WARPLAB

What am I looking at?

What you are looking at

The top half is one qubit: an arrow on the Bloch sphere that you turn with two angles or with gates, then measure, one copy at a time or ten thousand. The bottom half is a Bell test: an entangled pair goes to Alice and Bob, each turns a detector to an angle, and the CHSH value S counts how well their outcomes agree. The entangled pair plays beside the best local hidden-variable rule on the same game, so you can watch one pass 2 and the other stay at it.

Controls

On this device:

Mouse

  • Drag the sphere to turn the view, sideways or up and down; the state does not change.
  • Drag the polar angle θ and azimuth φ sliders to point the arrow; any change empties the histogram.
  • Click a gate (H, X, Y, Z, S, T) to turn the arrow about its axis; Reset to |0⟩ points it straight up.
  • Pick an axis, then +1, +10, +100 or +1000 to measure fresh copies, up to 10 000; Measure this qubit collapses the arrow.
  • Drag Alice's and Bob's angle sliders in 5° steps, or click Tsirelson's angles or All aligned; any change clears the rounds.
  • Click +1, +100 or +1000 to send pairs; Clear starts the count again.

Keyboard

  • Tab moves through every slider and button in page order; the sphere's view does not take keys.
  • On a slider, the arrow keys move one step (1° for θ and φ, 5° for a detector); Home and End jump to its ends.
  • Enter or Space presses the focused button: a gate, an axis, a number of copies, a preset or a number of pairs.
  • Shift+Tab goes back to the previous control.
  • H or ? opens this panel. Esc closes it.
Other devices

Touch

  • Drag sideways on the sphere to turn the view; it does not tilt up or down.
  • Drag the θ and φ sliders to point the arrow, or tap a gate to turn it.
  • Tap an axis, then a number of copies to measure; tap Measure this qubit to collapse it.
  • Drag the four detector sliders, or tap Tsirelson's angles or All aligned.
  • Tap +1, +100 or +1000 to send pairs, and Clear to start again.

Glossary

Qubit
The quantum version of a bit. Its state is α|0⟩ + β|1⟩, a mix of the two values that is only settled when you measure it.
Why it matters Every quantum computer is built from qubits like this one.
|0⟩ and |1⟩
The two basic states of a qubit, like 0 and 1 of an ordinary bit. On the sphere |0⟩ is the top and |1⟩ the bottom.
Amplitudes α and β
The two numbers, possibly complex, that say how much of |0⟩ and of |1⟩ the state holds. |α|² + |β|² = 1, and |α|² is the chance of +1 when you measure along z.
Why it matters A complex amplitude carries a phase, which no single measurement along z can see but a gate can turn into a difference.
Bloch sphere
A ball on which every state of one qubit is an arrow from the centre to the surface. Opposite points are opposite outcomes of one measurement.
Why it matters It turns two complex amplitudes into one picture you can rotate.
Polar angle θ
How far the arrow tips down from the top, from 0° (|0⟩) to 180° (|1⟩). At 90° it lies on the equator, an equal mix of |0⟩ and |1⟩.
Azimuth φ
How far the arrow is turned round the vertical axis, from −180° to 180°. At 0° it points along x, at 90° along y. It is the phase between α and β.
Bloch vector (x, y, z)
The arrow's three coordinates, a vector of length 1. Each one is the average outcome you would get measuring many copies along that axis.
Why it matters z = 1 means +1 every time along z; z = 0 means half and half.
Gate
An operation that turns the arrow about an axis by a fixed angle. It changes the state without measuring it, so nothing is lost.
H (Hadamard)
A half turn about the axis halfway between x and z. It swaps |0⟩ and |+⟩, so H on |0⟩ lands on the equator.
Why it matters It is the usual way to put a qubit into an equal mix of |0⟩ and |1⟩.
X, Y and Z
Half turns about the x, y and z axes. X swaps |0⟩ and |1⟩, the quantum NOT. Z flips the sign of |1⟩, which a measurement along z cannot see.
S and T
Smaller turns about z: S a quarter turn, T an eighth. S takes |+⟩ to |+i⟩. Both leave the chances along z as they were.
Measurement axis
The direction you measure along: x, y or z. The answer is +1 or −1; along z they mean |0⟩ and |1⟩, along x |+⟩ and |−⟩, along y |+i⟩ and |−i⟩.
Fresh copies
Qubits prepared the same way and each measured once. The histogram counts how many gave +1 and how many −1; the same copies give the same outcomes on every visit.
Born rule
The chance of +1 along an axis n is |⟨+n|ψ⟩|², set by the angle between the arrow and the axis. The tick on each bar marks it, and the measured share settles on it as copies add up.
Why it matters Quantum mechanics predicts chances, not single outcomes, and this is how.
Collapse
Measuring this qubit gives +1 or −1 with the Born rule's chances and moves the arrow onto that end of the axis. Measure again along the same axis and you get the same answer every time.
Entangled pair Φ⁺
Two qubits in the state (|00⟩ + |11⟩)/√2: neither has an answer of its own, but measured along the same direction they always agree. Alice's outcomes alone stay 50/50 whatever Bob does.
Why it matters The agreement shows only when the two lists are put side by side, so it cannot carry a message.
Settings and rounds
Each round a referee hands Alice and Bob a random setting, 0 or 1. They win when their outcomes agree, except when both settings are 1: then they must differ. Rounds counts the pairs sent so far.
Detector angles a, a′, b, b′
The direction each detector turns to in the x–z plane: a for Alice's setting 0, a′ for her setting 1, b and b′ for Bob's. 0° points along z, 90° along x.
Why it matters The angles, not the pair, decide how far the score can go.
Correlation E
For one pair of settings, how often the outcomes agree minus how often they differ, as a share of those rounds. It runs from 1 (always agree) to −1 (always differ); for Φ⁺ it is cos(a − b).
CHSH value S
S = E(a, b) + E(a, b′) + E(a′, b) − E(a′, b′), one number from the four correlations. S measured comes from the rounds sent; S expected is what the theory gives for these angles.
S ± σ
The measured S with one standard deviation σ, the spread that chance alone gives with this many rounds. Four times the rounds halves σ.
Win rate
The share of rounds won. It follows S: win rate = 1/2 + S/8, so S = 2 is 75% and S = 2√2 about 85%.
Classical limit 2
No local hidden-variable rule can make S larger than 2, so none wins more than 75% of rounds. Every fixed strategy scores exactly 2 or −2.
Why it matters Going past 2 rules out every explanation in which each particle carries its answers with it.
Quantum bound 2√2
The most S any quantum state can reach, 2√2 ≈ 2.83. The entangled pair reaches it exactly at Tsirelson's angles.
Tsirelson's angles
a = 0°, a′ = 90°, b = 45°, b′ = −45°: each pair of settings 45° apart, except a′ and b′, which are 135° apart. They give the entangled pair S = 2√2. All aligned puts every detector at 0°, where both sources give S = 2.
Local hidden-variable rule
Each pair leaves with a hidden direction, and each detector answers +1 if it points within 90° of it, on its own. Its correlation is 1 − 2d/π for detectors d apart, and it reaches exactly 2 at Tsirelson's angles.
Why it matters It is the best a local rule can do, so it is the fair rival.
Beats 2 by nσ
Shown when |S| sits more than 3σ above 2; a smaller excess counts as within the classical limit. Until σ falls under 0.5 the panel asks you to send more pairs.
Why it matters A local run can land a little above 2 by chance, so only a clear margin counts.

