Physics 211B : Solution Set #4

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1 Pysics 211B : olution et #4 1] Atomic pysics Consider an ion wit a partially filled sell of angular momentum J, and Z additional electrons in filled sells ow tat te ratio of te Curie paramagnetic susceptibility to te Larmor diamagnetic susceptibility is χ para χdia = g2 L JJ + 1 2Zk B T 2 m r 2 were g L is te Landé g-factor Estimate tis ratio at room temperature olution: We ave derived te expressions and were χ dia = Zne2 6mc 2 r2 χ para = 1 3 n g 2 JJ + 1 Lµ B k B T g L = LL + 1 2JJ + 1 and were µ B = e /2mc is te Bor magneton Te ratio is tus χ para χdia = g2 L JJ + 1 2Zk B T, 2 m r 2 If we assume r 2 = a 2 B, so tat 2 /m r ev, ten wit T = 300 K and k B T 1 40 ev, g L = 2, J = 2, and Z 30, te ratio is χ para /χ dia 450, 2] Adiabatic demagnetization In an ideal paramagnet, te spins are noninteracting and te Hamiltonian is N p H = γ i J i H i=1 were γ i = g i µ i / and J i are te gyromagnetic factor and spin operator for te i t paramagnetic ion, and H is te external magnetic field a ow tat te free energy F H, T can be written as F H, T = T ΦH/T If an ideal paramagnet is eld at temperature T i and field H i ẑ, and te field H i is adiabatically lowered to a value H f, compute te final temperature Tis is called adiabatic demagnetization 1

2 b ow tat, in an ideal paramagnet, te specific eat at constant field is related to te susceptibility by te equation s c H = T = H2 χ T H T Furter assuming all te paramagnetic ions to ave spin J, and assuming Curie s law to be valid, tis gives c H = 1 3 n gµb H 2 pk B JJ + 1, k B T were n p is te density of paramagnetic ions You are invited to compute te temperature T below wic te specific eat due to lattice vibrations is smaller tan te paramagnetic contribution Recall te Debye result c V = 12 5 π4 nk B T Θ D 3, were n = 1/Ω is te inverse of te unit cell volume ie te density of unit cells and Θ D is te Debye temperature Compile a table of a few of your favorite insulating solids, and tabulate Θ D and T wen 1% paramagnetic impurities are present, assuming J = 5 2 olution: Te partition function s a product of single-particle partition functions, and is explicitly a function of te ratio H/T : Z = i J i m= J i e mγ ih/k BT = ZH/T Tus, were Te entropy is F = k B T ln Z = T ΦH/T, N p sin Ji Φx = k B ln γ i x/k ] B sin γ i x/2k B i=1 = F T = ΦH/T + H T Φ H/T, wic is itself a function of H/T Tus, constant means constant H/T, and H f H i = T f T i T f = H f H i T i Te eat capacity is C H = T = x T H x = Φ x, 2

3 wit x = H/T Te isotermal magnetic susceptibility is Tus, Next, write χ 2 F = H 2 T C H = H2 T χ = 1 T Φ x C H = 1 3 n gl µ p k B JJ + 1 B H k B T T 3 C V = 12 5 π4 n k B Θ D 2 and we set C H = C V to find T Defining Θ H g L µ B H/k B, we obtain T = 1 π 5π 36 JJ + 1 n ] 1/5 p n Θ2 H Θ 3 D et J 1, g L 2, n p = 001 n and Θ D 500 K If H = 1 kg, ten Θ H = 0134 K For general H, find T 3 K H kg] 2/5 3] Ferrimagnetism A ferrimagnet is a magnetic structure in wic tere are different types of spins present Consider a sodium cloride structure in wic te A sublattice spins ave magnitude A and te B sublattice spins ave magnitude wit < A eg = 1 for te A sublattice but = 1 2 for te B sublattice Te Hamiltonian is H = J ij i j + g A µ 0 H i A z i + g B µ 0 H j B z j were J > 0, so te interactions are antiferromagnetic a Work out te mean field teory for tis model Assume tat te spins on te A and B sublattices fluctuate about te mean values A = m A ẑ, = m B ẑ and derive a set of coupled mean field equations of te form m A = F A βg A µ 0 H + βjzm B m B = F B βg B µ 0 H + βjzm A were z is te lattice coordination number z = 6 for NaCl and F A x and F B x are related to Brillouin functions ow grapically tat a solution exists, and fund te criterion for 3

