Too complicated to solve, so he replaced the physics with dice — and it worked
In the 1950s, physicists could fire slow neutrons at a heavy nucleus and measure, one by one, the energies at which it absorbed them. Each of those resonances is an excited state of a nucleus with a couple of hundred nucleons in it, and the experiments produced them by the dozen — long, precise lists of numbers that theory could not touch.
The reason theory could not touch them is worth being blunt about. Predicting a resonance energy means finding an eigenvalue of the nucleus’s Hamiltonian, and that Hamiltonian is a matrix on a space whose dimension counts the ways two hundred fermions can be arranged among their available states. That number is astronomical. Worse, even with unlimited computing you would still need the matrix entries, and the force between nucleons was — and to a useful precision still is — not known well enough to write them down.
Wigner’s response was to stop. Not to approximate the Hamiltonian better, but to give up on it entirely and ask a different question: if I knew nothing about this matrix except its symmetry, what could I still predict?
It is worth getting a feel for the size of the thing, because the usual word — “intractable” — undersells it. Put A nucleons into N available single-particle states. They are fermions, so no two share a state, and the number of arrangements is the binomial coefficient C(N, A). Sixteen particles among forty-four states already gives about 4×1011 states; twenty-four among sixty gives 3.6×1016.
Diagonalising this matrix would take about 1.08 × 1014 times the present age of the universe.
Every number here is computed from C(N, A), the count of ways to place A indistinguishable fermions in N single-particle states. That binomial coefficient, rather than NA, is Pauli exclusion doing its work. This is a counting model of the size of the problem, not a reproduction of any particular published nuclear structure calculation — the real ones impose angular momentum and separate protons from neutrons, which changes the constants and not the shape of the curve. Note too that the bar chart peaks in the middle: a half-filled shell is the worst case, which is exactly where the heavy nuclei Wigner cared about tend to sit.
Try it: Start with a small shell — say N = 20 and A = 8, which is 125,970 states and diagonalises in about two milliseconds — and then walk the sliders up. By N = 60, A = 24 the dense diagonalisation needs about 4.7×1049 operations, which on a machine doing a billion billion of them a second is roughly 1014 times the present age of the universe. The count peaks at half filling, which is where heavy nuclei live.
Modern nuclear structure theory does far better than dense diagonalisation, and the numbers above are a counting model rather than a description of anyone’s actual calculation. But the shape of the curve is the point, and no amount of cleverness with sparse solvers repeals it: the state space of a heavy nucleus grows faster than any hardware ever will.
Here is the idea, and it is genuinely strange the first time you meet it. Replace the nuclear Hamiltonian — the real one, with all its unknown matrix elements — with a matrix whose entries are random numbers. Not as an approximation to the true Hamiltonian; nobody claims the nucleus has random matrix elements. The claim is statistical: a nucleus is one draw from an ensemble of plausible complicated systems, and if a property is shared by almost all members of that ensemble, the nucleus will have it too.
That reframing changes what counts as a prediction. You give up on where the levels are — no theory of this kind will ever tell you the energy of the fourth resonance in a particular isotope. What you get in exchange is a prediction about the pattern the levels make: how often two of them sit close together, how the gaps are distributed. And that prediction is sharp, because the ensemble is concentrated: almost every member looks the same in this respect.
The nuclear Hamiltonian is real symmetric, because the nuclear force respects time-reversal symmetry. That fixes the class: β = 1. Wigner wrote down the resulting spacing distribution and presented it at a conference on neutron physics held in Gatlinburg, Tennessee, in 1956. It fits on a line:
p(s) = (π/2) s e−πs²/4
One formula, no adjustable parameters, no nuclear physics in it anywhere. Before looking at whether it works, look at what it is predicting — and see whether you can tell the two hypotheses apart by eye.
Two ladders of energy levels below, both drawn at the same average density. In one, the levels repel; in the other they are dropped independently. Before Wigner, the reasonable default would have been the independent one — a complicated system, a great many unrelated states, so surely the levels fall where they fall.
One of these two ladders has levels that repel one another; the other has levels dropped independently at random. Both have exactly the same average density. Which one would you bet is the nucleus?
