RJS-Vol-1-Sept-2006-4.dvi RUHUNA JOURNAL OF SCIENCE Vol. 1, September 2006, pp. 32–40 http://www.ruh.ac.lk/rjs/ issn 1800-279X ©2006 Faculty of Science University of Ruhuna. Variation of magnetic energy in oriented ultra-thin ferromagnetic films P. Samarasekara Department of Physics, University of Ruhuna, Matara, Sri Lanka, pubudus@phy.ruh.ac.lk The angle dependence of magnetic energy of ultra-thin films with N=1 to 5 layers has been investigated. The effect of stress and the demagnetization factor on the classical Heisenberg Hamiltonian has been taken into account in addition to other magnetic factors. Films of sc(001) and bcc(001) ferromagnetic indicate preferred in plane and perpendicular orienta- tions for N≤3 and N≥4, respectively. Easy and hard directions of all sc(001), bcc(001) and fcc(001) films with five layers are θ = 7.2°and θ = 114°respectively. According to our simu- lation, ultra-thin films with one or two layers can be easily oriented in θ = 279°direction. Key words : Spin, magnetic moment, magnetic anisotropy, ferromagnetic materials, thin films, Heisenberg Hamiltonian 1. Introduction. Although the experimental evidence of oriented and non-oriented ferromagnetic and ferrimagnetic films can be found in many early reports, the theoretical aspects of oriented ferrimagnetic films were not revealed in a considerable manner yet. The sputter synthesis of aligned strontium hexaferrites, SrO.6(Fe2O3), films can be given as an example of the deposition of oriented ferromagnetic film (Hegde et al. 1994a). The sputter synthesis of TbCu7 type Sm(CoFeCuZr) films with controlled easy axis orientation (Hegde et al. 1994b) and nitriding studies of aligned high anisotropy ThMn12-type NdFe11Co1−yMoyN film samples can be found in some experimental reports of oriented films (Navarathna et al. 1994). Easy axis oriented Nickel ferrite (Samarasekara et al. 1996) and Lithium mixed ferrite (Samarasekara and Cadieu 2001) thin films have been fabricated on single crystal substrates by means of pulse laser deposition technique. The magnetic properties of ferromagnetic films have been investigated using Bloch-spin wave theory (Martin and Robert 1951). Monte Carlo simulation of hys- teresis loops of FeF2(110)/Fe bilayers has been performed to study the exchange interaction between antiferromagnetic and ferromagnetic sublattices (Lederman et al. 2004). The Energy of oriented ferromagnetic thick films has been calculated using the Heisenberg Hamiltonian in one of our early report. The energy of oriented ultra-thin ferromagnetic films will be described in this report. The effect of spin exchange interaction, magnetic dipole interaction, anisotropies up to eighth order, demagnetization factor and stress induced anisotropy on Heisenberg Hamiltonian was studied. 32 Samarasekara: Variation of magnetic energy... Ruhuna Journal of Science 1, pp. 32–40, (2006) 33 2. Model and discussion The classical Heisenberg Hamiltonian for ferromagnetic thin films can be given as following. H = − J 2 ∑ m,n ~Sm.~Sn + ω 2 ∑ m 6=n ( ~Sm.~Sn r3mn − 3(~Sm.~rmn)(~rmn.~Sn) r5mn ) − ∑ m D (2) λm (Szm) 2 − ∑ m xD (4) λm (Szm) 4 − ∑ m D (6) λm (Szm) 6 − ∑ m D (8) λm (Szm) 8 − ∑ m,n [ ~H − (Nd ~Sn/µ0)].