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Part of Quantum Algorithms
Home/Quantum Physics/Lessons/Bell State Preparation
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Bell State Preparation

beginner2 qubits~5 min

Create maximal entanglement with just two gates

The question

How do you create a pair of qubits that are perfectly correlated, even though each one individually looks completely random?

Why this matters

Bell states are the simplest entangled states and the foundation of quantum information. They were proposed by physicist John Bell in 1964 to test whether quantum mechanics could be explained by hidden local variables. The answer was no — and Bell states became the cornerstone of quantum teleportation, superdense coding, quantum key distribution, and error correction. If you understand Bell states, you understand the core resource that makes quantum computing different from classical computing.

Intuition

Start with two qubits in ∣00⟩. Apply Hadamard to the first qubit to put it in superposition: now the state is (∣0⟩+∣1⟩)/2​⊗∣0⟩. Then apply CNOT with the first qubit as control. The ∣0⟩ branch leaves the target alone, giving ∣00⟩. The ∣1⟩ branch flips the target, giving ∣11⟩. The result is (∣00⟩+∣11⟩)/2​ — a state where the two qubits are perfectly correlated but individually random. This is entanglement in its purest form.

Key insight

Entanglement is created by applying a two-qubit gate (CNOT) to a qubit that is already in superposition. If the control qubit were definite (∣0⟩ or ∣1⟩, CNOT would just do a classical operation. Superposition + controlled interaction = entanglement.

Circuit
Open in simulator →
q0q1HCX
▶ Try it in the simulator
Step-by-step walkthrough
1

Initialize ∣00⟩

Both qubits start in the ∣0⟩ state. The system is completely unentangled.

∣00⟩
2

Apply Hadamard to qubit 0

Hadamard puts qubit 0 into an equal superposition. The two qubits are still independent — no entanglement yet.

2​∣0⟩+∣1⟩​⊗∣0⟩=2​∣00⟩+∣10⟩​
3

Apply CNOT (control: q0, target: q1)

CNOT creates entanglement. The ∣00⟩ term stays ∣00⟩ (control is 0, target unchanged). The ∣10⟩ term becomes ∣11⟩ (control is 1, target flips). Now the two qubits are linked.

2​∣00⟩+∣11⟩​=∣Φ+⟩
4

Measure both qubits

You get 00 or 11, each with 50% probability. You never see 01 or 10. The outcomes are perfectly correlated even though each individual outcome is random.

What the measurement tells you

Measuring 00 or 11 with equal probability (never 01 or 10) confirms the Bell state. The perfect correlation cannot be explained by each qubit independently deciding its value.

Full technical statement

The Bell state ∣Φ+⟩=(∣00⟩+∣11⟩)/2​ is a maximally entangled state of two qubits. It cannot be written as a product of individual qubit states (∣ψA​⟩ times ∣ψB​⟩, which is the mathematical definition of entanglement. If you mathematically remove one qubit from the description (a process called tracing out), the remaining qubit is in a completely mixed state -- meaning it carries zero information about the outcome. All the information lives in the correlations between the two qubits. There are four Bell states, and they form a complete basis for the two-qubit state space: |Φ±⟩ = (∣00⟩±∣11⟩/2​ and |Ψ±⟩ = (∣01⟩±∣10⟩/2​.

Classical approach

Classically, you could prepare two correlated coins (both heads or both tails) by looking at one and setting the other. But the coins had definite values all along.

Quantum approach

In a Bell state, neither qubit has a definite value before measurement. The correlation exists without either qubit having a pre-determined outcome. This is what Bell’s theorem proves cannot be classical.

Tempting but wrong

It is tempting to think entangled qubits can send messages instantly. That is not what happens. Each qubit’s local measurement outcome is completely random. The correlations only become visible when you compare results from both sides, which requires ordinary (speed-of-light-limited) classical communication.

Where this leads

Bell states are the starting point for quantum teleportation (transferring a state using entanglement + classical bits) and superdense coding (sending 2 classical bits using 1 qubit + entanglement).

Gates you should know first
H gate →CNOT gate →
Test your understanding

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

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