advanced32 min studyLesson 334 in Physics

Ising transfer matrix in one dimension

Solve the finite periodic nearest-neighbour Ising chain and extract its analytic free energy, correlation length, and zero-field susceptibility.

On this page

What you’ll understand

  • Construct a transfer matrix without double-counting the magnetic field.
  • Distinguish its finite trace from the dominant-eigenvalue thermodynamic limit.
  • Explain why a growing low-temperature correlation length does not imply a positive-temperature transition.

An exact solution is useful precisely where a plausible approximation can fail. The one-dimensional nearest-neighbour Ising ferromagnet is such an example: uniform mean-field reasoning suggests ordering, but the exact chain has no spontaneous magnetization at positive temperature and zero field. A two-by-two transfer matrix performs the entire configuration sum while preserving local correlations. Its two eigenvalues reveal both the thermodynamic free energy and the length over which a spin influences its neighbours.

Put each local weight on one bond

Take a ring with N spins, ferromagnetic coupling J, and field energy h per spin. Introduce dimensionless couplings K=J/(kBT)K=J/(k_BT) and H=h/(kBT)H=h/(k_BT). Assign half of each endpoint's field energy to each bond so every site field is counted exactly once around the ring. Matrix indices are classical spin values, not quantum states whose amplitudes interfere. Multiplication sums positive statistical weights over intermediate spin signs.

Periodic closure makes the partition sum the trace of the Nth matrix power. An open chain instead needs boundary vectors, which changes finite-size factors even though the bulk limit is the same. Diagonalizing the matrix gives two eigenvalues. At any finite positive temperature its entries are positive, and the largest eigenvalue is unique and smooth for finite real field. An eigenvalue crossing that could create a finite-temperature bulk singularity is absent in this nearest-neighbour chain.

Tss′=eKss′+H(s+s′)/2,T=(eK+He−Ke−KeK−H),ZN=Tr⁡TN=λ+N+λ−NT_{ss'}=e^{Kss'+H(s+s')/2},\qquad T=\begin{pmatrix}e^{K+H}&e^{-K}\\e^{-K}&e^{K-H}\end{pmatrix},\qquad Z_N=\operatorname{Tr}T^N=\lambda_+^N+\lambda_-^N
s,s′=±1, K and H are dimensionless, and ZN is dimensionless. J and h are energies. The formula assumes a classical nearest-neighbour ring with N bonds and periodic boundary conditions; finite-chain boundary conditions must not be changed silently.↔ Scroll wide equations horizontally.

A subleading eigenvalue carries the correlation length

At zero field the eigenvectors are symmetric and antisymmetric under reversal, with eigenvalues 2cosh⁡K2\cosh K and 2sinh⁡K2\sinh K. In the infinite ferromagnetic chain, the spin correlation at a separation of j bonds is (tanh⁡K)j(\tanh K)^j. Writing that decay as an exponential defines a correlation length measured in lattice spacings. At finite K, the ratio lies strictly between zero and one, so the correlation length is finite even when it is large.

The dominant eigenvalue alone determines the thermodynamic free energy per site, but it does not contain all finite-size information. The omitted contribution relative to the leading partition term is (λ−/λ+)N(\lambda_-/\lambda_+)^N. That correction can matter in a cold chain shorter than its correlation length. At infinite temperature the zero-field ratio vanishes and distinct spins are independent; as temperature tends to zero the ratio approaches one and the correlation length diverges.

λ+=2cosh⁡K,λ−=2sinh⁡K,C(j)=⟨s0sj⟩=(tanh⁡K)j,ξa=−1ln⁡(tanh⁡K),f=−kBTln⁡λ+\lambda_+=2\cosh K,\quad\lambda_-=2\sinh K,\quad C(j)=\langle s_0s_j\rangle=(\tanh K)^j,\quad \frac{\xi}{a}=-\frac1{\ln(\tanh K)},\quad f=-k_BT\ln\lambda_+
These zero-field formulas apply to the infinite ferromagnetic chain at T>0. a is lattice spacing in metres, xi is a length, C is dimensionless, and f is free energy per site in joules. At K=0, xi has zero limiting value and correlations at nonzero separation vanish.↔ Scroll wide equations horizontally.