4 broken symmetry solutions to exist wen H = 0, ie find T c Ten linearize, expanding for small m A, m B, and H, and solve for m A T and m B T and te susceptibility χt = 1 2 H g Aµ 0 m A + g B µ 0 m B in te region T > T c Does your T c depend on te sign of J? Wy or wy not? b Work out te spin wave teory and compute te spin wave dispersion You sould treat te NaCl structure as an FCC lattice wit a two element basis Assume a classical ground state N in wic te spins are up on te A sublattice and down on te B sublattice, and coose A ublattice + = a 2 A a a 1/2 B ublattice + = 2 b b 1/2 b = 2 A a a 1/2 a + = b 2 b b 1/2 z = a a A z = b b How does te spin wave dispersion beave near k = 0? ow tat te spectrum crosses over from quadratic to linear wen ka A / A olution: We apply te mean field Ansatz i = m A,B and obtain te mean field Hamiltonian H MF = 1 2 NJzm A m B + ga µ 0 H + zjm B i + gb µ 0 H + zjm A j i A j B Assuming te sublattice magnetizations are collinear, tis leads to two coupled mean field equations: m A x = F A βga µ 0 H + βjzm B were m B x = F B βgb µ 0 H + βjzm A, F x = x, and B x is te Brillouin function, B x = ctn x 1 2 ctn x 2 Te mean field equations may be solved grapically, as depicted in fig 1 Expanding F x = x + Ox3 for small x, and defining te temperatures k B T A,B 1 3 A,B A,B + 1 zj, we obtain te linear equations, m A T A T m B = g Aµ 0 zj H m B T m A = g Bµ 0 zj H, 4

5 Figure 1: Grapical solution of of mean field equations wit A = 1, = 2, g A = g B = 1, zj = 1, and H = 0 Top: T > T c ; bottom: T < T c wit solution Te susceptibility is m A = g AT A T g B T A T 2 T A m B = g B T g A T A T 2 T A µ 0 H zj µ 0 H zj wic diverges at χ = 1 N M H = 1 2 H g Aµ 0 m A + g B µ 0 m B = g2 A T A + g 2 B T 2g A g B T A T 2 T A µ 2 0 2zJ, T c = T A = A A z J 3k B Note tat T c does not depend on te sign of J Note also tat te signs of m A and m B may vary For example, let g A = g B g and suppose A > Ten < T A < T A and wile m A < 0 for all T > T c, te B sublattice moment canges sign from negative to positive 5

6 at a temperature > T c Finally, note tat at ig temperatures te susceptibility follows a Curie χ T 1 beavior Now let s work out te spin wave teory Consider a bipartite lattice, ieone formed from two interpenetrating Bravais lattices Te A sublattice sites are located at R and te B sublattice sites at R + δ For te NaCl structure, A and B are FCC lattices, and we may take a 1 = a ŷ + ẑ, a 2 = a ˆx + ẑ, a 3 = a ˆx + ŷ, δ = a ẑ, were a is te Na-Cl separation Immediately, we know our spin-wave spectrum will exibit two excitation brances We assume te classical ground state is a Néel state were all of te A sublattice spins are pointing down z = and all te B sublattice spins are pointing up z = + Te most general Hamiltonian wic is isotropic in spin-space and composed of bilinear operators is H = R,R J AA R R A R A R + J BB R R R R + J AB R R δ A R R + µ 0 H R g A AR z + g B BR z In our case, J AA = J BB = 0 and J AB R R δ = J δ R R δ,a, but it is instructive to consider te more general case in wic all pairs of spins potentially interact Even more generally, let s consider te anisotropic case, wit te field directed along te direction of anisotropy: H = J AA R R AR x AR x + AR y AR y + AA AR z AR z R,R + J BB R R BR x BR x + BR y BR y + BB BR z BR z + J AB R R AR x BR x + AR y BR y + AB AR z BR z + µ 0 H g A AR z + g B BR z R Writing i j ] i j i j + + i z z j, and expanding te radicals in te Bogoliubov transformation, we obtain A R A R ] = AA 2 A AA A a AA R a R + a R a R + A a R a R + a R a R + B R R ] = BB 2 B BB b BB R b R + b R b R + B b R b R + b R b R + A R R ] = AB A + AB B a AB R a R + A b R b R A a R b R + a R b R + 6

