advanced32 min studyLesson 336 in Physics

Real-space renormalization examples

Perform an exact zero-field Ising-chain decimation and distinguish preserved long-distance physics from lost microscopic information.

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What you’ll understand

  • Sum over alternating Ising spins to derive a renormalized coupling.
  • Track the change of lattice spacing and correlation length together.
  • Distinguish exact closure in one dimension from truncation in more general blocking schemes.

Renormalization in real space asks what statistical model remains after some microscopic variables are summed out. The goal is to preserve predictions for retained variables at larger physical separations while changing the resolution of the description. The zero-field nearest-neighbour Ising chain offers an unusually clean example: eliminating every other spin leaves another nearest-neighbour chain. That exact closure makes the transformation transparent, but should not be mistaken for a property of arbitrary blocking prescriptions.

Decimate one middle spin exactly

Consider retained spins s and s′ separated by one spin sigma. The two adjacent bond weights multiply to eKσ(s+s′)e^{K\sigma(s+s')}. Summing sigma over its two signs gives 2cosh⁡[K(s+s′)]2\cosh[K(s+s')]. This takes one value when the retained spins agree and another when they disagree. Represent those two weights as a common factor A times a new bond weight eK′ss′e^{K'ss'}, then solve their ratio to determine the effective coupling.

The common factor does not affect normalized probabilities of retained configurations, but it contributes to the partition function and free energy. Dropping it is acceptable only if the desired observable is insensitive to that normalization. For a periodic chain with even length, applying this local sum everywhere leaves half as many spins and twice the lattice spacing. A decimation step therefore changes both the coupling parameter and the unit used to measure separation.

∑σ=±1eKσ(s+s′)=AeK′ss′,K′=12ln⁡cosh⁡(2K),tanh⁡K′=(tanh⁡K)2,a′=2a\sum_{\sigma=\pm1}e^{K\sigma(s+s')}=A e^{K'ss'},\qquad K'=\frac12\ln\cosh(2K),\qquad \tanh K'=(\tanh K)^2,\qquad a'=2a
K=J/(kBT) and K′ are dimensionless ferromagnetic couplings; A is a positive normalization factor. a and a′ are lattice spacings in metres. The transformation is exact for zero field, nearest-neighbour interactions, and decimation of alternating sites in an even periodic chain.↔ Scroll wide equations horizontally.

The physical correlation length survives the change of units

Write q=tanh⁡Kq=\tanh K. The original infinite-chain correlation across j bonds is qjq^j. Retained nearest neighbours were separated by two original bonds, so their correlation is q2q^2. This directly explains the recursion q′=q2q'=q^2 without relying on the algebraic identity for hyperbolic functions. The renormalized chain reproduces all retained-spin correlations at the corresponding physical separations.

The correlation length expressed in new lattice spacings is half the old numerical value. Multiplying by the new spacing restores the same physical length, so the transformation does not make the material lose correlation in metres. It changes the resolution at which that correlation is described. Once the coarse spacing exceeds the physical correlation length, the effective coupling looks weak and the long-distance system appears almost uncorrelated.

ξ′a′=−1ln⁡q′=−12ln⁡q=12ξa,ξ′=ξas physical lengths\frac{\xi'}{a'}=-\frac1{\ln q'}=-\frac1{2\ln q}=\frac12\frac{\xi}{a},\qquad \xi'=\xi\quad\text{as physical lengths}
Here xi′/a′ is the correlation length in coarse lattice units, while xi′ and xi denote the same retained-correlation decay length in metres. The identity uses 0<q<1, the infinite-chain limit, and the exact doubling of physical spacing.↔ Scroll wide equations horizontally.

Fixed points classify long-distance outcomes

Repeated squaring drives every finite-temperature ferromagnetic value 0<q<10<q<1 toward zero. The only endpoint with perfect correlation is q=1, corresponding to zero temperature, and small deviations away from it grow under blocking. There is no intermediate finite-temperature critical fixed point in this exact one-dimensional model. This agrees with the transfer-matrix result, while the normalization factor accumulates the free energy discarded by looking at coupling flow alone.

In higher dimensions, summing or grouping spins typically generates longer-range and multispin interactions. Retaining only the original nearest-neighbour coupling then becomes an approximation in a larger space of Hamiltonians. A majority rule is not automatically an exact partition sum, and an attractive fixed point of a crude recursion is not sufficient evidence for quantitatively correct exponents. Universality explains why some discarded directions are harmless asymptotically, but their irrelevance must be established rather than assumed from convenience.

Decimate a correlation ratio of one half

An infinite zero-field Ising chain has q=tanh⁡K=1/2q=\tanh K=1/2 and spacing a. Decimate alternate sites. Find the new q, the new coupling, and the correlation length in both coarse lattice units and physical length.

  1. The retained nearest-neighbour correlation is the original two-bond correlation, so q′=q2=1/4q'=q^2=1/4. This corresponds to a weaker effective dimensionless coupling at the coarser resolution.

  2. Invert the hyperbolic tangent: K′=12ln⁡[(1+q′)/(1−q′)]=12ln⁡(5/3)≃0.2554K'=\tfrac12\ln[(1+q')/(1-q')]=\tfrac12\ln(5/3)\simeq0.2554. The original coupling was 12ln⁡3≃0.5493\tfrac12\ln3\simeq0.5493, but these numbers describe different lattice spacings.

  3. The coarse numerical length is ξ′/a′=1/ln⁡4≃0.7213\xi'/a'=1/\ln4\simeq0.7213. Since a′=2aa'=2a, the physical length is 1.4427a1.4427a, equal to the original a/ln⁡2a/\ln2. No physical shortening has occurred.

The result

Decimation gives q′=1/4 and K′≈0.2554; the correlation length is 0.7213 coarse spacings or 1.4427 original spacings.

Explore the model

Interactive lab

Interactive lab · markov

Squaring a subleading eigenvalue

For the first run, Reset and set BOTH Transition p=0.20 and Transition q=0.20, then take two Steps; predict P(state 1)=0.32. For the second run, Reset and set BOTH controls to 0.32, then take one Step; predict the same probability. These exact temporal Markov chains have subleading factors 0.60 and 0.36, respectively.

Squaring a subleading eigenvalueExact 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 forTwo applications square a matrix eigenvalue, paralleling the decimation algebra. The instrument does not sum Ising bonds, perform spatial blocking, or report a free energy.

Check your understanding

After exact alternating-site decimation, the correlation length measured in coarse lattice spacings halves. What happens to its physical length?

Try it yourself

Start with q=0.80 in the exact zero-field chain recursion. Apply three decimations. Find the final q, the spacing factor, and the ratio of final to initial correlation length measured in lattice units. Does this produce a finite-temperature fixed point?

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

Give me a hint

Every step squares q and doubles spacing. Three squarings raise the original q to power eight.

Compare with the solution

After three steps q3=q23=0.808=0.16777216q_3=q^{2^3}=0.80^8=0.16777216. The spacing is a3=8aa_3=8a. Since ln⁡q3=8ln⁡q\ln q_3=8\ln q, the numerical correlation length is one eighth its initial value, while the physical product with spacing remains unchanged. Repeating the map drives q toward zero for every initial value strictly between zero and one. The starting value is therefore in the disordered finite-temperature basin, not at an intermediate critical fixed point. Exact decimation preserves retained-spin predictions but does not reconstruct the removed microscopic configurations.

Take it with you

  • Decimation sums statistical weights and changes the resolution of the model.
  • A normalization factor matters for free energy even when it cancels from probabilities.
  • Exact one-dimensional closure does not justify arbitrary higher-dimensional truncations.

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