20 Magnetic Excitons Quantum field theory deals with systems possessing a large number of de- grees of freedom. This chapter presents a simple model, where we study the magnetic excitations of a long chain of coupled spins. We show that one can associate the excited states of the system with quasi-particles that propagate along the chain. We recall that, for any integer k: N n=1 e 2iπkn/N = N if k = pN, with p integer; = 0 otherwise. 20.1 The Molecule CsFeBr 3 Consider a system with angular momentum equal to 1, i.e. j = 1 in the basis |j, m common to ˆ J 2 and ˆ J z . 20.1.1. What are the eigenvalues of ˆ J 2 and ˆ J z ? 20.1.2. For simplicity, we shall write |j, m = |σ,whereσ = m =1, 0, −1. Write the action of the operators ˆ J ± = ˆ J x ± i ˆ J y on the states |σ. 20.1.3. In the molecule Cs Fe Br 3 ,theionFe 2+ has an intrinsic angular mo- mentum, or spin, equal to 1. We write the corresponding observable ˆ J ,and we note |σ the eigenstates of ˆ J z . The molecule has a plane of symmetry, and the magnetic interaction Hamiltonian of the ion Fe 2+ with the rest of the molecule is ˆ H r = D ¯h 2 ˆ J 2 z D>0 . What are the eigenstates of ˆ H r and the corresponding energy values? Are there degeneracies? 204 20 Magnetic Excitons 20.2 Spin–Spin Interactions in a Chain of Molecules We consider a one-dimensional closed chain made up with an even number N of Cs Fe Br 3 molecules. We are only interested in the magnetic energy states of the chain, due to the magnetic interactions of the N Fe 2+ ions, each with spin 1. We take {|σ 1 ,σ 2 , ··· ,σ N },σ n =1, 0, −1, to be the orthonormal basis of the states of the system; it is an eigenbasis of the operators { ˆ J n z } where ˆ J n is the spin operator of the n-th ion (n =1, ··· ,N). The magnetic Hamiltonian of the system is the sum of two terms ˆ H = ˆ H 0 + ˆ H 1 where ˆ H 0 = D ¯h 2 N n=1 ( ˆ J n z ) 2 has been introduced in 1.3, and ˆ H 1 is a nearest-neighbor spin–spin interaction term ˆ H 1 = A ¯h 2 N n=1 ˆ J n · ˆ J n+1 A>0 . To simplify the notation of ˆ H 1 , we define ˆ J N+1 ≡ ˆ J 1 . We assume that ˆ H 1 is a small perturbation compared to ˆ H 0 (A D), and we shall treat it in first order perturbation theory. 20.2.1. Show that |σ 1 ,σ 2 , ··· ,σ N is an eigenstate of ˆ H 0 , and give the cor- responding energy value. 20.2.2. What is the ground state of ˆ H 0 ? Is it a degenerate level? 20.2.3. What is the energy of the first excited state of ˆ H 0 ? What is the degeneracy d of this level? We shall denote by E 1 the corresponding eigenspace of ˆ H 0 , of dimension d. 20.2.4. Show that ˆ H 1 can be written as ˆ H 1 = A ¯h 2 N n=1 1 2 ( ˆ J n + ˆ J n+1 − + ˆ J n − ˆ J n+1 + )+ ˆ J n z ˆ J n+1 z . 20.3 Energy Levels of the Chain We now work in the subspace E 1 . We introduce the following notation |n, ± = |σ 1 =0,σ 2 =0, ··· ,σ n = ±1,σ n+1 =0, ··· ,σ N =0. Owing to the periodicity of the chain, we define |N +1, ± ≡ |1, ±. 20.3 Energy Levels of the Chain 205 20.3.1. Show that ˆ H 1 |n, ± = A(|n −1, ± + |n +1, ±)+|ψ n , where |ψ n is orthogonal to the subspace E 1 . Without giving the complete form of |ψ n , give an example of one of its components, and give the energy of the eigenspace of ˆ H 0 to which |ψ n belongs. 