7 Te spin-wave Hamiltonian is ten H = A R,R J AA + R R a R a R + a R a R AA a R a R AA a R a R + AA A R,R J BB R R b R b R + b R b R BB b R b R BB b R b R + BB + J AB R R δ A a R b R + a R b R R,R We now Fourier transform, wit and + AB a R a R + AB A b R b R AB A a R 1 e ik R a k, b R 1 e ik R+δ a k, N A N B k k Ĵ AA k = R Ĵ BB k = R Ĵ AB k = R J AA R e ik R J BB R e ik R J AB R + δ e ik R+δ Here, N A = N B = 1 2N is alf te total number of lattice sites Tis leads to H = E 0 + ω A k a k a k + ω Bk b k b k νk a k b k ν k a k b k k, wit ω A k = +g A µ 0 H + 2 A ĴAA k AA Ĵ AA 0 + AB Ĵ AB 0 ω B k = g B µ 0 H + 2 ĴBB k BB Ĵ BB 0 + AB A Ĵ AB 0 νk = A Ĵ AB k Note tat ĴAAk and ĴBBk are bot real, but not necessarily ĴABk All tree satisfy Ĵ k = Ĵ k, since tey are Fourier transforms of real functions JR 1 OK, now we do te Bogoliubov tang, and write a k = u k α k + vk β k b k = u k β k + vk α k a k = u k α k + v k β k b k = u k β k + v k α k 1 For te NaCl structure, Ĵ AB k as defined is real 7

8 Tis preserves te bosonic commutation relations: αk, α k ] = βk, β k ] = ak, a k ] = bk, b k ] = δkk ] αk, β k = αk, β ] ] k = ak, b k = ak, b ] k = 0 ubstituting into H, we find H = k H k, wit H k = ω A k u 2 k + ωb k v 2 k νk uk vk ν k u k v k α k α k + ω A k vk 2 + ωb k uk 2 νk uk vk ν k u k v k β k β k ωa + k + ω B k u k v k νk u 2 k ν k vk 2 α k β k + ωa k + ω B k u k v k ν k u k 2 νk v k 2 α k β k + const We now write νk νk e iϕk and use te freedom to coose u k, v k to eliminate te α k β k and α k β k terms in H k : were u k = cos θ k e iϕk/2 v k = sin θ k e +iϕk/2, tan 2θ k = 2 νk ω A k + ω B k Tis leads to te dispersions E + k = +ω k + ω+ 2 k ν2 k E k = ω k + ω+ 2 k ν2 k, were ω ± k 1 2 ω A k ± ω B k, and were E + k is te α-boson dispersion, and E k is te β-boson dispersion: H = k E + k α k α k + E k β k β k Let s ceck our dispersion in some simple cases For te NaCl structure, we take J AA R = J BB R = 0 and ĴABR + δ = δ R+δ,a Ten Ĵ AB k = 2J cos k x a + cos k y a + cos k z a 8

9 In order for te spin-wave Hamiltonian to be stable, we must ave E α,β k 0 for all k in te Brillouin zone Oterwise, te ground state energy can be lowered by adding α or β excitations, and at te level of te spin-wave Hamiltonian tere is noting to prevent us from adding an infinite number of suc excitations ie tere is no spin-wave repulsion in order to keep lowering te energy At te zone center, te energy gap is E g H = 0 = r ĴAB0, were r = 2 1 A AB 4 A 1 2 A AB Te gap vanises wen r = 0, wic occurs at te isotropic point AB = 1 For AB < 1, E g H = 0 < 0 and te spectrum is unstable Precisely at AB = 1, te spectrum is unstable in an infinitesimal field Te system would prefer to enter te spin flop pase, in wic te spins lie predominantly in te x y plane wit some small component parallel to te z-axis If we furter restrict AB = 1, ten te spin-wave dispersion for H = 0 becomes E ± k = ± 1 2 A ĴAB ĴAB0 A 2 + A k 2 R 2 + Ok 4, were R 2 R R2 J AB R d R J ABR, wit d te dimension of space For te NaCl structure, R = a/ 3 If bot A and are large, but teir difference is of order unity, ten a separation of scales develops For k A / A, te low-lying spin-wave branc disperses quadratically, as in te case of te ferromagnet For A / A k π/a, te dispersion is linear, as in te case of te antiferromagnet At very long wavelengts, ten, te ferrimagnet beaves as a ferromagnet, wit a net spin A per unit cell At somewat longer wavelengts but still large compared wit te lattice spacing, tis quadratic dispersion crosses over to a linear one, typical of an antiferromagnet 4] Tinking about singlets Consider te antiferromagnetic Heisenberg model on a bipartite lattice: H = J ij i j were J > 0 and te sum is over te bonds of te lattice a Break up te lattice into a dimer covering Tere are exponentially many suc dimer coverings, ie te number grows as e αn were N is te number of lattice sites Tink about tiling a cessboard wit wit dominoes Te analysis of tis problem was performed by H N V Temperley and M E Fiser, Pil Mag 6, Denote one sublattice as A and te oter as B You are to develop a mean field teory of interacting dimers in te presence of a self-consistent staggered field A = mẑ Te mean field Hamiltonian ten breks up into a sum over dimer Hamiltonians H dimer = J A + z 1J A + z 1J A = J A H s z A z B 9

