<div dir="ltr"><div class="gmail_default" style="font-family:arial,helvetica,sans-serif">Is it possible to calculate free energy in WIEN2k or we need additional calculations such as DFT+EDMFT or using PhonoPy(QE)?</div><div><div dir="ltr" class="gmail_signature" data-smartmail="gmail_signature"><div dir="ltr"><font face="georgia, serif">Best regards.</font><div><font face="georgia, serif">Thank you!!</font><div><font face="georgia, serif">Nileema Sharma</font></div></div></div></div></div><br></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Wed, May 20, 2020 at 12:06 PM Xavier Rocquefelte <<a href="mailto:xavier.rocquefelte@univ-rennes1.fr">xavier.rocquefelte@univ-rennes1.fr</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">And another comment. Applying the magnetization along a high symmetry <br>
direction does not mean that you will apply SOC in the direction for <br>
which the effect will be maximum... It is why, people study the <br>
magnetocrystalline anisotropy energy (MAE).<br>
<br>
Best regards<br>
<br>
Xavier<br>
<br>
Le 20/05/2020 à 08:07, Peter Blaha a écrit :<br>
> I think nobody really understands your question.<br>
><br>
> Obviously for a hexagonal system the c-axis is a very high symmetry <br>
> direction. This corresponds to 0 0 1<br>
><br>
> Of course, the a (or b) direction is also a special direction, but it <br>
> will break hexagonal symmetry as the a and b directions will no longer <br>
> be equivalent. So when you choose 1 0 0, in many cases the symmetry <br>
> will be automatically reduced by symmetso.<br>
><br>
><br>
> Am 20.05.2020 um 05:30 schrieb Nileema Sharma:<br>
>> Dear WIEN2k users and developers!!<br>
>> I am working on a hexagonal system, I would like to know in which <br>
>> direction(s) should I apply SOC so that it would represent high <br>
>> symmetry crystallographic direction, for the calculation of the <br>
>> energy in that particular direction?<br>
>> Best regards.<br>
>> Thank you!!<br>
>> Nileema Sharma<br>
>><br>
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