[Wien] spin polarization of valence states

Tue Dec 11 11:12:00 CET 2007

If you are talking about photoemission experiments, the question is more complicated.

You will detect spin polarized electrons even from non polarized initial states, what is basically known as Fano-effect (and its later refinements by Cherepkov).
This are truely relativistic effects that are not implemented in Wien2k.
Anyway, the spin polarization is in the final state and not in the initial state, it emerges from the optical selection rules (note the photons do not act on the spin but
on the total angular momentum). This means you populate the final states with different m_j projection numbers  differently even if the initial states had an equal occupation
of the m_j quantum numbers.
As an example, assume circularly polarised light (The details will become more complicated for linearly polarized ligh (see the work of Heinzmann et al.).
This will imply the selection rule Delta m_j=+1 for sigma^+ polarization.
If youre initial state is s_1/2,1/2 the final states is p_3/2,3/2 and for s_1/2,-1/2 you have p_1/2,1/2 and p_3/2,1/2
immidiately, you see that the final state has not longer an equal distribution of the m_j states.
The result is that you have spin polarized electrons, even so the initial state was not polarized at all and had no SO-splitting (|0+1/2| = |0-1/2|).
In the same way you will find in emission from the tungsten d-bands
polarised electrons, what does NOT mean that the bands are spin polarized anywhere in the BZ.
(Following Cherepkov you will find that the effect is even larger due to the SO-splitting of the d-states (indeed not possible for s-states).
However, if you calculate the cross section in detail, you find that you need the difference in the final state radial wave functions for p_1/2 and p_3/2,
otherwise the effect will vanish.

You may ask Jürgen Braun (braun at cs.uni-hildesheim.de) he did calculations for W or N.A. Cherepkov (State University of Aerospace Instrumentation, St. Petersburg, Russia)
that will give you much more details than I can give here in short. Both will be happy to help you with such questions, just remark that I did send you.

Ciao
Gerhard

________________________________________
Von: wien-bounces at zeus.theochem.tuwien.ac.at [wien-bounces at zeus.theochem.tuwien.ac.at] im Auftrag von Oleg Artamonov [arto at mail.nnz.ru]
Gesendet: Dienstag, 11. Dezember 2007 09:42
An: A Mailing list for WIEN2k users
Betreff: Re: [Wien] spin polarization of valence states

Dear Gerhard,

Perhaps, I ask my question not completely correct. Of course, the net spin
polarization  is absent in tungsten.
We measure in the experiment the asymmetry of the spin-up and spin-down
valence electron density in the chosen point of the BZ. This local
difference of the spin density is due to the SO interaction.
I would like to get from the calculations the SPIN ORIENTATION of the
valence electrons in the chosen point of the BZ. How is it possible to
extract this information (including the sign of the spin)?
Thanks,
                Oleg Artamonov.

----- Original Message -----
From: "Gerhard Fecher" <fecher at uni-mainz.de>
To: "A Mailing list for WIEN2k users" <wien at zeus.theochem.tuwien.ac.at>
Sent: Monday, December 10, 2007 11:18 AM
Subject: Re: [Wien] spin polarization of valence states

The SO interaction will split states according to their total angular
momentum into such with
j=|l+s| and j=|l-s|
this does NOT mean that those states are spin polarised.
To have a net spin polarisation of the density of states one needs an
unequal occupation of states with different projection quantum number m_s or
say better m_j=m_l+m_s.
In paramagnetic materials one has no magnetic exchange interaction.
Without exchange interaction - this is the case in W - all states with the
same absolut value of m_j are equally occupied
(for example the |l,j,m_j> state with |2,5/2,5/2> has always the same
occupation like the state with |2,5/2,-5/2>)
and therefore one has no spin polarisation anywhere in the BZ of a
paramagnetic material.

If the question was related to photoemission then check J. Braun, Rep. Prog.
Phys. 59 (1996) 1267–1338 (and references there) to see how it is still
possible to detect spin polarised "photo"electrons through the selection
rules.

Ciao
Gerhard

________________________________________
Von: wien-bounces at zeus.theochem.tuwien.ac.at
[wien-bounces at zeus.theochem.tuwien.ac.at] im Auftrag von Peter Blaha
[pblaha at theochem.tuwien.ac.at]
Gesendet: Sonntag, 9. Dezember 2007 15:57
An: A Mailing list for WIEN2k users
Betreff: Re: [Wien] spin polarization of valence states

Of course in tungsgten even with SO included, the net magnetic moment is
zero and there is no difference between spin-up and dn densities
(integrated over all states of the BZ).
You are also right, SO could give you some splitting of a state at a
certain k-point (leading to possible spin-polarization), but there will
be an equivalent k-point in the BZ where this SO effect is just the
opposite, so in summary the effect is zero.

Oleg Artamonov schrieb:
> Dear Wien users,
>
> Can anybody answer my question. I want to get information about the
> spin-polarization of the valence states. One can expect in the case of
> tungsten the significant energy shift between spin-up and spin-down
> particular states. How is it possible to extract the spin polarization
> state from the result of calculations with the spin-orbit interaction
> taken into account?. I use Wien2k_7.
> Thanks,
>             Oleg Artamonov.
>
>
> ------------------------------------------------------------------------
>
> _______________________________________________
> Wien mailing list
> Wien at zeus.theochem.tuwien.ac.at
> http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien
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