20.3.2. Consider the circular permutation operator ˆ T, and its adjoint ˆ T † , defined by ˆ T |σ 1 ,σ 2 , ··· ,σ N = |σ N ,σ 1 , ··· ,σ N−1 ˆ T † |σ 1 ,σ 2 , ··· ,σ N = |σ 2 ,σ 3 , ··· ,σ N ,σ 1 . Write the action of ˆ T and ˆ T † on the states |n, ±. 20.3.3. Check that, in the subspace E 1 , ˆ H 1 and A( ˆ T + ˆ T † )havethesame matrix elements. 20.3.4. Show that the eigenvalues λ k of ˆ T are the N-th roots of unity (we recall that N is assumed to be even): λ k =e −iq k q k = −π + 2kπ N k =0, ··· ,N − 1 . 20.3.5. We seek, in E 1 , the 2N eigenvectors |q k , ± of ˆ T , each corresponding to an eigenvalue λ k . Each |q k , ± is written |q k , ± = n c n (k) |n, ± . (20.1) (a) Write a recursion relation between the coefficients c n . (b) Show that c n (k)= 1 √ N e iq k n (20.2) is a solution of this recursion relation. (c) Show that the states |q k , ± defined using (20.1) and (20.2) are ortho- normal. (d) Show that the vectors |q k , ± are also eigenvectors of ˆ T † and ˆ T + ˆ T † ,and give the corresponding eigenvalues. (e) Calculate the scalar product n, |q k , (, = ±) and write the expan- sion of the states |n, ± in the basis |q k , ±. 20.3.6. We treat the Hamiltonian ˆ H 1 of Sect. 2 as a perturbation to ˆ H 0 . We limit ourselves to the first excited level of ˆ H 0 , and we want to calcu- late how the perturbation lifts the degeneracy of this level. We recall that, in the degenerate case, first order perturbation theory consists in diagonaliz- ing the restriction of the perturbing Hamiltonian in the degenerate subspace of the dominant term ˆ H 0 . 206 20 Magnetic Excitons (a) Explain why the results of questions 3.3 and 3.5 above allow one to solve this problem. (b) In first order perturbation theory, give the new energy levels which arise from the first excited state of ˆ H 0 , and the corresponding eigenstates. (c) Draw qualitatively the energies E(q k ) in terms of the variable q k which can be treated as a continuous variable, q k ∈ [−π,+π[, if N is very large. 1. What is the degeneracy of each new energy level? 20.4 Vibrations of the Chain: Excitons We now study the time evolution of the spin chain. 20.4.1. Suppose that at time t =0, the system is in the state |Ψ(0) = =± N−1 k=0 ϕ k |q k , with =± N−1 k=0 |ϕ k | 2 =1. Setting ω =2A/¯h, write the state |Ψ(t) at a later time t. 20.4.2. We assume that the initial state is |Ψ(0) = |q k , +. (a) Write the probability amplitude α n (t) and the probability P n (t)offinding at time t the n-th spin pointing upwards, i.e. σ n =+1andσ m =0for m = n. Show that P n (t) is the same for all sites of the chain. (b) The molecules of the chain are located at x n = na,wherea is the lattice spacing. Show that the probability amplitude α n (t) is equal to the value at x = x n of a monochromatic plane wave Ψ k (x, t)=Ce i(p(q)x−E(q)t)/¯h , where C is a constant, q = q k ,andx is the abscissa along the chain. Express p(q)intermsofq. (c) Show that Ψ k (x, t) is an eigenstate of the momentum operator ˆp x = (¯h/i)∂/∂x along the chain. Show that the value of p(q) ensures the periodicity of Ψ k (x, t), i.e. Ψ k (x + L, t)=Ψ k (x, t), where L = Na is the length of the chain. (d) Show that, for |q k |1,Ψ k (x, t) satisfies a Schr¨odinger equation for a particle of negative mass m, placed in a constant potential; give the value of m. 20.4.3. In a more complete analysis, one can associate quasi-particles to the magnetic excitations of the chain. These quasi-particles, which we call “mag- netic excitons”, have an energy E(q k ) and a momentum p(q k ). At very low temperatures, T ≈ 1.4 K, the chain is in the ground state of ˆ H 0 . If low energy neutrons collide with it, they can create excitons whose energy and momentum can be determined by measuring the recoil of the neutrons. The experimental result for E(q) as a function of q ∈ [−π,0] is given in Fig. 20.1. 