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 cos 2 (  )+MH cos(  -  0 ).  E/  =0;  2 E/  2  =0.  E/  = K sin 2(  )-MH sin(  -  0 ). K sin 2  =MH sin(  -  0 ).  2 E/  2  =2K cos 2(  )-MH cos(  -  0 ). 2K cos 2  =MH cos(  -  0 ).

4. Coherent rotation K sin 2  =MH c sin(  -  0 ). K cos 2  =MH c cos(  -  0 )/2. H c (  0 )=(2K/ M)[1-(tan  0 ) 2/3 +(tan  0 ) 4/3 ] 0.5 / (1+(tan  0 ) 2/3 ).

5. Special case:  0 =0 H c0 =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<H c H c =2K/M. In general, at the local energy maximum, cos  m =MH/2K. E max = -K cos 2  m +MH cos  m = (MH) 2 /4K. E 0 =E(  =0)=-K+MH For H c -H= , U=E max -E 0 =NM 2  2 /4K. Rate of switching, P = exp(-U/k B T) where is the attempt frequency

7. Special case:  0 =0, H c (T) H c0 =2K/M. For H c0 -H= , U=E max -E 0 =NM 2  2 /4K. Rate of switching, P = exp(-U/k B T). H c (T) determined by P  ¼ 1. We get H c (T)=H c0 -[4K k B T ln(  )/NM 2 ] 0.5 In general H c0 -H c (T)/ T . For  0 =0,  =1/2; for  0  0,  =3/2

8. Non-uniform magnetization: formation of domains due to the dipolar interaction E dipo =(  0 /8  ) s d 3 R d 3 R’ M(R)M(R’)  ia  jb (1/|R-R’|). After two integrations by parts and assuming that the surface terms are zero, we get E dipo =(  0 /8  ) s d 3 R d 3 R’  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 / V 2/3. MFig. 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 M 2 AL/8  0 M 2 V/8  MFig. 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 =dM x /dx +dM z /dz=-M+M=0. Thus the magnetic energy is small. M Fig. 1 x z

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 cos 2 (  ). Minimizing U, we get the equation –Jd 2  /dx 2 +2K sin(2  )=0. This can be written as -d 2  /dt 2 +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 d 2 y/dt 2 -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(  )|=2t;  =2 tan -1 exp(2t). t=-1,  =0; t=1,  = .

17. Non-uniform magnetization: Spin wave Rate of change of angular momentum, ~ dS i /dt is equal to the torque, [S i, H]/i where H is the Hamiltonian, the square bracket means the commutator. Using the commutation relationship [S x,S y ]=iS z  : [S, (S¢ A)]=iA£ S. For example x component [S x, S y A y +S z A z ] =iS z A y -iA z S y We obtain ~ dS i /dt=2J S i £   S j+ 

18. Ferromagnetic spin waves Consider a ferromagnet with all the spins line up in equilibrium. Consider small deviation from it. Write S i =S 0 +  S i, we get the linearized equation We get ~ d  S i /dt=-2J S 0 £   (  S i -  S i+  )

19. Ferromagnetic spin waves: ~ d  S i /dt=-2J S 0 £   (  S i -  S i+  ) Write  S i =A k exp(i  k t-k  r), we obtain the equation i ~  k A k =CA k £ S 0 ; C= 2J   (1-e ik   ). In component form (S 0 along z): i ~  k A kx =CA ky, i ~  k A ky =-CA kx For S 0 along z, A k =A(1, i, 0) and ~  k = 2J|S 0 |   (1-cos{k   }). For k  small,   k ~Dk 2 where D=JzS 0  2.

20. Spin wave energy gap At k=0,  k =0. Suppose we include an anisotropy term H a =-(K/2)  i S iz 2 =-(K/2)  i [S  -(  S  ) 2 ]. In terms of Fourier transforms H a =(K/2)  k (  S k ) 2 +constant. i ~  k A k =C’A k £ S 0 ; C’= 2J   (1-e ik   )+K. ~  k = 2J|S 0 |   (1-cos{k   })+K.  k=0 =K. This is usually measured by FMR

21. Magnon: Quantized spin waves a=S + /(2S z ) 1/2, a + =S - /(2S z ) 1/2. [a,a + ]~[S +,S - ]/(2S z )=1. aa + =S - S + /(2S z )=(S 2 -S z 2 -S z )/2S z =[S(S+1)- S z 2 +S z ]/2S z =[(S+S z )(S-S z )+S-S z ] /2S z. S-S z ~aa + H exch =-J  (S-a i + a i )(S-a j + a j )+(S i + S j - +S i - S j+ )/2 ~ constant-JS  (-a i + a i -a j + a j +a i a j + +a i + a j ) =  k   k n k