Appendix A Properties of Yttrium–Iron–Garnet (YIG) Structure and Physical Properties Chemical formula Y3Fe5O12 Crystal structure Cubic Number of formula units per unit cell 8 Lattice constant (25 ◦C) 12.376 A˚ Mass density (25 ◦C) 5172 kg/m3 Octahedral Sublattice 27 −3 Site density Na 8.441 × 10 m Angular momentum Ja 5/2 Land´e g factor ga 2 Magnetization at 0 K Ma (0 K) 391.5 kA/m Tetrahedral Sublattice 27 −3 Site density Nd 12.66 × 10 m Angular momentum Jd 5/2 Land´e g factor gd 2 Magnetization at 0 K Md (0 K) 587.2 kA/m Molecular Field Constants [1] Octahedral λaa 735.84 Tetrahedral λdd 344.59 Nearest-neighbor λad 1100.3 Phenomenological Exchange Constant [2] 2 −16 2 Defined by hex = λex∇ m (3.104) λex 3 × 10 m Macroscopic Magnetic Properties Magnetization at 0 K Mtot (0 K) 196 kA/m Magnetization at 298 K Mtot (298 K) 140 kA/m Curie temperature Tc 559 K 333 334 References Anisotropy Constants 3 First-order cubic (4.2 K) [3] Kc1 (4.2 K) −2480 J/m 3 First-order cubic (295 K) [3] Kc1 (295 K) −610 J/m 3 Second-order cubic (4.2 K) [4] Kc2 (4.2 K) −118.0J/m 3 Second-order cubic (273 K) [4] Kc2 (273 K) −26.0J/m Optical Properties [5–7] Refractive index (1.2 μm) n 2.2 Faraday rotation (1.2 μm) φF 240 deg/cm Absorption coefficient (1.2 μm) α 0.069 cm−1 μ 2 − × −4 Cotton–Mouton tensor elements (1–3 m) g44MS 1.14 10 2 × −5 ΔgMS 5.73 10 Cotton–Mouton rotation (1.14 μm) ΦCM −160 deg/cm References [1] E. E. Anderson, “Molecular field model and magnetization of YIG,” Phys. Rev., vol. 134, p. A1581, 1964. [2] E. H. Turner, “Interaction of phonons and spin waves in yttrium iron garnet,” Phys. Rev. Lett., vol. 5, p. 100, 1960. [3] P. Hansen, “Anisotropy and magnetostriction of gallium-substituted yt- trium iron garnet,” J. Appl. Phys., vol. 45, p. 3638, 1974. [4] P. Escudier, “L’anisotropie de l’aimantation; un param´etre important de l’´etude de l’anisotropie magn´etocrystalline,” Ann. Phys. (Paris), vol. 9, p. 125, 1975. [5] M. J. Weber, Ed., CRC Handbook of Laser Science and Technology, Vol. IV Optical Materials Part 2: Properties. Boca Raton, FL: CRC Press, 1986, p. 291. [6] R. V. Pisarev, I. G. Sinii, N. N. Kolpakova, and Y. M. Yakovlev, “Magnetic birefringence of light in iron garnets,” Sov. Phys. JETP, vol. 33, p. 1175, 1971. [7] G. A. Smolensky, R. V. Pisarev, I. G. Sinii, and N. N. Kolpakova, “Cotton- Mouton birefringence of ferrimagnetic garnets,” J. Physique Colloque, vol. C1 32, pp. c1–1048, 1971. Appendix B Currents in Quantum Mechanics B.1 Density of States The valence electrons in a metal can be modeled as free electrons contained in a box the size of the sample. Because of the boundary conditions at the faces of the box, the allowed values of wave number k are quantized. The actual conditions at the surface of a metal may be very complicated, but much of solid-state physics depends primarily on the fact that the modes are quantized, rather than on exactly what the boundary conditions of a particular sample happen to be. Consequently, it is common to make the simple assumption of periodic boundary conditions. In this assumption, the wave function of an electron at x = L should equal to that at x = 0, where L is the length of the sample along the x-direction. For this to be true, we have exp(ikL) = exp(0) = 1, which gives the discrete values of wave number k =2πn/L, where n is an integer. In k space, the total number of states N in a sphere of radius |k| is given by the volume of the sphere divided by the volume occupied by one state. Thus 4 1 4πk3L3 N =2 πk3 =2 , (B.1) 3 (2π/L)3 3(2π)3 where we have assumed the sample is a cube with length L on each side so that 3 the minimum volume in k space is ΔkxΔkyΔkz =(2π/L) . The factor of 2 accounts for the degeneracy in the energy of spin-up and spin-down electrons.1 In the free electron approximation, the Hamiltonian H is merely the kinetic energy operator (cf. Table 1.1) p2 2 H = = − ∇2 , (B.2) 2m 2m 1 When we consider ferromagnetic metals, this degeneracy is lifted by the inter- action with the internal exchange field causing a splitting of the energy band diagram. 