Semicircle law and Stieltjes transforms
Choose a consistent resolvent sign and recover a global spectral density from the correct analytic branch.
On this page
The question for today
What you’ll understand
- State a global semicircle law under explicit sufficient entry assumptions.
- Fix the sign, half-plane, and infinity normalization of a Stieltjes transform.
- Use inversion without treating finite atomic spectra as continuous densities.
The semicircle law identifies the limiting empirical distribution of a normalized Wigner spectrum. A Stieltjes transform provides an analytic encoding of that measure and often makes the limit easier to derive than a long list of moments. Its sign convention matters: two common definitions differ by a minus sign, so the imaginary half-plane and the selected square-root branch must be checked together.
A global law concerns normalized mass
For the real Wigner ensemble specified in the preceding lesson, with a fixed centered unit-variance off-diagonal law having all moments finite and an independent centered diagonal law with finite moments, the empirical spectral measures converge weakly in probability to the displayed semicircle measure. These are sufficient assumptions rather than a claim of minimal moment hypotheses. Weak convergence tests bounded continuous functions and describes a fraction of the eigenvalues. It does not alone control the largest eigenvalue or the spacing between neighbors.
The limiting density integrates to one, has support from minus two to two, and has the Catalan even moments found by the walk expansion. A finite number of outliers can lie far away while changing the empirical measure by only a vanishing amount. Likewise, a narrow interval whose width shrinks with dimension is outside the direct scope of a fixed-test-function weak limit. Global density, extremal eigenvalues, and local spacing laws are different statements even when they arise from the same ensemble.
The physical branch is analytically constrained
For a probability measure on the real line, the chosen transform has positive imaginary part when the argument does, and behaves as minus the reciprocal of the argument at infinity away from the real axis. For the semicircle law it satisfies a quadratic equation. The algebraic equation has two roots; the half-plane and infinity conditions select exactly one. A numerical quadratic solver returning the other root has not found another possible spectral law for the same normalized matrix model.
Write the square root of as the analytic branch on the complement of the interval from minus two to two that behaves like at infinity. In particular it is positive for real arguments greater than two and negative for real arguments less than minus two. This is a statement about the composite analytic square root, not a prescription to take the principal scalar square root separately at every point. That distinction prevents a sign discontinuity when moving across the upper half-plane.
Inversion is a limit of a smoothed measure
The imaginary part of the transform is an integral of when . Dividing by pi gives convolution with a Cauchy probability density. At continuity points of a limiting density, sending the positive smoothing parameter to zero recovers that density. Integrating over an interval recovers its mass, with half weights at boundary atoms in the symmetric inversion convention. Atoms themselves are recovered by .
At finite dimension the empirical measure is atomic, so a plotted smoothed density depends on the chosen positive smoothing width. Taking that width to zero before a matrix-size limit is different from first obtaining a continuous limiting law. The transform bound also deteriorates as smoothing vanishes. Local laws control this shrinking-scale regime with extra estimates; the global transform convergence theorem does not automatically allow arbitrarily small widths or certify every numerical histogram.
Work it through
Select a root at a purely imaginary argument
Evaluate the semicircle Stieltjes transform at twice i and verify both the branch sign and a basic magnitude bound.
Substitute into the quadratic equation. Its two roots are and . Only the first has positive imaginary part.
Thus , approximately . Equivalently, the selected square root of is , consistent with analytic continuation from the upper half-plane.
Its magnitude is less than , matching the general bound . At large imaginary arguments the same branch approaches minus the reciprocal of the argument, providing a second independent branch check.
The correct transform is positive imaginary; solving the quadratic without its analytic conditions would leave an invalid second root.
The living notebook
Explore the model
Interactive lab · root-finding
A branch choice uses more than an equation
Pause/Reset, set target 2 and initial upper bound 2. After two Steps the positive-square-root bracket has midpoint 1.25 and half-width 0.25. For a fresh comparison Pause/Reset, set both target 2 and upper bound 4; two Steps give midpoint 1.5 and half-width 0.5. Both brackets contain sqrt(2).
- Midpoint after 0 refinements
- 1.5
- Bracket error bound
- 1.5
- Computed residual · x² − c
- 0.25
Bisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. 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 forAn equation plus a specified bracket selects a root. This real bisection instrument can approximate the square root in the worked example, but it does not evaluate complex Stieltjes branches or random spectra.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the finite matrix with eigenvalues minus one and one, compute its empirical Stieltjes transform at twice i. Compare it with the worked semicircle value and explain the difference.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Average the reciprocals of minus one minus twice i and one minus twice i, using complex conjugates to simplify each term.
Compare with the solution
The two reciprocals are and . Their average is , different from . Both have the required sign, but they encode different probability measures: a two-atom finite empirical measure and a continuous limiting semicircle measure. A finite spectrum need not equal its proposed large-dimension limit. At positive smoothing width its transform is perfectly well defined even though it has no ordinary continuous density to invert pointwise.
Keep the idea
Take it with you
- Global weak convergence controls normalized spectral mass, not every extremal or microscopic statistic.
- The Stieltjes denominator, imaginary sign, and infinity normalization must agree.
- Density inversion involves smoothing limits, which must not be interchanged without justification.
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