 Home Mathematics  Solutions to Problems of Chapter 2

Problem 1. Ising antiferromagnet: order parameter

Answer: We take the case of a chain of N Ising spins. The order parameter is defined by Ms = ^ (— 1)' S',

(staggered magnetization) where 5, is the spin at the lattice site

Problem 2. Potts model:

(a) Order parameter of q-state 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.

• (b) The ground state and its degeneracy when J > 0: When J > 0, in the ground state all lattice sites have the same Potts value.
• (c) If / < 0: The interaction of two different values has lower energy (zero). If q = 2 we see that the ground state is a configuration of alternating spins: This is the ''antiferromagnetic ordering.” The degeneracy is 2 (permutation of the two values of q).

For q = 3 (/ = 1, 2, 3), the ground state is constructed by choosing sequences of diagonal lines 1-2-3-1-2-3 ... or 1-3-2-1-3-2 ..., 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 2l, 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 ground-state 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 ground-state configuration constructed in the way described in the text.

by such a way do not have a long-range order as seen in the example shown in Fig. 18.4 (right panel). Note that there are 2N 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 2w+“> N+a2N+- _ 2aNi (o < 1). It is much larger than the degeneracy 2N 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) S2 cos(ffjOj) 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 Ai-spin thickness, ф is я/N. If N is large as it is the frequent case, ф is small so that E ~ -2] S2(l — ф2} = —2] S2 + ) S2(/>2. The first term is the ferromagnetic-state energy and the second term is the energy of the spin deviation. The total energy of the wall is Д = NJ S22 = NJ S2(tt/IV)2 = JS2n2/N.

Problem 4. Bragg-Williams 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 up-up 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 down-down 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 mean-field 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 Tc = zj/kB. When T > Tc, one has X = 0. The entropy (18.27) is then S = kBNn2. 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 = 2N. The entropy is then S = kB In W = kBN 2.

Remark: The Bragg-Williams 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:

• (a) Since e > ф, the energy of a spin is lower when it is surrounded by neighbors of the other kind. This yields a perfect antiferromagnetic ordering at Г =0.
• (b) • One has 0 < N^,i = N[l+x)/4 < N/2; hence,-1 <

x < l.Whenx = 0, A/|,; = N/4: The system is in the disordered state. The number of f-spins occupying sites 11 is NbU = N/2 - Nb, = 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/2-NUn = N[ l-x)/4.

One considers the case x > 0 in the following:

• The probability for a f-spin 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,//) = Wt,„/(A//2) = (l-x)/2.

In the same way, for a 4,-spin one has

P(l, П = NUif{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 - x2)/2; hence, Wt,t = NP{f, t) = 7V(1 - x2)/2. In the same way, one has: = N[ 1 - x2)/2.

For a tT pair, the probability is Therefore, Nbi=NPtf, i)=N{l + x2).

• One has E = [Л/^ + Thus, • When E is given, x2 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 f-spins 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 = kB In £2.

• The temperature T is calculated by (see Appendix A): One gets This equation is of the mean-field 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 right-hand side is positive. This relation is satisfied if on the left-hand side one has 2(e - ф)/(квТ] - 1 > 0 namely T < 2(e — ф)/кв = Tc. Tc is the critical temperature.

If we return to the binary alloy language, we say we have an ordered binary alloy structure when T < Tc and a disordered structure for T > Tc.

Problem 6. Critical temperature of ferrimagnet:

Solution: We make an expansion of (2.73) and (2.74) when < SZA > and < Sg > are small in the same manner as for (2.20). We have where Replacing (18.38) in (18.37), we write For < SA 0, we can simplify it on both sides. The remaining equation is The right-hand side of this equation is positive; therefore, ас — 1 > 0. Replacing the coefficients a and c we obtain This means that non-zero solutions of < SZA > are found only below TN. Note that replacing (18.37) in (18.38) gives the same solution.

Problem 7. Improvement of mean-field 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 two-spin 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 = 0) with these neighbors by the mean-field 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 Sz = (5? + S?)/2. We have where and Hence, We see that < Sz >= 0 is a solution of this equation. An expansion around < Sz >= 0 gives The solution < Sz 0 is possible if Tc is obtained by solving— 3+2/Sc(Z-l)7 -e 2^cI = 0 where pc = (кдТсI-1. We obtain Problem 8. Interaction between next-nearest neighbors in mean- field treatment:

Solution:

• (a) All spins are parallel at Г =0. All interactions are fully satisfied.
• (b) The hypothesis of the mean-field theory: All neighboring spins of a spin are replaced by an average value which is used to calculate the value of the spin under consideration.
• (c) The energy of a spin at T = 0 is E = — Z) j — Z2where Zx and Z2 are the numbers of nearest neighbors and of next-nearest neighbors, respectively. For a body-centered cubic lattice, Zl = 8, Z2 = 6. E is the energy which maintains the spin ordering: The lower it is, the higher the temperature is needed to destroy the ordering. Thus, the stronger J2 is, the higher the transition temperature becomes.
• (d) The same calculation as that in the chapter by replacing CJ with ZJi + Z2/2 in Eqs. (2.5)-(2.7) and in the following equations to obtain the final mean- field equation.
• (e) The critical temperature is obtained by replacing CJ in Eq. (2.23) by ZJi + Z2/2.
• (f) Now we suppose /2 < 0. When |/2| 3>7i, it is obvious that the J2 interaction imposes the antiferromagnetic ordering to make the overall energy negative. The spins on the cube corners form an antiferromagnetic sublattice, the centered spins form another antiferromagnetic sublattice, independent of the first one. Since each spin has 4 up neighbors and 4 down neighbors (make a figure to convince yourself), its interaction energy with nearest neighbors is zero, independent of J. The energy of such a spin configuration is thus EAntif = 22 = -^21721If hJ i then the ferromagnetic state is more favorable. Its energy is Eperro = -ZJ - Z2/2 = -Zih+Z2J2.

The critical value of I/2I below which the ferromagnetic state is stable is determined by solving Eperr0 < EAntif’ We have |/2C| = 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

• (i) in question (a): no change
• (ii) in question (b): we have the inverse order of energies £4 < £3 < £2 < Ei, because J < 0. The remaining calculation is exactly the same. The four configurations embedded in the crystal give the following energies: When putting these energies in the calculation of the average < (Sf + SJ)/2 >, be careful to use the energy of the corresponding crystal-field 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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