<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><div dir="ltr">The first BVS number is for the lattice you used, the second is an approximate estimate for an expanded lattice with the typical PBE expansion. You can/should put into a file ".latcalib" (note the "." at the front) the linear expansion/contraction for the DFT optimized volume.<div><br></div><div>The numbers you have are a bit odd, particularly the Sr is rather high. Did you take the positions from some report of SrCoO2.5 or are you just calculating the metastable SrCoO3? If a phase is metastable (or unstable) you cannot expect it to yield "simple" results such as a Co4+ valence. The Sr-Co-O family is quite complex.</div></div><div class="gmail_extra"><br><div class="gmail_quote">On Thu, May 12, 2016 at 4:44 AM, 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">
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Dear Prof. Peter and Laurence,</div>
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I am really thankful for your valuable suggestions.</div>
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<i>Prof. Laurence::</i></div>
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These are the bond-valence sums, I am getting </div>
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Atom 1 equiv 1 Sr Bond-Valence Sum 2.40 2.58</div>
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Atom 2 equiv 1 Co Bond-Valence Sum 3.03 3.20</div>
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Atom 3 equiv 1 O O1 Bond-Valence Sum 2.25 2.39</div>
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Atom 4 equiv 1 O O2 Bond-Valence Sum 1.67 1.78</div>
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Please tell the two values in each line. Can I now say that these are the valence state for respective atom. If it is so, why for Co it is not 4!!!!</div>
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<i>Prof. Peter:: </i></div>
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<div class="gmail_default"><font color="#0b5394" face="georgia, serif">I tried to stabilize a "high-spin" state for Co by changing the 0.8 on spin-dn to 0.2, and the 0.59 to 0.9 in spin-up.</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif">For Co up spin</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.89437 0.00000 0.00000 0.00000 -0.04204</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.93434 0.00000 0.00000 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.91726 0.00000 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.00000 0.93434 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> -0.04204 0.00000 0.00000 0.00000 0.89437</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif">Total spin moment: 4.57467 </font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif">For Co dn spin</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.34200 0.00000 0.00000 0.00000 -0.05202</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.33947 0.00000 0.00000 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.34654 0.00000 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.00000 0.33947 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> -0.05202 0.00000 0.00000 0.00000 0.34200</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif">Total spin moment: 1.70947</font></div>
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<div class="gmail_default"><font color="#0b5394" face="georgia, serif">Total energy is 516 meV greater than the previous case. </font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif">SPIN MAGNETIC MOMENTS OF MIXED CHARGE DENSITY</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"><span style="white-space:pre-wrap"></span>INTERSTITIAL = 0.14578</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> <span style="white-space:pre-wrap">
</span>Sr = 0.00800</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"><span style="white-space:pre-wrap"></span>Co = 2.89880</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"><span style="white-space:pre-wrap"></span>O1 = 0.23087</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"><span style="white-space:pre-wrap"></span>O2 = 0.17575 </font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif">total moment in the CELL = 3.87383</font></div>
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</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif">Looking at the moments, can I say Co is in high spin state, as the moment is still low than its actual theoretical high spin moment (~5 muB).</font></div>
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Thanks and Regards</div>
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<div class="gmail_quote">On Tue, May 10, 2016 at 5:24 PM, Komal Bapna <span dir="ltr">
<<a href="mailto:komal.bapna@gmail.com" target="_blank">komal.bapna@gmail.com</a>></span> wrote:<br>
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<div class="gmail_default"><font color="#0b5394" face="georgia, serif">Dear Prof. Peter</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif">Thank you for your kind reply..</font></div>
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<div class="gmail_default"><font color="#0b5394" face="georgia, serif">The Density matrix for Co up spin is..</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.84607 0.00000 0.00000 0.00000 -0.08661</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.92206 0.00000 0.00000 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.59673 0.00000 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.00000 0.92206 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> -0.08661 0.00000 0.00000 0.00000 0.84607</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif">Total electrons: 4.13298 </font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> </font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif">and Density matrix for Co dn spin is:</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.20583 0.00000 0.00000 0.00000 0.06263</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.80064 0.00000 0.00000 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.25923 0.00000 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.00000 0.00000 0.00000 0.80064 0.00000</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif"> 0.06263 0.00000 0.00000 0.00000 0.20583</font></div>
<div class="gmail_default"><font color="#0b5394" face="georgia, serif">Total electrons: 2.27217 </font></div>
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</font></div>
<div class="gmail_default"><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"><font color="#0b5394" face="georgia, serif"> </font></div>
<div class="gmail_default"><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>
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<div class="gmail_default"><font color="#0b5394" face="georgia, serif">Thanks and Regards </font></div>
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<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>
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Sir, </div>
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<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>
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<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>
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<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>
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<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
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</blockquote></div><br><br clear="all"><div><br></div>-- <br><div class="gmail_signature"><div dir="ltr"><div><div dir="ltr"><div><div dir="ltr"><div dir="ltr"><div dir="ltr"><div dir="ltr"><div dir="ltr"><span style="font-size:12.8px">Professor Laurence Marks</span><br></div><div dir="ltr"><span style="font-size:12.8px">"Research is to see what everybody else has seen, and to think what nobody else has thought", </span><span style="font-size:12.8px">Albert Szent-Gyorgi</span><br><a href="http://www.numis.northwestern.edu" target="_blank">www.numis.northwestern.edu</a> ; <span style="font-size:12.8px">Corrosion in 4D: </span><a href="http://MURI4D.numis.northwestern.edu" style="font-size:12.8px" target="_blank">MURI4D.numis.northwestern.edu</a><div><span style="font-size:12.8px">Partner of the CFW 100% program for gender equity, </span><a href="http://www.cfw.org/100-percent" style="font-size:12.8px" target="_blank">www.cfw.org/100-percent</a></div><div>Co-Editor, Acta Cryst A</div></div></div></div></div></div></div></div></div></div></div>
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