~Sm − ∑ m Ks sin 2θm Above equation for oriented ferromagnetic thin films can be deduced to following form (Usadel and Hucht 2002). E(θ) = − 1 2 N ∑ m,n=1 (J Z|m−n| − ω 4 Φ|m−n|) + 3ω 8 cos 2θ N ∑ m,n=1 Φ|m−n| − cos2 θ N ∑ m=1 D(2)m − cos 4 θ N ∑ m=1 D(4)m − cos6 θ N ∑ m=1 D(6)m − cos 8 θ N ∑ m=1 D(8)m − N (Hin sin θ + Hout cos θ − Nd µ0 + K5 sin 2θ) Here J, D(2)m , D (4) m , D (6) m , D (8) m , N, Hin, Hout, Nd, K5, Z|m−n|, Φ|m−n|, ω and θ are spin exchange interaction, second, fourth, sixth and eight order anisotropy coefficients, number of layers of the film, in plane applied magnetic field, out of plane applied magnetic field, demagnetization factor, stress induced anisotropy coefficient, num- ber of nearest neighbors between layers, constants related to partial summation of dipole interaction, strength of long range dipole interaction and the angle between the spin and the film normal, respectively. For microscopic anisotropy, sixth and eighth order anisotropies are also induced. For an ultra thin film with few layers, it is reasonable to assume that D(2)m , D (4) m , D (6) m , D (8) m are constants. Because Zδ>1 = 0, ∑N m,n=1 Z|m−n| = N Z0 + 2(N − 1)Z1. After assuming that Φδ>1 = 0, ∑N m,n=1 Φ|m−n| = N Φ0 + 2(N − 1)Φ1. Then above equation can be reduced into following form, E(θ) = − J 2 [N Z0 + 2(N − 1)Z1] + ω 8 [N Φ0 + 2(N − 1)Φ1](1 + 3 cos 2θ) −N [D(2)m cos 2 θ + D(4)m cos 4 θ + D(6)m cos 6 θ + D(8)m cos 8 θ +Hin sin θ + Hout cos θ − Nd µ0 + K5 sin 2θ] (1) Samarasekara: Variation of magnetic energy... 34 Ruhuna Journal of Science 1, pp. 32–40, (2006) Figure 1 The graph between number of layers and the angle (radians) for minimum energy. For the minimum energy, ∂E ∂θ = 0, and following equation of N can be obtained. N = 1.5ωΦ1 sin 2θ K . Here, K = 0.75ω(Φ0 + 2Φ1) sin 2θ − 2D (2) m sin θ cos θ − 4D(4)m sin θ cos 3 θ − 6D(6)m sin θ cos 5 θ − 8D(8)m sin θ cos 7 θ + Hin cos θ − Hout sin θ + 2K5 cos 2θ K5 ω = 10, D(2)m ω = 30, D(4)m ω = 20, D(6)m ω = 10, D(8)m ω = 5, Hin ω = Hout ω = 10, and Nd µ0ω = 10 will be used for the simulation in this report. For sc(001) lattice, Φ0 = 9.0336, Φ1 = −0.3275 and N = − 0.4913 sin 2θ K . Here K = −23.716 sin 2θ − 20(4 + 3 cos2 θ + 2cos4θ) cos3 θ sin θ + 10(cos θ − sin θ) + 20 cos 2θ. The graph between N and angle (θ) in radians is given in Figure 1. The angles indicated in all the graphs are given in radians. According to this graph, films with N=1, 2 can be easily oriented in θ = 279°direction. Similarly N corresponding to the minimum energy of fcc(001) and bcc(001) can be found using above equation. When the summations of D(2)m , D (4) m , D (6) m , D (8) m are considered, E(θ) = − J 2 [N Z0 + 2(N − 1)Z1] + ω 8 [N Φ0 + 2(N − 1)Φ1](1 + 3 cos 2θ) Samarasekara: Variation of magnetic energy... Ruhuna Journal of Science 1, pp. 32–40, (2006) 35 − cos2 θ N ∑ m=1 D(2)m − cos 4 θ N ∑ m=1 D(4)m − cos 6 θ N ∑ m=1 D(6)m − cos 8 θ N ∑ m=1 D(8)m − N (Hin sin θ + Hout cos θ − Nd µ0 + K5 sin 2θ) When θ = 0, E(0) = − J 2 [N Z0 + 2(N − 1)Z1] + ω 2 [N Φ0 + 2(N − 1)Φ1] − N ∑ m=1 D(2)m − N ∑ m=1 D(4)m − N ∑ m=1 D(6)m − N ∑ m=1 D(8)m − N (Hout − Nd µ0 ) When θ = 90°, E(90°) = −J 2 [N Z0 + 2(N − 1)Z1] − ω 4 [N Φ0 + 2(N − 1)Φ1] − N (Hin − Nd µ0 ) When E(0) > E(90), 3ω 4 [N Φ0 + 2(N − 1)Φ1] + N (Hin − Hout) > N ∑ m=1 D(2)m + N ∑ m=1 D(4)m + N ∑ m=1 D(6)m + N ∑ m=1 D(8)m When E(0) < E(90), 3ω 4 [N Φ0 + 2(N − 1)Φ1] + N (Hin − Hout) < N ∑ m=1 D(2)m + N ∑ m=1 D(4)m + N ∑ m=1 D(6)m + N ∑ m=1 D(8)m According to above equations, the in plane or perpendicular orientation is pre- ferred depending on the marginal value of N ∑ m=1 D(2)m + N ∑ m=1 D(4)m + N ∑ m=1 D(6)m + N ∑ m=1 D(8)m given by following equation. N ∑ m=1 D(2)m + N ∑ m=1 D(4)m + N ∑ m=1 D(6)m + N ∑ m=1 D(8)m = 3ω 4 [N Φ0 + 2(N − 1)Φ1] + N (Hin − Hout) Samarasekara: Variation of magnetic energy... 36 Ruhuna Journal of Science 1, pp. 32–40, (2006) For sc(001) with Hin = Hout N ∑ m=1 D(2)m + N ∑ m=1 D(4)m + N ∑ m=1 D(6)m + N ∑ m=1 D(8)m = 6.284N + 0.491 The anisotropy energy of a thin film is given by following equation. E(0) − E(90°) = 3ω 4 [N Φ0 + 2(N − 1)Φ1] − N ∑ m=1 D(2)m − N ∑ m=1 D(4)m − N ∑ m=1 D(6)m − N ∑ m=1 D(8)m + N (Hin − Hout) For Hout = 0, E(0) − E(90) ω = 16.284N − 64.51 When N ≥ 4, E(0)>E(90°) and perpendicular orientation is preferred. When N ≤ 3, E(0)