Problem set — phase transitions and the Ising model

Problem set: phase transitions and the Ising model¶

Exam-style problems. Work them with a pen before touching a computer; problems 4 and 5 are meant to be finished in the laboratory, and problem 6 is a challenge. Solutions and marking rubrics live with the instructor material and are deliberately not on this site.

Energies are in units of the coupling $J$, temperatures in units of $J/\kB$ and fields in units of $J$, as in the module. The square lattice has $q = 4$ neighbours per site and Onsager’s critical temperature is $\kB T_c = 2J/\ln(1 + \sqrt{2}) = 2.2692\,J$.

Problem 1 — the acceptance rule is a choice, within limits¶

A Markov chain proposes to flip one spin, chosen uniformly at random, and accepts the proposal with probability $A(\Delta E)$, where $\Delta E$ is the energy change of the flip.

(a) Show that detailed balance with respect to $p_s \propto e^{-E_s/(\kB T)}$ holds if and only if $A(\Delta E)/A(-\Delta E) = e^{-\Delta E/(\kB T)}$ for every $\Delta E$. Why does the proposal step not appear in this condition?

(b) Verify that the Metropolis rule $A = \min(1, e^{-\Delta E/(\kB T)})$ and the Glauber rule $A = 1/(1 + e^{\Delta E/(\kB T)})$ both satisfy it.

(c) A student proposes $A' = \tfrac12 \min(1, e^{-\Delta E/(\kB T)})$, “to be safe”. Does it sample the right distribution? What does it cost? Explain why no rule can use a constant $c > 1$ in place of $\tfrac12$.

(d) For the square lattice at $h = 0$ and $\kB T = 2J$, list the possible values of $\Delta E$ for a single flip and the Metropolis and Glauber acceptance probabilities for each. Which rule accepts more often, and does that make it more correct?

(e) Show that the chain with acceptance $A$ leaves the Boltzmann distribution stationary, by summing detailed balance over the initial configuration. Then explain why stationarity does not say how many sweeps the chain needs to reach it.

Problem 2 — domain walls in one and two dimensions¶

(a) An open chain of $N$ spins at $h = 0$ is all up. Show that introducing one domain wall costs $\Delta E = 2J$ and gains entropy $\Delta S \approx \kB \ln N$, and hence that $\Delta F < 0$ for $N > e^{2J/(\kB T)}$. Evaluate this threshold at $\kB T = 0.5\,J$ and $\kB T = 0.2\,J$. What do you conclude about long-range order at any $T > 0$?

(b) The exact energy per spin of the infinite chain is $E/(NJ) = -\tanh(J/(\kB T))$. Show that this corresponds to a density of domain walls $n_w = (1 - \tanh(J/(\kB T)))/2$ per bond, and compare $1/n_w$ with the threshold of part (a) at $\kB T = 0.5\,J$.

(c) Now the square lattice, $L \times L$. A domain of reversed spins enclosed by a closed wall of length $\ell$ bonds costs $2J\ell$. The number of closed walls of length $\ell$ through a given bond is at most $3^{\ell}$. Show that the free energy of such walls is positive for $\kB T < 2J/\ln 3$. Evaluate the bound and compare it with Onsager’s $T_c$.

(d) Explain in two or three sentences why part (c) permits order in two dimensions but does not prove it, and why the same argument cannot be repeated in one dimension.

Problem 3 — mean field: what it gets right and how wrong it is¶

(a) Starting from a single spin in a field $h_{\mathrm{eff}}$, $\langle s \rangle = \tanh(h_{\mathrm{eff}}/(\kB T))$, derive the mean-field equation $m = \tanh((qJm + h)/(\kB T))$, naming the approximation precisely.

(b) Show graphically or by expansion that at $h = 0$ the equation has nonzero solutions only below $\kB T_c^{\mathrm{MF}} = qJ$.

(c) Expand $\tanh x \approx x - x^3/3$ and show that just below $T_c^{\mathrm{MF}}$, $m_0 \approx \sqrt{3(1 - T/T_c^{\mathrm{MF}})}$. What is the mean-field exponent $\beta$? What is the exact exponent on the square lattice?

