Following Ising himself, put the spins on a chain. The model can be solved exactly with a 2 × 2 transfer matrix, and the answer is that there is no phase transition at any finite temperature: the critical point sits at Tc = 0, Hc = 0, where the free energy has a cusp and the susceptibility and correlation length diverge. Watch the chain fluctuate, then follow the exact solution step by step.
Based on lecture notes by Sang Hoon Lee for Chapter 2, following K. Christensen and N. R. Moloney, Complexity and Criticality (2005). Units: kB = 1, and J = 1 unless stated. References
Top: the chain now (white up, black down). Below: its history, one row per frame with time running downward, so domains appear as vertical stripes and domain walls as their edges. Periodic boundaries, sN+1 = s1.
With periodic boundaries, E = −J Σi sisi+1 − H Σi si, and the field term can be split evenly between neighbors, H si → (H/2)(si + si+1). The Boltzmann weight then factorizes into one factor per bond, Tsisi+1 = exp[βJ sisi+1 + (βH/2)(si + si+1)], the entries of a real symmetric 2 × 2 matrix. Summing over each spin is a matrix product, so
In the thermodynamic limit only the larger eigenvalue survives, giving
At H = 0, f = −T ln(2 cosh βJ), which tends to −T ln 2 (pure entropy of random spins) as T → ∞ and to −J (pure energy of aligned spins) as T → 0. For any T > 0, m → 0 as H → 0, so m0(T) = 0; only at T = 0 is m0 = ±1. The free energy develops a cusp at (T, H) = (0, 0), a singular point: the critical point. Setting J = 0 recovers the non-interacting results, m = tanh βH and χ = β sech² βH.
At H = 0, m0 = 0 for T > 0, so g(r) = ⟨sisi+r⟩. Writing the couplings as Ji and differentiating Z = 2N Π cosh βJi once with respect to each bond between the two spins gives
ξ goes to 0 as T → ∞ and grows as ½e2βJ as T → 0+; at exactly T = 0 all spins are aligned and g = 0. Summing g over all sites recovers the susceptibility, Σj g = (1 + tanh βJ)/(1 − tanh βJ) = e2βJ = kBTχ(T, 0).
In zero field, T = 0 plays the role of the critical point: χ and ξ diverge as it is approached, together with the onset of spontaneous magnetization. So (Tc, Hc) = (0, 0). As in one-dimensional percolation, where pc = 1 and p ≤ 1, the critical point can be approached from one side only. The analogy is exact: with p ↔ tanh βJ the correlation functions and the susceptibilities have the same form. The only difference is the reference state. In percolation ξ measures fluctuations away from the empty lattice; in the Ising model, away from randomly oriented spins.
| 1D percolation | 1D Ising model, H = 0 | |
|---|---|---|
| Control parameters | p | T and H |
| Critical point | pc = 1, approached from below | (Tc, Hc) = (0, 0), approached from above |
| Correlation function | g(r) = pr | g(r) = (tanh βJ)r |
| Correlation length | ξ = −1/ln p | ξ = −1/ln tanh βJ |
| Sum rule | χ = (1 + p)/(1 − p) | kBTχ = (1 + tanh βJ)/(1 − tanh βJ) = e2βJ |
| ξ measures fluctuations away from | the empty configuration | randomly oriented configurations |
There is also a thermodynamic reason why an infinite aligned cluster survives only at Tc = 0. Compare a single domain of aligned spins with the same chain containing a “droplet” of flipped spins, bounded by two domain walls that each cost 2J:
The two walls can sit at N(N − 1) pairs of positions, which is entropy. So F2-dom − F1-dom ≈ 4J − 2kBT ln N, and the single domain is unstable whenever 2J/kBT < ln N, which for N → ∞ is every T > 0. Setting the difference to zero estimates the largest domain, N ≈ e2J/kBT, consistent with ξ ≈ ½e2βJ. There is no phase transition at any finite temperature in one dimension.
This links back to symmetry breaking: for a finite chain, ⟨M⟩ = 0 when H → 0 is taken first, and in one dimension even N → ∞ first does not help at T > 0, because domain walls destroy order on scales beyond ξ. The live chain at the top shows exactly this.
Key reference. K. Christensen and N. R. Moloney, Complexity and Criticality, Imperial College Press, London (2005), Chapter 2.