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Solutions to Problems of Chapter 2Problem 1. Ising antiferromagnet: order parameter Answer: We take the case of a chain of N Ising spins. The order parameter is defined by M_{s} = ^ (— 1)' S', (staggered magnetization) where 5, is the spin at the lattice site Problem 2. Potts model: Answer: (a) Order parameter of qstate Potts model:
where Mi is the number of spins in the state / (/ = 1, ■ • •, q) and N the number of spins in the system. In an ordered state, only one of the states is present, Af, is equal to N, so p = 1, and in the disordered state, all M, are equal (= N/q), so p = 0.
For q = 3 (/ = 1, 2, 3), the ground state is constructed by choosing sequences of diagonal lines 123123 ... or 132132 ..., namely any sequence of diagonal lines with no adjacent similar numbers. An example is shown in Fig. 18.3. There are 3 ways for choosing a number for the first diagonal line, 2 ways for each of the following lines. Hence, the number of configurations in the ground state (degeneracy) is 3 x (2)^{i_1} x 2 oc 2^{l}, L being the number of diagonal lines (equal to the linear lattice size). The last factor 2 is to take into account the fact that there are two diagonal lines in the square lattice. This construction generates a semiordering: There is an ordering on each diagonal line but no ordering on the second diagonal line perpendicular to the first one. There is another way to construct the ground state which is completely disordered: Let us consider the square lattice defined on the xy plane. Each lattice site is defined by two indices (/, j). We fill the lattice sites Figure 18.3 An example of the construction of a groundstate configuration which has an order on each diagonal line (discontinued lines) but no order on the perpendicular diagonal lines. line by line starting from j = 0, from / = 0 to / = L (from left to right, bottom to top). The lattice sites on the first x line (7 = 0) can be randomly filled with three values 1,2, 3, never similar values at two adjacent sites. On the next line (7 = 1), the lattice site at (/, j) has its neighbor at (/', j  1) (below, on the previous line): there are two ways to choose its value which should be different from that at (/, j  1). Once its value is chosen, we go next to its neighbor at (i + 1, j): Its value should be different from that at (/, j) and that at (/ + 1, j — 1) (see Fig. 18.4, left panel). If the value at (/, j) is equal to the value at (/' + 1, j — 1), then we have two possible values for (/' + 1, j). If they are different, then we have only one possible value left (no choice) for (;' +1, j). We go to the next lattice site on the same line, namely the site at (/ + 2, 7), and we proceed in the same manner: We do not have any problem which can stop the pursuit of the construction of the line. We go to the next line at 7 + 1 and continue the construction from left to right on the line. We see that the ground states obtained Figure 18.4 Left: indexation of lattice sites for description in the text. Right: example of a random groundstate configuration constructed in the way described in the text. by such a way do not have a longrange order as seen in the example shown in Fig. 18.4 (right panel). Note that there are 2^{N} to place the numbers on the first line in the way described above. On each of the following lines, a number of the sites have only one choice as said above: Only a, N sites (a, < 1) have two choices. So, the random ground states have a degeneracy is of the order of 2^{w+}“> N+a_{2}N+ _ 2^{aNi} (o < 1). It is much larger than the degeneracy 2^{N} of the ordered ground states shown in Fig. 18.3. (d) The Potts model is equivalent to the Ising model when q = 2: There are two states for each site of energy — J and 0 for Potts model, and ±J for Ising model. There is only a shift of energy in the calculation. Physical results are identical. Problem 3. Domain walls: Solution: The interaction between two neighboring spins is E = 2] Si ■ Sj = — 2) S^{2} cos(ffj — Oj) where 0, is the angle between the x axis and S,. In the ferromagnetic state, the angle ф = (в,  0j) = 0. In a domain wall which has Aispin thickness, ф is я/N. If N is large as it is the frequent case, ф is small so that E ~ 2] S^{2}(l — ф^{2}} = —2] S^{2} + ) S^{2}(/>^{2}. The first term is the ferromagneticstate energy and the second term is the energy of the spin deviation. The total energy of the wall is Д = NJ S^{2} 2 = NJ S^{2}(tt/IV)^{2} = JS^{2}n^{2}/N.Problem 4. BraggWilliams approximation: Solution: (a) Entropy: One has N+ = N[ 1 + X)/2, AL = N{ 1  X)/2, with 0 < X < 1. The number of configurations (microscopic states) is W = N/N_{+}NJ.. Using the Stirling formula (see Appendix A) Inn! ~ nlnn  n for large n, one has
(b) The probability to have an up spin at a lattice site: p+ = N+/N. The number of upup spin pairs is thus
because there are z bonds around each site. The factor 1/2 is to remove the double counting of each bond. In the same manner, the number of downdown spin pairs and the number of antiparallel spin pairs are
(c) The energy of the crystal is (d) The free energy
The minimum of F corresponds to dF/dX = 0. This gives
