Transcript Magnetism - University of Delaware

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-(tan0)2/3 +(tan0)4/3 ]0.5 /
(1+(tan0)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=NM22/4K.
• 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=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
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.
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 (zL/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
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
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
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  /[Ccos(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, =.
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(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.
Spin wave energy gap
• At k=0, k=0.
• Suppose we include an anisotropy term
2
2
2
Ha=-(K/2)i Siz =-(K/2)i[S -( Si) ]. 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)
=kk nk