Questions: The Ising Model and Magnetic Transitions

Question 1

The 1D Ising chain has no ferromagnetic phase transition at any finite temperature, while the 2D Ising model does. What is the fundamental physical reason for this difference?

• A. The 2D lattice has more sites, allowing longer-range correlations to develop and stabilize order
• B. Domain walls in 1D are single broken bonds (low energy, no entropy gain), so thermal fluctuations always proliferate them; in 2D, domain walls are extended line objects whose energy grows with length, making them costly enough to suppress at low T
• C. The coupling constant J has a different sign in 1D versus 2D models
• D. The 2D model has an external field h that stabilizes the ordered phase

Answer: (B)

The 1D argument: a domain wall between an up-region and a down-region costs energy 2J but gains entropy k ln N (it can be placed anywhere along the chain). For any T > 0 and large N, the free energy gain from entropy exceeds the energy cost — domain walls proliferate and destroy long-range order. In 2D, a domain wall is a closed loop whose energy scales with its perimeter, making it much more costly to create. Below T_c, this energy cost suppresses domain walls; above T_c, entropy wins. This is a dimension-dependent balance, not a difference in model parameters.

Question 2

Mean-field theory applied to the Ising model correctly predicts a ferromagnetic transition and that magnetization vanishes continuously at T_c, but gives the wrong critical exponent β = 1/2 instead of β = 1/8 (2D). Why?

• A. Mean-field theory uses an approximate partition function that omits nearest-neighbor pairs
• B. Mean-field theory replaces neighbors' spins with their average value, suppressing fluctuations that are actually large near T_c — the approximation fails worst where it matters most
• C. Mean-field theory only applies to models without an external field h, introducing systematic error
• D. Mean-field theory assumes a square lattice, which does not match the hexagonal structure of the 2D Ising model

Answer: (B)

Mean-field theory's key approximation is replacing σ_j with its mean ⟨σ_j⟩ = m, eliminating correlations between neighboring spins. Near T_c, fluctuations are actually enormous — the correlation length diverges — so the assumption that neighbors look like their average is maximally wrong exactly at the critical point. In high dimensions (d > 4, the upper critical dimension), fluctuations become small and mean-field exponents become exact. In 2D they fail badly, giving β = 1/2 versus the exact β = 1/8.

Question 3

In the Ising model with J > 0, the ferromagnetic phase at low temperature exists simply because energy usually wins over entropy.

• T. True
• F. False

Answer: False

Energy wins over entropy only below T_c. Above T_c, entropy dominates and the system is disordered (m = 0). The phase transition is precisely the temperature at which these competing tendencies balance. Saying 'energy always wins when J > 0' ignores the temperature-dependence of the free energy: at high T, the entropy of disordered configurations (which vastly outnumber ordered ones) overwhelms the energy benefit of alignment.

Question 4

The partition function Z of an N-spin Ising model contains exactly 2^N terms, because each spin independently takes one of two values.

• T. True
• F. False

Answer: True

Z = Σ_{all configs} exp(−βH), where the sum runs over all possible assignments of ±1 to each of the N spins. Since each spin has 2 choices and the choices are independent, there are exactly 2^N configurations. For macroscopic N (~10^23), this sum is astronomically large — which is why direct computation is impossible and techniques like the transfer matrix method, mean-field approximation, or Monte Carlo simulation are needed.

Question 5

Explain in terms of energy-entropy competition why the Ising model has a ferromagnetic phase transition at a finite critical temperature T_c (in d ≥ 2).

Model answer: At low temperature, the Boltzmann factor strongly weights low-energy (aligned) configurations, so the system orders: spins align, m ≠ 0. At high temperature, the entropy of the vast number of disordered configurations overwhelms the energy benefit of alignment, so m = 0. There is a critical temperature T_c at which these tendencies balance — below it, order is thermodynamically stable; above it, disorder is. This competition is encoded in the free energy F = E − TS: at low T, minimizing E dominates; at high T, maximizing S dominates.

The entropy argument is quantitative: there are vastly more disordered configurations than ordered ones (one perfectly aligned state vs. 2^N − 2 others). Only the energy penalty J for misaligned neighbors keeps order stable at low T. The transition at T_c is where the free energy first prefers disorder — a collective phenomenon that requires d ≥ 2 for the energetic cost of disorder (domain walls) to be large enough to sustain an ordered phase.