1 / 16

Homework

Homework. 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 ,

aislin
Download Presentation

Homework

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Homework

  2. 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(ik t-kr), Ak=A(1, i, 0) and ~k = 2J|S0| (1-cos{k }). For k small, k~Dk2 where D=JzS02

  3. Ferromagnetic spin waves  Si=Ak exp(ik 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

  4. 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。 • ~k = 2J|S0| (1-cos{k })+K

  5. 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

  6. Magnon heat capacity • <E>=kk<nk>=kk /(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/(D3/242)]0xm dx x3/2/(ex-1). • At low T aprroximate xm by . Then <E>/V T5/2; C T3/2

  7. 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).

  8. Antiferromagnetic magnons: physics related to superconductivity • H=J  SjSj+ -2BHASajz +2BHASbjz. • a=Sa+/(2Sz)1/2, a+=Sa-/(2Sz)1/2 ; b+=Sb+/(2Sz)1/2, b=Sb-/(2Sz)1/2 ; Sajz=S-aj+aj, Sblz=-S+bl+bl. • H=ek[k( ak+bk++akbk)+(ak+ak+bk+bk)]+ a(ak+ak+bk+bk)]; e=2JzS, k= exp(ik)/z, a= 20Ha. • H involves products of two creation operators!

  9. AF magnons: [ak+,H]=ek[ak+,akbk] + (e+a)[ak+,ak+ak] = -ekbk -(e+a)ak+ ; [bk,H]= ekak+ +(e+a)bk; • Define k= ukak-vkbk+ ; k=ukbk-vkak+. Look for solutions of the form k+ exp(i t). itk+=[k+,H]=-k+. • [k+,H]= uk [-ekbk -(e+a)ak+ ]-vk [ekak+ +(e+a)bk]=- ( ukak +-vkbk ). Get uk (e+a) +vkek = uk ;uk ek +vk (e+a)=- vk .

  10. uk (e+a) +vkek = uk ;uk ek +vk (e+a )= - vk • (e+a )2-2 =(ek )2 ; k2 = (e+a )2-(ek )2 • Long wavelength limit k= [(e+a+e) (e (1-k )+a)]0.5 ; • k=0= [(2e+a ) a ] 0.5 >> a; (FMR) • k (a=0)=e (1-k 2) 0.5 k. (For F, k2)

  11. Normalization [k,k+]=uk2[ak,ak+]+vk2[bk+,bk]=uk2-vk2=1. • Write u=cosh , v=sinh  • Homework : Is it true that tanh 2=-(e+a)/[e(1-k)]?

  12. Superconductivity and antiferromagnet • Superconductivity • k=ukck-vkc-k+; -k=ukc-k+vkck+ • AF-Magnon: • k= ukak-vkbk+ ; k=ukbk-vkak+.

  13. Ground state magnetization • ak=ukk+vkk+; bk=ukk+vkk+ • <Saz>=NS- ak+ak=NS- (uk2k+k+vk2kk++ off-diagonal terms). • At T=0, nk=0, NS-<Saz>= vk2=k sinh2(k)  ddk/k. Fluctuation is infinite in 1 dimension.

  14. 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.

More Related