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<div class="moz-cite-prefix">nspin is used to define whether you are
performing a <br>
non magnetic (nspin=1,default), collinear magnetic (nspin=2) or
non-collinear/fully relativistic (nspin=4) calculation.<br>
<br>
In the atomic case if you want to fix the total spin (the up/dw
density unbalance) you can use<br>
the total_magnetization variable (there use to be a redundant
multiplicity variable but i think it has been suppressed).<br>
<br>
In most cases you just run nspin=2 calculations specifying a
non-zero starting_magnetization to break the up/dw symmetry<br>
<br>
stefano<br>
<br>
On 03/10/2015 09:22, Venkataramana Imandi wrote:<br>
</div>
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cite="mid:CAEJJhnuG5_quvG9eEznNd4=jxCsUgLfSnyBnALRqPVaFcJ58cw@mail.gmail.com"
type="cite">
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Dear all<br>
<br>
I want to calculate ground state total energy of single Iridium
neutral gaseous atom.<br>
The electronic configuration of Iridium atom is
[Xe].4f^14.5d^7.6s^2 and in the 5d orbital, three unpaired
electrons are there.<br>
So, the resultant spin multiplicity is 4. Hence, in keywords
list, nspin=4 or noncolin=.true. is essential.<br>
I am asking that whether nspin means spin multiplicity or not ?.<br>
</div>
</blockquote>
<blockquote
cite="mid:CAEJJhnuG5_quvG9eEznNd4=jxCsUgLfSnyBnALRqPVaFcJ58cw@mail.gmail.com"
type="cite">
<div dir="ltr">Am i correct for determining total energy of
Iridium with nspin=4 or noncolin=.true. ?<br>
In case of bulk Iridium (total atoms=72), then can I skip
nspin=4 or noncolin=.true. ?<br>
In case of bulk Iridium electrons can get paired or not ?<br>
<br>
Finally I want to calculate cohesive energy of bulk Iridium.<br>
<br>
I am extremely say sorry, if questions are fundamental and
trivial.<br>
<br>
Any suggestions are appreciated and thanks in anticipated.<br>
<br>
<br>
<div class="gmail_signature">venkataramana<br>
</div>
<div class="gmail_signature">PhD student<br>
</div>
<div class="gmail_signature">IIT Bombay<br>
</div>
<div class="gmail_signature">Mumbai<br>
</div>
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<br>
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