1 / 22

5.0 引言 5.1 轨道 , 相互作用与自旋 5.2 原子和分子的磁矩 5.3 晶体的磁矩 5.4 晶体的磁各向异性 5.5 习题

5. 　磁性与电子态. 5.0 引言 5.1 轨道 , 相互作用与自旋 5.2 原子和分子的磁矩 5.3 晶体的磁矩 5.4 晶体的磁各向异性 5.5 习题. Outline. General remarks Uniaxial cases Cubic crystals Why success limited. Decoupling of Spin from Orbit. Even for a spin-dependent Exc, such as Von Barth and Hedin ( 1972),

lan
Download Presentation

5.0 引言 5.1 轨道 , 相互作用与自旋 5.2 原子和分子的磁矩 5.3 晶体的磁矩 5.4 晶体的磁各向异性 5.5 习题

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 5.　磁性与电子态 5.0 引言5.1 轨道,相互作用与自旋5.2 原子和分子的磁矩5.3 晶体的磁矩5.4 晶体的磁各向异性5.5 习题

  2. Outline General remarksUniaxial cases Cubic crystalsWhy success limited

  3. Decoupling of Spin from Orbit Even for a spin-dependent Exc, such as Von Barth and Hedin (1972), Vsxc = dExc/dr = Vxc0(r)+sVxcm(r,m) The spin-up and spin-down states are decoupled. Total energy depends only on the magnitude of spin polarization, m, but independent of its direction.

  4. Spin-orbit Coupling Causes Anisotropy Spin-orbit coupling Hsl = [(1/4c2r) ¶V/ ¶r]l(r)×s = x( r)l(r)×s Total energy variation Esl(s ) = E(H0+Hsl)- E(H0)

  5. First order energy E(s) = Sx <o|l×s |o> because <o|l|o>=0 Second (even) order energy, E(s )=Sx 2|<o|l×s |e>|2 / (e (o)-e (e)) + h.o.t Mostly between spin-down bands Perturbation Analysis

  6. Directional Dependence Due to orbital character of the o-e pairs near Fermi surface

  7. Perpendicular AnisotropyD.S.Wang et al, PRB47, 14932, 1993 Fe film: coupling <5|lz|5*>=<xz|lz|yz> causes perpendicular anisotropy Singularity occurs when |e (o)-e (e)| <»x

  8. In-plane AnisotropyD.S.Wang et al, JMMM 129, 344, 1994 Co film: coupling <5|ly|1>=<xz|ly|z2> <5*|lx|1>=<yz|lx|z2> causes in-plane anisotropy Singularity occurs when |e (o)-e (e)| <»x

  9. Experiments vs TheoryD.S.Wang et al. JMMM 140, 643, 1995

  10. Anisotropy vs Band Filling

  11. Anisotropy of X-Co-XD.S.Wang PRB48, 15886, 1993

  12. Ni Layers on Cu Substrate J.Henk et al, PRB59, 9332 (1999)

  13. Ni Layers on Cu Substrate J.Henk et al, PRB59, 9332 (1999) For fct, the bulk contribution is nearly correct, but contribution of the sub-surface layer seems wrong.

  14. Distorted Cubic Crystals T.Burkert et al. PRB 69, 104426 (2004)

  15. Cubic Crystals – Early Empirical Authors E(001)-E(111) in meV/atom Remarks bcc Fe fcc Co fcc Ni Experiments -1.4 1.8 2.7 Kondorskii et al /JETP36,188(1973) x x 1.3 Empirical Fritsche et al /J.Phys.F17,943(1987) 7.4 x 10.0

  16. Cubic Crystals - LSDA Authors E(001)-E(111) in meV/atom Remarks bcc Fe fcc Co fcc Ni Experiment -1.4 1.8 2.7 Daalderop et al /PRB41,11919(1990) -0.5 x -0.5 Strange et al /Physica B172,51(1991) -9.6 x 10.5 Trygg et al /PRL75,2871(1995) -0.5 0.5 -0.5 Razee at al /PRB56,8082(1997) -0.95 0.86 0.11 Halilov et al /PRB57,9557(1998) - 0.5 0.3 0.04 -2.6 2.4 1.0 x scaling

  17. Cubic Crystals – LSDA+OP Authors E(001)-E(111) in meV/atom Remarks bcc Fe fcc Co fcc Ni Experiment -1.4 1.8 2.7 Trygg et al /PRL75,2871(1995) -1.8 2.2 -0.5 OP Yang et al. /PRL87,216405(2001) U=1.2 x U=1.9 in eV J=0.8 x J=1.2 in eV Xie et al /PRB69,172404(2004) U=1.15 U=1.41 U=2.95 in eV J=0.97 J=0.83 J=0.28 in eV

  18. Ab Initio Attempt - Summary • Bulk uniaxial cases are good • Surface (interface) layers are fair • Cubic crystals are poor

  19. Uniaxial Case: Two <o|*|e>Pairs Reconsider the second order perturbation, E(s )=-Sx 2|<o|l×s |e>|2 / (e (e)-e (o)) It holds only when e(e)-e(o) > x , and S » 1 . For uniaxial cases, the regular part is in 2nd order (x 2/ D)! When e(e)-e(o) < x , degenerate perturbation applies, E(s )=-S |x <o|l×s |e>| and S » (x2 / | Ñke(o)×Ñke(e)| ). Singular at those k points. Total contribution is in 3rd order (x 3/ D2)!.

  20. Cubic Case: Two <o|*|e>Pairs The second order perturbation, E(s )=-Sx 2|<o|l×s |e>|2 / (e (e)-e (o)) is isotropic. For cubic case, the regular part of anisotropy goes to E(s ) »-Sx 4|<l×s >|4 / (e (e)-e (o))3 , and S » 1 . The contribution is in the 4th order (x2/ D3)! The singular part with, E(s )=-S |x <o|l×s |e>| and S » (x2 / | Ñke(o)×Ñke(e)| ) Singular at those k points. Total contribution is in 3rd order (x 3/ D2)!.

  21. Challenge in Cubic Case • Count the correlation in acceptable accuracy between the nearly degenerate pairs of empty and occupied states around Fermi surface!.

  22. Concluding Comment One can not claim understand unless he can calculate ! - J.C.Slater

More Related