Chapter Twelve Ferromagnetism and Antiferromagnetism
In some materials, there are important interactions between magnetic moments resulting in ordered magnetic phases in the absence of H.
M >> 0 M > 0 M = 0 Parallel alignment Anti-parallel Anti-parallel alignment Ferromagnetism Ferrimagnetism Anti-Ferromagnetism What are the interactions ? (1) Dipole-dipole interaction Not Important µ produces local field → align adjacent µ very weak (2) “Exchange” interaction QM interaction due to overlap of electron wavefunctions (Spin dependent) U = −2J∑Si •Sj ij Ferromagnetism M T=300K Ms
H Hc
The typical magnetization curve of the ferromagnet A hysteresis loop of magnetization is present.
Ms : saturation magnetization
Mr = M(0) residual magnetization
Hc coercive field M(± Hc)=0 χ Ms M 1 χ = ~ Curie-Weiss law H T −θ
Ferromagnet Paramagnet T Tc Curie temperature High T (T>Tc) : paramagnetic FCC Low T (T Mo 1073K 1043K (emu/g) s M 627K T(K) 273K Approaching Tc (Curie temperature) χ ∝ − −γ High T (T>Tc) : paramagnetic (T Tc ) β Low T (T Experimental data γ β Tc(K) Fe 1.33 0.34 1043 Ni Co 1.21 1388 Ni 1.35 0.42 627 Gd 1.3 292 CrO2 1.63 387 CuBr3 1.21 0.37 33 EuS 0.33 17 o Tc=354 C How will Ms vary with temperature ? What is Tc ? Recall Langevin function M /M M 1 s o = L(a) = coth(a)− Mo a µH where a = in the theory of k BT paramagnetism Weiss suggested that a “molecular field HM” in addition to H T/Tc is acting on material. : H + HM= H + γM µ µγ HM M Mo M k BT When H=0, a = = = a k BT k BT Mo Mo µγMo M/Mo is a linear function of a with a slope proportional to T. M k T = B a T T Three straight lines: T Mo µγMo 2 1 3 w/. T3>T2>T1 P o The intersection P gives spontaneous fractional magnetization Ms/Mo M/M achieved at that temperature. Ms decreases with increasing T and reaches 0 at T2. When T is higher than T2, the spontaneous magnetization vanishes a=µH/kBT implying that T2 is Tc. The Curie temperature Tc : k T 1 µγMo B c = T = c 3k µγMo 3 B M k T T M Hence, = B a = a intersects w/. = L(a) at P M Mo µγMo 3Tc o determining Ms/Mo. The law of corresponding states : All ferromagnetic materials, which naturally have different values of Mo and Tc have the same value of Ms/Mo for any particular value of T/Tc. Very nearly, but not exactly, correct M in the calculation is magnetic moment per unit volume, emu/cm3 . n: # of atoms per unit volume, changes with temperature. Exact statement : take M[emu/g] = M[emu/cm3] / ρ[g/cm3] instead of M. All materials have the same values of Ms/Mo for the same value of T/Tc. Μ k BT T = a = a M o µγρM o 3Tc o M / s The Weiss’s prediction of a law of M corresponding states is verified. But the shape of the curve is wrong. ∞ Weiss-Langevin : J= T/Tc As discussed in the previous chapter, Ch.11. QM need to be taken into accounted for the source of M. M 2J +1 (2J +1)a' 1 a' = BJ (a') = coth − coth M o 2J 2J 2J 2J gJµ H µ H where a'= B = H k BT k BT M k BT The straight line : = a' M o µHγρMo 1 J +1 The slope of B (a’) at the origin is : J 3 J µHγρM o J +1 g(J +1)µBγρM o Curie temperature Tc = = 3k B J 3k B Μ J +1 T Μ M / M = a' s = tanh s o M J 3T o c J=1/2 M o T/Tc in fairly good agreement with experiment J=1/2 implies that the magnetic moment is due entirely to spin, g=2 and there is no orbital contribution. Experimental data Ferromagnetism is due g* g+ essentially to electron spin with little or no contribution from Fe 2.12 2.08 orbital motion of the electrons. Co 2.22 2.18 At 0K, the spins on all the atoms are Ni 2.2 2.09 parallel, in one direction. At a higher temperature, a certain fraction of the Cu2MnAl 2.01 2.00 total, determined by the Brillouin Ni78Fe22 2.07 2.11 