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Solutions to Problems of Chapter 4Problem 1. Proofs of (4.13): Solution: We consider the following integral in the complex plane: where C is an integral contour chosen in the plane. This integral has a pole at z_{0} = — ie on the imaginary axis in the lower half plane. If t — t' > 0, we choose the contour in the lower half plane including the pole z_{0}: C = — R —>■ +R + C where ±R are on the real axis and C is the half circle going from R to —R in the lower half plane. Using the theorem of residues we write
where the minus sign come from the sense of the contour and the only pole lying inside the contour is z_{0}:
Taking the limit R > oo, we can show that the integral fc—юс ^{6}z+L' ’ g°^{es t0 zero} Therefore,
Now, if t — t' < 0, we choose the contour C in the upper half plane. The righthand side of (18.71) is zero because there is no pole inside C. We have thus
Combining (18.72) and (18.73) and using the Heavyside function 0(t — tf) = 1 if t — t' > 0, = 0 if t — t' < 0, we obtain the formula (4.13):
Problem 2. Demonstration of Eq. (4.22): We have (4.21) Noting that 0(t) = f^SftJdt so that dd(t)/dt = <5(t), we write
where we have used in the second equality a commutation relation for the first term and the equation of motion idO/dt = —Щ, 0] for the second term. We calculate now the commutator A = ['H, S',^{f}]. Using H of (3.31) without the factor 2 of у and without the applied field term, we have
where the prefactor  was added to remove the double counting due to the double sum and where we have used
Note that in the last line of Eq. (18.77) we gathered the sums by changing the dummy variables and permuted operators such as S,^{+} Sf, = Sf,Sf because the indices / and /' indicate two different neighboring sites [in the Hamiltonian, /' and m' are neighboring sites, one of these sites is equal to / because of the delta functions in the 4th equality of Eq. (18.77)]. Inserting (18.77) into (18.76) and putting Г = / + p where p are the vectors connecting / to its neighbors, we obtain Eq. (4.22):
Problem 3. Helimagnet by Green’s function method: Solution: We consider a crystal of simple cubic lattice with Heisenberg spins of amplitude 1/2. The interaction Ji between nearest neighbors is ferromagnetic. Suppose that along the у axis there exists an antiferromagnetic interaction J_{2} between next nearest neighbors, in addition to/i. (a) In the xz plane the only interaction is /1, so the spins in the plane are parallel. Along the у axis, the competition between in the ferromagnetic / j and the antiferromagnetic /_{2} can give rise to a helical structure (see Section 3.4). Let в be the helical angle between two nearest neighboring spins in they direction. The energy of a spin is written as
where the first term is the energy in the xz plane, the second and third terms are the energy in the у direction. Minimizing E with respect to # we have The solutions are
^  = a_{c}. This solution corresponds to the helimagnetic configuration. We can compare the energies of the three solutions
We see that E_{2} is a maximum, and
(b) Let # be the helical angle between two nearest neighboring spins in they direction. The Hamiltonian in terms of # is given by (3.124) (without the anisotropy term):
where R,_{y} is the distance vector between the two spins / and j, # is the vector of magnitude # perpendicular to the angle plane (xz). In the present problem, we have
Using the Fourier transforms and summing on neighbors with their corresponding angles mentioned above, we obtain where and where
in which, Z=4 is the number of nearest neighbors in the xz plane, в = arccos(/i/4/_{2}), and Ук — [2 cos(/c_{x}) + 2 cos(/c_{z})] /Z. Nontrivial solutions of (18.83) impose that
We have
We show in Fig. 18.7 the spin wave spectrum (18.88) as a function of the wave vector in the helical direction ky, for $ = Л/3 (/_{2}/7i = 0.5) at k_{x} = k_{y} = 0. We see that w = 0 at в.
