1 / 38

Theoretical description of electrons in single molecule magnets

Theoretical description of electrons in single molecule magnets. Ernest R Davidson Universities of Washington and North Carolina. [Fe 8 (C 6 H 12 N 3 H 3 ) 6 (OH) 12 O 2 ] +8 S=10. Note μ 3 O's. Fe III (d 5 ,S=5/2) 40 singly occupied orbitals. CASSCF or CASCI not feasible

auryon
Download Presentation

Theoretical description of electrons in single molecule magnets

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. Theoretical description of electrons in single molecule magnets Ernest R Davidson Universities of Washington and North Carolina

  2. [Fe8(C6H12N3H3)6(OH)12O2]+8 S=10 Note μ3 O's FeIII(d5,S=5/2) 40 singly occupied orbitals

  3. CASSCF or CASCI not feasible 40 singly occupied orbitals coupled as S=10 ~240 Slater determinants 847,660,528 with M=10 574,221,648 combinations with S=10, M=10 6,328 combinations with eight S=5/2 atoms coupled to give S=10, M=10 Would need to compute all S to show 10 was ground state

  4. Note assumed one-to-one correspondence between Heisenberg spin functions and Slater determinants: αAαAβBβB Det{(closed shells of A)φA1αφA2α (closed shells of B) φB1βφB2β}

  5. Reminder: Defines S Defines M This seems to be major source of confusion.

  6. Similarly a pair of d5 S = 5/2 centers like Mn…Mn αAαAαAαAαA βBβBβBβBβB is 0.4% S=5, 4% S=4, 14% S=3, 30% S=2, 36% S=1, and only 17% S=0 even though it is 100% M=0

  7. [Fe8(C6H12N3H3)6(OH)12O2]+8 S=10 Picture shows M = 10 NOT S=10!

  8. C. Cañada PhD Thesis 2002 S = 5 6 FeIII(d5,S=5/2) 30 singly occupied orbitals

  9. S=0

  10. Picture shows lowest energy spin orientation, not S2 eigenfunction! H=E0-2∑JABSA•SB ZINDO calculation

  11. D > 0

  12. Noodleman and Yamaguchi: For two magnetic centers, calculate spin-unrestricted energies and Ŝ2for SA ≥ SB, MA = SA, MB = ±SB, so M = |SA±SB| with values for SA and SB assumed. Substitute in averaged Heisenberg Hamiltonian H = E0– 2J ŜA•ŜB Ŝ = ŜA + ŜB 2 ŜA•ŜB = Ŝ2 – SA(SA+1) – SB(SB+1) and solve for J and E0. Use calculated value or M2+SA+SB for Ŝ2. Then the correct spin-eigenfunction energy levels are E = E0 – J [S(S+1) – SA(SA+1) – SB(SB+1)] for S= SA – SB to SA + SB in steps of 1.

  13. SA2 in black above atoms SA•SB in black above bonds SzA in blue Ideal S2 = M2 +SA+SB Mn2(OH)4(H2O)2 MnII d5 S=5/2

  14. Ligand separated MnII(d5S=5/2) A prototype dinuclear complex Lowest E for S =0 or 5 E=J/2[S(S+1)-35/2], S = 0,1,2,3,4,5

  15. For more than two spin centers follow Noodleman and Yamaguchi example: Substitute in averaged Heisenberg Hamiltonian H = E0 –ΣJABSA •SB from enough calculations to give E0 and all JAB. Then diagonalize H = E0–Σ JAB SA•SB in direct product basis of atomic spin eigenfunctions to get energy levels.

  16. Deliberately misuse DFT calculations. For Treat Kohn-Sham determinant as though it were a wave function to evaluate with projected operator. Or else use chemical intuition value in agreement with spin-space interpretation of Heisenberg Hamiltonian.

  17. Consider a molecule with 3 high-spin d5 transition metal atoms in a triangle. If all J were equal for d5 S=5/2 atoms in a triangle, H = -2J[ŜA•ŜB + ŜB•ŜC + ŜA•ŜC] E = -J[S(S+1) -105/4] S = ½ (2) ; S=3/2 (4) ; S=5/2 (6) ; S=7/2 (5) ; S=9/2 (4) ; S=11/2 (3) ; S=13/2 (2) ; S=15/2 (1) J derived from calculations with αA5αB5αC5 M=15/2 100% S=15/2 αA5αB5βC5 M=5/2 55% S=5/2, 30% S=7/2, 12% S=9/2 …

  18. Mn3O(OH)4(H2O)4 AIM charge

  19. For four or more magnetic centers we begin to get intermediate spin as the ground state. Many states of each spin. Maximum spin is the sum of the SA.

  20. Mn4O2(OH)4(H2O)6 MnII4

  21. MnII4 model

  22. Usually SA is assigned by chemical intuition! Unclear how to extract a value for SA from a wave function. J is obtained by least-squares fit to experimental data. Less clear how to obtain J from calculations. Unclear how accurate Heisenberg-Hamiltonian model is.

  23. where either Or else

  24. Basic assumption in extracting J from broken spin computations, DFT or UHF, is that the same energies are just mixed with different weights. This is easily tested for UHF by extracting the ES by spin projection (PUHF). Results show this assumption is NOT valid for UHF! Not testable for DFT because PDFT is not defined. Much debate in the literature about meaning of DFT with broken spin.

  25. Stretched H2 The prototype diradical

  26. H2 aug-cc-pvqz Localized, broken symmetry, broken spin for M=0, R > 1.8

  27. H2 aug-cc-pvqz H2 aug-cc-pvqz

  28. END The standard model is to use the Heisenberg Hamiltonian with parameters from fitting experiment or from DFT or UHF broken spin calculations. There is no check on accuracy of individual states derived in this way. There is evidence that broken spin calculations for different M differ in ESand not just in the weights. There is evidence that assumed site S values have meaning even though site charges do not match assumed oxidation numbers. Broken spin DFT calculations do not give the energy ES with S = M as often assumed for mono-radical and di-radical.

More Related