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

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

Hc

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).

Coherent rotation

• K sin 2=MHc sin(-0).

• K cos 2=MHc cos(-0)/2.

• Hc(0)=(2K/ M)[1-(tan0)2/3 +(tan0)4/3 ]0.5 / (1+(tan0)2/3).

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.

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=NM22/4K.

• Rate of switching, P = exp(-U/kBT) where is the attempt frequency

Special case: 0=0, Hc(T)

• Hc0=2K/M.

• For Hc0-H=, U=Emax-E0=NM22/4K.

• 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

Non-uniform magnetization: formation of domains due to the

dipolar interaction• Edipo=(0/8) s d3R d3R’ M(R)M(R’) iajb(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.

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

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/80 M2 V/8

M Fig. 1

Fig. 2

L/2-L/2

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.

M

Fig. 1

x

z

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

xy

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

Domain wall configuration

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.

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, =.

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+

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+ )

Ferromagnetic spin waves:

• ~ d Si/dt=-2J S0£ ( Si- Si+ ) • Write Si=Ak exp(ik 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=JzS02.

Spin wave energy gap

• At k=0, k=0.

• Suppose we include an anisotropy term Ha=-

(K/2)i Siz2=-(K/2)i[S-( S)

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

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) =kk nk