10 were te effective staggered field is H s = z 1Jm, and z is te lattice coordination number Find te eigenvalues of te dimer Hamiltonian wen = 1 2 b Define x = 2/J Wat is te self-consistent equation for x wen T = 0? Under wat conditions is tere a nontrivial solution? Wat ten is te self-consistent staggered magnetization? How does it compare wit te result of spin-wave teory? olution: Te mean field Hamiltonian, H = J A H s z A z B, is written in matrix form for = 1 2 as 1 4 J H = J H 1 s 2 J J 1 4 J + H s Clearly te states and are eigenstates of H wit eigenvalues 1 4 J Te oter two eigenvalues are easily found to be λ ± = 1 4 J ± H 2 s J 2, Te ground state eigenvector is ten Ψ 0 = α + β, wit β α = x 1 + x 2, wit x = 2H s /J Te staggered moment is ten m = z A 1 α 2 β 2 = 2 α 2 + β 2 = ince m = 1 2 x/z 1, we ave 1 + x 2 = z 1, or 4 J x x 2 m = zz 2 2z 1 for te staggered magnetization For z = 4 square lattice we find m = , wic is 94% of te full moment = 1 2 pin wave teory gives m , wic is only 62% of te full moment for = 4 2 5] Let s all do te spin flop In real solids crystal field effects often lead to anisotropic spin-spin interactions Consider te anisotropic Heisenberg antiferromagnet in a uniform magnetic field, H = J ij x i x j + y i y j + z i z j + i z i 10

11 were te field is parallel to te direction of anisotropy Assume δ 0 and a bipartite lattice a Tink first about classical spins In a small external field, sow tat if te anisotropy is not too large tat te lowest energy configuration as te spins on te two sublattices lying predominantly in te x, y plane and antiparallel, wit a small parallel component along te direction of te field Tis is called a canted, or spin-flop structure Wat is te angle θ c by wic te spins cant out of te x, y plane? Wat do I mean by not too large? You may assume tat te lowest energy configuration is a two sublattice structure, rater tan someting nasty like a four sublattice structure or an incommensurate one b Now work out te quantum spin wave teory To do tis, you ll ave to rotate te quantization axes of te spins to teir classical directions Tis means taking x cos θ x + sin θ z y y z sin θ x + cos θ z wit θ = ±θ 0, depending on te sublattice in question How is θ 0 related to θ c above? Tis may seem like a pain in te neck, but really it isn t so bad Besides, you souldn t complain so muc And stand up straigt you re sloucing And brus your teet c Compute te spin wave dispersion and find under wat conditions te teory is unstable olution: We start by assuming a two-sublattice structure in wic te spins lie in te x z plane Any two-sublattice structure is necessarily coplanar Let te A sublattice spins point in te direction θ = θ A, φ = 0 and let te B sublattice spins point in te direction θ = θ B, φ = 0 Te classical energy per bond is ten εθ A, θ B = J 2 sin θ A sin θ B + J 2 cos θ A cos θ B z cos θa + cos θ B Note tat in computing te energy per bond, we must account for te fact tat for eac site tere are 1 2z bonds, were z is te coordination number Te total number of bonds is tus N bonds = 1 2Nz, were N is te number of sites Note also te competition between and Large makes te spins antialign along ẑ, wile large prefers alignment along ẑ Let us first assume θ A = θ B = θ c and determine θ c Let eθ A, θ B εθ A, θ B /J 2 : eθ c eθ A = θ c, θ B = θ c = sin 2 θ c + cos 2 θ c 2 cos θ c e = sin θ θ c 21 + cos θ c 2 c Tus, te extrema of eθ c occur at sin θ c = 0 and at cos θ c =

12 Te latter solution is present only wen > / 1 Te energy of tis state is e = per bond To assess stability, we ll need te second derivatives, 2 e 2 e θ 2 = θa =θ A c θ 2 = sin 2 θ θa =θ c cos 2 θ c + B c cos θ c θ =θ B c θ =θ B c 2 e θ A θ = cos 2 θ θa =θ c + sin 2 θ c, B c θ =θ B c from wic we obtain te eigenvalues of te Hessian matrix, λ + = cos 2 θ c + cos θ c 2 = λ = 1 + cos θ c = Assuming > 0, we ave tat λ + > 0 requires > 1, wic is equivalent to cos 2 θ c < 1, and λ > 0 requires 2 < 1 + Tis is te meaning of not too large Te oter extrema occur wen sin θ c = 0, ie θ c = 0 and θ c = π Te eigenvalues of te Hessian at tese points are: θ c = 0 : λ + = λ = 1 + θ c = π λ + = 1 + λ = 1 12

