IFF 10 p. 1 Magnetism in low dimensions from first principles Atomic magnetism Gustav Bihlmayer Institut für Festkörperforschung, Quantum Theory of Materials Gustav Bihlmayer Institut für Festkörperforschung Institute for Advanced Simulation Forschungszentrum Jülich
IFF 10 p. 2 Atomic-like states in oxides Magnetite: Fe3O4 Fe 3+ Fe 3+ Fe 2+ 2-4O Oxygen (fcc) tetrahedral Fe octahedral Fe magnetic moment: A2 (here) influence of crystal field: A3 (next talk) electronic & magnetic interaction: A4 and A6 (tomorrow) reduced hopping between Fe sites: - atomic-like behaviour (A2) - correlation effects (A5, A8...)
IFF 10 p. 2 Atomic-like states in oxides 2+ Fe in octahedral field 6 d-electrons 3+ Fe in tetrahedral field t 2g 5 d-electrons e g electronic levels filled according to Hund s rules
IFF 10 p. 3 Electronic states in an atom Intraatomic exchange interaction Atoms in a magnetic field Spin-Orbit coupling Hund s rules in a solid
IFF 10 p. 4 Atomic Schrödinger equation Non-relativistic, N particles, nuclear number Z: i 1 2 2 i Z r i + 1 2 j i 1 r ij Ψ(r 1,r 2,...,r N ) = EΨ(r 1,r 2,...,r N ) Rotationally invariant Hamiltonian: angular momentum (L) is a good quantum number. orbital (angular) momentum operator: L = i h(r ) Eigenvalues of L 2 : h 2 L(L + 1) Eigenvalues of L z : hm L
IFF 10 p. 5 Single electron quantum numbers single electron with spin σ : ( 1 2 Z r ) ψ(r) = Eψ(r) Transformation r r, ϑ, φ Separation: ψ(r) = R n,l (r)θ l,ml (ϑ)φ ml (φ)σ ms (σ) Principal quantum number n defines energy eigenvalue Azimuthal quantum number l orbital angular momentum Magnetic quantum number m l orientation of l w.r.t. an axis Spin quantum number m s spin (m s = ± 1 2 ) w.r.t. an axis
IFF 10 p. 6 Many (two) electron states Single electron states m (1) l, m (1) s and m (2) l, m (2) s E.g. 2p states: l = 1 therefore m l = 1, 0, 1 and m s =,
IFF 10 p. 6 Many (two) electron states Single electron states m (1) l, m (1) s and m (2) l, m (2) s E.g. 2p states: l = 1 therefore m l = 1, 0, 1 and m s =, Possible 2-electron states: 15 determinants m (1) l m (1) s, m (2) l m (2) s 36 6 2 = 15 (Pauli principle, indistinguishable particles)
IFF 10 p. 6 Many (two) electron states Single electron states m (1) l, m (1) s and m (2) l, m (2) s E.g. 2p states: l = 1 therefore m l = 1, 0, 1 and m s =, Possible 2-electron states: 15 determinants m (1) l m (1) s, m (2) l m (2) s 36 6 2 = 15 (Pauli principle, indistinguishable particles) Group according to M L = i m(i) l E.g. 1, 1 has M L = 2 and M S = 0 and M S = i m(i) s
