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Fields, scales and critical phenomena

FR01 · L3 · Chapter 7

Testing a convincing data collapse

Derive a finite-size scaling curve and expose a false critical point with held-out sizes.

Before you start: Chapters 5–6, standard errors and dimensionless ratios.

You will be able to: Separate fitted and held-out sizes and explain why a free-field exponent is not an interacting exponent.

Calculated ideal referenceAnalytic reference curves, with axes and parameter definitions in the caption.Susceptibility divided by squared length0.00.00.0224.00.0458.00.06812.00.09016.0Dirichlet continuum referenceSquared length times mass parameter
Ideal analytic reference; dimensionless reference units. χ/L2=[1−2tanh⁡(x/2)/x]/x,x=rL2\chi/L^2=\left[1-2\tanh(\sqrt{x}/2)/\sqrt{x}\right]/x,\quad x=rL^2The value at the origin is one twelfth. The Lab also shows finite-lattice and sampled results.

A larger system can expose a critical-point error that several smaller systems hide. The useful test is a prediction: choose a scaling law using the small sizes, then ask where the larger sizes should land. A neat overlap of curves becomes more informative when a size you kept aside joins it.

Prerequisites: Euclidean lattices and errors establishes covariance and source response; Keeping low-energy physics separates exact elimination from a truncated model. Here you will derive a finite-interval scaling curve, estimate it from saved samples and diagnose a misleading collapse.

Why the length enters the answer

Consider a one-dimensional field on an interval. The field is zero at both ends. The interior has NN equally spaced sites, spacing aa, and physical length L=(N+1)aL=(N+1)a. The coordinate increases from the left endpoint to the right. The parameter rr has units of inverse length squared. We use a dimensionless Euclidean action and measure length in a fixed reference unit:

S=12∑j=0N(ϕj+1−ϕj)2a+ar2∑j=1Nϕj2,ϕ0=ϕN+1=0.S=\frac12\sum_{j=0}^{N}\frac{(\phi_{j+1}-\phi_j)^2}{a} +\frac{ar}{2}\sum_{j=1}^{N}\phi_j^2, \qquad \phi_0=\phi_{N+1}=0.

The difference term penalizes abrupt changes between neighbors. The mass term penalizes a large field even when it varies slowly. On an infinite line with positive rr, correlations decay over the length ξ=1/r\xi=1/\sqrt r. On a finite interval, the endpoints also matter. When ξ\xi becomes comparable with LL, increasing the correlation length cannot be understood without mentioning the interval.

There are two lengths and one independent ratio:

Lξ=Lr,x=rL2.\frac{L}{\xi}=L\sqrt r, \qquad x=rL^2.

This suggests comparing systems at the same xx. Doubling the interval then requires dividing rr by four. The fields have the same correlation range relative to their own length, although their unscaled responses can be very different.

A response that fluctuations can measure

Apply a weak uniform source hh. Its action term is −ha∑jϕj-ha\sum_j\phi_j, so positive hh favors positive field values. Differentiate the mean integrated field with respect to hh, then divide by LL. For a zero-mean Gaussian field the result is

χL=a2L∑i,j⟨ϕiϕj⟩=⟨q2⟩,q=aL∑jϕj.\chi_L=\frac{a^2}{L}\sum_{i,j}\langle\phi_i\phi_j\rangle =\langle q^2\rangle, \qquad q=\frac{a}{\sqrt L}\sum_j\phi_j.

The quantity χL\chi_L is the susceptibility: the response per interval length to a small uniform source. You can obtain it by applying a source, or by recording source-free fluctuations and averaging q2q^2. The second route supplies the saved samples used here. Each recorded projection comes from an entire field configuration.

The continuum response gives an independent reference. Write u(z)u(z) for the derivative of the mean field at position zz with respect to the source. The Gaussian field equation gives

−u′′(z)+ru(z)=1,u(0)=u(L)=0.-u''(z)+ru(z)=1,\qquad u(0)=u(L)=0.

The constant on the right is the uniform source. For r>0r>0, a constant particular solution is 1/r1/r. A symmetric combination of exponential solutions adjusts it to vanish at both endpoints:

u(z)=1r[1−cosh⁡ ⁣(r(z−L/2))cosh⁡(rL/2)].u(z)=\frac1r\left[ 1-\frac{\cosh\!\left(\sqrt r(z-L/2)\right)} {\cosh(\sqrt rL/2)} \right].

Average this response over the interval. Integrating the hyperbolic cosine gives

χL=1r[1−2tanh⁡(rL/2)rL].\chi_L=\frac1r\left[ 1-\frac{2\tanh(\sqrt rL/2)}{\sqrt rL} \right].

Dividing by L2L^2 leaves a function of xx alone:

χLL2=F(x),F(x)=1−2tanh⁡(x/2)/xx.\frac{\chi_L}{L^2}=F(x), \qquad F(x)=\frac{1-2\tanh(\sqrt x/2)/\sqrt x}{x}.

The plotted dashed curve is this continuum reference. The colored points come from saved lattice samples. Their small difference includes both sampling error and finite-spacing corrections.

At zero mass the expression looks singular, but the response remains finite on the interval. Solve the differential equation directly: u(z)=z(L−z)/2u(z)=z(L-z)/2. Its spatial average is L2/12L^2/12, giving F(0)=1/12F(0)=1/12. The endpoints have supplied the finite length scale.

Work through one saved record

Use N=31N=31, a=1a=1, L=32L=32 and r=0.00390625r=0.00390625. These values give x=4x=4 and ξ/L=1/2\xi/L=1/2. The exact continuum response is 61.031961.0319. The discrete lattice calculation gives 60.954260.9542.

