[Wien] spin polarization of valence states
Oleg Artamonov
arto at mail.nnz.ru
Tue Dec 11 13:42:49 CET 2007
Dear Gerhard,
We measure the spin-asymmetry of the inelastic scattering of the
spin-polarized low energy electron from the valence electron on W single
crystal. The more probable origin of the measured spin-asymmetry is the
exchange interaction between primary and valence electrons. The SO in the
valence states may give the energy splitting in some points of the BZ. For
this reason we expect the asymmetry of the spin density of the valence
electron in
thisi points. To conform this explanation I would like to get from the
calculations the SPIN ORIENTATION of the
valence electrons in the chosen point of the BZ.
For the normal incidence the measured spin-asymmetry changes the sign for
symmetry points in the BZ (relative to Г point) and that confirms the idea
of the SO origin of the spin-asymmetry in the initial state.
Best regards,
Oleg.
----- 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: Tuesday, December 11, 2007 1:12 PM
Subject: Re: [Wien] spin polarization of valence states
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.
>
>
> ------------------------------------------------------------------------
>
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> Wien mailing list
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