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Bell-State Entanglement Visualizer

Quantum

$H$ on $q_0$ followed by $\operatorname{CNOT}(q_0\to q_1)$ . Toggling the initialization ports selects which Bell pair is prepared; amplitudes and the correlation index are derived, and a measure button accumulates real Born-rule shots.

6 components · 5 connections · state-vector + Born sampler

the aha
Measure one qubit and you instantly know the other — not because a signal raced across, but because they were never really two things.

why it works

Start with

|00\rangle

, apply

H

to the first qubit, then a CNOT, and you reach

|\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt 2}

. Notice what is missing: there is no

|01\rangle

and no

|10\rangle

. The two qubits have no state of their own — only the pair has a state. A skeptic says "they were secretly agreed in advance," like two gloves in sealed boxes. Quantum mechanics answers with the CHSH test: local hidden agreements can never push the correlation witness

S

above

2

, yet this state reaches

S = 2\sqrt 2 \approx 2.83

. Watching

S

climb past

2

is watching local realism fail in front of you.

governing model

|\Phi^+\rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2}} \;\cdot\; \langle Z \otimes Z \rangle \in [-1, +1]

structural inventory

$2\times$ qubit register line
Hadamard gate
CNOT control/target
measurement ports

boundaries

$4$ Bell states
amplitudes $\pm 1/\sqrt{2}$
CHSH witness $S\in[0,2\sqrt{2}]$

fault relay CNOT entangler offline

Prepare $|\Phi^+\rangle$ and measure ×64 repeatedly.

Only $00$ and $11$ ever appear, roughly half each — perfectly correlated outcomes from a state with no definite value beforehand.
Run the CHSH witness.

The empirical $S$ settles near $2\sqrt 2$ . Any glove-in-a-box explanation is capped at $2$ ; you are past it. That gap is the whole of quantum non-locality.
Take the entangler offline.

Skip the CNOT and you get a plain product state — the correlation collapses back below $2$ . No entanglement, no violation.

causal signal trace paused

Prepare $∣ 00 ⟩$ Both qubits begin separable and deterministic.

t_{stage} = 0.00 s

step 1 / 4 Prepare |00⟩

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