advanced32 min studyLesson 338 in Physics

Wilson–Fisher fixed-point calculation

Find the interacting scalar fixed point near four dimensions and extract exponents consistently to first order in epsilon.

Before you begin

On this page

The question for today

What you’ll understand

Balance engineering growth against the one-loop quartic correction.
Distinguish fixed-point coordinates from universal linearized exponents.
Re-expand derived exponents consistently to the perturbative order retained.

Below four dimensions, a positive quartic coupling initially grows away from the Gaussian fixed point. Its one-loop correction opposes that growth, creating an interacting fixed point when the dimension is only slightly below four. The small parameter is epsilon, the difference between four and the spatial dimension. This construction explains how non-Gaussian critical exponents can be computed systematically, while also showing why an unqualified numerical substitution at epsilon equal to one is not a controlled high-precision calculation.

Solve the weakly interacting fixed point

Keep the previous scalar action with quartic coefficient u divided by four factorial and dimensionless coupling $g=K_d u\Lambda^{d-4}$ . For small positive epsilon, assume the fixed-point coupling is of order epsilon. To determine it at leading order, set the mass-dependent denominators to one in the quartic equation. The tree term and quadratic loop term then balance at a nonzero positive coupling proportional to epsilon.

The mass equation must also vanish. Its leading solution is negative and of order epsilon because it compensates the positive shell correction. This bare mass coordinate depends on the cutoff convention; it is not an observable inverse susceptibility. Higher-order operators and higher loops shift the coordinates. Their influence must be organized by perturbative order rather than solved as if the truncated flow equations were exact for all coupling values.

0=\epsilon g_* -\frac32g_*^2+O(\epsilon^3),\qquad g_* =\frac{2\epsilon}{3}+O(\epsilon^2),\qquad \bar r_*=-\frac{g_*}{4}+O(\epsilon^2)=-\frac\epsilon6+O(\epsilon^2)

These dimensionless coordinates use the previous sharp-shell scalar convention g=Kd u Lambda^(d−4), with increasing ell toward the infrared. They describe the weak-coupling fixed point for small positive epsilon=4−d, not a controlled arbitrary-dimension exact solution.↔ Scroll wide equations horizontally.

Linearization identifies the thermal scaling field

Perturb both couplings around the fixed point and diagonalize the linearized flow. The thermal direction is generally a mixture of bare mass and quartic deviations, rather than the mass coordinate alone. Its eigenvalue is positive, so one must tune the corresponding temperature-like field to remain critical. The leading quartic correction direction is negative and describes attraction toward the fixed point within the symmetry-preserving critical surface.

A conjugate ordering field is another relevant perturbation, even though it has been set to zero in this even-field subspace. Saying the fixed point has only one relevant direction without that qualification would be misleading. The thermal eigenvalue determines nu by its reciprocal. Expanding that reciprocal to the same order as the calculated eigenvalue gives a consistent epsilon series; treating the reciprocal as an exact rational formula silently includes uncontrolled higher powers.

y_t=2-\frac\epsilon3+O(\epsilon^2),\qquad y_{\mathrm{irr}}=-\epsilon+O(\epsilon^2),\qquad \nu=\frac1{y_t}=\frac12+\frac\epsilon{12}+O(\epsilon^2),\qquad \eta=O(\epsilon^2)

yt and yirr are eigenvalues of the infrared flow generator in the reversal-even scalar subspace, not physical frequencies. nu is the thermal correlation-length exponent. The leading irrelevant exponent magnitude is epsilon at this order; the field anomalous dimension eta has no term linear in epsilon.↔ Scroll wide equations horizontally.

Derive other exponents without manufacturing accuracy

Using Fisher's relation with eta zero to first order gives gamma equal to one plus epsilon over six. Hyperscaling in dimension four minus epsilon gives alpha equal to epsilon over six after the product is expanded consistently. The magnetization exponent is one half minus epsilon over six. These relationships refer to the conventional short-range scalar continuous transition and include nonuniversal amplitudes that this calculation does not determine.

