Homework

Ferromagnetic spin waves

• Consider a ferromagnet with all the spins line up in equilibrium. Consider small deviation from it. Write Si=S0+ Si,

• Si=Ak exp(ik t-k r), Ak=A(1, i, 0) and ~k = 2J|S0| (1-cos{k }). For k small, k~Dk2 where D=JzS02

Ferromagnetic spin waves

Si=Ak exp(ik t-k r), Ak=A(1, i, 0). Take the real part. At t=0, S is along x at r=0 and along y at k r=/2. When t=/2, S is along y at r=0 and along –x at k r=/2

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-Sz

2-Sz)/2Sz=[S(S+1)-Sz

2+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。

• ~k = 2J|S0| (1-cos{k })+K

Quantization: Magnons are Bosons

• Eigenvalue of n=a+a is quantized with eigenfunction |n>=a+|n-1>/n0.5. (the conjugate is a|n>=n0.5|n>.

• First prove that the normalization is correct:<n|n>=<n-1|aa+|n-1>/n=<n-1|(a+a+1)|n-1>/n

=[(n-1)<n-2|n-2>+1]/n=1.Finally a+a|n>=a+n0.5|n-1>=n|n>. Thus the

energy of the system changes by integer multiples of k

Magnon heat capacity

• <E>=k k<nk>=kk /(e/kBT-1)

• For T<J, only magnons with small k is excited. If T>K, can neglect the gap. <E>=[V/(2)3] d3k Dk2/(eDk2/kBT-1).

• <E>/V=[(kBT)5/2/(D3/242)]0xm dx x3/2/(ex-1).

• At low T aprroximate xm by . Then <E>/V T5/2; C T3/2

Refresher for Bose Statistics

• <n>=k=0 e-kx k/Z where x= /kBT.

• Z=k e-kx =1/(1-e-x).

• <n>=-x lnZ=e-x/(1-e-x)=1/(ex-1).

Antiferromagnetic magnons: physics related to superconductivity• H=J SjSj+ -2BHASa

jz +2BHASbjz.

• a=Sa+/(2Sz)1/2, a+=Sa-/(2Sz)1/2 ; b+=Sb+/(2Sz)1/2, b=Sb-/(2Sz)1/2 ; Sa

jz=S-aj+aj, Sb

lz=-S+bl+bl.

• H=ek[k( ak+bk

++akbk )+(ak+ak+bk

+bk)]+ a(ak

+ak+bk+bk)]; e=2JzS, k= exp(ik)/z,

a= 20Ha.• H involves products of two creation

operators!

AF magnons:

[ak+,H]=ek[ak

+,akbk] + (e+a)[ak+,ak

+ak] = -ekbk -(e+a)ak

+ ; [bk,H]= ekak+ +(e+a)bk;

• Define k= ukak-vkbk+ ; k=ukbk-vkak

+. Look for solutions of the form k

+ exp(i t). it k

+=[k+,H]=- k

+.

• [k+,H]= uk [ -ekbk -(e+a)ak

+ ]-vk [ekak+ +

(e+a)bk]=- ( ukak +

-vkbk ). Get uk (e+a) +vk ek = uk ; uk ek +vk (e+a)=- vk .

• uk (e+a) +vk ek = uk ; uk ek +vk (e+a )= - vk

• (e+a )2 -

2 =(ek )2 ; k 2 = (e+a )2

-(ek )2

• Long wavelength limit k = [(e+a+ e) (e

(1-k ) +a)]0.5 ;

k=0 = [(2e+a ) a ] 0.5 >> a ; (FMR)

k (a=0)=e (1-k 2) 0.5 k. (For F, k2)

• Normalization [k,k+]=uk

2[ak,ak+]

+vk2[bk

+,bk]=uk2-vk

2=1.

• Write u=cosh , v=sinh • Homework : Is it true that tanh 2=-

(e+a)/[e(1-k)]?

Superconductivity and antiferromagnet

• Superconductivity

• k=ukck-vkc-k+; -k=ukc-k+vkck

+

• AF-Magnon:

• k= ukak-vkbk+ ; k=ukbk-vkak

+.

Ground state magnetization

• ak=ukk+vkk+; bk=ukk+vkk

+

• <Saz>=NS- ak

+ak=NS- (uk2k

+k+vk2kk

+

+ off-diagonal terms).

• At T=0, nk=0, NS-<Saz>= vk

2=k sinh2(k) ddk/k. Fluctuation is infinite in 1 dimension.

Magnons: Holstein-Primakoff transformation

• Define spin wave operators a, a+ by S+/(2S)1/2=(1-a+a/2S)1/2a; S-/(2S)1/2= a+(1-a+a/2S)1/2 a; Sz=S-a+a

• Assume a+a/2S<<1, Sz~S; then [S+,S-] =2Sz=2S[a,a+]=2S if [a,a^+]=1. a behaves like a boson destruction operator.