<div dir="ltr"><font><font size="4"><div class="im">Dear Prof. Blaha,<br><br>I have two other questions:<br><br>>><i> 4) If the forces calculated with spin-orbit coupling are indeed incorrect and meaningless,<br></i>>><i> can we use PORT option in case.inM when doing atomic relaxation (min_lapw)? This<br>
</i>>><i> question arises because I read in the wien2k user guide that "It (PORT) minimizes the<br></i>>><i> total energy and NOT the forces (using the forces as derivative of E vs. atomic positions).".<br>
</i>>><i> As I understand, the PORT method will use "total energy", instead of "force on nuclei",<br></i>>><i> as a criterion to find the equilibrium positions of atoms, i.e., the PORT method will find<br>
</i>>><i> an energy minimum by a real displacement of atoms, rather than by find a structure in<br></i>>><i> which forces on atoms are zero. This means that the forces on nuclei is NOT the essential<br></i>>><i> ingredient in the PORT method, and PORT method is still valid (as long as the total<br>
</i>>><i> energy are calcuated correctly) even in the presence of spin-orbit coupling. Is my<br></i>>><i> ratiocination correct?<br></i><br></div>> No, you can't use PORT. PORT uses the "gradiant of E-tot", i.e. the forces in order to move<br>
> atoms "in the proper direction" and if they do not fit with the resulting E-tot, it will give up.<div class="im">===============================================<br>Take a material, in which only one internal freedom x exists, as an example.<br>
If I do a series of total energy calculations with difference x and find a <br>energy surface with respect to x, then I can obtain a force by taking a<br>energy derivative with respect to x. Is the force obtained by this method<br>
meaningful (even in the presence of spin orbit coupling)?<br></div><br><br>> PS: As mentioned in UG, you can switch off SO for light atoms (eg. oxygens), and then the forces for<br>> those atoms are still ok and meaningful.<br>
====================================================<br>Does this mean that the spin-orbit term of Pulay correction for the<br>investigated atom, eg. oxygen, comes from the oxygen itself only, <br>and is irrelevant to those atoms around the oxygen?</font></font><font size="4"><font color="#888888"><br>
</font><br>Thank you very much.<br><br>Best Regards,<br>Zhiyong Zhu<br></font></div>