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<p class="MsoNormal"><span style="font-size:11.0pt">Dear QE users,<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt">I’m quite new to QE and was trying to get the energy difference among different spin state (low spin, intermediate spin and high spin) in the LaCoO3 compound. I started with a simple rhombohedral primitive
cell which has 2 Co, 2 La and 6 O in it. To depict the high spin state, there will be 4 net spins up, so I set the tot_magnetization=8. And for the intermediate spin state, I set tot_magnetization=4.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt">The output has the total magnetization equals exactly to the number I set, and absolute value is very close to that (e.g., 4.09 for IS state), which seems OK to me. But if I check the magnetization value on
each atom, for HS state, O has 0.29 value and both Co has 2.82. I would imagine it should be 0 for O and 4 for Co each. And for IS state, the two Co has different values, one is 0.98 and the other is 2.28. I was expecting the value to be 2 for each Co. I’m
wondering if these are reasonable results? Should I expect or is there a way to get the ideal spin state in the real calculation?<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt">Thank you so much for your help!<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt">Sincerely,<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt">Shenli<o:p></o:p></span></p>
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