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Ferromagnetism

Ferromagnetism. At the Curie Temperature Tc, the magnetism M becomes zero. Tc is mainly determined by the exchange J. As T approaches Tc, M approaches zero in a power law manner (critical behaviour). M. Tc. Coercive behaviour.

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Ferromagnetism

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  1. Ferromagnetism • At the Curie Temperature Tc, the magnetism M becomes zero. • Tc is mainly determined by the exchange J. • As T approaches Tc, M approaches zero in a power law manner (critical behaviour). M Tc

  2. Coercive behaviour • Hc, the coercive field, is mainly determined by the anisotropy constant (both intrinsic and shape.) Hc

  3. Coherent rotation model of coercive behaviour • E=-K cos2 ()+MH cos( -0). • E/=0; 2E/2=0. • E/= K sin 2()-MH sin( -0). • K sin 2=MH sin( -0). • 2E/2=2K cos 2()-MH cos( -0). • 2K cos 2=MH cos( -0).

  4. Coherent rotation • K sin 2=MHc sin( -0). • K cos 2=MHc cos( -0)/2. • Hc(0)=(2K/ M)[1-(tan0)2/3 +(tan0)4/3 ]0.5 / (1+(tan0)2/3).

  5. Special case: 0=0 • Hc0=2K/M. • This is a kind of upper limit to the coercive field. In real life, the coercive field can be a 1/10 of this value because the actual behaviour is controlled by the pinning of domain walls.

  6. Special case: 0=0, finite T, H<Hc • Hc=2K/M. • In general, at the local energy maximum, cos m=MH/2K. • Emax= -K cos2m +MH cos m= (MH)2/4K. • E0=E(=0)=-K+MH • For Hc-H=, U=Emax-E0=NM22/4K. • Rate of switching, P = exp(-U/kBT) where  is the attempt frequency

  7. Special case: 0=0, Hc(T) • Hc0=2K/M. • For Hc0-H=, U=Emax-E0=NM22/4K. • Rate of switching, P = exp(-U/kBT). • Hc(T) determined by P ¼ 1. We get Hc(T)=Hc0-[4K kB T ln()/NM2]0.5 • In general Hc0-Hc(T)/ T. For 0=0, =1/2; for 0 0, =3/2

  8. Non-uniform magnetization: formation of domains due to the dipolar interaction • Edipo=(0/8) s d3R d3R’ M(R)M(R’) iajb(1/|R-R’|). • After two integrations by parts and assuming that the surface terms are zero, we get • Edipo=(0/8) s d3R d3R’M(R)M(R’)/|R-R’| where the magnetic charge M=r¢ M.

  9. Non-uniform magnetization: formation of domains • For the case of uniform magnetization in fig. 1, there is a large dipolar energy proportional to the volume V • For fig. 2, the magnetic energy is very small. But there is a domain wall energy / V2/3. M Fig. 1 Fig. 2

  10. Uniform magnetization: magnetic energy • The magnetic charge at the top is M(z=L/2)=Md (z-L/2)/dz=M (z-L/2); similarly M(z=-L/2)=-M (z+L/2). The magnetic energy is 0 M2 AL/8= 0 M2 V/8. M Fig. 1 Fig. 2 L/2 -L/2

  11. Magnetic charge density is small for closure domains • For the closure domain, as one crosses the domain boundary, the magnetic charge density is M=dMx/dx +dMz/dz=-M+M=0. Thus the magnetic energy is small. z M x Fig. 1

  12. Domain walls • Bloch wall: the spins lie in the yz plane. The magnetic charge is small. • Neel wall: the spins lie in the xz plane. The dipolar energy is higher because the magnetic charge is nonzero here. z x y

  13. Domain wall energy • Because the exchange J is largest, first neglect the dipolar copntribution. • Assume that the angle of orientation  changes slowly from spin to spin. • The exchange energy is approximately Js (d/dx)2

  14. Domain wall configuration  

  15. Domain wall energy • Energy to be minimized: U=J s (d/dx)2-Ks cos2(). • Minimizing U, we get the equation • –Jd2/dx2+2K sin(2)=0. This can be written as • -d2/dt2+2 sin (2)=0 where t=x/l; the magnetic length l=(J/K)0.5. This looks like the same equation for the time dependence of a pendulum in a gravitational field : m d2y/dt2-m g sin y=0.

  16. Domain wall energy • From the ``conservation of energy’’, we obtain the equation (d/dt)2+ cos (2)=C where C is a constant. • From this equation, we get s d  /[C-cos(2)]0.5= t. To illustrate, consider the special case with C=1, then we get the equation s d/sin()=t. Integrating, we get ln|tan(/2)|=2t; =2 tan-1 exp(2t). • t=-1, =0; t=1, =.

  17. Non-uniform magnetization: Spin wave Rate of change of angular momentum, ~ dSi/dt is equal to the torque, [Si, H]/i where H is the Hamiltonian, the square bracket means the commutator. • Using the commutation relationship [Sx,Sy ]=iSz: [S, (S¢ A)]=iA£ S. For example x component [Sx, SyAy+SzAz] =iSzAy-iAzSy • We obtain ~ dSi/dt=2J Si£Sj+

  18. Ferromagnetic spin waves • Consider a ferromagnet with all the spins line up in equilibrium. Consider small deviation from it. Write Si=S0+ Si, we get the linearized equation • We get ~ d Si/dt=-2J S0£(  Si- Si+ )

  19. Ferromagnetic spin waves: • ~ d Si/dt=-2J S0£(  Si- Si+ ) • Write  Si=Ak exp(ik t-kr), we obtain the equation • i ~ k Ak =CAk £ S0; C= 2J(1-eik ). In component form (S0 along z): i ~ k Akx =CAky , i ~ k Aky =-CAkx • For S0 along z, Ak=A(1, i, 0) and ~k = 2J|S0| (1-cos{k }). For k small, k~Dk2 where D=JzS02.

  20. Spin wave energy gap • At k=0, k=0. • Suppose we include an anisotropy term Ha=-(K/2)i Siz2=-(K/2)i[S2-( Si)2]. In terms of Fourier transforms Ha=(K/2)k ( Sk)2+constant. • i ~ k Ak =C’Ak £ S0; C’= 2J(1-eik )+K. • ~k = 2J|S0| (1-cos{k })+K. • k=0=K. This is usually measured by FMR

  21. Magnon: Quantized spin waves • a=S+/(2Sz)1/2, a+=S-/(2Sz)1/2. • [a,a+]~[S+,S-]/(2Sz)=1. • aa+=S-S+/(2Sz)=(S2-Sz2-Sz)/2Sz=[S(S+1)-Sz2+Sz]/2Sz=[(S+Sz)(S-Sz)+S-Sz] /2Sz. • S-Sz~aa+ • Hexch=-J (S-ai+ai)(S-aj+aj)+(Si+Sj-+Si-Sj+)/2 ~ constant-JS (-ai+ai-aj+aj+aiaj++ai+aj) =kk nk

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