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Bell State

Entangle two qubits so they are perfectly correlated

intermediate2 qubits·~2 min

The question

How can two qubits be perfectly correlated when each one alone looks random?

Before you start

You will use two qubits and two gates. First, a Hadamard puts qubit 0 into superposition. Then a CNOT gate links qubit 1 to qubit 0 so that they always agree. The result is called a Bell state, the simplest example of quantum entanglement.

What you will see

After the Hadamard, qubit 0 is in a 50/50 superposition and qubit 1 is still $∣0 ⟩$ . The CNOT gate then says: if qubit 0 is $∣1 ⟩$ , flip qubit 1. Because qubit 0 is in superposition, CNOT creates a state where both qubits are $∣0 ⟩$ together OR both are $∣1 ⟩$ together, with nothing in between. Measuring either qubit alone gives a random result, but whenever you measure both, they always match. This correlation is stronger than anything classical physics allows.

The circuit

Circuit

Open in simulator →

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Step-by-step walkthrough

Apply Hadamard to qubit 0

H puts qubit 0 into superposition while qubit 1 stays in $∣0 ⟩$ . At this point the two qubits are still independent. The histogram shows equal weight on $∣00 ⟩$ and $∣10 ⟩$ . Look at the Bloch spheres: qubit 0 is on the equator, qubit 1 is still at the north pole.

\frac{∣0 ⟩ + ∣1 ⟩}{2} \otimes ∣0 ⟩ = \frac{∣00 ⟩ + ∣10 ⟩}{2}

Apply CNOT (control: q0, target: q1)

CNOT flips qubit 1 when qubit 0 is $∣1 ⟩$ . The $∣00 ⟩$ branch stays $∣00 ⟩$ and the $∣10 ⟩$ branch becomes $∣11 ⟩$ . Now the histogram shows only $∣00 ⟩$ and $∣11 ⟩$ , with zero weight on $∣01 ⟩$ or $∣10 ⟩$ . Both Bloch sphere arrows move inward, showing that each qubit individually is mixed, even though the pair is in a sharp, well-defined state.

\frac{∣00 ⟩ + ∣11 ⟩}{2} = ∣ Φ^{+} ⟩

What to notice

The histogram shows only two bars (00 and 11) with equal height. The bars for 01 and 10 are completely absent.
Each single-qubit Bloch sphere arrow shrinks inward, indicating a mixed local state.
The joint state is perfectly defined even though the individual parts look random.
Run many shots: the two qubits always agree, but which value they agree on is random each time.

Tempting but wrong

It is tempting to think one qubit secretly carries a hidden value that the other one copies. That is not what happens. Before measurement, neither qubit has a definite value. The correlation is built into the joint quantum state, not into hidden local properties. This distinction is what Bell's theorem proves.

Expected result

You see only 00 and 11 in the output, each with roughly 50% frequency. You never see 01 or 10. Over 1000 shots, the counts for 00 and 11 should each be near 500.

Connection to the theory

This experiment is a hands-on companion to the Entanglement lesson. That lesson explains why the Bell state cannot be written as a product of two separate qubit states, what it means for each qubit to be locally mixed, and how entanglement differs from classical correlation. After running this experiment, the formula $∣Φ + ⟩ = (∣00 ⟩ + ∣11 ⟩) / 2$ should feel like something you have seen in action.

Read the full lesson →

Test your understanding

In a Bell state, you measure qubit 0 and get |1⟩. What is the state of qubit 1?

▶ Load in simulator ↗ Nielsen and Chuang, Quantum Computation and Quantum Information ↗ MIT OCW 8.06: entanglement notes

Interference

GHZ State

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