Domain walls explain the missing transition

An isolated reversal boundary costs a finite bond energy rather than an energy proportional to the length of the reversed region. At every positive temperature there is a nonzero density of such domain walls in an infinite chain, breaking it into finite correlated domains. Their density becomes exponentially small at low temperature, so a short experimental chain can look ordered over its whole length without representing an infinite-system ordered phase.

The connected-correlation sum gives the zero-field susceptibility per site to h as χ=(kBT)−1(1+q)/(1−q)\chi=(k_BT)^{-1}(1+q)/(1-q) with q=tanh⁡Kq=\tanh K, hence χ=e2K/(kBT)\chi=e^{2K}/(k_BT). It grows sharply as temperature falls but remains finite at each positive temperature. This exact result exposes the failure of mean-field independence assumptions in one dimension. Long-range couplings, infinitely many local states, or a different model can evade the simple nearest-neighbour conclusion, so its assumptions belong to the result.

A chain with correlation ratio one half

Set zero field and K=12ln⁡3K=\tfrac12\ln3 for an infinite ferromagnetic chain of lattice spacing a. Find the correlation across three bonds, the correlation length, and the susceptibility per site to the energy field h.

  1. Since e2K=3e^{2K}=3, the identity tanh⁡K=(e2K−1)/(e2K+1)\tanh K=(e^{2K}-1)/(e^{2K}+1) gives q=(3−1)/(3+1)=1/2q=(3-1)/(3+1)=1/2. This is the subleading-to-leading transfer-eigenvalue ratio.

  2. The three-bond correlation is C(3)=q3=1/8C(3)=q^3=1/8. The length is ξ/a=−1/ln⁡(1/2)=1/ln⁡2≃1.443\xi/a=-1/\ln(1/2)=1/\ln2\simeq1.443, so a long chain loses correlation over only a few sites.

  3. The per-site susceptibility is (kBT)−1(1+1/2)/(1−1/2)=3/(kBT)(k_BT)^{-1}(1+1/2)/(1-1/2)=3/(k_BT). Its thermal-energy factor is required because h, rather than H, is the differentiation variable.

The result

The three-bond correlation is 1/8, the correlation length is 1.443a, and the per-site susceptibility is 3/(kBT), finite at this positive temperature.

Explore the model

Interactive lab

Interactive lab · markov

A positive two-state matrix

For each run, Reset and set BOTH Transition p and Transition q. Compare p=q=0.25 with p=q=0.50. Predict that the first chain approaches equal stationary probabilities gradually, while the second reaches them after one Step. The model starts in state 0 and shows exact Markov probabilities, not sampled spins or Ising Boltzmann weights.

A positive two-state matrixExact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA two-state matrix can have a dominant stationary eigenvector and a decaying subleading component. This Markov instrument is an analogy, not a transfer-matrix solver or spatial spin chain.

Check your understanding

For the finite periodic nearest-neighbour Ising chain, which expression is the exact partition function?

Try it yourself

A four-site periodic chain has zero field and K=12ln⁡3K=\tfrac12\ln3. What fraction of the leading partition term does the subleading term contribute? Find the exact factor relating Z4 to the dominant-eigenvalue estimate, and explain why a finite correction does not imply a phase transition.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the eigenvalue ratio q=1/2, then raise it to the number of bonds. Compare the correction with the leading term before normalizing by the full sum.

Compare with the solution

The correction relative to the leading term is q4=(1/2)4=1/16=0.0625q^4=(1/2)^4=1/16=0.0625. Hence Z4=(17/16)λ+4Z_4=(17/16)\lambda_+^4, whereas the leading-only estimate is λ+4\lambda_+^4. The omitted term is 1/17 of the exact partition function, a different fraction from 1/16. If desired, λ+=4/3\lambda_+=4/\sqrt3 gives Z4=272/9Z_4=272/9. These are finite positive sums with analytic temperature dependence at positive temperature. A measurable finite-size correction therefore does not identify a singularity or a positive-temperature transition.

Take it with you

  • The transfer matrix sums classical weights with explicit boundary conditions.
  • Its subleading eigenvalue controls correlations and finite-size corrections.
  • The short-range one-dimensional Ising chain has no positive-temperature ordering transition.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.