2021 at 1:57 PM Fabian Ducry <fabia... at gmail.com> wrote:
> Dear Sumit
>
> A supercell of (2x2x2) has to contain 2*2*2=8 times the number of atoms of
> the unit cell. So It can never have an odd number as in your case 63. You
> should follow a tutorial on how to prepare such a structure, e.g. with ASE.
> Here, for example, you find instruction on how to build a crystal:
> http://exciting-code.org/lithium-atomic-simulation-environment. Once you
> have the ASE Atoms object building the supercell is as easy as
> atoms*(2,2,2). When you have a correct crystal, make sure that the &CELL
> section in the cp2k input matches the lattice dimensions of the crystal.
> You cannot just use an arbitrary number.
>
> CELL_OPT will optimize both the atom positions as well as the lattice
> dimensions.
>
> Fabian
> On 12.05.2021 06:19, sumit agrawal wrote:
>
>
> Dear Lucas Lodeiro and Fabian,
>
> Thanks for your nice suggestions. I am new to cp2k. I have just
> started learning this. I have some query regarding my procedure for
> preparation for crystal structure. I am following below procedure to
> prepare the crystal structure:
>
> First, I have prepared the crystal structure (2x2x2) with the lattice
> dimension a=5.2229 A, b=5.2229 A and c=5.2229 A. It contains 63 atoms.
> After that I removed some atoms from x, y and z directions so that it
> become periodic in x, y and z direction. After this, it contains 32 atoms.
> Now I save this lattice in XYZ format. This will be my
> initial coordinate system. Am I correct?
>
> After this I took this coordinate file and ran for the cp2k calculation.
>
> Now in cp2k calculations for this lattice structure what will be my cube
> size? (Here you mention it will not be 20 A. )
>
> One more query, here I am starting with a lattice dimension with 5.2229 A
> . So after Cell_OPTIMIZATION, will the geometry of my crystal change or not?
>
> Yes, dispersion correction must be required for this type of calculation,
> I will do that.
>
> Here I am attaching my crystal structure images.
>
>
>
> Thanks
> Sumit
>
>
>
> On Tue, May 11, 2021 at 11:02 PM Lucas Lodeiro <eluni... at gmail.com>
> wrote:
>
>> Hi Sumit,
>>
>> First of all, you cannot find the lattice constant with this type of
>> calculation. You ran an ENERGY calculation. If you want to do it, you need
>> to run a CELL_OPT calculation.
>> In your current calculation, you use a 20 A lattice constant cube. Also,
>> I think your initial cell dimension is not correct, since the lattice
>> constant of Argon is 5.26 A, then a 2x2x2 supercell is not 20 A sized.
>> Also, your system has 63 atoms... ¿?.
>>
>> Other related topics are: a 2x2x2 supercell is not sufficient to compute
>> a periodical Argon system... You need a bigger supercell, or use kpoints.
>> Also, as this system just interacts by dispersion, probably the PBE
>> functional is not sufficient at all. You need to account the dispersion
>> terms by means of Grimes D3 or other.
>>
>> Regards
>>
>> El mar, 11 may 2021 a las 14:08, sumit agrawal (<
>> sumitag... at gmail.com>) escribió:
>>
>>> Hii everyone,
>>>
>>> I want to calculate a lattice constant for the Ar crystal of (2*2*2 unit
>>> cell). I am attaching my input and out files. Where I will find the
>>> lattice constant ?
>>>
>>> Please help me!!
>>>
>>>
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