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Solutions to Problems of Chapter 5Problem 1. Chain of Ising spins by exact method: We write the Hamiltonian as
with where p = J/к_{в}Т. Since a„a_{n+}i = ±1, we have the following identity (by verification):
Equation (18.117) becomes
Except the first and the last terms, all other terms of the sum in the square brackets [• • • ] are zero because in each term there is one a which appears once in the factor giving rise, when summed up, two terms of opposite signs. The first term in (18.119) does not depend on a, it gives 2^{w}(cosh p)^{N}. The last term yields 2^{w}(sinh p)^{N} because each a appears twice in its factor. We have
For T Ф 0 (p Ф oo), we have cosh p > sinh p. With N 1, we can neglect (sinh p)^{N} compared to (cosh p)^{N}. Thus,
The free energy is
The average energy is thus [see (A.10)]
The heat capacity is Figure 18.9 Energy E (top) and heat capacity C (bottom) per spin versus temperature T.J/кв = 1 has been used. The energy and the heat capacity per spin are shown in Fig. 18.9. We see that C has a maximum but no divergence. Thus, there is no phase transition at finite T. Problem 2. Chain of Ising spins by microcanonical method:
ttt'l'ttt • • or a block of spins is reversed ■ • • tt1444ttt • • • • The energies of the two states are identical with two "unsatisfied bonds" of energy +/. The energy of the system in the first excited state is thus E{2) = J (JV  2) + 2J = ) [N  2 x 2) where 2 is the number of unsatisfied bonds. The degeneracy of the first excited state is equal to the number of ways to choose two unsatisfied bonds among N bonds of the system. This is given by g{2) = 2Cj/ where the factor 2 results from the global reversal of the whole system. One deduces that the energy E (2n) of a state in which there are 2n unsatisfied bonds is £(2/7) = J[N 2/7) + 2nJ = —J [N — 2 x 2n). The degeneracy is 9(2n) = 2 Cj?. The energy of the system is maximum when all bonds are unsatisfied, namely 2n = N so that E_{max} = 0. The degeneracy of this state is g = 2C$ = 2.
= k_{B}[ 2 + In ЛП — ln(2n)!  ln[N  2n)!] ~ k_{B}[ 2 + N In N  2n In 2/7  (IV  2n) In[N  2n)] where the Stirling formula has been used. The micro canonical temperature is
from which,
where x is the percentage of unsatisfied bonds x = One obtains
At low temperatures, x *■ 0: All spins are parallel (the system is ferromagnetic). At high temperatures, x *■ 1/2: Half of the bonds are unsatisfied, the system is disordered (paramagnetic). Problem 3. Chain of Ising spins by canonical method: One considers again the system defined in Problem 2 but in the canonical situation: It is maintained at the temperature T. (a) The partition function is given by
In combining the two following Newton relations, one has
since all odd terms in n are canceled out. Putting и = exp(—02)), and using (18.128), one rewrites (18.127) as
This relation of Z is exact (see Problem 1). (b) The system energy is thus
(c) The average percentage x of unsatisfied bonds is
At low temperatures, x ■ 0. At high temperatures, x ■ 1/2. One finds again here the results using the microcanonical method found in Problem 2. The curves E and the calorific capacity C are plotted in Problem 1 (Fig. 18.9). Problem 4. Low and hightemperature expansions of the Ising model on the square lattice: Solutions: We have
where the sum is performed over nearest neighbors and ^{a} ''0) = ^{±1} (a) The partition function
where К = J /(k_{B}T). In the ground state (GS), we have E_{0} = —2JN = N_{b}J and Z = 2exp[N_{b}K) where the factor 2 comes from the GS degeneracy (reversing all spins), N_{b} = 2N is the total number of links. Figure 18.10 Real square lattice (solid lines) and its dual lattice (broken lines). Let us construct the dual lattice by drawing the links (broken lines) perpendicular to the real links (solid lines) as shown in Fig. 18.10. The case of the square lattice is special: Its dual lattice formed by the broken lines is also a square lattice. This is not the case in general: For example, the dual lattice of the triangular lattice is the honeycomb lattice. (b) Lowtemperature expansion: ^{• [1]} Figure 18.11 Graphs (heavy lines) crossing broken links for one, two three reversed spin clusters. The reversed spins are shown by black