20.4 Vibrations of the Chain: Excitons 207 Fig. 20.1. Experimental measurement of the excitation energy E(q) as a function of q between −π and 0. The energy scale is in meV (10 −3 eV) (a) Deduce from that data approximate values for D and A. (b) What do you think of the approximation D A and of the comparison between theory and experiment? How could one improve the agreement between theory and experiment? (c) Is it justified to assume that the chain is in its ground state when it is at thermal equilibrium at 1.4 K? We recall the Boltzmann factor: N(E 2 )/N (E 1 )=exp[−(E 2 − E 1 )/kT], with k =8.6 ×10 −5 eV K −1 . 20.4.4. Consider, at time t = 0, the state |Ψ(0) = N−1 k=0 ϕ k |q k , + with N−1 k=0 |ϕ k | 2 =1. We assume that N 1, that the coefficients ϕ k have significant values only in a close vicinity of some value k = k 0 , or, equivalently, q ≈ q 0 , and that, to a good approximation, in this vicinity, E(q)=E(q 0 )+(q − q 0 )u 0 ,u 0 = dE dq q=q 0 . Show that the probability P n (t)offindingσ n = +1 at time t isthesameas the probability P n (t ) of finding σ n = +1 at another time t whose value will be expressed in terms of t and of the distance between the sites n and n . Interpret the result as the propagation of a spin excitation wave along the chain. Calculate the propagation velocity of this wave and give its numerical value for a = 0.7 nm and q 0 = −π/2. 20.4.5. We now assume that the initial state is |Ψ(0) = |n =1, +. 208 20 Magnetic Excitons (a) Write the probability P m (t)offindingσ m = +1 at a later time t? (b) Calculate the probabilities P 1 (t)andP 2 (t), in the case N =2, and inter- pret the result. (c) Calculate P 1 (t) in the case N =8. Is the evolution of P 1 (t) periodic? (d) For N 1, one can convert the above sums into integrals. The prob- abilities are then P m (t) ≈|J m−1 (ωt)| 2 where the J n (x) are the Bessel functions. These functions satisfy |J n (x)| 2 =1andJ n =(−) n J −n . For x 1wehaveJ n (x) ≈ 2 πx cos(x −nπ/2 − π/4) if x>2|n|/π, and J n (x) ≈ 0ifx<2|n|/π. Which sites are appreciably reached by the probability wave at a time t such that ωt 1? (e) Interpret the result as the propagation along the chain of a probability amplitude (or wave). Calculate the propagation velocity and compare it with the result obtained in question 4.4). 20.5 Solutions Section 20.1: The Molecule CsFeBr 3 20.1.1. The results are: ˆ J 2 :2¯h 2 , ˆ J z : m¯h ; m =1, 0, −1. 20.1.2. One has: J + |1 =0 J + |0 =¯h √ 2|1 J + |−1 =¯h √ 2|0 J − |1 =¯h √ 2|0 J − |0 =¯h √ 2|−1 J − |−1 =0. 20.1.3. The eigenstates are the states |σ. The state |0 corresponds to the eigenvalue E =0,whereas|+ and |−, which are degenerate, correspond to E = D. Section 20.2: Spin–Spin Interactions in a Chain of Molecules 20.2.1. It is straightforward to see that ˆ H 0 |σ 1 ,σ 2 ···σ N = D N n=1 (σ n ) 2 |σ 1 ···σ N , the corresponding eigenvalue being E = D σ 2 n . 