IFF 10 p. 6 Many (two) electron states Group according to M L = i m(i) l E.g. 1, 1 has M L = 2 and M S = 0 and M S = i m(i) s 1,1 1 D M L 1,0 1,0 1,0 1,0 1, 1 1, 1 0,0 1, 1 1, 1 1 S 0, 1 0, 1 0, 1 0, 1 1, 1 3 P M S
IFF 10 p. 7 Electronic multiplets Multiplet term symbol: M L where M = 2M S + 1 and L = S, P, D, F,... for 2p states: 1 D, 3 P, 1 S. Other example with d-states: 3d 2 1 G 1 G 3 F 1 D 3 P M L 1 S density matrix (m,m ) 3F 2 m m 1 1 G D + + 3 F M S 3F 4
IFF 10 p. 8 Hund s rules Which multiplet is the ground state? select largest spin multiplicity M select largest orbital multiplicity L compatible with M less than half-filled shells: M S antiparallel to M L more than half-filled shells: M S parallel to M L
IFF 10 p. 9 Electronic states in an atom Intraatomic exchange interaction Atoms in a magnetic field Spin-Orbit coupling Hund s rules in a solid
IFF 10 p. 10 Example: 2 particles Schrödinger equation (H 1 + H 2 + H 12 )Ψ(r 1,r 2 ) = EΨ(r 1,r 2 ) where H 12 = 1 r 12 and H i = 1 2 2 i Z r i with H i ψ ν (r i ) = ǫ ν ψ ν (r i ) Ansatz for Ψ (spatial part): Ψ(r 1,r 2 ) = aψ 1 (r 1 )ψ 2 (r 2 ) + bψ 1 (r 2 )ψ 2 (r 1 ) = a 1 1 2 2 + b 1 2 2 1
IFF 10 p. 10 Example: 2 particles Schrödinger equation (H 1 + H 2 + H 12 )Ψ(r 1,r 2 ) = EΨ(r 1,r 2 ) where H 12 = 1 r 12 Ansatz for Ψ (spatial part): Ψ(r 1,r 2 ) = aψ 1 (r 1 )ψ 2 (r 2 ) + bψ 1 (r 2 )ψ 2 (r 1 ) = a 1 1 2 2 + b 1 2 2 1 leads to H 12 (a 1 1 2 2 +b 1 2 2 1 ) = (E ǫ 1 ǫ 2 )(a 1 1 2 2 +b 1 2 2 1 ) = ǫ 12 (a 1 1 2 2 +b 1 2 2 1 ) or a( 1 1 2 2 H 12 1 1 2 2 ǫ 12 ) + b 1 1 2 2 H 12 1 2 2 1 = 0 a 1 2 2 1 H 12 1 1 2 2 + b( 1 2 2 1 H 12 1 2 2 1 ǫ 12 ) = 0
IFF 10 p. 11 Coulomb & exchange integrals Coulomb integral: J = 1 2 2 1 H 12 1 2 2 1 = ψ 1 (r 1 ) 2 ψ 2 (r 2 ) 2 r 12 dr 1 dr 2 Exchange integral: K = 1 2 2 1 H 12 1 1 2 2 = ψ1 (r 1)ψ2 (r 2)ψ 1 (r 2 )ψ 2 (r 1 ) r 12 dr 1 dr 2 rewrite in matrix form: ( (J ǫ 12 ) K K (J ǫ 12 ) ) ( a b ) = 0
IFF 10 p. 11 Coulomb & exchange integrals ( (J ǫ 12 ) K K (J ǫ 12 ) ) ( a b ) = 0 has solutions ǫ 12 = J K : a = b = 1 2 and ǫ 12 = J+K : a = b = 1 2 Since K > 0, for the lowest energy the spatial part of Ψ is antisymmetric, therefore the spin part has to be symmetric: 1 2 or 1 2 or 1 2 1 2 + 2 1 This describes a triplet state.
