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<div class="moz-cite-prefix">dear Venkataramana Imandi<br>
nspin=2 + starting_magnetization/=0 should be ok for your case i
think.<br>
<br>
the value of the starting magnetization should not be very
important. Its role is to break the up/down symmetry in the first
iteration and then the code should reach self consistency.<br>
Unless the system has many competing solutions with different
magnetizations the final scf value of the magnetization should be
the same. <br>
The initial value may affect the number of iterations needed to
reach self-consistency but should not affect the final self
consistent results. You may want to explore this dependence if you
plan to make many calculations with different supercells to study
defect or other properties.<br>
Starting magnetization can be used to explore different magnetic
configurations (ferromagnetic, AFM, more complex configurations if
many inequivalent atoms/sites are present).<br>
In bulk Iridium i think a simple FM solution should be enough.<br>
<br>
best<br>
<br>
stefano<br>
<br>
On 03/10/2015 11:29, Venkataramana Imandi wrote:<br>
</div>
<blockquote
cite="mid:CAEJJhnsg_mhGhKCJQBPBAGfkfH7YyYa5wzgNoVGNu00hFrLW9w@mail.gmail.com"
type="cite">
<div dir="ltr">
<div>Dear prof. <b>STEFANO DE GIRONCOLI</b><br>
<br>
Many thanks for spontaneous reply. On the basis of your
answer, If I understood correctly, I can use nspin=2 for
atomic Iridium and bulk Iridium<br>
(since Iridium is paramagnetic from literature data). However,
I have to specify starting_magnetization in the input file in
the both atomic Iridium and bulk Iridium<br>
input files. The keywords list information says that values
range between -1 (all spins down for the valence electrons of
atom type 'i') to 1 (all spins up). Iridium has three unpaired
electrons in the spin up.<br>
1. It indicates can I use starting_magnetization(1)=1 along
with nspin=2.<br>
2. If not that value, what value I have to use, I don't know,
please suggest me.<br>
3. I am not using full relativistic pseudopotential for
Iridium, so, I can skip nspin=4, am I correct ?.<br>
<br>
Please verify my assumptions.<br>
</div>
The reply of previous thread of clean stop of running job, now
I got clean stop during running job.<br>
</div>
<div class="gmail_extra"><br>
<div class="gmail_quote">On Sat, Oct 3, 2015 at 12:52 PM,
Venkataramana Imandi <span dir="ltr"><<a
moz-do-not-send="true"
href="mailto:venkataramana.imandi@gmail.com"
target="_blank"><a class="moz-txt-link-abbreviated" href="mailto:venkataramana.imandi@gmail.com">venkataramana.imandi@gmail.com</a></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"><br clear="all">
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>
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>venkataramana<br>
</div>
<div>PhD student<br>
</div>
<div>IIT Bombay<br>
</div>
<div>Mumbai<br>
</div>
</div>
</blockquote>
</div>
<br>
<br clear="all">
<br>
-- <br>
<div class="gmail_signature">venkataramana</div>
</div>
<br>
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