Exact solution via transfer matrix: T = [[e^(J/kT), e^(-J/kT)], [e^(-J/kT), e^(J/kT)]]
The 1D Ising model is exactly solvable. The transfer matrix T connects adjacent spins: Z = Tr(T^N).
Eigenvalues λ± = e^(J/kT) ± e^(-J/kT) give correlation length ξ = −1/ln(tanh J/kT).
No phase transition at finite T in 1D — thermal fluctuations always disorder the chain.
Top: current spin configuration (blue=up, red=down). Bottom: exact thermodynamic quantities vs temperature.