IFF 10 p. 12 Example: the He atom Both electrons in 1s level: Spin part has to be antisymmetric: ) 1 2 E 1 = 2ǫ 1s + J + K Ψ 1 = 1s 1 1s 2 ( 1 2 2 1
IFF 10 p. 12 Example: the He atom Both electrons in 1s level: Spin part has to be antisymmetric: ) 1 2 E 1 = 2ǫ 1s + J + K Ψ 1 = 1s 1 1s 2 ( 1 2 2 1 Electrons in 1s and 2s level: Spin part can be symmetric (triplet): E 2 = ǫ 1s +ǫ 2s +J K Ψ 2 = 1 1s 1 2s 2 1s 2 2s 1 2 1 2 1 2 1 2 + 2 1 1 2
IFF 10 p. 12 Example: the He atom Both electrons in 1s level: Spin part has to be antisymmetric: ) 1 2 E 1 = 2ǫ 1s + J + K Ψ 1 = 1s 1 1s 2 ( 1 2 2 1 Electrons in 1s and 2s level: Spin part can be symmetric (triplet): E 2 = ǫ 1s +ǫ 2s +J K Ψ 2 = 1 1s 1 2s 2 1s 2 2s 1 2 1 2 1 2 1 2 + 2 1 1 2 E 3 = ǫ 1s + ǫ 2s + J + K singlet is higher in energy
IFF 10 p. 13 Two p-electrons carbon atom: configuration 1s 2 2s 2 2p 2 ; possible multiplets: 1 D, 3 P, 1 S 1 D term: M L = 2, M S = 0, e.g. E( 1 D) = E( 1, 1 ) 3 P term: M L = 1, M S = 1, e.g. E( 3 P) = E( 1, 0 ) 1 S term: M L = 0, M S = 0 (contains also 1 D and 3 P ). 1,1 1 D M L 1,0 1,0 1,0 1,0 1, 1 1, 1 0,0 1, 1 1, 1 1 S 0, 1 0, 1 0, 1 0, 1 1, 1 3 P M S
IFF 10 p. 14 Coulomb & exchange integrals (2) Problem: compare different Coulomb & exchange integrals Simplification: factorize (atomic-like) wavefunctions: Ψ α (r 1 ) = 1 r 1 R(r 1 ; n α l α )Θ(ϑ 1 ; l α m α l )Φ(φ 1 ; m α l )Σ(σ 1 ; m α s ) rewrite J and K in angular (a, b) and radial (F, G) part: α 1 β 2 1 α 1 β 2 r 12 α 1 β 2 1 β 1 α 2 r 12 = = a (k) (l α m α l, l β m β l )F (k) (n α l α, n β l β ) k=0 b (k) (l α m α l, l β m β l )G(k) (n α l α, n β l β ) k=0
Ordering of the multiplets With the help of these Slater parameters, F (k) J.C.Slater, Phys.Rev. 34, 1293 (1929) we can write the energies of the terms: E( 1 S) = 2E 2p + F 0 + 10 25 F 2 E( 1 D) = 2E 2p + F 0 + 1 25 F 2 E( 3 P) = 2E 2p + F 0 5 25 F 2 8. Mar. 2010 Since F 0 = F 0 (21, 21) > 0 and F 2 = F 2 (21, 21) > 0, the triplet is again the ground state. IFF 10 p. 15
IFF 10 p. 16 Electronic states in an atom Intraatomic exchange interaction Atoms in a magnetic field Spin-Orbit coupling Hund s rules in a solid
IFF 10 p. 17 Schrödinger equation + magnetic field total momentum of charged particle in external field: p q c A (minimal coupling) interaction of a magnetic moment µ with a magnetic field: V Z = µ B where µ = µ B g s S For an electron of unit charge and g s = 2, we can introduce Lorentz-force and Zeeman term in the Schrödinger equation: i ( 1 2 ( p(r i ) 1 ) ) 2 c A(r i) + V (r i ) + 1 2c σ i B(r i ) + 1 2 j i 1 r ij Ψ = EΨ
IFF 10 p. 18 Magnetic field and vector potential Vector potential A and magnetic field: B = A Coulomb gauge: A = 0 Magnetic field in z-direction: A = 1 2 r 0 0 B z = 1 2 yb z xb z 0 therefore ( p 1 c A ) 2 = p 2 + 2i c A + 1 c 2A2 = p 2 + 1 c B L + 1 4c 2B2 z(x 2 + y 2 )
