Solutions to Problems of Chapter 3

Problem 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 Sk2 (a = 1) and integrating, we obtain (3.51) and then (3.52).

Problem 2. Chain of Heisenberg spins:


(a) One has

  • (Z = 2, number of neighbors)
  • (b) If/2 < 0,

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) = ^ (y2 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:


(a) co = 2J SZ(1 — Yk) where Z = 4 (number of nearest neighbors in a square lattice), yk = (cos(/txa) + cos(/c,,a))/2. со —> 2J S(ka)2 when к 0.

  • (b) <5г> = 1/2 - A fZB И: constant, see (3.43)-
  • (3.43)]. The most important contribution to the integral comes from the small к region where w —>■ 2J S(ka)2. We have Z> ~ 1/2— л Izb i+efs^y-1 — 1/2 —A fZB This integral

diverges at к = 0; hence, < Sz > is not defined if T Ф 0. There is no long-range order for T Ф 0 in 2D (see the rigorous theorem of Mermin-Wagner in Ref. [231]).

Note: In 3D, we replace in the integral 2:rkdk by 4rck2dk. The integral does not diverge at к = 0. The long-range ordering exists at Г ^ 0 in 3D.

Problem 4. Demonstration of Eqs. (3.131)-(3.133): Demonstration: We have

where we can show that ak 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

Left: Union-Jack lattice

Figure 18.5 Left: Union-Jack lattice: diagonal, vertical and horizontal bonds denote the interactions Ju J2 and J3, 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. "Union-Jack" 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 — 02 = $2 — =

в3в = 2тг/3. One can also write

The minimum of E corresponds to Si + S2 + S3 = 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 S2 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


We deduce

To calculate the angle between two spins, for instance Si and S4, we write


We find in the same manner,

We have

Note that |012| = 3|0|. These solutions exist if | cosв < 1, namely r/ > r]c = 1/3. When r/ = 1, we have в = ,т/4, 012 = Зтг/4.

Examples of frustrated spin systems. Left

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:


  • (a) To follow the method of the chapter.
  • (b) Yes, because the integral does not diverge any more at к = 0 in the presence of d.

Problem 9. Commutation relations of Holstein-Primakoff 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 Holdstein-Primakoff 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, S7n] = -S78lm.

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 next-nearest 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 J2 affects modes near kx = ky = kz = n/a. To increase J2 makes the frequencies of these modes decrease.

The first mode to become zero occurs at

, _ , _ 2 1-M

t — tc — з 1+|a|

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