1. Counting. How many configurations does a 10×1010 \times 10 Ising lattice have? How long would it take to enumerate them at 10910^9 configurations per second? Compare with the age of the Universe, 1.4×10101.4 \times 10^{10} years.

  2. Fluctuations. Derive Eqs. (3) and (4) from ZZ.

  3. Independent spins. For J=0J = 0, compute ZZ, mm, the energy and the specific heat per spin. Show that χ=1/kBT\chi = 1/k_BT at h=0h = 0.

  4. Specific heat of the chain. Show that the specific heat (7) is maximum when Ktanh⁡K=1K \tanh K = 1, and find the corresponding temperature.

  5. Transfer matrix. Check the eigenvalues λ±\lambda_\pm of T\mathcal{T}, and derive the magnetization (10) from m=∂ln⁡λ+/∂(βh)m = \partial \ln\lambda_+ / \partial(\beta h). What are its limits for T→0T \to 0 and T→∞T \to \infty?

  6. Mean field. (a) Derive m≃3(1−T/Tc)1/2m \simeq \sqrt{3}(1 - T/T_c)^{1/2} and the Curie–Weiss law from Eq. (12). (b) Show that the mean-field specific heat jumps from 32kB\frac32 k_B to 0 at TcT_c.

  7. Duality. Solve tanh⁡K=e−2K\tanh K = e^{-2K} and check Eq. (15).

  8. Scaling relations. Check that the exact 2D exponents and the mean-field exponents (in d=4d = 4) satisfy the four scaling relations of Section 5.4.

  9. Detailed balance. Show that the Metropolis acceptance (23) satisfies detailed balance (21). Do the same for the heat-bath probability.

  10. Acceptance rates. On the square lattice at h=0h = 0, list the possible values of ΔE\Delta E and the corresponding Metropolis acceptance probabilities at T=1T = 1 and T=3T = 3.

  11. Wolff algorithm. Consider a cluster CC with n1n_1 parallel bonds on its boundary before the flip and n2n_2 after. Show that the energy changes by ΔE=2J(n1−n2)\Delta E = 2J(n_1 - n_2), and that the ratio of the probabilities to build CC in the forward and backward moves is (1−p)n1−n2(1-p)^{n_1-n_2}. Deduce that the choice p=1−e−2βJp = 1 - e^{-2\beta J} satisfies detailed balance with an acceptance of 1.

  12. Lattice gas. With si=2ni−1s_i = 2n_i - 1, show that the Ising Hamiltonian becomes H=−4J∑⟨ij⟩ninj−μ∑ini+constH = -4J\sum_{\langle ij\rangle} n_i n_j - \mu \sum_i n_i + \text{const}, and express μ\mu in terms of hh and JJ (on the square lattice).

  13. With the code. (a) Using ising_2D.py, option 3, measure the height of the specific heat peak for L=8L = 8, 16, 32 and 64, and show that it grows like ln⁡L\ln L. (b) Using option 5, estimate TcT_c from the crossings of the Binder cumulant.

Answers

  1. 2100≈1.3×10302^{100} \approx 1.3 \times 10^{30}; 1.3×10211.3 \times 10^{21} s =4×1013= 4 \times 10^{13} years, three thousand times the age of the Universe.

  2. ∂βln⁡Z=−⟨H⟩\partial_\beta \ln Z = -\left\langle H \right\rangle and ∂β2ln⁡Z=⟨H2⟩−⟨H⟩2\partial_\beta^2 \ln Z = \left\langle H^2 \right\rangle - \left\langle H \right\rangle^2; with ∂T=−kBβ2∂β\partial_T = -k_B\beta^2 \partial_\beta, C=kBβ2(⟨H2⟩−⟨H⟩2)C = k_B \beta^2 (\left\langle H^2 \right\rangle - \left\langle H \right\rangle^2). Similarly with derivatives with respect to βh\beta h.

  3. Z=(2cosh⁡βh)NZ = (2\cosh\beta h)^N, m=tanh⁡βhm = \tanh\beta h, e=−htanh⁡βhe = -h \tanh \beta h, c=kB(βh)2/cosh⁡2βhc = k_B (\beta h)^2 / \cosh^2 \beta h.

