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Dear Stefaan,<br>
I have been watching this very interesting discussion and I found
myself that you brought me a bee in my mind. What would be "true
many-body ground state" for this isolated C atom?<br>
Regards,<br>
Jose<br>
<br>
<br>
Stefaan Cottenier wrote:
<blockquote cite="mid46A0BFED.3020406@fys.kuleuven.be" type="cite">
<blockquote type="cite">
<pre wrap="">Just to be complete, hoping not to add confusion. In 'real world' C
isolated atom is diamagnetic. That is found when applying the
Russel-Saunders coupling (two Hund's rules + spin orbit coupling).
That's not possible to reach this diamagnetic ground state with a
single-electron approach.
</pre>
</blockquote>
<pre wrap=""><!---->Completely right, Thomas, of course. A free Carbon atom has a 3P_0
ground state, hence S=1 (2S+1=3, S=1 corresponds to 2 mu_B spin moment),
L=1 (whence P) and J -- which is the many-body 'sum' of L and S -- is
zero. WIEN2k will correctly find that S=1, because it will find
:MMTOT=2.000. If one would calculate the orbital magnetic moment (by
lapwdm, 1-3), one would find -1.0 mu_B. This corresponds to one
particular orientation of L with respect to S (antiparallelly aligned,
due to the third Hund's rule), not to the true many-body ground state.
Still, this is a better approximation to the ground state energy of the
free carbon atom than taking e.g. a non-magnetic atom.
</pre>
<blockquote type="cite">
<pre wrap="">A question : can someone help me to install Wien2k (ifc and mkl too) on
my Ubuntu 6.06 system ?
</pre>
</blockquote>
<pre wrap=""><!---->Not me, I'm sorry ;-)
Stefaan
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</pre>
</blockquote>
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