Universality here means that models flowing to the same stable critical theory share suitable asymptotic exponents and scaling functions after metric normalizations. Their microscopic couplings, transition temperatures, and crossover scales can differ. Extra symmetry-breaking interactions, long-range tails, disorder, or a first-order transition can change the outcome. Higher-order epsilon calculations and resummation improve quantitative predictions in three dimensions, but agreement of an arbitrary smooth fit with one leading-order exponent is not evidence of the Wilson–Fisher mechanism.

Work it through

A small-epsilon scalar fixed point

Use epsilon=0.30 as an illustrative expansion parameter. To first order, calculate the nonzero fixed-point coupling, bare mass coordinate, correlation-length exponent, and susceptibility exponent. Keep the normalization from the previous lesson.

The quartic fixed point is $g_*=2(0.30)/3=0.20$ to the retained order. The mass coordinate is $\bar r_*=-0.30/6=-0.050$ , a scheme-dependent coordinate rather than a negative physical response.
The thermal eigenvalue is $y_t=2-0.30/3=1.90$ plus omitted terms. The consistently expanded exponent is $\nu=1/2+0.30/12=0.525$ to first order.
Since eta has no linear term, $\gamma=(2-\eta)\nu=1+0.30/6=1.05$ at this order. Using $1/1.90=0.5263$ as though exact would add a different, uncontrolled collection of terms beyond first order.

The result

At first order, g*=0.20, r-bar*=−0.050, nu=0.525 and gamma=1.05, with order-epsilon² corrections not determined by these numbers.

The living notebook

Explore the model

Interactive lab

Interactive lab · optimization

Attraction in one linearized direction

For each run, Reset and set BOTH Step size η and Initial position x₀=1. Compare η=0.20 and η=0.50 after four Steps, predicting positions 0.4096 and 0.0625. The lab performs gradient descent on x²/2, with each Step one iteration; interpret contraction only as an analogy to an irrelevant RG perturbation.

ObjectiveDiscrete iterate

Position x0: 2
Objective x²/2: 2
Iteration behavior: Stable step

Step size η0.3Initial position x₀2

n = 0 · t = 0 s

Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. 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 stable direction loses memory of its initial deviation. This optimizer does not calculate a Wilson–Fisher fixed point, thermal relevant direction, or critical exponent.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

A reciprocal generates infinitely many powers of epsilon. Only its constant and linear terms are justified by a first-order eigenvalue; omitted corrections to the eigenvalue change the higher-order coefficients.

Your turn

Try it yourself

At first order for the same scalar theory, use $\alpha=\epsilon/6$ , $\beta=1/2-\epsilon/6$ , and $\gamma=1+\epsilon/6$ . With epsilon=0.60, compute the three exponents and check the Rushbrooke scaling combination. State the approximation limitation.

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

Give me a hint

Substitute into each truncated series separately, then form alpha+2beta+gamma. An exact cancellation in that combination does not remove uncertainty in each exponent.

Compare with the solution

The values are $\alpha=0.10$ , $\beta=0.40$ , and $\gamma=1.10$ . Their combination is $0.10+2(0.40)+1.10=2.00$ , as the scaling identity requires through the retained order. The consistency check tests the algebra and shared scaling assumptions, not the numerical accuracy of leading-order estimates at epsilon 0.60. Every individual exponent still has an uncomputed correction of order epsilon squared in this calculation. The beta here is a magnetization exponent, not inverse thermal energy, and the chosen fractional dimension is an expansion device rather than a specified real material.

Keep the idea

Take it with you

The interacting fixed point balances quartic relevance against loop corrections near four dimensions.
Universal exponents come from scaling directions, not the numerical bare coordinates.
Perturbative order must be respected when combining or evaluating exponent series.

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