13 Witout loss of generality we may assume 0, in wic case te θ c = π solution is always unstable Tis is obvious, since te spins are anti-aligned wit te field For θ c = 0, te solution is stable provided < /zj 1 For general, te stability condition is < /zj 1 Te oter possibility is tat is so large tat neiter of tese solutions is stable, in wic case we suspect θ A = 0 and θ B = π or vice versa Tus, for < zj1 +, te solution wit θ c = cos 1 /zj1 + is stable Te Hessian matrix in tis case is 2 e 2 e θa 2 θ A θ B + 1 = 1 θ =0 A 2 e θ B θ A 2 e θ 2 B θ B =π wose eigenvalues are 2 λ ± = ± 1 + Tus, tis configuration is stable only if > We will perform te spin-wave analysis for te case 2 1 < < 1 + We begin te analysis by rotating tey spins in eac sublattice by an angle ±θ in te x z plane Tus, In terms of te ± operators, x θ = cos θ x + sin θ z y θ = y z θ = sin θ x + cos θ z + θ = 1 2 cos θ cos θ 1 + sin θ z θ = 1 2 cos θ cos θ sin θ z z θ = 1 2 sin θ sin θ + cos θ z Te Holstein-Primakoff transformation is given by + = a 2 a a 1/2, = 2 a a 1/2 a, z = a a, 13

14 in wic case we ave + θ = 2 cos θ + 1 a + θ = 2 cos θ 1 a + z θ = 2 sin θ a 2 cos θ 1 a + sin θ a a + 2 cos θ + 1 a + sin θ a a + 2 sin θ a + cos θ a a +, were te stands for terms of iger order in te expansion of 2 a a 1/2 Hamiltonian may be written as a sum over links, Te H = J ij H ij, were H ij = i j i j + + z i j z zj z i + j z We assume tat i A and j B Remembering tat te spins are rotated by ±θ on alternate sublattices, we obtain H ij = cos 2 θ sin 2 θ a i a i a j a j sin θ cos θ a i a i a j + a j a j a j a i + a i cos2 θ sin 2 θ a i + a i a j + a j 1 2 a i a i a j a j + zj 2 sin θ a i + a i a j a j zj cos θ a i a i + a j a j 2 Te spin-wave Hamiltonian is obtained by dropping terms wic contain more tan two boson operators: H ij = sin 2 θ cos 2 θ zj cos θ a i a i + a j a j cos2 θ sin 2 θ a i + a i a j + a j 1 2 a i a i a j a j /2 sin θ 1 + cos θ + a i + a i a j a j + 2 cos 2 θ sin 2 θ + 2 cos θ Te tird line above contains bare a and a operators, and is also formally of order 3/2 We can eliminate it by coosing θ suc tat cos θ = Note te minus sign, wic is due to te fact tat te vacuum state 0 for te bosons, prior to rotation, is directed along ẑ 14

15 We next substitute for θ and obtain te bond Hamiltonian H ij = a 1 + i a j + a i a j a i 1 + a j + a j a i + a i a i + a j a j Note tat te last term is te classical ground state energy Fourier transforming, ij a i a i + a j a j = 1 2 z a k a k + a k a k k i a j + a ij a i a j = 1 2 z γ k a k a k + a k a k k ij a i a j + a j a i = 1 2 z γ k a k a k + a k a k, k were γ k = 1 z is a sum over nearest neigbor vectors δ On te square lattice, we ave γ k = 1 2 coskx a + cosk y a We ten obtain H = 1 2 J2 k δ e ik δ ω k a k a k + a k a k + ν k a k a k + a k a k, wit ω k = 1 + γ k 1 2 r γ k ν k = 1 2 r γ k r We now invoke te Bogoliubov transformation, 2 a k = cos β k α k sin β k α k a k = cos β k α k sin β k α k, wit tan2β k = ν k ω k 15

16 to obtain te spin-wave Hamiltonian H = J 2 k E k a k a k J2 k E k ω k, wit E k = ω 2k ν2k = 1 + γ k 1 + γ k r γ k Te second constant term in H is te sift of te ground state energy due to quantum fluctuations Tis term is negative, since E k ω k Te spin-wave teory is stable provided E k is real and nonnegative for all k ince γ k 1, 1 ], we ave instabilities at r = 0 zone corner, γk = 1 and at r = 2 zone center, γ k = +1 Tese are precisely te classical instabilities we found earlier: r = 0 = 1 2 r = 2 =