ANOMALIES EXPLAINED · 8 OF 12

Tsirelson’s bound

In the CHSH form of a Bell test, the quantum score stops at 2√2, though a ban on faster-than-light signals alone would allow 4. One shared state, measurements whose order across the stations does not matter, and the Born rule’s squares set that ceiling.

This answer is proved in the paper (a preprint, posted 14 November 2026, not peer reviewed) for one system split into two blocks, taking as a premise that each probability for these measurements is the squared length of a part of the state. Owed, not yet proved: that two separate laboratories are two such blocks, and two statements, one about what a probability may depend on and one about what a click is and that one occurs (all three are named in the three lines below).

The ceiling, 2√2

2.828427…

Closest of the measurements the paper cites: 2.82759 ± 0.00051 (Poh et al. 2015). Ceiling for a table that only forbids signals: 4. The test on this page computes the sum live, from the matrices of the cube of on/off switches in section 02 and from the rounds it simulates.

Figure 1 simulates the test live, in your browser. The run on top draws every round from the cube of section 02; the arrows below are station B’s arrows b0 and b1 of section 02.

Why Not Four? · film, 3 min 59 s

The paradox in three lines

  1. The puzzle

    Two stations far apart, A and B, each choose one of two settings and get a result, +1 or −1. A correlation is how often the two results agree minus how often they differ: +1 if they always agree, −1 if they always differ. Add three of the four correlations and subtract the fourth. In quantum mechanics that score, S, stops at 2√2 ≈ 2.83. Yet the rule that seems to govern distant stations, that neither station’s choice can change the odds of the other’s results, allows S = 4. Why does it stop at 2√2?

  2. On the record

    The record is reality’s ledger of events: every event is a posting, an entry in it, and no posting is erased. On it, the ceiling comes from three conditions: both stations act on one shared state; each act has two outcomes, +1 and −1, and is its own undoing: done twice, it gives back the state it started from, so asked twice it agrees with itself (the textbook’s observable with values +1 and −1), and every act at one station commutes with every act at the other, which for acts means that on every state they give the same odds in either order; and each read (the textbook’s probability, once the total is set to one) is a square: the squared length of the part of the state that matches that outcome. With these, S is at most 2√2, the combined length of a unit square’s two diagonals, and suitable acts on one cube of on/off switches (section 02) reach it exactly; one table of probabilities that merely does not signal reaches 4. The ceiling, that one cube reaches it, and the table that reaches 4 are proved in the paper’s Section 7 (the theorems “Three ceilings” and “One cube reaches the middle ceiling”, Theorems 27 and 28; sections 02 and 03 below).

  3. What comes back, and what is owed

    Every number the record gives for this paradox, starting with the ceiling 2√2 = 2.828427…, is the one ordinary textbook quantum mechanics gives. The closest of the measurements the paper cites, 2.82759 ± 0.00051 (Poh et al. 2015), sits 0.00084 under the ceiling, 1.6 times its quoted uncertainty; no measurement has passed it.

    Owed, and not yet proved: that two separate laboratories are two blocks of one cube, sharing one complex unit i (composition). This paradox is the composition law itself, so it is not resolved until composition is proved. For the acts used here, squared weights are a premise; for reads of a register’s positions they are proved, given five premises named on the Born rule page, one of them another owed statement: that a read weighs each position by the value held there and by nothing else (the position read). Also owed: what an apparatus and its click are on the record, and that one outcome is posted (attachment).

01 · The history

From 2, to 2√2, to 4.

  1. 1964

    John Bell showed that if each particle carried its results in advance, the correlations between two distant measurements would obey a limit that quantum mechanics breaks.

  2. 1969

    John Clauser, Michael Horne, Abner Shimony and Richard Holt put the limit in the form most tests since have used (CHSH, from their initials): for results carried in advance, S is at most 2. Experiments pass 2; that story, with the loophole-free tests, is on the Bell’s inequality page.

  3. 1980

    Boris Tsirelson (published as B. S. Cirel’son) proved that quantum mechanics passes 2 only so far: for any state and any two-outcome measurements on each side, S is at most 2√2 ≈ 2.828, the value an entangled pair at the best settings was already known to give.

  4. 1994

    Sandu Popescu and Daniel Rohrlich asked whether 2√2 is simply the most that correlations can do without a signal. It is not. They wrote down a table of probabilities in which neither station’s choice changes the other’s odds, and its S is 4. If the ban on signals allows 4, something else holds quantum correlations at 2√2. (The same table had appeared in 1985, in work by Peter Rastall and by Leonid Khalfin and Tsirelson; principles proposed since to rule it out without assuming quantum mechanics include information causality, Pawłowski and colleagues, 2009.)