circles, other spins are not shown. straight trimer along у axis, and four perpendicular trimers), E = E_{0} + 16/. Calculation of Z with the first excited states of energies E_{0} + 8J, Eо + 12J, Eо + 16/ : To do that, we have to find all excited states for each level. The first two levels concern one and two reversed neighboring spins as found above. However, the level Eq + 16/ corresponds not only to the excited trimer shown above but also to two other cases: a cluster of four sites forming a square, two disconnected reversed spins. Both cases have 8 broken links. The first case has a degeneracy of N (the number of choices of the first site of the square), the second case has a degeneracy of N(N — 5) (the number of choices of the first reversed spin is N, the number of choices of the second disconnected reversed spin is N — 5 where 5 is the number of spins concerned by the first reversed spins which are to be avoided). We finally have (c) We draw a path P which encircles each cluster of reversed spins: This path crosses the broken links around the cluster. For such a closed path, we can verify that it crosses an even number of broken links as follows. Imagine a rectangular path, the number of broken links is even. Now, including an additional site anywhere around the path will add two additional broken links. Excluding a site inside the path will reduce the number of broken links by 2. Such a construction shows clearly that any closed path crosses an even number of broken links. Let t(P) be the number of broken links crossed by the path. Since the variation of energy when breaking a link is AE = J  (—J) = 2J. A path crossing £(P) broken links corresponds to AE = +2£(P) J, so that the system energy is £(P) = E_{0} + AE = E_{0} + 2t(P) J. The partition functions is thus
(d) Hightemperature expansion: Using Eq. (18.118) we write the partition function as
Since there are N_{b} links, there are N_{b} factors in the product. We draw a link between two nearest sites in each factor (this link is in the real space). We expand the product, we see that if a given spin appears an odd number of times in a term of the resulting polynomial, then when summing on its values, this term gives two opposite values yielding a zero contribution. Each nonzero term contains an even number of times of each spin, so that the spin factor in front of tanh К is equal to 1. The power of tanh К for this term is nothing but the number of links of spins in front of it. Consider, for example, the term x = (Ti
Figure 18.12 Graphs linking nearest sites: The lower left graph represents the term a^a_{L} right one represents the term where 2^{N} comes from the sum V_{tJi}__{±1} and f(P) is the number of links in the closed graph P. (e) Duality: The partition functions Z in (18.133) and (18.134) have the same structure: Since the prefactors are nonsingular, the summations over the closed paths deter?mine the singularity of Z. Note that £(P) in (18.133) corresponds to the path drawn in the dual lattice (see Fig. 18.11). While, £(P) in (18.134) corresponds to the path drawn in the real lattice (see Fig. 18.12). Nevertheless, these two kinds of path have the same structure with an even number of links in each path. Therefore, we can connect the two Z by fixing
where K* corresponds to the lowГ phase and К to the highГ phase. The above relation (18.136) is called the "duality” condition which connects the low and the highГ phases. We deduce from (18.134) and (18.133) the following relation between the hightemperature Z (К) and the lowtemperature Z (К*):
where £]p(tanh /Г)‘(^{р}) has been replaced by ^{e}~^{2K e}^ and then by Z(K*)/[2e^{NbK}') = Z(/6*)/(e^{iV}'’^{,r}) from (18.133) (the factor 2 in the denominator is neglected because N is large). (f) The critical temperature of the Ising model on the square lattice is obtained by using the duality. We have
from which This relation is symmetric with respect to К and K*: when K* increases, К decreases, and vice versa. If the system undergoes a single phase transition, it should undergo at the same point of К and K*, namely K* = K_{c}. To satisfy (18.139), we should have
Thus,
Therefore, k_{B}T_{c}/J = 1/K_{C} ~ 2.27 which is the exact Onsager's solution. Problem 5. Critical temperatures of the triangular lattice and the honeycomb lattice by duality: Solutions: We have
