[Wien] spin polarization of valence states

Tue Dec 11 20:30:33 CET 2007

The SO interaction will split (for example) the d-bands into
d_3/2 and d_5/2 bands
both of them will be completely degenerate with respect to m_s and therefore they will have no net spin polarisation the expectation value
of m_s is Zero for both d_3/2 and d_5/2.
Therefore, I assume you like to distinguish the bands with different j.

To see which band belongs to j=3/2 (l-s) or j=5/2 (l+s), or say better those with different double group symmetry,  you would need a band character plot with respect to the total angular momentum and its projection j, m_j.
as far as I know, the band character plots in Wien2k are only with respect to l, m_l so they do not help.
With the density of states it  may work if you neglect the small components of the relativistic wave functions, as you can transform the |l,m_l>|s, m_s> states by Glebsch-Cordan coefficients easily into |j, m_j> states.
Actually, I use fully relativistic programs if I need the double group representations of the bands or densities, as they automatically provide the j-dependent radial functions as well as their large and small components.

Maybe, Peter knows a solution for Wien2k.

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 19:03
An: A Mailing list for WIEN2k users
Betreff: Re: [Wien] spin polarization of valence states

Dear Gerhard,

I work in the cooperation with J. Kirshner and R. Feder. But, please come
back to my primary question. Let's suppose that due to SO coupling we see
band splitting in the paramagnetic crystal. Is it possible using Wien2k to
distinguish which sing of the spin is predominant in each of the bands?
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 8:37 PM
Subject: Re: [Wien] spin polarization of valence states

> In that case you should consult some basic works:
> Joachim Kessler "Polarized electrons" (this group did a lot of work on
> scattering of spin polarized electrons)
> J¨¹rgen Kirschner "Polarized electrons at surfaces" (who invented the spin
> detector based on scattering of low energy electrons at tungsten)
> both published by Springer
> or
> R. Feder (ed.) Polarized Electron in Surface Physics, World Scientific
> Singapore
>
> 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 13:42
> An: A Mailing list for WIEN2k users
> Betreff: Re: [Wien] spin polarization of valence states
>
> 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 §¤ 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 J¨¹rgen 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¨C1338 (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
>> Wien at zeus.theochem.tuwien.ac.at
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