<html xmlns:v="urn:schemas-microsoft-com:vml" xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:w="urn:schemas-microsoft-com:office:word" xmlns:m="http://schemas.microsoft.com/office/2004/12/omml" xmlns="http://www.w3.org/TR/REC-html40"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8"> <meta name="Generator" content="Microsoft Word 15 (filtered medium)"> <style></style> </head> <body lang="EN-US" link="blue" vlink="purple"> <div class="WordSection1"> Dear Katarina<o:p></o:p> <o:p> </o:p> you have indeed to approach each spin channel in the BS section from a “closed shell” point of view. The slides <a href="https://pc2.uni-paderborn.de/fileadmin/pc2/cp2k_workshop/Implicit_Solvent_Methods_and_DFT_U.pdf#page=22"> 22 and 23</a> from the last <a href="https://pc2.uni-paderborn.de/teaching/trainings/cp2k-tutorial"> CP2K tutorial in Paderborn</a> show an example for a BS section for UO2. The BS section is not perfectly intuitive.<o:p></o:p> <o:p> </o:p> HTH<o:p></o:p> <o:p> </o:p> Matthias<o:p></o:p> <o:p> </o:p> From: cp...@googlegroups.com <...@googlegroups.com> On Behalf Of katarína stanciaková Sent: Donnerstag, 8. November 2018 14:36 To: cp2k <...@googlegroups.com> Subject: Re: [CP2K:10917] Re: Initial guess of the density matrix<o:p></o:p> <o:p> </o:p> <div> <div> Dear Matthias,<o:p></o:p> </div> <div> thank you for the suggestion. Could you, please, explain why NEL in alpha channel is 1? <o:p></o:p> </div> <div> Is it becasue neutral Cu has 4s1, so we add one electron and then divide by 2? What if the result would be even number?<o:p></o:p> </div> <div> <o:p> </o:p> </div> <div> Thanks!<o:p></o:p> </div> Dňa štvrtok, 8. novembra 2018 11:49:26 UTC+1 Matthias Krack napísal(-a):<o:p></o:p> <blockquote style="border:none;border-left:solid #CCCCCC 1.0pt;padding:0cm 0cm 0cm 6.0pt;margin-left:4.8pt;margin-top:5.0pt;margin-right:0cm;margin-bottom:5.0pt"> <div> <div> Dear Katarina<o:p></o:p> <o:p></o:p> You may try for Cu1 alpha<o:p></o:p> <o:p></o:p> NEL 1 0<o:p></o:p> L 0 2<o:p></o:p> N 4 3<o:p></o:p> <o:p></o:p> instead of<o:p></o:p> <o:p></o:p> NEL 0 0<o:p></o:p> L 0 2<o:p></o:p> N 4 3<o:p></o:p> <o:p></o:p> Hope that works.<o:p></o:p> <o:p></o:p> Matthias<o:p></o:p> <o:p></o:p> From: <a href="javascript:" target="_blank">cp...@googlegroups.com</a> <<a href="javascript:" target="_blank">cp...@googlegroups.com</a>> On Behalf Of katarína stanciaková Sent: Donnerstag, 8. November 2018 11:13 To: cp2k <<a href="javascript:" target="_blank">cp...@googlegroups.com</a>> Subject: Re: [CP2K:10913] Re: Initial guess of the density matrix<o:p></o:p> <o:p></o:p> <div> Dear all, I am trying to set up calculation for my Cu+ system, however, even after reading this thread I cannot understand how the electrons are assigned to channels in CP2K. I wanna test two possible spin states (just for testing purposes) of Cu+: Cu+ with electron configuration 3d9 4s1 (Cu1) and 3d10 4s0 (Cu2). I tried to build my input sections: &KIND Cu1 ELEMENT Cu BASIS_SET DZVP-MOLOPT-GTH POTENTIAL GTH-BLYP-q11 &BS &ALPHA &ALPHA NEL 0 0 L 0 2 N 4 3 &END &BETA NEL -1 -2 L 0 2 N 4 3 &END &END and &KIND Cu2 ELEMENT Cu BASIS_SET DZVP-MOLOPT-GTH POTENTIAL GTH-BLYP-q11 &BS &ALPHA &ALPHA NEL -1 0 L 0 2 N 4 3 &END &BETA NEL -1 0 L 0 2 N 4 3 &END &END Is this correct?<o:p></o:p> <div> <o:p></o:p> </div> <div> Also,regarding previous example on Cl1 - what is the meaning of half-electrons in alpha/beta channel? <o:p></o:p> </div> <div> <o:p></o:p> </div> Thank you very much for the help Katarina Dňa streda, 10. februára 2016 11:53:26 UTC+1 S Ling napísal(-a):<o:p></o:p> <blockquote style="border:none;border-left:solid #CCCCCC 1.0pt;padding:0cm 0cm 0cm 6.0pt;margin-left:4.8pt;margin-top:5.0pt;margin-right:0cm;margin-bottom:5.0pt"> <div> You cannot have half an electron on a p orbital?