where the sum is performed over nearest neighbors on the triangular lattice and = ±1. The dual lattice by construction is a honeycomb lattice shown by the broken lines in Fig. 18.13. First, we follow the same method as for the square lattice in Problem 4 above: Writing the partition functions of the triangular lattice using lowtemperature expansion with graphs on links of the dual lattice and hightemperature expansion with graphs on links of the real lattice, we obtain a relation similar to Eq. (18.137) Figure 18.13 Triangular lattice (solid lines) and its dual honeycomb lattice (broken lines). where Z_{t}[K) denotes the lowtemperature partition function of the triangular lattice, Z_{h}{K*) the hightemperature dual (honeycomb) lattice and N_{b} = 3N is the total number of links. Now, we can relate the two partition functions in another relation as follows. For convenience, the dual lattice is shifted as shown in Fig. 18.14. We consider the plaquette defined by three sites 1,2 and 3 with a site at the center. We calculate the following quantity:
where we used the remark before Eq. (18.134) to expand the product of the first equality. Now, we consider the plaquette defined by three sites 1, 2 and 3 of the triangular lattice. We calculate the following quantity:
If we set the factors of (cti a_{2} + a_{2} <73 + 03 or) in (18.144) and (18.145) to be equal, then we have
so that
This relation connects the two dual lattices. To find the full partition functions, it suffices to sum over all spins of the plaquette and to take the product over all plaquettes on each side of the above equation, we then have Figure 18.14 Triangular lattice (solid lines) and its dual honeycomb lattice (broken lines) shifted for convenience: The four sites are numbered from 0 to 3 for calculating the partition function. Replacing Z_{h}[K*) given by (18.148) in (18.143), we have Using (18.146) to eliminate K* in the above equation, and after some algebra, we get
Let us calculate the critical temperature of the triangular lattice. We note that by graph constructions for low and hightemperatures, we obtained Eq. (18.135) which is veiy general, independent of the lattice structure: It was established using the square lattice, but all arguments leading to it are also valid for the triangular lattice. Using Eq. (18.135) to eliminate K* in Eq. (18.146), and after some algebra, we obtain
As before in the case of the square lattice, this relation shows that if К increases, K^{+} decreases, and vice versa. The transition should occur at the same critical temperature K_{c} = K+. The solution of Eq. (18.151) at K_{c} is thus
To find the critical temperature of the honeycomb lattice, we follow the same method as above: We obtain Solutions to Problems of Chapter 8Problem 1. Surface magnon: Solution: In the ferromagnetic case, we just write the equation of motion for S+. We use next the Fourier transform in the xy plane. We obtain then, for n > 2,
and, for the first two layers, where E = e = 4^{1}, and JS h
To study bulk modes, we replace U„±i = U_{n} exp(±/k_{2}a/2) in Eq. (18.152) to obtain the energy of the bulk mode of wave vector (кц, k_{z}):
For surface modes, we replace U_{n}±i = Ui4>^{n} in Eqs. (18.152)(18.154) to obtain surfacemode energy E and the damping factor ф. Problem 2. Critical nextnearestneighbor interaction: Solution: If and J_{2} are both ferromagnetic (>0), the ferromagnetic state is stable. However, if J_{2} becomes negative, the ferromagnetic state becomes unstable beyond a critical value of e = J_{2}/Ji For an infinite crystal, the critical value is obtained by setting the energy E of the lowest magnon mode equal to zero. This corresponds to the instability due to a soft mode. The stable state is no more ferromagnetic for e > e_{c} with e_{c} = , namely J_{2} < — /i Problem 3. Uniform magnetization approximation: Solution: If we replace all < Sjj > by a unique value M, we see that all elements of the matrix M in Section 8.4 are proportional to M. The eigenvalues £, obtained by solving detM = 0 is, therefore, proportional to M. Problem 4. Multilayers: critical magnetic field Solution: Without applied field, the spin configuration is Л(ир spin)—B(down spin)C(up spin). In the very strong field applied along z, the В spins all turn up. We calculate the critical field beyond which the spin configuration is that state. The inplane exchange energy for a В spin does not change whether the В film is in up or down state. We consider a column of spins in the z direction. The energy of the spins of В when they are antiparallel to H is