IFF 10 p. 19 Perturbation theory Extract magnetic part of Hamiltonian H mag = 1 2c B (L i + σ i ) + 1 i 8c 2B2 z (x 2 i + yi 2 ) i and consider these terms as a perturbation: E = E (0) 0 + 0 H mag 0 + n 0 first order in B: paramagnetic term 0 H mag n 2 E (0) 0 E n (0) second order in B: diamagnetic and van-vleck paramagnetic term
IFF 10 p. 20 Paramagnetism With µ B = 1 2c and µ 0 tot = µ B 0 i (L i + 2S i ) 0 first order term: µ B B 0 L i + 2S i 0 = B µ 0 tot i exists only, if there is a moment in the ground state! magnet liquid oxygen for 1µ B and 1 Tesla 10 5 ev paramagnetism of oxygen: from demo.physics.uiuc.edu
IFF 10 p. 21 Diamagnetism + 1 8c 2B2 z 1 8c 2B2 z 0 i 0 i (x 2 i + yi 2 ) 0 ri 2 sinϑ 2 i 0 = induced moment antiparallel to field exists also without permanent moment! magnitude 10 10 ev levitation of pyrolytic carbon in an external field (from en.wikipedia.org)
Paramagnet in a B-field MS = +3/2 Without orbital moment: 2µ B B 0 S i 0 = i 2µ B B z 0 S z,i 0 i Level splitting according to M S! with orbital moment: total angular momentum J = L+S S = 3/2 without field E 4 S MS = +1/2 MS = -1/2 MS = -3/2 with field How do L and S couple? 8. Mar. 2010 B IFF 10 p. 22
IFF 10 p. 23 Electronic states in an atom Intraatomic exchange interaction Atoms in a magnetic field Spin-Orbit coupling Hund s rules in a solid
IFF 10 p. 24 Lorentz Transformation Electric and magnetic fields in a moving reference-frame (primed): E = γ (E v B) B = γ 1 with γ = 1 v2 /c 2 ( B v E ) c 2 ( ) γ 1 (E v)v v 2 ( ) γ 1 (B v)v For v << c and B = 0 we get B 1 c 2 (v E) v 2
IFF 10 p. 25 The spin-orbit coupling (SOC) term Realtivistic effects: Dirac equation! Up to order (v/c) 2 : [ E V (r) + 1 2 2 + 1 2c 2 (E V (r))2 i c A(r) 1 2c 2A2 (r) + i (2c) 2E(r) p 1 (2c) 2 σ (p E(r)) + 1 2c σ B(r) ] ψ = 0 potential near the nucleus B p σ electron in an atom E
IFF 10 p. 26 Atomic SOC σ (E(r) p) = σ ( V (r) p) = 1 r dv (r) σ (r p) = 1 dr r dv (r) (σ L) dr coupling between spin and orbital momentum spin and orbital moment are antiparallel Coulomb-like potential for small r like V (r) = Z r ξ = 1 r dv (r) dr Z coupling strength scales with ψ ξ ψ, i.e. approx. Z 2 large for heavy atoms, splittings up to 1.5 ev (Bi)
IFF 10 p. 27 Hydrogenic atom with SOC E = E NR + E soc = 1 2 Z 2 n 2 α2 Z 4 2n 3 ( 1 j + 1 2 3 ) 4n n l s j E NR [ Z2 2 ] E soc[ α2 Z 4 2 ] 1 0 1/2 1/2 1 1/4 0 n = 3 n = 2 3D 5/2 3P 3/2,3D 3/2 3S 1/2,3P 1/2 2 0 1/2 1/2 1/4 5/64 2P 3/2 2S 1/2,2P 1/2 2 1-1/2 1/2 1/4 5/64 2 1 1/2 1/2 1/4 1/64 3 0 1/2 1/2 1/9 9/324 3 1-1/2 1/2 1/9 9/324 3 1 1/2 3/2 1/9 3/324 n = 1 3 2-1/2 3/2 1/9 3/324 3 2 1/2 5/2 1/9 1/324 1S 1/2
IFF 10 p. 28 More than 1 electron & SOC Small SOC: J = S + L with S = i s i and L = i l i Large SOC: J = i j i with j i = s i + l i µ tot = µ orb + µ spin = µ B (L + 2S) total moment is not collinear to J
IFF 10 p. 28 More than 1 electron & SOC Small SOC: J = S + L with S = i s i and L = i l i Large SOC: J = i j i with j i = s i + l i µ tot = µ orb + µ spin = µ B (L + 2S) total moment is not collinear to J MJ = +3/2 Small B-field: MJ = +1/2 µ tot = g J µ B J(J + 1) J = 3/2 MJ = -1/2 (µ tot ) z = g J µ B M J with g J = 1 + J(J+1)+S(S+1) L(L+1) 2J(J+1) without field MJ = -3/2 with field E=g µ B J B