function, flip over into the other direction; the value of that fraction determines the Ni79Mo5Fe16 2.12 2.10 value of Ms. Reduction of Ms (T≠0) is due At 0K to excitation of spin waves. At T≠0 In the calculation shown above, we only consider the effect of the molecular field and put the applied field H equals to zero. µ (H + H ) µ (H + γρM ) Now we apply a magnetic field H, a'= H M = H k BT k BT Μ k BT k BT µHH k BT H = a'− = a'− M o µHγρM o µHγρM o k BT µHγρM o γρM o B(a’) Curve is the Brillouin J=1/2 function for J=1/2. o M / Lines 2 & 4 represent the M molecular field alone. The dashed lines 2’ & 4’ represent the molecular and a applied field. H a’ γρM o Above the Curie temperature T>Tc For instance, T=1.2Tc, the effect of the applied field move the intersection point of the field line and the magnetization curve from the origin to point B. Μ J +1 Near the origin = a' M o 3J Μ µH Μ o (J +1)/ 3k BJ C' Susceptibility χ = = = H T − µHγρΜ o (J +1)/ 3k BJ T −θ µ Μ (J +1) µ γρΜ (J +1) C'= H o and θ = H o 3k BJ 3k BJ θ is the temperature at which susceptibility becomes infinite. Tc is the temperature at which the spontaneous magnetization becomes 0. µHγρM o J +1 Tc = identical 3k B J Below the Curie temperature T For instance, T=0.5Tc, the effect of the applied field shift the intersection point of the field line and the magnetization curve from point P to point P’. Since the magnetization curve is nearly flat in this region, the increase in relative magnetization from P to P’ is very slight. “molecular field HM” : γρMs spin k T H = γρM = B c M M s µ M s H o J=1/2 ferromagnetic paramagnetic For Fe, µo=221.9emu/g o -16 M (1.38×10 erg/K )(1043K) / = -20 (218emu/g ) M (2.06×10 erg/Oe )(221.9emu/g ) = 6.9×106 Oe Co H =11.9×106 Oe T(K) M 6 Ni HM =14.7×10 Oe These fields are very much larger than any continuous field yet produced in the laboratory. However, the molecular field is in no sense a real field, but rather a force tending to make adjacent atomic moments parallel to one another. Ferromagnetic N M [emu/g] µ [µ ] M o = μ H o H B A Fe 221.9 2.22 N : Avogadro’s number Co 162.5 1.72 A : Atomic mass Ni 57.5 0.60 −20 µB = 0.927×10 erg/Oe The huge difference between a ferromagnetic and paramagnetic is due to the degree of alignment achieved and not to any large difference in the size of moment per atom. Exchange energy Play an important part of the total energy of many molecules and of the covalent bond in many solids. Heisenberg showed that it also played a decisive role in ferromagnetism. Ferromagnetism is due essentially to electron spin and reduction of Ms (T≠0) is due to excitation of spin waves. Quantized spin waves : Magnons In general, consider N spins N where J is called exchange integral and U = −2J Si •Sj ∑ spin angular momentum of atom i : Si i,j=1 If J is positive, U is a minimum when the spins are parallel (φij=0) and a maximum when they are anti-parallel (φij=π). If J is negative, U is a maximum when the spins are parallel (φij=0) and a minimum when they are anti-parallel (φij=π) As we have already seen, ferromagnetism is due to the alignment of spin moments on adjacent atoms. J>0 for ferromagnetism The ground state of a simple ferromagnet has all spins parallel N = − • Treat Si as 2 U 2J∑Si Si+1 classical vector Uo = −2JNS i=1 What is the energy of the first excited state? One particular spin is reversed. U = U + 8JS2 OR o we can form an excitation of much lower energy if all spins share the reversal The elementary excitations of a spin system have a wave like form and called magnons. Analog to lattice vibrations or phonons A spin wave on a line of spins Side