<o:p></o:p> <div> <o:p></o:p> </div> <div> In your case of Cl1, you have one extra electron in the alpha spin channel, and by doing that, you break the symmetry of alpha and beta spin channels, i.e. you MAY end up with different occupations in the alpha and beta spin channels after the calculation. Of course, this also depends on your specific system, that a symmetry-broken solution is energetically more stable.<o:p></o:p> </div> <div> <o:p></o:p> </div> <div> SL<o:p></o:p> </div> <div> <o:p></o:p> </div> </div> <div> <o:p></o:p> <div> On 9 February 2016 at 07:49, tao liu <<a href="mailto:liu...@gmail.com">liu...@gmail.com</a>> wrote:<o:p></o:p> <blockquote style="border:none;border-left:solid #CCCCCC 1.0pt;padding:0cm 0cm 0cm 6.0pt;margin-left:4.8pt;margin-top:5.0pt;margin-right:0cm;margin-bottom:5.0pt"> <div> Dear All, I can understand the &BS setting for Cu1, but still confused on Cl and Cl1. &KIND Cl BASIS_SET DZVP-MOLOPT-GTH POTENTIAL GTH-BLYP-q7 &BS &ALPHA NEL 2 L 1 N 3 &END &BETA NEL 2 L 1 N 3 &END &END &END &KIND Cl1 ELEMENT Cl BASIS_SET DZVP-MOLOPT-GTH POTENTIAL GTH-BLYP-q7 &BS &ALPHA NEL 2 L 1 N 3 &END &BETA NEL 0 L 1 N 3 &END &END &END As suggested by Marcella, for 'Cl' in the input file, Alpha is (5+2)/2=3.5, and same for Beta, in the end 'Cl' should have 3s-2 3p-7 for 'Cl1' in the input file, Alpha is (5+2)/2=3.5, and (5+0)/2 =2.5 for Beta, in the end 'Cl1' should have 3s-2 3p-6 Could anyone tell me if I understand correctly or not ? Thanks, Tao<o:p></o:p> </div> -- You received this message because you are subscribed to the Google Groups "cp2k" group. To unsubscribe from this group and stop receiving emails from it, send an email to <a href="mailto:cp2k+u...@googlegroups.com">cp2k+u...@googlegroups.com</a>. To post to this group, send email to <a href="mailto:cp...@googlegroups.com">cp...@googlegroups.com</a>. Visit this group at <a href="https://groups.google.com/group/cp2k" target="_blank">https://groups.google.com/group/cp2k</a>. For more options, visit <a href="https://groups.google.com/d/optout" target="_blank">https://groups.google.com/d/optout</a>.<o:p></o:p> </blockquote> </div> <o:p></o:p> </div> </blockquote> </div> -- You received this message because you are subscribed to the Google Groups "cp2k" group. To unsubscribe from this group and stop receiving emails from it, send an email to <a href="javascript:" target="_blank">cp2k+u...@googlegroups.com</a>. To post to this group, send email to <a href="javascript:" target="_blank">cp...@googlegroups.com</a>. Visit this group at <a href="https://groups.google.com/group/cp2k" target="_blank"> https://groups.google.com/group/cp2k</a>. For more options, visit <a href="https://groups.google.com/d/optout" target="_blank"> https://groups.google.com/d/optout</a>.<o:p></o:p> </div> </div> </blockquote> </div> -- You received this message because you are subscribed to the Google Groups "cp2k" group. To unsubscribe from this group and stop receiving emails from it, send an email to <a href="mailto:cp2k+unsu...@googlegroups.com">cp2k+unsu...@googlegroups.com</a>. To post to this group, send email to <a href="mailto:cp...@googlegroups.com">cp...@googlegroups.com</a>. Visit this group at <a href="https://groups.google.com/group/cp2k">https://groups.google.com/group/cp2k</a>. 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