<p dir="ltr">Well, one might in principle perform a non-scf calculation starting from a potential that is not the scf one. Not sure why would anybody want to do so. Paolo</p>
<div class="gmail_extra"><br><div class="gmail_quote">Il 01/nov/2016 02:36 PM, "dario rocca" <<a href="mailto:roccad@gmail.com">roccad@gmail.com</a>> ha scritto:<br type="attribution"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr">Dear Q.J. Wang,<div>No, it is not possible.<br><div>In a nscf calculation you need to "fix" your Hamiltonian which is determined from a previous scf calculation.   </div><div>Best,</div><div>Dario Rocca</div></div></div><div class="gmail_extra"><br><div class="gmail_quote">On Tue, Nov 1, 2016 at 2:06 PM, Q.J.Wang <span dir="ltr"><<a href="mailto:wangqj1@126.com" target="_blank">wangqj1@126.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div style="line-height:1.7;color:#000000;font-size:14px;font-family:Arial">Dear QE developer,<br>I want to know whether QE can perform a nscf calculation without a scf calculation?<br><br><br><br><div style="zoom:1">--<br><div>Best regards</div><div> </div><div>Q.J.Wang</div><div> </div><div>XiangTan University </div><div style="clear:both"></div></div></div><br><br><span title="neteasefooter"><p> </p></span><br>______________________________<wbr>_________________<br>
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