This equation is equivalent to the meanfield equation (2.12) in the case where S = ±1. (e) Using an expansion of tanh[^A'] when X *■ 0 up to third order in X and proceeding in the same manner as for Eq. (2.23), one obtains T_{c} = zj/k_{B}. When T > T_{c}, one has X = 0. The entropy (18.27) is then S = k_{B}Nn2. This result can be directly obtained by the simple following argument: In the disordered phase, each spin is independent with two states ±1. There are N sites so the total number of system spin configurations is just W = 2^{N}. The entropy is then S = k_{B} In W = k_{B}N 2. Remark: The BraggWilliams approximation was initially used for binary alloys where an A atom is represented by an up spin, and a В atom by a down spin. The mixing of A and В atoms is favored if J is negative: This is equivalent to an antiferromagnetic ordering where an A atom (up spin) is surrounded by В atoms (down spins) at low temperatures. Problem 5. Binary alloys by spin language: Solution:
x < l.Whenx = 0, A/,; = N/4: The system is in the disordered state. The number of fspins occupying sites 11 is N_{bU} = N/2  N_{b}, = N{ 1  x)/4. For down spins ( В atoms), one has N^u + N^u = N/2; hence, Л/;,// = Л/(1+х)/4. One deduces Nj.,/ = N/2N_{U}n = N[ lx)/4. One considers the case x > 0 in the following: • The probability for a fspin to be at a site of the type I is P(t, /) = N^j/[N/2) = (l+x)/2, and that at a site of the type II is P(t,//) = W_{t},„/(A//2) = (lx)/2. In the same way, for a 4,spin one has P(l, П = N_{U}if{N/Z) = (1  x)/2 and P(j, II) = N,,n/{N/2) = (1 + x)/2. • Let N^f, N[^, and be the numbers of ft, 44 and t4 pairs. The probability to have aft pair is P(t, t) = P(t, 7)P(t,//) + P(t,//)P(t4) = (1  x^{2})/2; hence, W_{t},_{t} = NP{f, t) = 7V(1  x^{2})/2. In the same way, one has: = N[ 1  x^{2})/2. For a tT pair, the probability is
Therefore, N_{bi}=NPtf, i)=N{l + x^{2}). • One has E = [Л/^ + Thus,
• When E is given, x^{2} is determined. One takes x > 0. Let £2(£) be the number of microscopic states of energy equal to E. To calculate £2(E), one calculates the number of ways to choose N[ 1 + x)/4 fspins among N/2 sites of the type 1 and at the same time to choose 1V(1 + x)/4 sites among N/2 sites of the type II to place them. One has
Using the Stirling formula, one writes The entropy is given by 5 = k_{B} In £2. • The temperature T is calculated by (see Appendix A): One gets This equation is of the meanfield equation type [see (2.12)]. We have x = 1 at T =0 and x = 0 at T = oo. Between these limits, there exists a temperature below which x is not zero. An expansion at small x gives
If x Ф 0, one can simplify the two sides to get
Since 2(e  ф)/(квТ)> 0, the righthand side is positive. This relation is satisfied if on the lefthand side one has 2(e  ф)/(квТ]  1 > 0 namely T < 2(e — ф)/к_{в} = T_{c}. T_{c} is the critical temperature. If we return to the binary alloy language, we say we have an ordered binary alloy structure when T < T_{c }and a disordered structure for T > T_{c}. Problem 6. Critical temperature of ferrimagnet: Solution: We make an expansion of (2.73) and (2.74) when < S^{Z}A > and < Sg > are small in the same manner as for (2.20). We have
where Replacing (18.38) in (18.37), we write
For < S_{A} >ф 0, we can simplify it on both sides. The remaining equation is
The righthand side of this equation is positive; therefore, ас — 1 > 0. Replacing the coefficients a and c we obtain
This means that nonzero solutions of < S^{Z}A > are found only below T_{N}. Note that replacing (18.37) in (18.38) gives the same solution. Problem 7. Improvement of meanfield theory: Solution: (a) We write
The states of two spins 1/2 are
To calculate we use
We obtain a matrix 4 x 4. A simple diagonalization gives the following eigenvalues for the twospin cluster: (b) We put the cluster of two spins S, and Sy in a lattice: it has (Z  1) neighbors. We treat the interaction of the 4 cluster configurations found above in zero field (В = 0) with these neighbors by the meanfield theoiy. The energy of the cluster in the crystal depends on the embedded cluster spin configurations, they are in increasing energies: We consider the cluster of two spins as a superspin with the z component S^{z} = (5? + S?)/2. We have
where
and
Hence,
We see that < S^{z} >= 0 is a solution of this equation. An expansion around < S^{z} >= 0 gives
The solution < S^{z} >ф 0 is possible if T_{c} is obtained by solving— 3+2/S_{c}(Zl)7 e ^{2}^^{cI} = 0 where p_{c} = (кдТ_{с}I^{1}. We obtain
Problem 8. Interaction between nextnearest neighbors in mean field treatment: Solution:
The critical value of I/2I below which the ferromagnetic state is stable is determined by solving Ep_{err0} < EAntif’ We have /_{2}^{C} = or J{ = f£. When I/2I < J21 (^{or} /2 > J2) ^{we have the} ferromagnetic ordering. Otherwise, we have the antiferromagnetic one. Problem 9. We repeat Problem 7 in the case of an antiferromagnet: Solution: With J < 0: the changes with respect to the ferromagnetic case are
When putting these energies in the calculation of the average < (Sf + SJ)/2 >, be careful to use the energy of the corresponding crystalfield spin configuration in the argument of the exponential and to use the correct sign of each neighboring spin. Make a draw to help. We will have
It is the same as the ferromagnetic result, bearing in mind that J < 0 here. 
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