The saved chain uses seed 353159 and contains 16,384 projections. It gives

χ^L=60.8456±1.5105,χ^L/L2=0.0594196±0.0014751.\widehat\chi_L=60.8456\pm1.5105, \qquad \widehat\chi_L/L^2=0.0594196\pm0.0014751.

The uncertainty is one estimated standard error, calculated from batches of successive samples. Both references lie comfortably within it. The gap between the two references is much smaller than this chain's sampling uncertainty.

Why use batches? Each new Gaussian mode retains 80 percent of its previous value and receives a fresh Gaussian contribution. The squared projection therefore remembers earlier draws. Treating all draws as independent gives a standard error of only 0.67730.6773. Splitting the trace into 128 batches of 128 draws captures that memory in the spread of batch averages. The estimated effective sample count is about 3,294.

For this sampling algorithm the squared projection's lag correlation is 0.82k0.8^{2k}. Its exact asymptotic variance inflation is (1+0.82)/(1−0.82)≈4.56(1+0.8^2)/(1-0.8^2)\approx4.56. The generated-data tests compare the batch estimate with longer batches. In an unfamiliar interacting system, you would also investigate equilibration and slow modes.

How the wrong critical point passes a small test

A proposed collapse uses

xtrial=(r−rc)L1/ν.x_{\mathrm{trial}}=(r-r_c)L^{1/\nu}.

The exact Gaussian benchmark has rc=0r_c=0 and ν=1/2\nu=1/2. Keep that exponent and shift the proposed critical point to rc=−0.0003r_c=-0.0003. The horizontal displacement becomes 0.0003L20.0003L^2. It is 0.07680.0768 for length 16, but 4.91524.9152 for length 128.

The small curves move little compared with the range displayed. Their standardized RMS difference rises only from about 1.01 to 1.10. The two held-out sizes tell a clearer story: their RMS rises from about 0.94 to 7.14. These numbers compare the recorded response with interpolated small-size curves and include both uncertainties. The RMS diagnoses disagreement. Shared interpolation errors prevent treating it as a significance probability.

Finite spacing provides another reason to compare sizes carefully. At zero mass the discrete response can be summed exactly:

χL=a2N(N+2)12,χLL2=112(1−a2L2).\chi_L=\frac{a^2N(N+2)}{12}, \qquad \frac{\chi_L}{L^2}=\frac1{12}\left(1-\frac{a^2}{L^2}\right).

A systematic drift proportional to a2/L2a^2/L^2 is therefore expected even with the right critical point. Refining the lattice at fixed physical length separates that cutoff correction from a change in volume.

This benchmark approaches the massless Gaussian limit from nonnegative rr. A negative quadratic coefficient would make its infinite-volume action unstable without an additional stabilizing interaction. Interacting critical points require their own scaling functions and exponents.

Try the prediction before uncovering the data

Open the finite-size scaling Lab. Choose “One curve, five sizes” and explain why the unscaled responses separate. Then choose “A tempting wrong fit”. The two larger sizes are hidden. Sketch where you expect them to appear before choosing “Ask the larger lattices”.

Now change the proposed exponent under “Adjust the proposed scaling law”. Predict whether this stretches every size equally. Record the small-size and held-out residuals, together with the number of compared points. Moving data outside the comparison range can reduce that count; it should prompt a closer look at the remaining test.

Exercises

  • At fixed x=4x=4, what mass parameter should accompany a length change from 32 to 64?
Answer and reasoning

Since r=x/L2r=x/L^2, the initial value 0.003906250.00390625 becomes 0.00097656250.0009765625. The correlation length doubles with the interval, so their ratio stays unchanged.

  • Use the zero-mass lattice result to predict how the scaled-response error changes when aa is halved at fixed LL.
Answer and reasoning

The difference from 1/121/12 is −a2/(12L2)-a^2/(12L^2). Halving aa reduces its magnitude by four. This refinement changes the cutoff while preserving the physical volume.

  • A colleague obtains an excellent overlap after choosing all fit settings by looking at all five sizes. Propose a new test.
Answer and reasoning

Freeze those settings and generate a previously unused larger size with independent seeds. Choose the comparison range before uncovering it. An existing size can no longer serve as a fresh held-out test after it helped select the settings.

  • Transfer the method to periodic boundaries. Would the same dashed curve apply at zero mass?
Answer and reasoning

Periodic boundaries admit a uniform zero mode. Its quadratic stiffness vanishes at zero mass, so the unconstrained Gaussian integral diverges. The finite Dirichlet result relies on fixed endpoints. A periodic benchmark must keep positive mass or explicitly constrain the zero mode, then derive the response for that choice.

The Gaussian action, source response and correlation-length argument follow David Tong's Statistical Field Theory, sections 2.2–2.3. The finite-interval reference above is derived for this boundary condition. The Lab's raw downloads preserve the projections, matching field snapshots, seeds and model version needed to recompute each estimate.

Try the physics

Finite-size Scaling →Compare Gaussian fields on five lattice sizes. Try a convincing fit on the small lattices, then uncover the larger samples and see whether the prediction survives.

Sources and reading scope

  1. David Tong, Statistical Field Theory, Chapter 2

    Gaussian integrals, inverse-kernel covariance, correlation length and the Euclidean quantum-mechanical interpretation. Finite-chain normalizations are derived for this course.

  2. David Tong, Statistical Field Theory, Chapter 3, §3.5, Figure 30

    One-loop Wilson–Fisher flow near four dimensions, pp. 90–91, Figure 30. The Lab reconstructs an equation-derived phase portrait, with its parameters and convergence checks saved separately.

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