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Solutions to Problems of Chapter 3Problem 1. Demonstration of (3.51)(3.52): Demonstration: In (3.50), by replacing Ц — 2J S(ka)^{2} and using (3.46), we have
Putting x = lfi2J Sk^{2} (a = 1) and integrating, we obtain (3.51) and then (3.52). Problem 2. Chain of Heisenberg spins: Solution: (a) One has
We plot w versus k. We see that со is strongly affected by у 2 when к > 0. Analytically, we take the derivative of со with respect to k, we have
This derivative is zero at к = 0 (uniform mode) and at cos(ka) = ^ (y_{2} c 0). This second case is called "soft mode" because the slope (stiffness) ofa>is zero at this value of k. We have a helimagnetic ordering for у _{2} < —У i/4 (see Sections 3.4 and 5.7.3). Problem 3. Heisenberg spin systems in two dimensions: Solution: (a) co = 2J SZ(1 — Yk) where Z = 4 (number of nearest neighbors in a square lattice), y_{k} = (cos(/t_{x}a) + cos(/c,,a))/2. со —> 2J S(ka)^{2} when к 0.
diverges at к = 0; hence, < S^{z} > is not defined if T Ф 0. There is no longrange order for T Ф 0 in 2D (see the rigorous theorem of MerminWagner in Ref. [231]). Note: In 3D, we replace in the integral 2:rkdk by 4rck^{2}dk. The integral does not diverge at к = 0. The longrange ordering exists at Г ^ 0 in 3D. Problem 4. Demonstration of Eqs. (3.131)(3.133): Demonstration: We have
where we can show that a_{k} and «_{k} obey the boson commutation relations (see similar demonstration in Problem 10). Replacing these expressions in the Hamiltonian (3.125), we have
The Hamiltonian is diagonal if the second term in the curly brackets {} is zero, namely
Omitting the constant term, we have where the energy of the magnon of mode к is Figure 18.5 Left: UnionJack lattice: diagonal, vertical and horizontal bonds denote the interactions J_{u} J_{2} and J_{3}, respectively. Right: Phase diagram of the ground state shown in the plane (a = /2//1, P = Js/Ji) Heavy lines separate different phases and spin configuration of each phase is indicated (up, down and free spins are denoted by +, — and 0, respectively). The three kinds of partially disordered phases and the ferromagnetic phase are denoted by 1,11, III and F, respectively. Problem 5. "UnionJack" lattice: Solution: We write the energy expression for each kind of configuration. Then, we compare two by two to determine the frontier between them. The result is shown in Fig. 18.5. See details in Ref. [67]. Problem 6. Ground state of the triangular antiferromagnet with XY spins: Solution: In the case of the triangular plaquette, suppose that spin S, (/ = 1, 2, 3) of magnitude S makes an angle 0, with the Ox axis. Writing E and minimizing it with respect to the angles 0,, one has
A solution of the last three equations is Q — 0_{2} = $2 — = в_{3} — в = 2тг/3. One can also write
The minimum of E corresponds to Si + S_{2} + S_{3} = 0 which yields the 120° structure. This is also true for Heisenberg spins. Problem 7. Ground state of Villain’s model: Solution: The energy of a plaquette of the 2D Villain's model with XY spins defined in Fig. 18.6 with Si and S_{2 }linked by the antiferromagnetic interaction >j, is written as
where (S,)^{2} = 1. The variational method gives By symmetry, Ti = Хг = Л, Лз = Л4 ~ д. We have
Hence,
We deduce To calculate the angle between two spins, for instance Si and S_{4}, we write
Hence,
We find in the same manner,
We have
Note that 0_{12} = 30. These solutions exist if  cosв < 1, namely r/ > r]_{c} = 1/3. When r/ = 1, we have в = ,т/4, 012 = Зтг/4. Figure 18.6 Examples of frustrated spin systems. Left: antiferromagnetic triangular lattice with vector spins (XY or Heisenberg spins), Right: Villain's model with XY spins. Problem 8. Uniaxial anisotropy: Answer:
Problem 9. Commutation relations of HolsteinPrimakoff operators: Solution: The operators a^{+} and a defined in the Holstein Primakoff approximation respect rigorously the commutation relations between the spin operators:
Demonstration: Replacing the spin operators by the HoldsteinPrimakoff operators, one has
If / = m, one has
If / ф m, one obtains in the same manner [S,^{+}, S“] = 0. For the second relation, when I = m one has
If фт, one obtains [Sf, S,^] = 0. Similarly, one has [Sf, S7_{n}] = S78_{lm}. Remark: One has used o+a/ = /a+a in the above demonstration of (18.67) because Problem 10. Operators defined in (3.74)(3.77) obey the commutation relations: Demonstration: We have
The same demonstration is done for the other relations. Problem 11. Magnon soft mode: Demonstration: The magnon spectrum (3.113) becomes unstable when the interaction between nextnearest neighbors defined in e, Eq. (3.107), is larger than a critical constant. The spectrum becomes unstable when one of its frequencies tends to zero: This mode is termed as "soft mode.” Numerically, we plot (3.113) versus к for various values of e and determine its critical value. Analytically, we see that interaction J_{2} affects modes near k_{x} = k_{y} = k_{z} = n/a. To increase J_{2} makes the frequencies of these modes decrease. The first mode to become zero occurs at , _ , _ 2 1M t — t_{c} — з _{1+}_{a} 
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