20.2.2. The ground state of ˆ H 0 corresponds to all the σ n equal to zero, so that E = 0. This ground state is non-degenerate. 20.2.3. The first excited state corresponds to all the σ’s being zero except one: σ n equal to ±1. The energy is D, and the degeneracy 2N , since there are N possible choices of the non-vanishing σ n , and two values ±1ofσ n . 20.2.4. J ± = J x ± iJ y . A direct calculation leads to the result. 20.5 Solutions 209 Section 20.3: Energy Levels of the Chain 20.3.1. The action of the perturbing Hamiltonian on the basis states is, set- ting = ±: ˆ H 1 |n, = A (|n −1, + |n +1,) +A n =n (|0, ···0,σ n = , 0 ···0,σ n = −1,σ n +1 =+1, 0 ···0 +|0 ···0,σ n = , 0 ···0,σ n =+1,σ n +1 = −1, 0 ···0) . The vector |ψ = |σ 1 =1,σ 2 = −1, 0 ···0,σ n = , 0 ···0 belongs to this latter set; it is an eigenvector of ˆ H 0 with energy 3D. 20.3.2. The definition of ˆ T , ˆ T † and |n, ± implies: ˆ T |n, ± = |n +1, ±; ˆ T † |n, ± = |n −1, ± . 20.3.3. We therefore obtain A( ˆ T + ˆ T † )|n, ± = A(|n −1, ± + |n +1, ±) . Since ˆ H 1 |n, ± = A(|n −1, ± + |n +1, ±)+|ψ n where n , ±|ψ n =0, ˆ H 1 and A( ˆ T + ˆ T † ) obviously have the same matrix elements in the subspace E 1 . 20.3.4. Since ˆ T N = ˆ I, an eigenvalue λ k satisfies λ N k = 1, which proves that each eigenvalue is an N-th root of unity. Conversely, we will see in the following that each N-th root of unity is an eigenvalue. 20.3.5. (a) The corresponding eigenvectors satisfy |q k , ± = n c n |n, ± ˆ T |q k , ± = λ k |q k , ± therefore one has n c n |n +1, ± = λ k n c n |n, ± . Hence the recursion relation and its solution are λ k c n = c n−1 c n = 1 λ n−1 k c 1 =e iq k (n−1) c 1 . 210 20 Magnetic Excitons (b) The normalization condition n |c n | 2 =1givesN|c 1 | 2 = 1. If we choose c 1 =e iq k / √ N, the eigenvectors are of unit norm and we recover the solution given in the text of the problem. (c) The scalar product of |q k , and |q k , is easily calculated: q k , |q k , = δ , 1 N n e 2iπn(k−k )/N = δ , δ k,k . (d) The vectors |q k , ± are eigenvectors of ˆ T † with the complex conjugate eigenvalues λ ∗ k . Therefore they are also eigenvectors of ˆ T + ˆ T † with the eigen- value λ k + λ ∗ k =2cosq k = −2 cos(2kπ/N). (e) From the definition of the vectors, we have n, |q k , = 1 √ N e iq k n δ and (directly or by using the closure relation) |n, ± = 1 √ N N−1 k=0 e −iq k n |q k , ± . 20.3.6. The restriction of ˆ H 1 to the subspace E 1 is identical to A( ˆ T + ˆ T † ) (question 3.3). In E 1 , the operator A( ˆ T + ˆ T † ) is diagonal in the basis |q k , ±. Therefore the restriction of ˆ H 1 is also diagonal in that basis. The energy levels are E(q k )=D +2A cos(q k ) , (20.3) corresponding to the states |q k , ±. As far as degeneracies are concerned, there is a twofold degeneracy for all levels (the spin value may be +1 or −1). In addition, for all levels except q = −π and q = 0, there is a degeneracy q k ↔ −q k (symmetry of the cosine). Therefore, in general, the degeneracy is 4. Section 20.4: Vibrations of the Chain: Excitons 20.4.1. At time t the state of the chain is (cf. (20.3)): |Ψ(t) =e −iDt/¯h k ϕ k e −iωt cos q k |q k , . 