view Top view A classical derivation of the magnon dispersion relation th For the p spin − 2JSp • (Sp−1 + Sp+1 ) The magnetic moment at site p : µ = - gµ S p B p µp 2J − − • + = −µ • − + 2J (Sp−1 Sp+1 ) p (Sp−1 Sp+1 ) gµB gµB Effective magnetic field or exchange th Classical mechanics field on the p spin Bp. dS p = µ × B dt p p dSp 2J = −gµBSp ×− (Sp-1 + Sp+1 )= 2J(Sp ×Sp-1 + Sp ×Sp+1 ) dt gµB x y z In Cartesian coordinate Assuming that Sp << S, Sp << S, and Sp = S x 2JS dSp 2J y z y z z y z y = y − y − y = (S S + S S −S S −S S ) (2Sp Sp-1 Sp+1 ) dt p p-1 p p+1 p p-1 p p+1 y dSp 2J z x z x x z x z 2JS x x x = (SpSp-1 + SpSp+1 −SpSp-1 −SpSp+1 ) = − (2S −S −S ) dt p p-1 p+1 dSz 2J p = (SxSy + SxSy −SySx −SySx ) = 0 dt p p-1 p p+1 p p-1 p p+1 x Sp = u exp[i(kpa −ωt)] y A trial solution set Sp = vexp[i(kpa −ωt)] z Sp = S 2JS − 4JS − iωu = (2 − e ika − eika )v = (1− cos(ka))v 2JS − 4JS − iωv = − (2 − e ika − eika )u = − (1− cos(ka))u 4JS iω (1− cos(ka)) For a set of solution = 0 4JS − (1− cos(ka)) iω 2 2 4JS x −ω + (1− cos(ka)) = 0 Sp = u cos(kpa −ωt) y Sp = u sin(kpa −ωt) 4JS ω = (1− cos(ka)) and u = −iv z Sp = S 2 At long wavelength (ka) ω ≅ 4JS = (2JSa 2 ) k 2 ka<<1 2 Dispersion relation for magnons in a ferromagnet in 1D w/. nearest-neighbor interactions ω 2.0 =1− cos(ka) 4JS 1.5 k2 may be determined accurately by neutron scattering or by spin /4JS 1.0 ω wave resonance in thin films. 0.5 At long wave length limit, 0.0 ω = (2JSa 2 ) k 2 = Dk 2 0.00 0.25 0.50 0.75 1.00 ka/π Fe Co Ni D 281 500 364 meVÅ2 T= 295K by Neutron scattering Magnon spectrum obtained by neutron scattering experiment ' k n k n k 2 2 2 '2 k n k n = + ωk 2M n 2M n 1 Quantization of spin waves ε = n + ω k k 2 k In thermal equilibrium the average value of the number of magnons 1 n k = Planck distribution exp(ωk / k BT)−1 3 1 The number of magnons from ω to ω+dω D(ω)dω = 4πk 2dk 2π 1 3 k 2 1 dk 2 1 ω D(ω) = 4πk 2 = 2π = π ω 2JSa 2 π 2 2 2 2 d 2k 2 2JSa 4JSa 3/ 2 1 ω = (2JSa 2 ) k 2 ω = 2 2 4π 2JSa dω 2JSa 2 = 2k dk 3/ 2 ∞ ∞ 1 ω Total number of dωD(ω) n = dω magnons ∫ k ∫ 2 2 0 0 4π 2JSa exp (ω / k BT)−1 ∞ ω x 2 Let x = and ∫dx x = 0.0587(4π ) k BT 0 e −1 3/ 2 3/ 2 ∞ 1 k T ∞ dω ω / k T dωD(ω) n = B B ∫ k 2 2 ∫ 0 4π 2JSa 0 k BT exp (ω / k BT)−1 3/ 2 k T 1 = 0.0587 4π 2 = 0.0587 B ( ) 2JS a3 SC BCC FCC The number N of atoms per unit volume: 1/a3 2/a3 4/a3 Q/a3 Fractional change of magnetization 3/ 2 ∆M ∑n 0.0587 k T = k = B M(0) NS SQ 2JS Bloch T3/2 law In agreement with experimental data Tc=627K Ni Mo=510 gauss At T=60K ~ 0.1Tc ∆M ≈ 2×10−3 Mo -6 -3/2 ∆M 3/2 A=(7.5±0.2)x10 K Ni 3/2 = AT Bloch T law -6 -3/2 Mo A=(3.4±0.2)x10 K Fe This application of band theory to magnetic problems was made by Stoner, Mott, and Slater in 1933. (the collective-electron theory) Why µH(0K) are 2.22, 1.72, and 0.6 µB for Fe, Co, and Ni, respectively? Electron distributions in free atoms K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn 3d 0 0 1 2 3 5 5 6 7 8 10 10 4s 1 2 2 2 2 1 2 2 2 2 1 2 3d+4s 1 2 3 4 5 6 7 8 9 10 11 12 The occupation of energy levels is in accordance with the Pauli exclusion principle. When atoms are brought close together to form a solid, the position of energy levels are profoundly modified. overlapping of electron clouds Splitting of levels When d=do, the 3d levels are spread into a band extending from B to C, and the 4s levels are spread into a much wider band extending from A to D. Why does the band spread wider in 4s than 3d? A: 4s electrons are farther from the nucleus. Interatomic distance d An important and difficult problem of the band theory is to calculate the “shape” of energy bands, i.e., DOS : N(E) for the band. Density of states N(E)=dN/dE The states are full of electrons, up to the Fermi energy at 0K. The total number of spins in the subband with spins up is different from that in the subband with spins down. Net magnetization Molecular field or exchange interaction The nonintegral values of the magnetic moments are easily explained by the complex shapes of the bands. Ni dN dN dN dN dE dE dE dE Co Fe 4s and 3d bands in Cu Fermi surface 2.2eV 7.1eV 3.46eV 3d 3d↑ 3d↓ 4s Filled-10electrons 5 electrons 5 electrons 4s and 3d bands in Ni below Tc above Tc 0.27 hole 0.54 hole 0.54 electron 0.54 5 ↑ 4s 3d↑ 3d↓ electron 3d 3d↓ electron 4.73 electron 4.46 electron n = number of 3d + 4s electrons per atom x = number of 4s electrons per atom n − x = number of 3d electrons per atom The magnetic moment per atom for this series of transition metal is predicted to increase as number of n-x of 3d electrons increase up to n-x=5 and then decrease to zero as number reach 10. At saturation (T=0K), the net moment is the difference between spin-up states (n-x=5) and the spin-down states, n-x-5: µ = [5 − (n − x − 5)]µB = [10 − n + x]µB For Ni, µ=0.6µB, x=0.6 Hence, µ = [10.6 − n]µB The agreement is generally good for Fe, Co, and Ni. Magnetic alloys : Alloys of the transition metals or rare earth metals ↑ ↓ Chemical valence Z = N + N Valence electrons ↑ ↓ ↑ Magnetization of the atom µ = (N − N )µB µ = (2N − Z)µB For the transition metals, N↑ is determined from the fact that the spin-up d bands lie entirely above or below the Fermi level. ↑ Magnetic valence Zm = 2Nd − Z ↑ Other contribution from the 4s band (actually 4sp hybrid band) Ns = 0.3µB ↑ The magnetic moment per atom µ = (Zm + 2Ns )µB = (Zm + 0.6)µB Ex. Fe0.8Co0.2 alloy : Zm = 2×0.8 +1×0.2 =1.8 µ = (1.8 + 0.6)µB = 2.4µB The average magnetic valence concept can then be used to predict the resulting magnetic moment when transition elements are alloyed. The Slater-Pauling curve : The curve of magnetic moment per atom versus electron-to-atom ratio. Assuming 0.3µB for the sp band spin-up electrons. + Magnetic valence Zm=2Nd -Z Band model contribution to magnetic moment per atom Ericksson et al. considered the contributions of the 3d and 4sp hybrid bands and obtained the magnetic moment resulting from the excess holes (Nh) in the spin-up and spin-down 3d bands.. Parameter Fe Co Ni Excess holes (Nh) 3.4 2.5 1.5 µspin3d 2.26 1.64 0.64 µspin4sp -0.07 -0.07 -0.02 µspintotal 2.19 1.57 0.62 µorbit3d 0.09 0.14 0.07 µ total 2.28 1.71 0.69 Experimental value 2.20 1.72 0.6 Criteria for the existence of ferromagnetism in a metal : The electrons responsible must lie in partially filled bands in order that there may be vacant energy levels available for electrons with unpaired spins to move into. Ruling out inner core electrons The density of states in the band must be high, so that the increase in energy caused by spin alignment will be small. Ruling out valence electrons The atoms must be the right distance apart so that exchange force can cause the d-electron spins in one atom to align the spins in the neighboring atom. Only Fe, Co, and Ni of transition metals are ferromagnetic. Many of rare earths are ferromagnetic below room temperature due to spin imbalance in their 4f bands. 