The expression (18.88) is reduced to the dispersion relation (4.32) for ferromagnets w = Z7i Figure 18.7 Spin wave spectrum versus the wave vector k_{y} in the simple cubic lattice with a helical structure in the у axis, in the case в = л/Ъ (namely I/2I//1 = 0.5), and k_{x} = k_{z} = 0. (ii) when cos <9 = 1 (collinear antiferromagnets), for /2 = 0, we have, by returning to (18.81) (18.82) to calculate A and В (J1 < 0),
Eq. (18.88) then gives the dispersion relation (4.59) of antiferromagnets: со = ±Z'Ji{S^{z}) v/T^WF Problem 4. Green's function method for a system of Ising spins ±1/2 in an applied magnetic field in one dimension: Solution: We consider the following Hamiltonian:
where S^{z} = ±1/2 is the z component of the spin given by the Pauli matrices. 2S^{Z} = a^{z} = ±1 [Eqs. (1.3)—(1.5)]. The Ising model corresponds to the assumption that only the z component is taken into account. For simplicity, we often take S_{z} = ±1, the results are, however, identical to a constant factor. Consider the site /. We define the following Green's function:
The equation of motion of G,(t) is (/; = 1)
This equation generates the following Green’s function: Equation (18.92) becomes
The equation of motion of G,+i(t) generates the following Green’s function:
We write the equation of motion of G_{l+2}(t) but we shall neglect Green's functions of higher orders. We have
The Fourier transforms С&(£) = J^°_{oa}Gi_{<}[E)e~'^{Et}dE (к = i, i + 1, / + 2) yield, putting Е' = Е дц_{в}Н, where
The solutions of these equations are The spectral theorem [see Eq. (4.39)] gives We expand these expressions at small values of H. At the first order in H, we have
where x'_{2} =< Sf__{t}Sf_{+1} >. We have three equations for four unknowns x_{b} x_{2}, x_{3} and x’_{2}. It is not possible to solve them. However, when H = 0, we have xi = x_{3} = 0, namely there is no ordering at T Ф 0 in one dimension (remember Xi =< Sf >), as we will show by the renormalization group in Section 5.4.2. To go further, we calculate the correlation function between the spin at the site 0 and the spin at the site n: We consider the following Green's functions:
Following the same method as above, we obtain where x_{li(1} =< S^{Z}0S^{Z} >, 2x_{2},„ =< + Sf) >, x_{3},„ =< czcz czcz ^ Jo^{5}—> When W = 0, we have Xi,„ = 4x_{3},„. Hence,
We take a solution of the form < S^S^{Z} >= AX^{n}~^{m} where A is a constant: when m = n, we have < S^{Z}S^{Z} >= A = 1/4. Setting n = 0, m = — 1, m = 1 and m = 0 in (18.99) we have
The solution of this equation is X = A tanh Ц, namely
We see that, for n = 0, we have < >= 1/4 as expected for a spin 1/2. For n = 1, we have
Substituting this in (18.96) and (18.97) we obtain The susceptibility is thus
This result is also obtained by the transfer matrix method [see (5.90)]. Note that the argument of the exponential is PJ because of the factor 2 in the Hamiltonian and spins 1/2: To find the argument of the exponential of x in Problem 5.6 [Eq. (5.90)] of Section 5.9, we divide / by 2 and multiply by 4 (inverse of the square of spin magnitude). Problem 5. Effect of nextnearest neighbor interaction: Guide: We follow the same calculation in Section 4.2 in adding the interaction /2 in the equations (4.22), (4.26) (4.33). We obtain, instead of (4.29), where Z is the number of nearest neighbors and Z_{2} that of nextnearest neighbors. We have
where
and
Problem 6. Magnon spectrum in Heisenberg triangular antiferro magnet: Green’s function method Solution: Using the ground state of a triangular lattice determined in Problem 6 of Section 3.6 for the angle e_{kik}' between two neighboring spins, the Fourier transformations of Eqs. (4.71)(4.72) can be written as
where
and
where in which, Z = 6 is the number of nearest neighbors, в = 2n/3 the angle between two neighboring spins, and yu = ^2 cos (/c_{x}a) + 4cos [k_{x}a/2) cos ^/q,av/3/2^J /Z. Nontrivial solutions of (18.109) impose
We have
We show in Fig. 18.8 the spin wave frequency w versus k_{y }in the first Brillouin zone for k_{x} = 0. Figure 18.8 Spin wave spectrum versus the wave vector k_{y} in the case of a triangular antiferromagnet (S = /2,k_{x} = 0, J = 1). Note that when cos в = 1, the expression (18.113) is reduced to w = ZJ (S^{z}) (1 — j^) which is the dispersion relation (4.32) for ferromagnets. When cos0 = 1, Eq. (18.113) becomes w = ZJ {S^{z}) Jl — which is precisely the dispersion relation (4.59) of collinear antiferromagnets. Using the formula (3.97) for T = 0, we have
where AS = sinh^{2} 0* is independent of T. is given by (3.83)
Writing sinh 0* as a function of tanh(20/(), one can calculate numerically the sum in AS using the wave vectors in the first Brillouin zone. We have the result AS ~ 0.21, namely the spin length at Г = 0 is reduced from 5 = 1/2 to 0.29 (compared to 0.303 for a square antiferromagnet). This strong zeropoint contraction is due to the frustration of the triangular antiferromagnet. Other more precise methods taking into account higherorder fluctuations should yield a smaller value for the spin length at T = 0. Remark: The integral should be performed inside the first Brillouin zone which is a hexagon. 
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