All help and questions

Qubit and Bell test

Turn a qubit on the Bloch sphere and measure it. Then share an entangled pair and watch the CHSH value beat the classical limit of 2.

One qubit

A qubit's state is α|0⟩ + β|1⟩, and its arrow on the Bloch sphere holds all of it. Turn the arrow with a gate or the two angles, then measure.

Drag the sphere to turn the view.

Amplitude of |0⟩ α
0.866
Amplitude of |1⟩ β
0.250 + 0.433i
Bloch vector (x, y, z)
(0.43, 0.75, 0.50)

Prepare the state

A gate turns the arrow about an axis. Try H on |0⟩: it lands on the equator, an equal mix of |0⟩ and |1⟩.

0%50%100%+1 |0⟩−1 |1⟩
MeasuredBorn rule
Born rule, +1
75.0%
Measured, +1
79.0%
Copies measured
100

Each copy is prepared the same way and measured once. One copy gives +1 or −1, never anything between; the share of +1 settles on |⟨+n|ψ⟩|² as copies add up.

Bell test

Two qubits in the entangled state Φ⁺ = (|00⟩ + |11⟩)/√2 fly apart to Alice and Bob. Each round a referee hands each of them a random setting, 0 or 1; each turns a detector to the angle chosen for that setting and measures. They win the round when their outcomes agree, except when both settings are 1: then they must differ.

aa′bb′

Rounds0

Entangled pair

S measured
—
S expected
2.828
Rounds won
—
Win rate expected
85.4%

Send more pairs

Quantum mechanics: E(a, b) = cos(a − b).

Local hidden variables

S measured
—
S expected
2.000
Rounds won
—
Win rate expected
75.0%

Send more pairs

Each pair leaves with a hidden direction; each detector answers +1 if it is within 90° of it, on its own. The best a local rule can do.

012322√2|S|
Classical limit 2Quantum bound 2√2
−10104590135180Angle between detectors (°)E
Entangled pairLocal hidden variablesMeasured

S = E(a, b) + E(a, b′) + E(a′, b) − E(a′, b′), where E is how often a setting pair agrees minus how often it differs. Any local hidden-variable rule keeps S ≤ 2, which caps the win rate at 75%; the entangled pair reaches 2√2 ≈ 2.83 and about 85% at Tsirelson's angles. The ± is one standard deviation from the number of rounds.

Alice's own outcomes stay 50/50 whatever Bob's detector does: the correlation shows only when the two lists are put side by side.

Sources

Analytic model. This module uses no instrument data.

Physics benchmarks

  • Measured frequencies converge on the Born rule |⟨+n|ψ⟩|²: with 10 000 copies per state and axis, the χ² over 150 cases is what chance gives, and the mean error falls as 1/√N. Every gate is unitary and turns the Bloch vector about its own axis.
  • For the maximally entangled state Φ⁺, S = 2√2 exactly at Tsirelson's angles and no angles give more; 40 000 sampled pairs give 2√2 within 3σ and 0.05, and beat 2 by over 10σ.
  • Local hidden variables: all 16 deterministic strategies score S = ±2, so no mixture exceeds 2; the hidden-direction model never exceeds 2 for any angles on a 15° grid, and 200 runs of 4000 pairs never beat 2 by more than 4σ.

A release ships only when these checks pass.