<div dir="ltr">Dear Fabian,<br><div><br></div><div>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? </div><div><br></div><div>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. </div><div><br></div><div>Where will I find the lattice constant in the output file?</div><div><br></div><div>One more query: will the lattice constant for 1x1x1 unit cell and nxnxn unit cell be same or different?</div><div><br></div><div><br></div><div><br></div><div><br></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Wed, May 12, 2021 at 1:57 PM Fabian Ducry <<a href="mailto:fabia...@gmail.com">fabia...@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
<div>
<p>Dear Sumit</p>
<p>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:
<a href="http://exciting-code.org/lithium-atomic-simulation-environment" target="_blank">http://exciting-code.org/lithium-atomic-simulation-environment</a>.
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.</p>
<p>CELL_OPT will optimize both the atom positions as well as the
lattice dimensions.</p>
<p>Fabian<br>
</p>
<div>On 12.05.2021 06:19, sumit agrawal
wrote:<br>
</div>
<blockquote type="cite">
<div dir="ltr"><br>
Dear Lucas Lodeiro and Fabian,<br>
<div><br>
</div>
<div>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:</div>
<div><br>
</div>
<div>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. </div>
<div>Now I save this lattice in XYZ format. This will be my
initial coordinate system. Am I correct?</div>
<div><br>
</div>
<div>After this I took this coordinate file and ran for the cp2k
calculation.</div>
<div><br>
</div>
<div>Now in cp2k calculations for this lattice structure what
will be my cube size? (Here you mention it will not be 20 A. )</div>
<div><br>
</div>
<div>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?</div>
<div><br>
</div>
<div>Yes, dispersion correction must be required for this type
of calculation, I will do that.</div>
<div><br>
</div>
<div>Here I am attaching my crystal structure images.</div>
<div><br>
</div>
<div><br>
</div>
<div><br>
</div>
<div>Thanks </div>
<div>Sumit</div>
<div><br>
</div>
<div><br>
</div>
</div>
<br>
<div class="gmail_quote">
<div dir="ltr" class="gmail_attr">On Tue, May 11, 2021 at 11:02
PM Lucas Lodeiro <<a href="mailto:eluni...@gmail.com" target="_blank">eluni...@gmail.com</a>> wrote:<br>
</div>
<blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
<div dir="ltr">Hi Sumit,
<div><br>
</div>
<div>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.</div>
<div>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... ¿?.</div>
<div><br>
</div>
<div>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.</div>
<div><br>
</div>
<div>Regards</div>
</div>
<br>
<div class="gmail_quote">
<div dir="ltr" class="gmail_attr">El mar, 11 may 2021 a las
14:08, sumit agrawal (<<a href="mailto:sumitag...@gmail.com" target="_blank">sumitag...@gmail.com</a>>)
escribió:<br>
</div>
<blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
<div dir="ltr">Hii everyone,<br>
<div><br>
</div>
<div>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
?</div>
<div><br>
</div>
<div>Please help me!!</div>
<div><br>
</div>
<div><br>
</div>
</div>
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