[Wien] Orbital angular momentum at Fermi surface

[fecher]

Dear Matsui,
Seems you forgot somehow the dipole selection rules for photoemission.
You see most probably how the final state is alligned after photoexcitation 
assuming you performed angular resolved photoemission.

As a simple example: If you excite an s electron (spherical) with linearly 
polarized photons then youre final state is a p-electron aligned along the 
axis of the electric field vector. As px, py and pz are equally distributed 
in the final state band of the fcc lattice, you see just the allignment of 
youre polarization, that selects which partial wave it likes to have.

If youre initial state is the p-state with spherical distribution, then youre 
possible final state is composed of an spherical part (s final state, p-s 
transition, l'=l-1) and / or a nonspherical part (d final state, p-d 
transition, l'=l+1), and you have to check which transition is allowed from 
which initial to what final state band at given photon energy.
In approximation, starting from a free atom, the angular distribution will 
depend on the so called angular asymmetry parameter (depending on the final 
state energy through the matrix elements and phases). Again you may see just 
the alignment of the polarization vector that determines via the selection 
rules the final state partial waves. However, now the major intensity may be 
either along or perpendicular to that vector more or less with a p-like or 
cartwheel-like,  or donath-like shape.
Say youre final state is dz2 and the polarization was z, then the initialstate 
was pz from the selection rule m=m' and l'=l+1.

For more details check some work on angular resolved photoemission.

Ciao Gerhard

Am Mittwoch, 26. Januar 2005 15:41 schrieb matui at ms.naist.jp:
> Thank you for a quick reply.
>
> > In fcc Cu the symmetry is cubic and for this symmetry, px,py and pz must
> > be equal. For this reason, WIEN2k gives only a p-charge (and it is clear
> > from the above that px=1/3p). The d-charge is splitted into eg and t2g
> > according to cubic symmetry.
> >
> > WIEN2k will always give you an m-splitting according to group theory.
>
> I understand and agree with this point.
>
> What I am interested in is "the axis" of p orbital in k space.
> I have performed a photoelectron spectroscopy measurement.
> In the X [100] direction, p orbital seems to have its axis oriented to
> [100] direction, while in the X [010] direction, p orbital seems to have
> its axis oriented to [010] direction. In the X [001] direction, p orbital
> seems to have its axis oriented to [001] direction. I am wondering wheather
> the Fermi surface at a specific X point such as in [100] direction have
> composition of px:py:pz=1/3:1/3:1/3 or 1:0:0.
>
> Should I modify fcc Cu just a little to break the symmetry and check it?
>
> Thank you.
>
> Fumihiko Matsui
>
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