[Pw_forum] Relation between the magnetic field and the Fermi energies

[Matthias Timmer]

Dear Madams and Sirs,

I am a PhD student from Germany, and at the moment I treat spin 
relaxation effects.
I am trying to figure out how the relation between the magnetic field 
(bfield(3) in weights, was in sum_band before) and the Fermi energies 
(in a constrained calculation) is justified, which is given there. The 
given relation is bfield(3) = 0.5D0*( ef_up - ef_dw ), which I 
understand is just a simple model, i.e. twice the Zeeman energy is equal 
to the difference in the Fermi energies.
However, the Bohr magneton should be present, if I assume that the 
magnetic field is given in Rydberg atomic units. The Bohr magneton is 
sqrt(2) in Rydberg a.u. (Bohr magneton = e \hbar / (2 m_electron) = 
sqrt(2) *1 / (2 * 0.5)). That  means I expect the formula bfield(3) = 
0.5D0*( ef_up - ef_dw )/sqrt(2) should be there. When I take a look at 
"electrons.f90", however, the output for bfield(3) says that it is 
already in Rydberg atomic units (format 9071).
Now I ask myself: Am I missing something (a line of code, or a line of 
thought, or a factor somewhere), or is there something wrong with the 
output (should it say "bohr" instead of "Rydberg" for bfield(3))?
I am sorry if this has been answered before, but at least I did not find 
this in the archive.
Thank you in advance for your answers!

Best regards,

Matthias Timmer
Universitaet Duisburg-Essen
Fachbereich Physik
Lotharstrasse 1
47057 Duisburg
-------------- next part --------------
A non-text attachment was scrubbed...
Name: Matthias.Timmer.vcf
Type: text/x-vcard
Size: 290 bytes
Desc: not available
Url : http://www.democritos.it/pipermail/pw_forum/attachments/20080717/db05abcd/attachment.vcf