1839 1803 1885 1885 1945 1879 1896 1880 1878 1886 1879 1843 1828 1878 1907 Bethe-Slater curve: postulated variation of J with the ratio ra/r3d. ra :the radius of an atom r3d : the radius of its 3d shell of electrons Exchange forces are responsible for magnetism, as well for antiferromagnetism, and ferrimagnetism. Antiferromagnetism M = 0 Spins are ordered in an antiparallel arrangement with zero net moment. Below TN :Néel temperature A small positive susceptibility present at all temperatures but varies in a peculiar way with temperature. a weak cusp at T=TN 1904-2000 1970 Anomalous paramagnetic C T > T χ ∝ N, T +θ like Curie temperature w/. Tc = -TN Paramagnetic T < TN, anti-ferromagnetic order Some antiferromagnetic materials material TN(K) θ(K) χ(0)/χ(TN) MnO 122 610 0.69 FeO 198 570 0.78 CoO 293 280 NiO 523 3000 0.67 Fe2O3 950 2000 Cr2O3 307 1070 0.76 FeS 613 857 FeCl2 24 -48 <0.2 FeF2 79 117 0.72 MnO2 84 116 0.93 α-Cr 310 Most are ionic compounds :oxides, sulphides, chlorides, and the like. Manganese Oxide MnO: Ordered arrangements of spins of the Mn2+ ions. A B Antiferromagnetic arrangement of A and B sublattices in 2D A B Considering only the nearest neighbor interaction AB HmA = −γMB Molecular field acts on the ion A due to magnetization of B HmB = −γMA Molecular field acts on the ion B due to magnetization of A Μ C MAT = C' ρ(H −γMB ) Above TN, χ = = ρH T MBT = C' ρ(H −γMA ) Curie law (MA + MB )T = C' ρ(2H −γ (MA + MB )) MT = 2C' ρH − C' ργM (T + C' ργ )M = 2C' ρH Μ 2C' Therefore, χ = = ρH T + C' ργ When a field is applied above TN, each sublattice becomes magnetized in the same direction as the field, but each sublattice set up a molecular field in the opposite direction to the applied field, tending to reduce both MA and MB. Hence, the susceptibility χ is smaller than that of an ideal paramagnetic in which the molecular field is zero. Below T , Each sublattice spontaneously magnetized, in zero applied N field, by the molecular field created by the other sublattice. M = MA + MB = 0 = − ργ where H=0 At T=TN, MATN C' MB MA C' ργ = − TN = TN MB The Néel temperature at which χ(T) is the maximum equals to θ. Below TN, each sublattice is spontaneously magnetized to saturation just as a ferromagnetic is. M A µH = BJ, Brillouin function M oA k BT For the spontaneous magnetization in the absence of applied field, HmA = −γMB = γMA = γρM A M A µHγρM A = BJ, M oA k BT M A MnF2 M oA T TN M − B M oB The net spontaneous magnetization is zero below TN. However, an applied field can produce a small magnetization. χ χ depends on the angle bet. the applied field and its spin axis. A B Spin H H A axis B χ║ χ┴ Field at right angle to spin axis M M Assuming that the applied field turn each B M sublattice magnetization away from spin axis α A H by a small angle α. This rotation immediately creates a HmA magnetization M in the direction of H. HmB The spins will rotate until that HmA + HmB cancel the applied field H. Hm 2HmA sin α = H = 2γρM A sin α = γρ and M = 2M A sin α H M Therefore, M 1 C χ⊥ = = = a constant H γρ 2θ Independent of temperature Field parallel with spin axis M Assuming that the applied field H B MA increase the zero-field value of the A- ∆ sublattice magnetization by MA and ∆MB decreases the corresponding value of H ∆MA the B-sublattice by ∆MB. A net magnetization in the direction of H M = M A − M B = ∆M A + ∆M B B(J,a’) nearly ∆M = ∆a'M B' (J, a ') linear A oA o ∆M ∆M A B µ M oA H o M oB ∆a'= (H a −γρ∆M B ) M / k BT M µH = (H a −γρ∆M A ) k BT 2 ng µH ∆M A = (H a −γρ∆M A )B' (J, a o ') a’=µHH/kBT 2k BT 2 M 2∆M 2ng µHB' (J, a o ') χ = = A = Therefore, 11 2 Ha Ha 2k BT + ng µHγρB' (J, a o ') temperature dependence C χ = It becomes 2 θ at T=TN. C It reduces to χ ∝ at high temperatures. T +θ It approaches to zero as T approaches 0. Theoretical calculation for J=1 For a powder specimen, there is no preferred orientation of the crystals. 