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

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Presentation transcript:

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 Fe III (d 5,S=5/2) 40 singly occupied orbitals Note μ 3 O's

CASSCF or CASCI not feasible 40 singly occupied orbitals coupled as S=10 ~2 40 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

Note assumed one-to-one correspondence between Heisenberg spin functions and Slater determinants: α A α A β B β B  Det{(closed shells of A)φ A1 α φ A2 α (closed shells of B) φ B1 β φ B2 β}

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

Similarly a pair of d 5 S = 5/2 centers like Mn…Mn α A α A α A α A α A β B β B β B β B β B 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

[Fe 8 (C 6 H 12 N 3 H 3 ) 6 (OH) 12 O 2 ] +8 S=10Picture shows M = 10 NOT S=10!

C. Cañada PhD Thesis 2002 S = 5 6 Fe III (d 5,S=5/2) 30 singly occupied orbitals

S=0

S=5 S=0 Canada-Vilalta, O’Brien, Christou, et al

Picture shows lowest energy spin orientation, not S 2 eigenfunction! H=E 0 -2∑J AB S A S B ZINDO calculation

D > 0

Noodleman and Yamaguchi: For two magnetic centers, calculate spin-unrestricted energies and  Ŝ 2  for S A ≥ S B, M A = S A, M B = ±S B, so M = |S A ±S B | with values for S A and S B assumed. Substitute in averaged Heisenberg Hamiltonian  H  = E 0 – 2J  Ŝ A Ŝ B  Ŝ = Ŝ A + Ŝ B 2  Ŝ AŜ B  =  Ŝ 2  – S A (S A +1) – S B (S B +1) and solve for J and E 0. Use calculated value or M 2 +S A +S B for  Ŝ 2 . Then the correct spin-eigenfunction energy levels are E = E 0 – J [S(S+1) – S A (S A +1) – S B (S B +1)] for S= S A – S B to S A + S B in steps of 1.

 S A 2  in black above atoms  S AS B  in black above bonds  S zA  in blue Ideal  S 2  = M 2 +S A +S B Mn 2 (OH) 4 (H 2 O) 2 Mn II d 5 S=5/2

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

For more than two spin centers follow Noodleman and Yamaguchi example: Substitute in averaged Heisenberg Hamiltonian  H  = E 0 – Σ J AB  S A S B  from enough calculations to give E 0 and all J AB. Then diagonalize H = E 0 – Σ J AB S A S B in direct product basis of atomic spin eigenfunctions to get energy levels.

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.

Consider a molecule with 3 high-spin d 5 transition metal atoms in a triangle. If all J were equal for d 5 S=5/2 atoms in a triangle, H = -2J[Ŝ A Ŝ B + Ŝ B Ŝ C + Ŝ A Ŝ C ] 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 α A 5 α B 5 α C 5 M=15/2 100% S=15/2 α A 5 α B 5 β C 5 M=5/2 55% S=5/2, 30% S=7/2, 12% S=9/2 …

Mn 3 O(OH) 4 (H 2 O) 4 AIM charge

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 S A.

Mn 4 O 2 (OH) 4 (H 2 O) 6 Mn II 4

Mn II 4 model

Usually S A is assigned by chemical intuition! Unclear how to extract a value for S A 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.

where either Or else

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 E S 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.

Stretched H 2 The prototype diradical

H 2 aug-cc-pvqz Localized, broken symmetry, broken spin for M=0, R > 1.8

H 2 aug-cc-pvqz

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 E S and 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 E S with S = M as often assumed for mono-radical and di-radical.