IFF 10 p. 29 Strong magnetic field Single electron, weak field: strong field: E B = g J µ B m J B z E B = µ B (m l + 2m s )B z Example: 3 P 3/2 : m l m s m j m l + 2m s 1 1/2 3/2 2 0 1/2 1/2 1 0 1/2 1/2 1 1 1/2 3/2 2 (Paschen-Beck effect)
IFF 10 p. 30 Electronic states in an atom Intraatomic exchange interaction Atoms in a magnetic field Spin-Orbit coupling Hund s rules in a solid
IFF 10 p. 31 Hund s rules (1) select largest spin multiplicity M (2) select largest orbital multiplicity L compatible with M (3a) less than half-filled shells: M S antiparallel to M L, i.e. J = L S (3b) more than half-filled shells: M S parallel to M L, i.e. J = L + S (1) and (2) are Coulomb (and exchange) effects (3a) and (3b) follow from spin-orbit coupling S=1L=-5 S=5/2 L=0 S=3/2 L=5
IFF 10 p. 31 Hund s rules (1) select largest spin multiplicity M (2) select largest orbital multiplicity L compatible with M (3a) less than half-filled shells: M S antiparallel to M L, i.e. J = L S (3b) more than half-filled shells: M S parallel to M L, i.e. J = L + S Ions in a solid : rule (1) survives orbital moments get quenched by the crystal field if sizable orbital moments arise, rule (3) holds
IFF 10 p. 32 Example: free rare earth (RE 3+ ) ions 8 6 S L J 4 2 0 La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
IFF 10 p. 33 RE 3+ ions in a solid: garnets www.luminesce.ca www.larsenglish.com www.gemtec.co.kr RE 3 Fe 5 O 12, better: 3 RE 2 O 3 5 Fe 2 O 3 Fe 3+ atoms: 10µ B RE 3+ ions: 6(L + 2S)µ B couple antiferromagnetically
IFF 10 p. 33 RE 3+ ions in a solid: garnets RE 3 Fe 5 O 12, better: 3 RE 2 O 3 5 Fe 2 O 3 Fe 3+ atoms: 10µ B RE 3+ ions: 6(L + 2S)µ B couple antiferromagnetically 50 40 mag. moment (µ B ) 30 20 10 0-10 exp. moment 12S - 10 6(L+2S)-10 6(L/3+2S)-10 Y Gd Tb Dy Ho Er Tm Yb Lu L is quenched in the solid, but S survives!
IFF 10 p. 34 Once more: magnetite (Fe 2+ O 2 )(Fe 3+ 2 O2 3 ) small µ orb anisotropy = 2 10 4 J/m 3 d m = L -2-1 0 1 2 "e g " e g t 2g "t " 2g m = 0 L cubic trigonal spherical atomic spin-orbit splitting Fe 2+ Fe 2+ Fe 2+
IFF 10 p. 34 Once more: magnetite (Fe 2+ O 2 )(Fe 3+ 2 O2 3 ) small µ orb anisotropy = 2 10 4 J/m 3 (Co 2+ O 2 )(Fe 3+ 2 O2 3 ) large µ orb anisotropy 10 6 J/m 3 d m = L -2-1 0 1 2 "e g " e g t 2g "t " 2g m = 0 L cubic trigonal spherical atomic spin-orbit splitting Co 2+ Co 2+ Co 2+
IFF 10 p. 35 Summary transition metal atoms in oxides carry an atomic-like spin-moment S but have a quenched orbital moment L that couples due to SOC to form a total angular moment J these moments behave according to Hund s rules first maximizing the spin multiplicity then maximizing a compatible orbital moment total moment determined by SOC depending of filling: J = L S or J = L + S talk available at: http://iffwww.iff.kfa-juelich.de/ gbihl/publ/g_bihlmayer_pub.html#b016