Why is it the square?

Paper 19A · One mixer suffices · Read on Zenodo

Quantum mechanics has a rule everybody uses and nobody quite explains. To get the chance of an outcome, you take the amplitude and you square it. Not cube it, not take it as it is. Square it. Generations of physicists have tried to derive that square from something more basic, and each derivation works by granting a fairly large package of assumptions first. This note asks the accountant's question instead: how small can the package get before the square stops being forced? The answer, for one sharply fenced family of rules, is very small indeed. One mixing operation is enough.

The rule, and the puzzle

Think of a quantum system as having several possible outcomes, and call them routes. Each route carries an amplitude, a number that can be positive, negative, or complex. The Born rule says: to turn amplitudes into probabilities, take the size of each amplitude, square it, and normalise so the probabilities add to one. It works perfectly. It has never failed a measurement. And it looks arbitrary. Why the second power?

The existing answers all buy the square with something. Gleason bought it with the full geometry of the space. Others bought it with decision-theoretic axioms, or with symmetry under every possible quantum operation at once, or with operational assumptions about how measurements compose. Each is a real result. But in every case a substantial amount of structure goes in before the square comes out, which makes it hard to say what the square actually costs.

Turning the question around

So instead of asking how fast the square follows from the whole symmetry group, this note asks the reverse: what is the smallest mixing assumption that still forces it?

To make that question answerable you have to fence off a family of candidate rules. The family here is the plainest one imaginable: each route gets a weight computed from its own amplitude alone, by one and the same function, and the weights are added up. Nothing crosses between routes. Call these coordinate-additive rules. Inside that family, the ordinary square is one candidate among infinitely many; you could just as well try the cube, or the fourth power, or something with no formula at all.

Then you ask which of those candidates can survive a symmetry. A symmetry here means an operation that genuinely mixes two routes into each other, the way a beamsplitter mixes two beams of light, and leaves the total weight unchanged.

The result: one beamsplitter is enough

The answer is sharp, and it comes in three parts.

Ask for one genuine mixer, and you get the square. If there are at least three routes, and the rule is unchanged by a single operation that really blends two of them, then the function has to be the square. Not approximately, not up to some family of exceptions. Exactly the square, on the whole range. One beamsplitter's worth of symmetry does the entire job.

Ask for shuffling only, and you get nothing. If the operations you allow merely relabel the routes and twiddle their phases, without blending anything, then every candidate function survives. The requirement is empty. It fixes nothing at all.

There is no middle ground. Between "shuffling, which constrains nothing" and "genuine mixing, which forces the square" there is no intermediate symmetry that would pin the function down halfway, at least in the regimes the theorems cover. The symmetry landscape has no interior: the family of allowed operations jumps straight from the small one to the largest one, with nothing in between.

And the geometry comes free

Here is the part that makes it more than a curiosity about exponents. Once the function is forced to be the square, the quantity the symmetry preserves is just the ordinary squared length of the quantum state. And there is a classical piece of mathematics (polarisation, from Jordan and von Neumann in 1935) which says that if you know squared lengths, you can recover angles: the full inner product can be reassembled from four length measurements.

So asking for one mixer does not only give you the square. It gives you the entire geometric structure of quantum states, the inner product and all. Inside this family, "demand one beamsplitter" and "demand the whole quadratic structure of quantum mechanics" turn out to be the same demand.

How the proof works, in words

The argument is elementary, and it has three moves worth seeing.

The first is a phase sweep. Put two routes side by side with a third route standing by as a spectator (this is why you need at least three). When the mixer acts on the pair, how the weight redistributes depends on the relative phase between the two amplitudes. Because the rule has to hold for every state, including every phase, that one operation quietly hands you an entire continuum of equations rather than a single one. No separate phase symmetry has to be assumed; the "for every state" does that work by itself.

The second is a centring move. Choose the phase so the mixer takes any lopsided pair and replaces it with a less lopsided one, with the same total. Repeat, and the imbalance shrinks geometrically. Any pair can be walked in toward the balanced middle.

The third is the midpoint equation. What survives is the statement that the function's value at the midpoint of any two points equals the average of its values at those points. A function with that property has to be a straight line, and a classical theorem says mere measurability is enough to conclude it, with no smoothness needed. Translate back out of squared variables and the straight line is precisely the square.

Where it stops

The bounds matter as much as the result, and the paper states them flatly.

It is a characterisation, not a derivation. The mixer premise is the assumption that quantum evolution is unitary, and it gets no independent justification here. Inside this family, the question "why is the exponent two?" becomes exactly the question "why does any genuine mixing symmetry hold at all?" That is the unitarity question, which the reconstruction programmes also leave open. The puzzle is relocated, not reduced, and the paper says so in those terms.

It fails for two routes. With only two routes there is no spectator, the sweep never gets going, and an infinite family of non-square candidates slips through. This is structural rather than a gap in the argument, and it mirrors the same dimension threshold that appears in Gleason's theorem.

It is proven for two-route mixers only. Operations that blend three or more routes at once are checked numerically and left open. And the family restriction is a strong assumption in its own right: the paper notes that the coordinate-additive family may contain no physical theory besides quantum mechanics itself.

Read the paper

The full note is freely available on Zenodo (concept DOI 10.5281/zenodo.21471576):

Read on Zenodo → · Technical version · Dansk version

Related on this site:

• Paper 19B — How long does the losing answer stay ahead? — the sibling note: the same Born probabilities, seen from the dynamics of a measurement rather than the algebra of the rule.
• What is a race? — the picture of competing routes this paper's vocabulary borrows, though none of the mathematics depends on it.