  5. 2015

    Hou Shun Poh, Siddarth Joshi, Alessandro Cerè, Adán Cabello and Christian Kurtsiefer aimed a test with pairs of photons at the bound itself (their paper is titled “Approaching Tsirelson’s bound in a photon pair experiment”) and made the closest approach among the measurements the paper cites: the diamond in Figure 1’s magnifier. It counted only pairs in which both photons arrived, so it was not loophole-free; the loophole-free tests the paper cites sit lower, at 2.42 ± 0.20 (Hensen et al. 2015) and 2.0747 ± 0.0033 (Storz et al. 2023).

02 · The ceiling

Why the sum stops at 2√2: a square’s two diagonals.

A state is a list of values, one for each configuration of the system, and it can be drawn as an arrow of length 1: the squares of the sizes of its values add up to 1. An act turns it into another arrow of length 1. Station A’s two acts give arrows a0 and a1; station B’s give b0 and b1.

Write E(x, y) for the correlation when station A uses setting x and station B setting y. It is the weight of agreeing minus the weight of differing, and when the acts commute across the stations and the weights are squares (for these acts, a premise), that difference comes out as the overlap of two arrows: E(x, y) = ⟨ax, by⟩. The overlap is 1 when the arrows point the same way, −1 when they point opposite ways, and the cosine of the angle between them in general.

Group the sum by station A’s arrows: S = ⟨a0, b0 + b1⟩ + ⟨a1, b0 − b1⟩. The overlap of an arrow of length 1 with another arrow is at most the other arrow’s length, so S is at most |b0 + b1| + |b0 − b1|, the lengths of the sum and of the difference of station B’s arrows (the Cauchy–Schwarz inequality).

Those two lengths are tied. The sum and the difference of two arrows are the two diagonals of the parallelogram the arrows span, and a parallelogram’s two diagonals, squared, add up to its four sides squared: here 1 + 1 + 1 + 1 = 4, whatever the angle between b0 and b1 (the parallelogram law). If two lengths have squares that add up to 4, together they come to at most 2√2, and they reach it only when they are equal, √2 each (the Cauchy–Schwarz inequality again). That happens when b0 and b1 are at right angles. Then they are two sides of a unit square, and their sum and their difference are its two diagonals. Figure 1, at the top of the page, draws this with arrows you can turn, and lays the two lengths at a right angle inside a square of side 2, whose diagonal is the same 2√2.

One cube reaches the ceiling exactly.

Take four on/off switches, two at each station, with sixteen combinations: what the paper calls a cube of four binary axes (the textbook’s four qubits). Station A’s two switches form one block of the cube, station B’s the other. The state holds the value ½ on each of the four combinations in which station B’s switches match station A’s, and 0 on the other twelve. The cube comes with its own flips: each flips one switch and carries the sign of every switch before it (the construction of Jordan and Wigner, 1928). Any two of these flips anticommute: applied one after the other to an arrow, the two orders give arrows pointing opposite ways. That each station’s own two flips anticommute is what lets the score reach the ceiling. Station A’s acts, g0 and g1, are the flips of its two switches. Station B uses h0 and h1, the flips of its own two switches with station A’s sign taken off, so that each touches only station B’s switches (that two laboratories’ acts are taken this way is part of the owed composition statement, section 04); its two acts are (h0 + h1)/√2 and (h0 − h1)/√2.

Then station B’s two arrows come out at right angles, and station A’s point along their sum and their difference: the best case above. Three correlations are 1/√2 and the fourth is −1/√2, so S = 4/√2 = 2√2 exactly. That is the paper’s theorem “One cube reaches the middle ceiling” (Theorem 28), and it holds on a cube of n + n switches for every n of at least 2. Reaching the ceiling with each station’s two acts anticommuting, on a state that pairs the two blocks, is Tsirelson’s own method of 1980.

Figure 2, in exact arithmetic: pick a pair of settings to see its correlation worked cell by cell, and open any of the four acts as its 16 × 16 matrix.

03 · Break it

Drop the state, keep only “no signal”, and the sum reaches 4.

Throw away the state and the acts, and keep one demand: neither station’s choice may show in the other’s results. Popescu and Rohrlich’s table meets it. For every pair of settings each result is +1 or −1 with even odds; the two results always agree in three of the four pairings and always differ in the fourth. Each station alone sees a fair coin whatever the other chose, so no signal can pass, and S = 1 + 1 + 1 − (−1) = 4. Four is the most any table can give, since each correlation lies between −1 and +1: forbidding signals takes nothing away.