  4. dc/dK=0dc/dK = 0 gives Ktanh⁡K=1K\tanh K = 1, K=1.200K = 1.200, kBT=0.834 Jk_B T = 0.834\,J, c=0.439 kBc = 0.439\,k_B.

  5. det⁡(T−λ)=0\det(\mathcal{T} - \lambda) = 0 gives λ2−2λeKcosh⁡βh+2sinh⁡2K=0\lambda^2 - 2\lambda e^K \cosh \beta h + 2\sinh 2K = 0. For T→0T \to 0, m→sign⁡(h)m \to \operatorname{sign}(h); for T→∞T \to \infty, m≃tanh⁡(βh)m \simeq \tanh(\beta h), free spins.

  6. (a) See Section 4. (b) Below TcT_c, e=−qJ2m2≃−3qJ2(1−T/Tc)e = -\frac{q J}{2} m^2 \simeq -\frac{3qJ}{2}(1 - T/T_c), so c=3qJ2Tc=32kBc = \frac{3qJ}{2T_c} = \frac32 k_B just below TcT_c, and c=0c = 0 above.

  7. With x=e−2Kx = e^{-2K}, (1−x)/(1+x)=x(1 - x)/(1 + x) = x gives x2+2x−1=0x^2 + 2x - 1 = 0, x=2−1x = \sqrt{2} - 1, Kc=12ln⁡(1+2)=0.4407K_c = \frac12 \ln(1 + \sqrt 2) = 0.4407.

  8. 2D: 0+14+74=20 + \frac14 + \frac74 = 2; 18×14=74\frac18 \times 14 = \frac74; 1×(2−14)=741 \times (2 - \frac14) = \frac74; 2×1=2−02 \times 1 = 2 - 0. Mean field in d=4d = 4: 0+1+1=20 + 1 + 1 = 2; 12×2=1\frac12 \times 2 = 1; 12×2=1\frac12 \times 2 = 1; 4×12=24 \times \frac12 = 2.

  9. If ΔE>0\Delta E > 0, W(σ→σ′)=e−βΔEW(\sigma\to\sigma') = e^{-\beta\Delta E} and W(σ′→σ)=1W(\sigma'\to\sigma) = 1, whose ratio is P(σ′)/P(σ)P(\sigma')/P(\sigma). Heat bath: the ratio of p+p_+ and p−p_- is e2β(J∑sj+h)e^{2\beta(J\sum s_j + h)}, the ratio of the Boltzmann weights.

  10. ΔE=−8,−4,0\Delta E = -8, -4, 0: always accepted. ΔE=4\Delta E = 4: e−4=0.018e^{-4} = 0.018 at T=1T = 1, e−4/3=0.26e^{-4/3} = 0.26 at T=3T = 3. ΔE=8\Delta E = 8: 3.4×10−43.4\times10^{-4} and 0.0690.069.

  11. Bonds inside the cluster are unchanged; boundary bonds switch between parallel and antiparallel. The boundary bonds that are parallel must all have been rejected: (1−p)n1(1 - p)^{n_1} forward and (1−p)n2(1-p)^{n_2} backward. With 1−p=e−2βJ1 - p = e^{-2\beta J} the ratio is e−2βJ(n1−n2)=e−βΔEe^{-2\beta J(n_1-n_2)} = e^{-\beta\Delta E}.

  12. H=−4J∑ninj−(2h−8J)∑ni+constH = -4J\sum n_i n_j - (2h - 8J)\sum n_i + \text{const} (each site has 4 neighbours), so μ=2h−8J\mu = 2h - 8J.

  13. (a) Kaufman's exact values of the maximum of cc are 1.191.19, 1.551.55, 1.901.90 and 2.252.25 for L=8L = 8, 16, 32 and 64: each doubling of LL adds about 0.350.35, close to (8Kc2/π)ln⁡2=0.343(8K_c^2/\pi)\ln 2 = 0.343, so cmax⁡≃0.49ln⁡L+constc_{\max} \simeq 0.49 \ln L + \text{const}. The simulated points agree with these exact curves (Figure 7c). The positions of the maxima, 2.3622.362, 2.3182.318, 2.2942.294 and 2.2822.282, approach TcT_c like 1/L1/L, as expected for ν=1\nu = 1. (b) The crossings are within about 0.010.01 of Tc=2.269T_c = 2.269.