<div dir="ltr"><div>Dear Matthias,</div><div>thank you for the suggestion. Could you, please, explain why NEL in alpha channel is 1? <br></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?</div><div><br></div><div>Thanks!<br> </div>Dňa štvrtok, 8. novembra 2018 11:49:26 UTC+1 Matthias Krack napísal(-a):<blockquote class="gmail_quote" style="margin: 0;margin-left: 0.8ex;border-left: 1px #ccc solid;padding-left: 1ex;">
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<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1f497d" lang="EN-GB">Dear Katarina</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1f497d" lang="EN-GB"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1f497d" lang="EN-GB">You may try for Cu1 alpha</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Consolas;color:#1f497d" lang="EN-GB"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Consolas;color:#1f497d" lang="EN-GB">NEL 1 0</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Consolas;color:#1f497d" lang="EN-GB">L 0 2</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Consolas;color:#1f497d" lang="EN-GB">N 4 3</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1f497d" lang="EN-GB"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1f497d" lang="EN-GB">instead of</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1f497d" lang="EN-GB"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Consolas;color:#1f497d" lang="EN-GB">NEL 0 0</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Consolas;color:#1f497d" lang="EN-GB">L 0 2</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Consolas;color:#1f497d" lang="EN-GB">N 4 3</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:Consolas;color:#1f497d" lang="EN-GB"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1f497d" lang="EN-GB">Hope that works.</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1f497d" lang="EN-GB"> </span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1f497d" lang="EN-GB">Matthias</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1f497d" lang="EN-GB"> </span></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">From:</span></b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"> <a href="javascript:" target="_blank" gdf-obfuscated-mailto="y93GMi4GBQAJ" rel="nofollow" onmousedown="this.href='javascript:';return true;" onclick="this.href='javascript:';return true;">cp...@googlegroups.com</a> <<a href="javascript:" target="_blank" gdf-obfuscated-mailto="y93GMi4GBQAJ" rel="nofollow" onmousedown="this.href='javascript:';return true;" onclick="this.href='javascript:';return true;">cp...@googlegroups.com</a>>
<b>On Behalf Of </b>katarína stanciaková<br>
<b>Sent:</b> Donnerstag, 8. November 2018 11:13<br>
<b>To:</b> cp2k <<a href="javascript:" target="_blank" gdf-obfuscated-mailto="y93GMi4GBQAJ" rel="nofollow" onmousedown="this.href='javascript:';return true;" onclick="this.href='javascript:';return true;">cp...@googlegroups.com</a>><br>
<b>Subject:</b> Re: [CP2K:10913] Re: Initial guess of the density matrix</span></p>
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<p class="MsoNormal">Dear all,<br>
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+:<br>
<br>
Cu+ with electron configuration 3d9 4s1 (Cu1) and 3d10 4s0 (Cu2). I tried to build my input sections:<br>
<br>
&KIND Cu1<br>
ELEMENT Cu<br>
BASIS_SET DZVP-MOLOPT-GTH<br>
POTENTIAL GTH-BLYP-q11<br>
&BS<br>
&ALPHA<br>
&ALPHA<br>
NEL 0 0<br>
L 0 2<br>
N 4 3<br>
&END<br>
&BETA<br>
NEL -1 -2<br>
L 0 2<br>
N 4 3<br>
&END<br>
&END<br>
<br>
and <br>
<br>
&KIND Cu2<br>
ELEMENT Cu<br>
BASIS_SET DZVP-MOLOPT-GTH<br>
POTENTIAL GTH-BLYP-q11<br>
&BS<br>
&ALPHA<br>
&ALPHA<br>
NEL -1 0<br>
L 0 2<br>
N 4 3<br>
&END<br>
&BETA<br>
NEL -1 0<br>
L 0 2<br>
N 4 3<br>
&END<br>
&END<br>
<br>
Is this correct?</p>
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<p class="MsoNormal"> </p>
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<p class="MsoNormal">Also,regarding previous example on Cl1 - what is the meaning of half-electrons in alpha/beta channel?
</p>
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<p class="MsoNormal"> </p>
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<p class="MsoNormal">Thank you very much for the help<br>
Katarina<br>
<br>
<br>
Dňa streda, 10. februára 2016 11:53:26 UTC+1 S Ling napísal(-a):</p>
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<p class="MsoNormal">You cannot have half an electron on a p orbital?</p>
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<p class="MsoNormal"> </p>
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<p class="MsoNormal">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.</p>
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<p class="MsoNormal"> </p>
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<p class="MsoNormal">SL</p>
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<p class="MsoNormal"> </p>
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<p class="MsoNormal"> </p>
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<p class="MsoNormal">On 9 February 2016 at 07:49, tao liu <<a>liu...@gmail.com</a>> wrote:</p>
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<p class="MsoNormal" style="margin-bottom:12.0pt">Dear All,<br>
<br>
I can understand the &BS setting for Cu1, but still confused on Cl and Cl1.<br>
<br>
&KIND Cl<br>
BASIS_SET DZVP-MOLOPT-GTH<br>
POTENTIAL GTH-BLYP-q7<br>
&BS<br>
&ALPHA<br>
NEL 2<br>
L 1<br>
N 3<br>
&END<br>
&BETA<br>
NEL 2<br>
L 1<br>
N 3<br>
&END<br>
&END<br>
&END<br>
&KIND Cl1<br>
ELEMENT Cl<br>
BASIS_SET DZVP-MOLOPT-GTH<br>
POTENTIAL GTH-BLYP-q7<br>
&BS<br>
&ALPHA<br>
NEL 2<br>
L 1<br>
N 3<br>
&END<br>
&BETA<br>
NEL 0<br>
L 1<br>
N 3<br>
&END<br>
&END<br>
&END<br>
<br>
As suggested by Marcella, <br>
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<br>
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<br>
<br>
Could anyone tell me if I understand correctly or not ?<br>
<br>
Thanks,<br>
Tao</p>
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