Figure 3. The slider sets how strongly the table tilts towards agreeing or differing in each pairing; at full tilt it is Popescu and Rohrlich’s table. The eight signalling checks step through cell by cell, and the line below compares the table’s sum with the most that one state, acts that commute across the stations and squared weights can give.

The reason, for one state with acts that commute across the stations and squared weights: a correlation of +1 needs the two arrows to coincide. Three of them make a0 = b0, a0 = b1 and a1 = b0, so all four arrows are one arrow, and the fourth correlation, −1, would then need a1 = −b1: an arrow equal to its own opposite. Section 02 says how far out of reach the table is: no choice of state and acts goes past 2√2.

Figure 4. Trying states one at a time could never show that none goes higher, so the figure computes the largest value over every state at once: the top eigenvalue of one matrix (16 × 16 for the cube’s acts). Set the angle between each station’s two acts, or draw 3,000 random sets of acts that commute across the stations.

Let the acts fail to commute across the stations, and on some state there is no single table to score: doing one act first changes the other’s odds there, so the table depends on which station reads first. The paper proves that acts can be read together exactly when they commute (Theorem 24); in the textbook this is Lüders’ theorem.

04 · What it means

One record, read from two places.

Nothing holds the table of section 03 under 2√2, because there is no state under it for both stations to act on. Once both act on one state, with acts that commute across the stations and weights that are squares, the sum can go no higher than 2√2. On the record, the ceiling is what one shared state looks like when it is read from two places.

The ceiling itself is a theorem of 1980, and, given one state, two-outcome acts and squared weights, it needs only that one station’s acts commute with the other’s. The textbook gets that commuting from a rule it takes as given: that two systems together live in the product of their two spaces, each system’s acts acting on its own factor alone. On the record the product is proved for one cube split into two blocks of switches. For two blocks of switches of one cube, the product is simply how the cube’s combinations divide, and an act that touches only one block commutes with every act that touches only the other, with no further condition (the theorem “Blocks of one cube”, Theorem 26). The argument needs nothing to travel between the blocks.

The cube’s own flips do not come sorted into two systems. Each carries the signs of the switches before it, so station B’s raw flips carry station A’s sign and do not commute with station A’s acts. Take that sign off, and station B’s acts touch only its own block; that each station’s acts are taken this way is part of the owed composition statement. There is one record: on the cube, with station B’s acts taken this way, the two blocks are two parts of one system, read on one shared state, whose acts stay separate.

The same parallelogram law appears in the Born rule, in a different role. There the premises of a read come to it, and it forces the read of a register’s positions to be the sum of the squares of the values held there, given five premises, one of them the owed position read (the Born rule page); here it is a property of the lengths that squares give.

On the record’s picture of two registers, one part of the owed composition statement meets experiment: that two separate laboratories share one complex unit, which the textbook’s product rule says too. In a test with three stations and two separate sources, designed to tell real numbers from complex ones (Renou et al. 2021), a score of its own, built from more correlations than S, reaches at most about 7.66 for blocks built on real numbers alone when their sources share nothing (a bound found numerically, not known to be tight), and 6√2 ≈ 8.49 for blocks sharing one complex unit. The measurements gave 8.09 ± 0.01 on one chip (Chen et al. 2022) and 7.83 ± 0.03 with the sources and the measurements kept apart (Wu et al. 2022). Real blocks whose sources may share a suitable state reach the complex values too: in this network one shared entangled bit is needed and is enough (Weilenmann, Gisin and Sekatski 2025), and when the sources’ independence is defined by what can be observed, every correlation of the complex theory in a finite network is reached (Hoffreumon and Woods, two preprints of 2026). So the test does not by itself tell them apart: on the record’s picture of two registers, it is evidence that registers held in two laboratories agree on their unit, and so for composition over one complex unit; it is not a measurement of composition (the why complex numbers page).

05 · Hard questions

What an expert asks first.

Isn’t this known?
Yes. Tsirelson proved the ceiling in 1980 and reached it with acts of the kind in section 02; the table that reaches 4 appeared in 1985, and Popescu and Rohrlich drew its lesson in 1994. The proof in section 02 is a standard one: the Cauchy–Schwarz inequality twice and the parallelogram law. What the record adds is where the split into two stations, and the commuting of their acts, comes from: on one cube the two stations are two blocks, the product is simply how the cube’s combinations divide, and the cube’s own flips, which carry the signs of the switches before them, can be read together across the stations only once station A’s sign is taken off station B’s flips (section 04). That two laboratories are such blocks, with their acts taken this way, on one shared state, is owed.
Is any of this different from ordinary quantum mechanics?
No number is: every exact value on this page is the textbook’s, and none of these experiments can tell the record from the textbook. What differs is what is taken as given: the product of two spaces is a postulate in the textbook; on the record it is a theorem for two blocks of one cube, and owed for two separate laboratories.
What is not proved?
Why two separate laboratories stop at 2√2. That needs composition, with the position read and attachment alongside it (the third of the three lines), and it takes squared weights for these acts as a premise. The three-station tests of section 04 exclude real blocks whose sources share nothing; on the record’s picture of two registers they are evidence for composition over one complex unit, as they are for the textbook’s product rule. They are not a measurement of composition.
What would break it?
A correlation sum reliably above 2√2, in a test that counts every round and keeps each station’s choice from reaching the other. It would show that the two stations are not one state read with acts that commute across them and squared weights, and it would break the textbook too.

