[Wien] Spin Magnetic susceptibility
pieper
pieper at ifp.tuwien.ac.at
Tue Oct 3 16:41:45 CEST 2017
My two cent on this:
Spins of itinerant electrons or, more specifically, electrons in the
immediate vicinity of the Fermi surface usually will dominate the
magnetic field response at T=0 (Pauli susceptibilty) - if there is a
Fermi surface, i.e. if you have a metal, and if you do not consider
magnetic order.
However, I would not jump to the conclusion that this is the 'only' spin
contribution. Basically chi=M/H. The spin part of the magnetization M is
the field (H) induced occupation difference between spin-up and -down
channel. It occures because of the Zeeman interaction between electron
spin and field. Wether or not the electrons are localized doesn't
matter. An example might be the large spin susceptibility of
antiferromagnetic insulators.
Best regards,
---
Dr. Martin Pieper
Karl-Franzens University
Institute of Physics
Universitätsplatz 5
A-8010 Graz
Austria
Tel.: +43-(0)316-380-8564
Am 02.10.2017 18:44, schrieb karima Physique:
> Thank you very much and as you said, we are still waiting for a
> confirmation from Prof. P. Blaha and Wien2k users.
>
> 2017-09-30 21:53 GMT+02:00 Yundi Quan <yquan at ucdavis.edu>:
>
>> Hi,
>>
>> As I understand it, the susceptibility you obtain by shifting DOS is
>> the Pauli susceptibility, i.e. itinerant electrons near the Fermi
>> level. But I might be wrong.
>>
>> On Sat, Sep 30, 2017 at 8:40 AM, karima Physique
>> <physique.karima at gmail.com> wrote:
>>
>>> Dear Prof. P. Blaha and Wien2k user;
>>>
>>> Is the spin part of the magnetic susceptibility due to the
>>> contribution of delocalized electrons only?
>>>
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