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Chapter12 Ferromagnetism

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Chapter12 Ferromagnetism 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<Tc) : ferromagnetic 1473K 1388K 1273K 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<Tc) : ferromagnet Ms ∝ (Tc − T) In mean field approximation : γ=1 and β=1/2. 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<Tc 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/.
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