The energy of В spins when they are parallel to H is
[J_{s} is negative). We see that £> < E_{AB} when H > H_{c} = (we have taken Ji = /_{2} = /3). Problem 5. Meanfield theory for thin films: Solution: We assume the Hamiltonian ~K = — / XLy/ S/Sy where S, = ±1 (Ising spin at the lattice site /). We suppose a simple cubic lattice with a (001) surface. Using the mean field theory we have the average values of the spins in the three layers
where ft = l/(k_{B}T) and Z = 4 the number of nearest neighbors in the xy plane. By symmetry < Si >=< S3 >; therefore, we have only two equations to solve. Numerically, it can be easily done in a selfconsistent manner. Problem 6. HolsteinPrimakoff method: Guide: We can modify the equations (3.41)(3.43) for a semiinfinite crystal: We write the equation of motion for a spin in each layer. We obtain a set of difference equations. We use next the Fourier transform in the xy plane. We can use (3.41)(3.43) to make some comments: The lower the energy e_{k} is, the larger < n_{k} > becomes [see (3.41)]. As a consequence, M becomes smaller [see (3.43)]. Thus, the lowenergy surface modes lower the layer magnetization near the surface with a stronger effect for surface magnetization because of the lack of neighbors. Problem 7. Frustrated surface: surface spin rearrangement Solution: If there is only the surface, then the surface spins form a planar 120° configuration (see Section 5.7.3). We suppose this structure lies in the xy plane. Now, when the beneath layer acts on the surface spins with a ferromagnetic interaction, the surface spins on a triangle turn into the z direction to satisfy partially the ferromagnetic interaction: beta be the projection on the (xy) plane the angle between two neighboring surface spins and f) be the angle between a surface spin and its neighbor in the second layer (see Fig. 18.15). We have
The energy of a cell formed by a surface triangle and the beneath triangle is, for spins 1/2,
The minimization of this energy gives
Hence,
This solution is possible for J_{s} < —J /9. For ]_{s} > —J /9, the stable state is ferromagnetic. Note that we can use the steepestdescent method described in Chapter 9 to determine numerically the ground state of classical spin systems. Problem 8. Ferrimagnetic film: Solution: Let us take into account the difference between the average layer magnetizations, namely < S^{z}m >=< Sf > if m belongs to the first layer, < Sf >=< S > if / belongs to the second layer, etc. Modifying (8.5)(8.6) for the first two layers, we have Figure 18.15 Spin configuration of the ground state. The projection on the (xy) plane of the angle between two surface spins is a = 120°. The angle between a surface spin and the beneath spin is /3. We can write, in the same manner, the two equations for layers 3 and 4 as
where the notations are defined as
Note that < S* > (n: layer n) is positive for n = 2, 4 (Л sublattice) and negative for n = 1, 3 [B sublattice). Other notations in the above equations were defined after (8.6). We can put the above equations under a matrix form
where x represents a nonzero elements. Nontrivial solutions impose that the determinant is zero. Solving numerically det  • • •  = 0 we obtain the energy eigenvalues, and by replacing them into the above matrix equation we get the spin wave amplitudes. If Ui = U_{2} = U3 = U_{4}, then the corresponding energy is a bulk mode. It is a surface mode otherwise. Note that the two surfaces of the 4layer film are not symmetric. We should have four distinct modes. For e = 0 and k_{x} = k_{y} = 0, we have ^(кц) = cos(^) cos(^) = 1 and хг(к) = i[cos(k_{A}o) + cos(/q,a)] = 1. We write for this case the matrix as follows:
where nonzero elements are
This matrix can be diagonalized without difficulty. For the case k_{x} = k_{y} = я/a, we proceed in the same manner.

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