(d) Above $T_c^{\mathrm{MF}}$, linearize in small $h$ and show that $\chi = \partial m/\partial h = 1/(\kB(T - T_c^{\mathrm{MF}}))$: the Curie–Weiss law. What is the mean-field exponent $\gamma$?

(e) By what percentage does mean field overestimate the square lattice’s transition temperature? What does it predict for the chain ($q = 2$), and what is the truth there?

(f) At $\kB T = 2.0\,J$, mean field gives $m_0 = 0.958$ and Onsager–Yang give 0.911; at $1.5\,J$, 0.990 and 0.987. Explain why mean field is better at the lower temperature.

Problem 4 — locating T_c from the susceptibility peak¶

Use the laboratory’s temperature sweep, or thermolab.ising.temperature_scan.

(a) For $L = 8$, 16 and 32, scan $\kB T$ from $2.20\,J$ to $2.80\,J$ in steps of $0.02\,J$, with at least 10⁴ sweeps per temperature after burn-in. Record the temperature of the peak of $\chi' = N\,\mathrm{Var}(|m|)/(\kB T)$ and its height. Repeat with three more seeds and quote each peak position with an error.

(b) Finite-size scaling predicts $T_{\mathrm{peak}}(L) = T_c + a\,L^{-1/\nu}$ with $\nu = 1$ for the square lattice. Fit your three peak positions against $1/L$ and extrapolate to $L \to \infty$. Compare with Onsager’s $T_c$, and state whether the difference is within your error.

(c) Plot the peak height against $L$ on logarithmic axes and fit a power. The theory says $\gamma/\nu = 7/4$; what do you find, and why might three small lattices disagree?

(d) A classmate fits $T_{\mathrm{peak}}(L)$ against $1/L^2$ instead and gets a different extrapolation. What does the choice of the exponent assume, and which of your conclusions is independent of that assumption?

Problem 5 — measuring an exponent, and why the answer is biased¶

(a) At $L = 64$, measure $\langle |m| \rangle$ at $\kB T = 1.8$, 1.9, 2.0, 2.1, 2.15, 2.2 and $2.25\,J$, with an error bar on each corrected for the autocorrelation time: $\sigma/\sqrt{N_{\mathrm{eff}}}$ with $N_{\mathrm{eff}} = n/(2\tau)$. Report $\tau$ at each temperature. Where is it largest?

(b) Fit $\ln\langle|m|\rangle$ against $\ln(T_c - T)$, using Onsager’s $T_c$. What exponent do you find? Compare with the exact $1/8$.

(c) Repeat the fit using only the three temperatures furthest from $T_c$, then only the three closest. Explain the difference: name one bias that dominates far from $T_c$ and one that dominates close to it.

(d) For your point at $2.25\,J$, compare the naive error $\sigma/\sqrt{n}$ with the corrected one. By what factor were you about to overstate your precision?

(e) Run the $2.25\,J$ point with four different seeds. Do the four values agree within their corrected error bars? Within the naive ones? What does this say about which error bar to quote?

Problem 6 — the paramagnet hidden inside the Ising model (challenge)¶

(a) Set $J = 0$. Show that the spins become independent, that the partition function factorizes into module 12’s single-spin form, and that $\langle m \rangle = \tanh(h/(\kB T))$ exactly.

(b) Take the exact chain magnetization from the module, $m = \sinh(h/(\kB T))/\sqrt{\sinh^2(h/(\kB T)) + e^{-4J/(\kB T)}}$, and show that it reduces to the result of part (a) as $J \to 0$. Then show that for small $h$ at fixed $J > 0$, $\chi = \partial m/\partial h|_{h=0} = e^{2J/(\kB T)}/(\kB T)$, and explain physically why coupling makes the chain more susceptible than free spins.

(c) For $J = 0$, verify the fluctuation identity $\chi = N\,\mathrm{Var}(m)/(\kB T)$ directly, computing $\mathrm{Var}(m)$ for $N$ independent spins.

(d) At $h = 0$ and $J > 0$ on the square lattice, explain why the identity of part (c) still holds for every finite $N$, but the estimator the laboratory uses below $T_c$ replaces $m$ by $|m|$. What would go wrong, in a run of 10⁴ sweeps at $\kB T = 2.0\,J$ on an $8 \times 8$ lattice, if you used $m$?