20.4.2. We now consider an initial state |q k , ±, evolving as e −iE(q)t/¯h |q k , ±. (a) We therefore obtain an amplitude α n (t)= 1 √ N e i(q k n−E(q k )t/¯h) and a probability P n (t)=|α n | 2 = 1 N , which is the same on each site. 20.5 Solutions 211 (b) In the expression α n (t)= 1 √ N e i(q k x n /a−E(q k )t/¯h) , we see that α n (t)isthevalueatx = x n of the function Ψ k (x, t)= 1 √ N exp[i(px −Et)/¯h] with E(q)=D +¯hω cos q and p(q)=¯hq/a. (c) The function Ψ k (x) is an eigenstate of ˆp x with the eigenvalue ¯hq k /a. Since N is even, we obtain: e iq k L/a =e iNq k =e 2πik =1, which proves the periodicity of Ψ k . (d) For |q k |1, cos q k =1−q 2 k /2 . Therefore E = E 0 + p 2 /2m with E 0 = D +2A and m = − ¯h 2 2Aa 2 = − ¯h ωa 2 . Ψ k then satisfies the wave equation i¯h ∂ψ ∂t = − ¯h 2 2m ∂ 2 ψ ∂x 2 + E 0 ψ, which is a Schr¨odinger equation for a particle of negative mass (in solid state physics, this corresponds to the propagation of holes and in field theory, to the propagation of anti-particles). 20.4.3. (a) With the data of the figure which resemble grosso modo the E(q) drawn in Fig. 20.2, one finds D +2A ∼ 3.2 × 10 −3 eV, and D − 2A ∼ 0.4 ×10 −3 eV. Therefore: D ∼ 1.8 × 10 −3 eV A ∼ 0.7 ×10 −3 eV . (b) The approximation D A is poor. The theory is only meaningful to order (A/D) 2 ∼ 10%. Second order perturbation theory is certainly necessary to account quantitatively for the experimental curve which has a steeper shape than a sinusoid in the vicinity of q = −π. (c) For T =1.4K,kT ∼ 1.2 × 10 −4 eV, exp(−(D − 2A)/kT ) ∼ 0.04. To a few % , the system is in its ground state. 20.4.4. Approximating E(q)=E(q 0 )+(q − q 0 )u 0 in the vicinity of q 0 ,we obtain α n (t)= 1 √ N e i(q 0 n−ω 0 t) k ϕ k e i(q k −q 0 )(n−u 0 t/¯h) . 212 20 Magnetic Excitons Fig. 20.2. Energy levels Since the global phase factor does not contribute to the probability, one has P n (t)=P n (t ) with t = t +(n − n) ¯h u 0 . This corresponds to the propagation of a wave along the chain, with a group velocity v g = u 0 a ¯h = a ¯h dE dq q=q 0 = − 2aA ¯h sin q 0 . For q 0 = −π/2anda =0.7 nm, we find v g ∼ 1500 ms −1 . One can also evaluate u 0 ∼ 1.2 meV directly on the experimental curve, which leads to v g ∼ 1300 ms −1 . 20.4.5. If |Ψ(0) = |n =1, +,thenϕ + k =e −iq k / √ N and ϕ − k =0. (a) The probability is P m (t)=|m, +|Ψ(t)| 2 ,where m, +|Ψ(t) = e −iDt/¯h N k e iq k (m−1) e −iωt cos q k . (b) N=2: There are two possible values for q k : q 0 = −π and q 1 = 0. This leads to P 1 (t)=cos 2 ωt, P 2 (t)=sin 2 ωt. These are the usual oscillations of a two- state system, such as the inversion of the ammonia molecule. (c) N=8: q k −π − 3π 4 − π 2 − π 4 0 π 4 π 2 3π 4 cos(q k ) −1 − 1 √ 2 0 1 √ 2 1 1 √ 2 0 − 1 √ 2 The probability P 1 of finding the excitation on the initial site is P 1 (t)= 1 4 cos 2 (ωt/2) + cos(ωt/ √ 2) 2 . The system is no longer periodic in time. There cannot exist t = 0 for which . basis of the states of the system; it is an eigenbasis of the operators { ˆ J n z } where ˆ J n is the spin operator of the n-th ion (n =1, ··· ,N). The magnetic Hamiltonian of the system is the sum. P n (t) is the same for all sites of the chain. (b) The molecules of the chain are located at x n = na,wherea is the lattice spacing. Show that the probability amplitude α n (t) is equal to the value at. calcu- late how the perturbation lifts the degeneracy of this level. We recall that, in the degenerate case, first order perturbation theory consists in diagonaliz- ing the restriction of the perturbing