<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><div dir="ltr"><div class="gmail_default" style=""><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif">Dear Prof. Peter</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif">Thank you for your kind reply..</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"><br></font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif">The Density matrix for Co up spin is..</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> 0.84607 0.00000 0.00000 0.00000 -0.08661</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> 0.00000 0.92206 0.00000 0.00000 0.00000</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.59673 0.00000 0.00000</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.00000 0.92206 0.00000</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> -0.08661 0.00000 0.00000 0.00000 0.84607</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif">Total electrons: 4.13298 </font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> </font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif">and Density matrix for Co dn spin is:</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> 0.20583 0.00000 0.00000 0.00000 0.06263</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> 0.00000 0.80064 0.00000 0.00000 0.00000</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.25923 0.00000 0.00000</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.00000 0.80064 0.00000</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> 0.06263 0.00000 0.00000 0.00000 0.20583</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif">Total electrons: 2.27217 </font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"><br></font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif">This gives me the net spin moment of 1.86 muB on Co.</font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"> </font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif">I understood by changing the occupancies we can change the spin state. Still I am confused whether I can use these occupation numbers of electrons in the respective orbitals to find the spin state of Co.Here, total number of electrons is 6.4, 1.4 electrons higher than the expected value for Co4+ ion. I may consider the extra electrons due to the bonding of Co-O bonds. But, I am really unable to calculate the spin state from the present configuration. Please suggest how to calculate spin state for the given configuration, so that I can modify further. </font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif"><br></font></div><div class="gmail_default" style=""><font color="#0b5394" face="georgia, serif">Thanks and Regards </font></div></div></div><div class="gmail_extra"><br><div class="gmail_quote">On Mon, May 9, 2016 at 3:59 PM, Komal Bapna <span dir="ltr"><<a href="mailto:komal.bapna@gmail.com" target="_blank">komal.bapna@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div class="gmail_default" style="font-family:georgia,serif;color:rgb(11,83,148)">Sir, </div><div class="gmail_default" style="font-family:georgia,serif;color:rgb(11,83,148)"><br></div><div class="gmail_default" style="font-family:georgia,serif;color:rgb(11,83,148)"><pre style="font-family:courier,'courier new',monospace;font-size:14px;white-space:pre-wrap;word-wrap:break-word;margin:0em;color:rgb(0,0,0);line-height:19.6px">Its true that we can not generate "ionic" electron density with lstart/dstart. When I tried to modify .inst file according to the Co4+ ionic state, it showed error.</pre><pre style="font-family:courier,'courier new',monospace;font-size:14px;white-space:pre-wrap;word-wrap:break-word;margin:0em;color:rgb(0,0,0);line-height:19.6px">My query is that how can we generate spin state configuration for such an ionic state, it spin states for Co and Co4+ are different and accordingly the magnetic moments.</pre><pre style="font-family:courier,'courier new',monospace;font-size:14px;white-space:pre-wrap;word-wrap:break-word;margin:0em;color:rgb(0,0,0);line-height:19.6px"><br></pre><pre style="font-family:courier,'courier new',monospace;font-size:14px;white-space:pre-wrap;word-wrap:break-word;margin:0em;color:rgb(0,0,0);line-height:19.6px">Please suggest.</pre><pre style="font-family:courier,'courier new',monospace;font-size:14px;white-space:pre-wrap;word-wrap:break-word;margin:0em;color:rgb(0,0,0);line-height:19.6px"><br></pre><pre style="font-family:courier,'courier new',monospace;font-size:14px;white-space:pre-wrap;word-wrap:break-word;margin:0em;color:rgb(0,0,0);line-height:19.6px">Thanks</pre></div></div><div class="gmail_extra"><div><div class="h5"><br><div class="gmail_quote">On Fri, May 6, 2016 at 3:18 PM, Komal Bapna <span dir="ltr"><<a href="mailto:komal.bapna@gmail.com" target="_blank">komal.bapna@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div class="gmail_default" style="font-family:georgia,serif;color:rgb(11,83,148)"><pre style="font-family:courier,'courier new',monospace;font-size:14px;white-space:pre-wrap;word-wrap:break-word;margin-top:0px;margin-bottom:0px;color:rgb(0,0,0);line-height:19.6px">Dear Wien users
I am working on Sr2CoO4. Here I wanted to study the system with different spin state configuration of Co4+, which is known to be valence state of Co in this system. I could understand how to create:
(a) High-spin configuration
(b) Intermediate spin configuration
(c) Low-spin configuration
for the given Co atoms in the .inst file as</pre><pre style="font-family:courier,'courier new',monospace;font-size:14px;white-space:pre-wrap;word-wrap:break-word;margin-top:0px;margin-bottom:0px;color:rgb(0,0,0);line-height:19.6px">Co
Ar 3
3, 2,2.0 N
3, 2,2.0 N
3,-3,3.0 N
3,-3,0.0 N
4,-1,1.0 N
4,-1,1.0 N (for HS state)
and
Co
Ar 3
3, 2,2.0 N
3, 2,2.0 N
3,-3,2.0 N
3,-3,1.0 N
4,-1,1.0 N
4,-1,1.0 N (for IS state)
But my query is that .inst file takes Co as neutral atom (9
electrons:3d74s2) and accordingly its spin state. As if Co were in 4+ state, I would have 5 electrons in 3d state (3d54s0) rather 9 electrons as is revealed from .inst file now.
Please suggest me how to give spin state for Co4+ for this system.
Thanks
</pre><span><font color="#888888"><div><br></div></font></span></div><span><font color="#888888"><div><br></div>-- <br><div><div dir="ltr"><div><div dir="ltr"><div><font color="#0b5394"><b>Komal</b></font></div></div></div></div></div>
</font></span></div>
</blockquote></div><br><br clear="all"><div><br></div></div></div><span class="HOEnZb"><font color="#888888">-- <br><div><div dir="ltr"><div><div dir="ltr"><div><font color="#0b5394"><b>Komal</b></font></div></div></div></div></div>
</font></span></div>
</blockquote></div><br><br clear="all"><div><br></div>-- <br><div class="gmail_signature"><div dir="ltr"><div><div dir="ltr"><div><font color="#0b5394"><b>Komal</b></font></div></div></div></div></div>
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