Exact & formalBasic circuitsBell / entanglement2026-08-12

Bell state measurement

A two-qubit entanglement example with a return contract and distribution check.

entanglementintroverification

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The Bell state is compact enough for a first experiment and rich enough to show that measurement outcomes can be correlated without choosing a classical bit in advance.

Circuit & simulation

q[0]

HCX

q[1]

0050%

1150%

What this takes and returns

TakesNothingWhat joins here

Nothing goes in, and that is what this entry is: a pipeline starts here. There is no upstream to choose — what you pick is what comes after.

Nothing in the Atlas meets this end.

Returns2 qubitsWhat joins here

Returns a 2-qubit register a next stage can take. This is the end that joins.

12 entries meet this end. Named below.

Prepares a state. Nothing goes in, and what comes out is a register another stage can take. See all 12 →

Fits after this 12

Where the map uses this

This record is an instance of an object the map names, so these are the processes that consume or produce one. None of them is about this record in particular.

State you can prepare 5 of 33 processes

Quantum linear solve hands one back
Prepare an input state hands one back
Amplify a success branch hands one back
Estimate an observable takes one
Minimise the objective over the parameters hands one back

How it works

The Bell state $\frac{1}{\sqrt2}(|00\rangle+|11\rangle)$ is prepared by $H$ on qubit 0 followed by $\mathrm{CX}_{0\to1}$ : starting from $|00\rangle$ , $H\otimes I$ gives $\frac{1}{\sqrt2}(|00\rangle+|10\rangle)$ , and CX then copies the control's value onto the target, $\frac{1}{\sqrt2}(|00\rangle+|11\rangle)$ — exactly the statevector this record's exact-state check compares against. Measuring both qubits in the computational basis yields $00$ or $11$ each with probability $1/2$ and never $01$ or $10$ : the two outcomes are perfectly correlated even though neither qubit's individual outcome is determined before measurement, the operational signature of entanglement (this state violates a Bell/CHSH inequality; it cannot be reproduced by any local hidden-variable model tossing independent coins). Because it is the smallest circuit exhibiting genuine quantum correlation, the Bell state anchors teleportation, superdense coding, and entanglement-verification protocols throughout this catalog.

Implementation

FrameworkNative

bell_state.py

from qiskit import QuantumCircuit

qc = QuantumCircuit(2, 2)
qc.h(0)
qc.cx(0, 1)
qc.measure([0, 1], [0, 1])

FINAL_CIRCUIT = qc

Quantum vs classical

Classical baseline

Use a classical state-vector or matrix simulation at the same width, precision, and measurement objective.

Quantum claim

The quantum record demonstrates a state or operator behavior; it does not make classical simulation or communication costs disappear.

How to compare

Compare fidelity, samples, gate depth, noise, memory, and the cost of preparing and reading the state.

Declared gaps

Nobody has reviewed this record for gaps yet.

Literature & references

On the Einstein Podolsky Rosen Paradox1964 · John S. Bell

Introduces the Bell inequality that distinguishes entangled correlations, such as the Bell state, from any local hidden-variable model.

doi.org/10.1103/physicsphysiquefizika.1.195 ↗

Quantum Computation and Quantum Information: 10th Anniversary Edition2010 · Michael A. Nielsen and Isaac L. Chuang

Standard graduate textbook covering the single- and two-qubit gate definitions, matrix identities, and circuit primitives this record states.

doi.org/10.1017/cbo9780511976667 ↗