A system's two-point correlation function ⟨σ(r)σ(0)⟩ behaves as exp(−|r|/ξ) for large |r|, with ξ finite. What does this indicate about the system's phase?
AThe system is in an ordered phase with long-range order characterized by correlation length ξ
BThe system is in a disordered phase — correlations decay to zero beyond a few correlation lengths, with no long-range coherence
CThe system is exactly at its critical point, where correlations decay as a power law
DThe system has partial long-range order, with ξ setting the range of ordering
Exponential decay to zero means distant points are statistically independent — the system has no global coherence. This is the disordered phase (e.g., a paramagnet above its Curie temperature). Long-range order requires ⟨σ(r)σ(0)⟩ to approach a non-zero constant as |r| → ∞. Option C (the critical point) is also wrong: at criticality, ξ diverges and correlations decay as a power law, not exponentially. A finite ξ means you are in a disordered phase, not at criticality.
Question 2 Multiple Choice
In a ferromagnet below the Curie temperature with order parameter m = ⟨σ⟩ ≠ 0, why does the two-point function ⟨σ(r)σ(0)⟩ approach m² rather than zero at large |r|?
ABecause at large distances, σ(r) and σ(0) become statistically independent conditional on the global order, so ⟨σ(r)σ(0)⟩ → ⟨σ(r)⟩⟨σ(0)⟩ = m·m = m²
BBecause ordered phases have no fluctuations, so every spin exactly equals m
CBecause m² is the theoretical maximum of the correlation function
DBecause the correlation function must equal 1 at all distances in the ordered phase
When |r| is large enough that the two spins are statistically independent, their joint expectation factorizes: ⟨σ(r)σ(0)⟩ → ⟨σ(r)⟩⟨σ(0)⟩ = m². This is non-zero precisely because the system has broken symmetry: m ≠ 0 means each spin has a non-zero mean even when uncorrelated with any other specific spin. In the disordered phase, m = 0, and the same factorization gives zero — which is why the correlation function decays to zero there.
Question 3 True / False
Exactly at the critical point of a second-order phase transition, the two-point correlation function decays exponentially with a very large but finite correlation length.
TTrue
FFalse
Answer: False
Exactly at the critical point, the correlation length ξ *diverges* to infinity. The two-point function then cannot decay exponentially (exponential decay requires a finite ξ); instead it follows a power law: ⟨σ(r)σ(0)⟩ ~ |r|^(−(d−2+η)), where η is a critical exponent. This scale-invariant power-law behavior is the signature of criticality and underlies universality — the same exponents appear in seemingly different physical systems. 'Very large ξ' describes near-critical behavior approaching the transition, not the critical point itself.
Question 4 True / False
Crystals, ferromagnets, and superconductors all exhibit long-range order, but the physical quantity that becomes long-range correlated differs between them.
TTrue
FFalse
Answer: True
All three share the same mathematical signature — a two-point function that approaches a non-zero constant at large separation — but the quantity being correlated differs. In ferromagnets it is the spin ⟨σ(r)σ(0)⟩; in crystals it is the density ⟨ρ(r)ρ(0)⟩ oscillating at lattice periodicity; in superconductors and superfluids it is the off-diagonal element of the one-particle density matrix ⟨ψ†(r)ψ(0)⟩ (off-diagonal long-range order, ODLRO). The mathematical criterion for long-range order is universal; the physical interpretation varies.
Question 5 Short Answer
Why is a non-zero order parameter m = ⟨σ⟩ equivalent to long-range order in the two-point correlation function?
Think about your answer, then reveal below.
Model answer: When two spins are very far apart, they become statistically independent. Their correlation then factorizes: ⟨σ(r)σ(0)⟩ → ⟨σ⟩² = m². If the system has broken symmetry and m ≠ 0, this limit is non-zero — which is the definition of long-range order. If m = 0 (no broken symmetry), the factorized limit is zero, and the correlation decays to zero, meaning no long-range order.
The connection reveals why spontaneous symmetry breaking and long-range order are two sides of the same phenomenon. A non-zero order parameter means every spin independently carries a preferred direction. Statistical independence at large distance doesn't destroy the correlation because each spin is individually biased toward m, so their product averages to m² rather than zero. This equivalence between m ≠ 0 and non-zero limiting correlation is what makes the order parameter a complete diagnostic of the ordered phase.