Basic circuits · Bell / entanglement
Bell state measurement
A two-qubit entanglement example with a return contract and distribution check.
Exact & formalUpdated
A small worked example, computed in your browser. It is not a result from this record's papers.
A Bell pair
2 qubits, preparing (|00⟩ + |11⟩) / √2.
Step 1 of 2 · H
Probability of each outcome after this step
- 0050%
- 0150%
Puts qubit 0 into an equal superposition of |0⟩ and |1⟩.
Probability spreads from 1 outcome to 2. All 2 are equally likely.
Entangles qubit 1 with qubit 0, so the two only ever agree.
Probability moves: 1 outcome appears that had none, 1 drops to zero, 2 carry probability now. Up: 11 0.0% → 50.0%. Down: 01 50.0% → 0.0%. This is the step that entangles the qubits: from here on no single qubit has a state of its own, only the register as a whole does.
Readout Measuring both qubits always gives 00 or 11, each about half the time, and never 01 or 10.
Open in Studio- Steps2
- Wires2
- Qubits2
- Depth2 gates
- Outcomes00 / 11
- Papers2
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.
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 13 →
The Bell state is prepared by on qubit 0 followed by : starting from , gives , and CX then copies the control's value onto the target, — exactly the statevector this record's exact-state check compares against. Measuring both qubits in the computational basis yields or each with probability and never or : 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.
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 13 →
- Qubits
- 2
- Depth
- 2 gates
- Outcomes
- 00 / 11
- State
- (|00⟩ + |11⟩)/√2
- Ideal P(00)
- 0.5
- Ideal P(11)
- 0.5
Hon q[0]CXon q[0], q[1]
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.
Nobody has reviewed this record for gaps yet.
from qiskit import QuantumCircuit
qc = QuantumCircuit(2, 2)
qc.h(0)
qc.cx(0, 1)
qc.measure([0, 1], [0, 1])
FINAL_CIRCUIT = qcWhere 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 34 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
Related entries
- Hadamard gate
Gates
- GHZ state preparation
Basic circuits
References
- On the Einstein Podolsky Rosen Paradox
John S. Bell · 1964
Introduces the Bell inequality that distinguishes entangled correlations, such as the Bell state, from any local hidden-variable model.
- Quantum Computation and Quantum Information: 10th Anniversary Edition
Michael A. Nielsen and Isaac L. Chuang · 2010
Standard graduate textbook covering the single- and two-qubit gate definitions, matrix identities, and circuit primitives this record states.