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# Solutions to Problems of Chapter 4

Problem 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 z0 = — 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 z0: 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 z0:

Taking the limit R -> oo, we can show that the integral fc—юс 6z+L' ’ g°es t0 zero- Therefore,

Now, if tt' < 0, we choose the contour C in the upper half plane. The right-hand 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 pre-factor | 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 J2 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

• (i) sin# = 0, namely # = 0 (solution 1), # = n (solution 2)
• (ii) cos# = (solution 3) if-1 < у < 1, namely

^ | = ac. This solution corresponds to the

helimagnetic configuration.

We can compare the energies of the three solutions

We see that E2 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

• (i) cos(0 • R,y) = 0 for nearest neighbors i and j belonging to the xz plane,
• (ii) cos(0 ■ Rjj) = cosd for nearest neighbors / and j on they axis (lattice constant=l),
• (iii) cos(0 • Rjj) = cos(26) for next nearest neighbors / and j on they axis (distance R,j = 2).
• (c) We define two Green’s functions by (4.69)-(4.70) which lead to, using the RPA decoupling scheme,

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(/cx) + 2 cos(/cz)] /Z.

Non-trivial 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 kx = ky = 0. We see that w = 0 at в.

• (d) Note that
• (i) when cos0 = 1 (ferromagnet), taking J2 = 0, we have

The expression (18.88) is reduced to the dispersion relation (4.32) for ferromagnets w = Z7iz>(l - yL) where Z' = 6 and yL = [2 cos(kx) + 2 cos ky + 2 cos(/tz)] /Z' [we have in this case В = 0 in (18.88) ].

Figure 18.7 Spin wave spectrum versus the wave vector ky 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 kx = kz = 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{Sz)

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 Sz = ±1/2 is the z component of the spin given by the Pauli matrices. 2SZ = az = ±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 Sz = ±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 Gl+2(t) but we shall neglect Green's functions of higher orders. We have

The Fourier transforms С&(£) = J^°oaGi<[E)e~'EtdE (к = i, i + 1, / + 2) yield, putting Е' = Е- дцвН,

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_tSf+1 >.

We have three equations for four unknowns xb x2, x3 and x’2. It is not possible to solve them. However, when H = 0, we have xi = x3 = 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 xli(1 =< SZ0SZ >, 2x2,„ =< + Sf) >, x3,„ =<

czcz czcz ^

Jo-5—>

When W = 0, we have Xi,„ = 4x3,„. Hence,

We take a solution of the form < S^SZ >= AXn~m where A is a constant: when m = n, we have < SZSZ >= 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 next-nearest 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 Z2 that of next-nearest 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 ekik' 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 (/cxa) + 4cos [kxa/2) cos ^/q,av/3/2^J /Z. Non-trivial solutions of (18.109) impose

We have

We show in Fig. 18.8 the spin wave frequency w versus ky in the first Brillouin zone for kx = 0.

Figure 18.8 Spin wave spectrum versus the wave vector ky in the case of a triangular antiferromagnet (S = /2,kx = 0, J = 1).

Note that when cos в = 1, the expression (18.113) is reduced to w = ZJ (Sz) (1 — j^) which is the dispersion relation (4.32) for ferromagnets. When cos0 = -1, Eq. (18.113) becomes w = ZJ {Sz) Jl — which is precisely the dispersion relation (4.59) of collinear antiferromagnets. Using the formula (3.97) for T = 0, we have

where AS = sinh2 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 zero-point contraction is due to the frustration of the triangular antiferromagnet. Other more precise methods taking into account higher-order 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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