2 2 χp = χ11cos θ + χ⊥ sin θ 1 2 = χ + χ⊥ 3 11 3 Ferrimagnetism Ferrimagnetic materials exhibit the phenomena of magnetic saturation and hysteresis, like ferromagnetics. The most important ferrimagnetic materials are certain double oxides of iron and another metal, called ferrites. The ferrites were developed into commercially useful materials. 1933 Si0.03Fe0.97 (Silicon ferrite) …. Until 1948, ferrimagnetics were separated from ferromagnetics. Néel provided the theoretical key to understand the ferrites. They fall mainly into two groups with different crystal structures : Cubic : MO•Fe2O3 where M is a divalent metal ion, Mn, Ni, Co, Fe,… Ironferrite FeO •Fe2O3 is the oldest magnetic material to man. Hexagonal : BaO •6Fe2O3 Barium ferrite Cubic ferrite Hexagonal ferrite metal ion Oxygen ion metal ion Tetrahedral A site Octahedral B site Barium ferrite Key: Lattice with bases of two magnetic atoms Moments anti-align, but do not cancel Eg. Magnetite Fe3O4=FeO • Fe2O3 parallel Fe2+ Fe3+ antiparallel M ≠ 0 Why is it different from ferromagnetic ? How to know this fact ? 3+ Fe ions have a spin of 5/2 and should contribute 5µB 2+ Fe ions have a spin of 2 and should contribute 4µB expectation Effective moment (2×5+4) µB = 14 µB at T=0K 2+ results It show only 4.1 µB only from Fe ions confirmed by Neutron diffraction Magnetization : complex behaviors FeO • Fe2O3 T > Tc, paramagnetic MA + MB χ = Let CA and CB be Curie constants for ions A and B. ρH MAT = CA ρ(H − γMB ) MBT = CBρ(H −γMA ) For non-zero (CA + CB )T − 2ργCACB solutions MA and MB T = ργ C C χ = c A B 2 − 2 At H=0 T Tc Below T c MO • Fe2O3 Cubic ferrite (emu/g) M T(oC) o M / M Below Tc Iron Garnets M3+ is a trivalent metal ion. 3+ M3 Fe5O12 3 Fe ion on tetrahedral site d 2 Fe3+ ion on octahedral site a 3 M3+ ion on site c Y3+ is a diamagnetic ion At T=0K, resulting in 5µB M decreases w/. increasing T and reaches to zero at T=Tc. Rare earth M3+ :are paramagnetic and magnetized (c) opposite to the resultant of the Fe3+ ions (a+d). M drops rapidly w/. increasing T The compensation temperature Due to weak c-a and c-d coupling. at which the magnetization crosses zero. Then M passes through zero and increase again due to Fe3+. T/Tc T/Tc Neutron magnetic scattering An X-ray photon see the spatial distribution of electronic charge, whether the charge density is magnetized or unmagnetized. A neutron sees two aspects of a crystal: the distribution of nuclei and the distribution of electronic magnetization. Neutron A neutron can be inelastically scattered by the ’ magnetic structure, kn kn With creation or annihilation of a magnon. magnon spectra 2 2 2 '2 k n k n = + ωk Magnon 2Mn 2Mn Neutrons are uncharged, easily penetrate electron cloud, and are scattered only by nucleus. If the scattering atom or ion has a net magnetic moment, that moment will interact with neutron beam, because the neutron has a small magnetic moment, ~0.3µB. Neutrons M=1.67x10-27kg 0.28 λ(A ) = When E=80meV , λ=1Å E(eV) -4 µ=5.4×10 µB Body-centered tetragonal BCT structure MnF2 T=23K T=300K MnF2 TN=67K T=80K TN=122K MnO T=293K MnO The first substance to be clearly recognized as antiferromagnetic in 1938. However, the first direct evidence came from Neutron diffraction experiment by Shull and Smart in 1949. Magnetic Domains Ferromagnetic materials are not uniformly magnetized, but break up into regions called domains. The magnetization for each domain will, in general, have a different orientation. Origin of domains – magnetic, exchange, and anisotropy energies Magnetic field energy Introducing more domains µ E = o H2dV decreasing the magnetic field energy 2 ∫ However, it increases the wall energy. For a 180o magnetization reversal in one step, an exchange