335 336 Appendix B Currents in Quantum Mechanics and when we assume an electron wavefunction of the form ψ ∼ eik·r,the Schr¨odinger equation Hψ = Eψ yields 2k2 = E. (B.3) 2m Combining (B.1) and (B.3), the number of states with energy less than E can be written 4πL3 2m 3/2 N =2 E3/2. (B.4) 3(2π)3 2 The density of states is defined as the number of states per unit energy, or dN g(E)= . (B.5) dE Thus, the density of states is given by 4πL3 2m 3/2 3 g(E)=2 E1/2 (B.6) 3(2π)3 2 2 L3 2m 3/2 √ = E. (B.7) 2π2 2 The Fermi energy is defined as the highest occupied energy state by an electron, at absolute zero temperature. Interactions at higher temperatures 2 2 typically occur near the Fermi energy, EF = kF/(2m), since electrons at the Fermi energy have nearby empty states. Consequently, the current is often calculated at the Fermi energy while assuming the parabolic band approxi- mation g(E) ∝ E1/2. Multiplying and dividing (B.7) by E and comparing the result with (B.4), we see that the relationship between the Fermi energy and the density of states can be expressed 3 N g(EF)= . (B.8) 2 EF At finite temperature, some of the electrons will be excited to states above the Fermi level. In this case, the density of states is related to the number of particles by ∞ N = g(E)f(E)dE , (B.9) 0 where − − 1 f(E)= 1+e(E EF)/(kBT ) (B.10) is the Fermi–Dirac distribution describing the probability of finding an elec- −23 tron with energy E at temperature T and kB =1.38 × 10 J/Kisthe Boltzmann constant. B.2 Electric and Spin Current Densities 337 Note that the above expressions assume degenerate spin-up and spin-down subbands. In a ferromagnetic metal, the molecular field interacting with the electron spin splits the bands as illustrated in Figure 10.1. If the magnetization is oriented upwards, then the Hamiltonian with the effects of the molecular field is (cf. Eq. (2.30)) −2 d 2 H = + μ μ σH , (B.11) 2m dx 0 B m where Hm is the molecular field and σ = ±1. Defining Em = μ0μBσHm,the density of states for the spin subbands can be expressed 1 g (E)= g (E − σE ) , (B.12) σ 2 m and the total density of states is given by g(E)= gσ(E) (B.13) σ = g↑(E)+g↓(E) (B.14) 1 = [g (E − E )+g (E + E )] . (B.15) 2 m m (B.16) B.2 Electric and Spin Current Densities Consider ψ to be a solution of the 1D Schr¨odinger equation ∂ψ 2 i = − ∇2ψ. (B.17) ∂t 2m Multiplying (B.17) by ψ∗ and subtracting the complex conjugate of the re- sulting equation, we obtain ∂ −i (ψ∗ψ)+ ∇·[ψ∗∇ψ − ψ∇ψ∗]=0. (B.18) ∂t 2m In obtaining (B.18), we have used the vector identity ∇·(f∇g)=∇f ·∇g + f∇2g. (B.19) If we define the probability density of finding an electron as ψ∗ψ,wecan express the charge density as ρ = qψ∗ψ. We can also define a current density j that satisfies the conservation equation ∂ρ + ∇·j =0. (B.20) ∂t 338 Appendix B Currents in Quantum Mechanics Multiplying (B.18) by the electronic charge q and comparing with (B.20), we make the identification −iq j = [ψ∗∇ψ − ψ∇ψ∗] 2m (B.21) q = Im(ψ∗∇ψ). m √ Choosing ψ =1/ keikx for the 1D case ensures the normalization dψ Im ψ∗ =1. (B.22) dx Under the two-band approximation, we treat the spin-up and spin-down carriers as being independent. Under this two-channel approximation, we can borrow standard techniques from quantum mechanics to solve for the prob- ability of carrier transmission across an interface, or even through a barrier with multiple interfaces. B.3 Reflection and Transmission at a Boundary Consider an electron that is incident on a planar interface between two media as illustrated in Figure B.1. The electron will be reflected with probability Medium 1 Medium 2 t r x Fig. B.1. Geometry for calculating the reflection and transmission coefficients of a particle at a boundary between two materials. amplitude r, and transmitted with probability amplitude t. In medium 1, the wave function is ψ1 = ψinc + ψref (B.23) 1 − = √ eik1x + re ik1x , (B.24) k1 (B.25) while in medium 2, the wave function is B.4 Tunneling Through a Barrier 339 t ik2x ψ2 = ψtr = √ e . (B.26) k2 The reflection and transmission coefficients can be found by requiring both the wave function and its derivative to be continuous at the boundary x =0. − + Requiring ψ1(0 )=ψ2(0 ) gives 1 1 √ (1 + r)=√ t.