Not claimed by the paper

That this test needs complex numbers. Real values reach 2√2 as well; the correlation test cannot tell real from complex.

06 · Read it

The paper and the sources.

The Five Postulates of Quantum Mechanics on the Record: Theorems, Premises, and What Is Owed

Jonathan Washburn · Recognition Physics Research Institute

Preprint, posted 14 November 2026; not peer reviewed. Tsirelson’s bound is paradox 8 of the twelve in its Section 8; the theorems used here are in its Sections 5, 6 and 7.

Read the preprint (PDF), the institute’s library

  • P. Jordan and E. Wigner, Über das Paulische Äquivalenzverbot, Z. Phys. 47, 631 (1928). doi:10.1007/BF01331938
  • G. Lüders, Über die Zustandsänderung durch den Meßprozeß, Ann. Phys. (Leipzig) 443, 322 (1951). doi:10.1002/andp.19504430510
  • J. S. Bell, On the Einstein Podolsky Rosen paradox, Physics 1, 195 (1964). doi:10.1103/PhysicsPhysiqueFizika.1.195
  • J. F. Clauser, M. A. Horne, A. Shimony and R. A. Holt, Proposed experiment to test local hidden-variable theories, Phys. Rev. Lett. 23, 880 (1969). doi:10.1103/PhysRevLett.23.880
  • B. S. Cirel’son (Tsirelson), Quantum generalizations of Bell’s inequality, Lett. Math. Phys. 4, 93 (1980). doi:10.1007/BF00417500
  • P. Rastall, Locality, Bell’s theorem, and quantum mechanics, Found. Phys. 15, 963 (1985).
  • L. A. Khalfin and B. S. Tsirelson, Quantum and quasi-classical analogs of Bell inequalities, in Symposium on the Foundations of Modern Physics, P. Lahti and P. Mittelstaedt, eds. (World Scientific, Singapore, 1985).
  • S. Popescu and D. Rohrlich, Quantum nonlocality as an axiom, Found. Phys. 24, 379 (1994). doi:10.1007/BF02058098
  • M. Pawłowski, T. Paterek, D. Kaszlikowski, V. Scarani, A. Winter and M. Żukowski, Information causality as a physical principle, Nature 461, 1101 (2009).
  • B. Hensen et al., Loophole-free Bell inequality violation using electron spins separated by 1.3 kilometres, Nature 526, 682 (2015). doi:10.1038/nature15759
  • H. S. Poh, S. K. Joshi, A. Cerè, A. Cabello and C. Kurtsiefer, Approaching Tsirelson’s bound in a photon pair experiment, Phys. Rev. Lett. 115, 180408 (2015). doi:10.1103/PhysRevLett.115.180408
  • M.-O. Renou et al., Quantum theory based on real numbers can be experimentally falsified, Nature 600, 625 (2021). doi:10.1038/s41586-021-04160-4
  • M.-C. Chen et al., Ruling out real-valued standard formalism of quantum theory, Phys. Rev. Lett. 128, 040403 (2022). doi:10.1103/PhysRevLett.128.040403
  • D. Wu et al., Experimental refutation of real-valued quantum mechanics under strict locality conditions, Phys. Rev. Lett. 129, 140401 (2022). doi:10.1103/PhysRevLett.129.140401
  • S. Storz et al., Loophole-free Bell inequality violation with superconducting circuits, Nature 617, 265 (2023). doi:10.1038/s41586-023-05885-0
  • M. Weilenmann, N. Gisin and P. Sekatski, Partial independence suffices to rule out real quantum theory experimentally, Phys. Rev. Lett. 135, 180201 (2025).
  • T. Hoffreumon and M. P. Woods, Quantum theory based on real numbers cannot be experimentally falsified, preprint, arXiv:2603.19208 (2026).
  • T. Hoffreumon and M. P. Woods, Clarifications on the experimental status of real quantum theory, preprint, arXiv:2610.03644 (2026).