energy per wall area has to be overcome: 2 W = JS2 ex a 2 For a 180o magnetization reversal in N steps, an exchange energy per wall area is reduced: 2 N π W = JS2 ex a 2 N a: lattice constant Domain wall orientations Bloch Wall Néel Wall Anisotropy Energy The magnetic moments in a magnetic material tend to line up preferentially along certain crystallographic direction. The exchange energy depends on the orbital overlap of electronic wavefunctions between electronic orbits on different sites. With crystal anisotropy the rotation away from the easy axis 2 2 2 2 2 2 2 2 2 costs extra energy UK = K1(α1 α 2 +α 2α3 +α3α1 )+ K 2α1 α 2α3 where k1 and k2 are the magnetic anisotropy constants N spins in a Bloch wall with lattice constant a UK = KNa 2 N π W = JS2 ex a 2 N The energy per unit area of the wall 2 N π σ = W + U ≈ JS2 + KNa W ex K a 2 N 2 ∂σ W 2 π Minimizing σW, = −JS + Ka = 0 ∂ N Na 2 π 2JS2 KJS N = and σ W = 2π Ka3 a 2 o For Iron, σW ≈ 1 erg/cm (180 wall, N≈300) Magnetic hysteresis Saturation Remanent magnetization magnetization Coercive field Saturation magnetization Demagnetized state 1/χ 1/χ T T Diamagnetism Paramagnetism Ferromagnetism χ Ms M 1 χ = ~ Curie-Weiss law H T −θ Ferromagnet Paramagnet T Tc Curie temperature A B A B Anti-ferromagnetism TN Néel temperature 1/χ A Ms B A Ferrimagnet B Paramagnet T Tc Ferrimagnetism Curie temperature Magnetic Devices Permanent Magnets Transformers Magnetic Amplifiers Data Storage Spin + Electronics Spin Magneto- Spin Electronics Photonics a multidisciplinary field including magnetism, semiconductor physics, optics, mesoscopic physics, superconductivity Quantum and new connections to other fields Electronics The central theme is how to manipulate the spin degree of freedom which interact with the solid-state environment. Spin-dependent transport Conventional transistors make use of current or voltage to control the transmitted current. Electronics Nature 404,918 (2000) Use the spin configurations to control current Spintronics The most commonly built structures for spin-dependent transport make use of (1) Giant magnetoresistance effect (GMR) (2) Tunneling magnetoresistance effect (TMR) M.N. Baibich et al., Phys. Rev. Lett. 61, 2472 (1988). Tunneling magnetoresistance effect (TMR) Tunneling conductance across the barrier is proportional to the product of density of states on both sides. G ∝ N1,↑ N2,↑ + N1,↓ N2,↓ R R parallel < anti− parallel GaMnAs/AlAs/GaMnAs tunneling junction M. Tanaka and Y. Higo, Phys. Rev. Lett. 87, 026602 (2001). Datta-Das Spin field-effect transistor Datta and Das, Appl. Phys. Lett. 56, 665 (1990). gate How to maintain spin coherence ? MRAM IBM MRAM images Magnetic RAM chips use magnetic rather than electrical structures to store information, so they do not need to be constantly powered to retain data, like current RAM technologies. much faster and less expensive 1989 - IBM scientists made a string of key discoveries about the "giant magnetoresistive" effect in thin-film structures. 2000 - IBM and Infineon established a joint MRAM development program. 2002 - NVE Announces Technology Exchange with Cypress Semiconductor. 2003 - A 128K bit MRAM chip was introduced, manufactured win 0.18 technology. 2004 June - Infineon unveiled a 16M bit prototype based on 0.18 September - MRAM becomes a standard product in Freescale, which has began sampling MR NEC, Toshiba claim Magnetorestitive RAM could replace flash memory and DRAM by as early as 2010 One issue involves the size of MRAM cells, which tend to be bigger than those of other memory types. New Technique could allow them to develop 256M bit MRAMs by early 2006