Both ladders are simulated. The repelling one is the spectrum of a β = 1 random matrix; the other is a sequence of independent exponential gaps. Neither contains measured resonance energies — this demo is about what the two hypotheses look like, not about any particular nucleus. Real resonance ladders are obtained by firing slow neutrons at a target and reading off the sharp peaks in the absorption cross-section, and they are compared with theory in exactly these units, rescaled so the mean spacing is 1.
Try it: Guess, reveal, then run a few more pairs. The statistics under each ladder make the difference concrete: in 44 gaps, the independent sequence typically has three or four times as many near-collisions and a conspicuously larger biggest gap. The repelling ladder is not regular — it is irregular in a very specific, measurable way.
Both sequences on that diagram are simulated, and deliberately so. The point being made is about what the two hypotheses look like, and a simulated ladder makes it without dressing invented numbers up as measurements.
Sample a large number of spacings from real symmetric random matrices and they fall onto Wigner’s curve — the L1 distance between the histogram and the surmise drops to about 0.025 with 34,000 spacings, while the distance to Poisson sits at 0.55 and stays there no matter how much data you collect.
Sampling… 0 matrices, 0 spacings.
What this does and does not show. Every spacing here came from a random matrix, so this establishes that the theory is internally consistent: real symmetric random matrices really do produce the β = 1 curve, and the agreement improves rather than plateaus. The claim that measured neutron resonances follow the same curve is a claim about experimental data, made in the lesson text and resting on published compilations — it is cited there, not reproduced here.
Try it: Watch the two numbers as the sample fills in. The distance to the β = 1 curve falls — roughly 0.11 at a thousand spacings, 0.038 at twelve thousand, 0.025 at thirty-four thousand — while the distance to Poisson holds at 0.55 throughout. A law that is merely undersampled improves with data; a law that is wrong does not.
Be careful about what that establishes. Everything in the demo above came out of a random matrix, so what it demonstrates is self-consistency: the theory does what the theory says it does. The empirical claim is separate, and it is this. Measured neutron resonance spacings in heavy nuclei follow the same curve. Individual isotopes give sequences too short to be decisive on their own, so the compelling test came from pooling resonances from many different nuclei into a single body of data — the so-called nuclear data ensemble, assembled in the early 1980s from over a thousand measured levels. Analysed together, they follow the β = 1 prediction and are decisively inconsistent with independent levels. That is an experimental result, cited here rather than reproduced.
The obvious objection is that this should not work. A nucleus is not random. It has a definite Hamiltonian, definite forces, definite structure — and a random matrix knows none of it. So why does the answer come out right?
Because complication itself is a kind of averaging. When a system is intricate enough that its eigenstates involve every degree of freedom at once, the specific values of the matrix elements stop leaving a fingerprint on the local statistics. What survives is only the constraints that no amount of complication can wash out — and the constraints are the symmetries. Below, three knobs change details and one knob changes symmetry. Only one of them moves the answer.
Change any of these and watch the number at the bottom refuse to move.
The only knob on this panel that changes the answer.
Sampling… 0 spacings.
The sample budget is fixed at 8,000 spacings however large the matrices are, so the comparison is not smuggling in a difference in statistics. With β held at 1, the distance wanders between about 0.04 and 0.06 across every setting of the left-hand panel — that is noise. Switch to β = 2 and it jumps to about 0.21; to β = 4 and about 0.50. The system’s size, its particular matrix elements, and which levels you happen to have measured are all invisible. Its symmetry is not.
Try it: Change the matrix size by a factor of eight, redraw the matrix, look at a narrow slice of the spectrum or almost all of it. The distance to the β = 1 curve stays between about 0.04 and 0.06 throughout — noise. Then click β = 2 and watch it leap to about 0.21, and β = 4 to about 0.50. That is the whole thesis in one number.
Once stated that way the idea stops being about nuclei at all. Nothing in it mentions the nuclear force, or fermions, or energy. It says: for a sufficiently complicated self-adjoint operator, the local spacing statistics depend on the symmetry class and on nothing else. If that is true, the same curve should turn up in every complicated system with the same symmetry — in a chaotic billiard, in a disordered conductor, and, as it eventually turned out, in the zeros of the Riemann zeta function, which have nothing whatsoever to do with physics.