Entanglement

The circuit is two gates. H puts q0 into (|0⟩+|1⟩)/√2 while q1 sits in |0⟩. Then CX acts on each branch separately: in the q0 = 0 branch it does nothing; in the q0 = 1 branch it flips q1. The result is (|00⟩ + |11⟩)/√2 — a superposition of both qubits zero and both qubits one, and nothing else.

Read the amplitudes literally. The components |01⟩ and |10⟩ have amplitude exactly zero — not small, zero. Mismatched outcomes aren't unlikely; in the ideal state they are impossible. Yet each qubit on its own is a fair coin: measure either one alone and you get 50/50.

This state cannot be written as 'q0's state' times 'q1's state'. The qubits no longer have individual states, only a joint one. That is what entangled means.

A fair objection: two classical coins glued to always match would produce the same histogram. If you could only ever measure in the computational basis, a Bell pair and a bag of pre-agreed bit pairs would be indistinguishable.

The difference appears when you change the measurement basis. Add an H to each qubit just before measuring — this measures along a rotated axis. The Bell state (|00⟩+|11⟩)/√2 is left exactly unchanged by that pair of H's, so the outcomes are still only 00 and 11. Pre-agreed classical pairs measured the same way would scatter uniformly across all four outcomes. Correlation that holds in more than one basis at once is the genuinely quantum part — it's what Bell-test experiments quantify.

There are four maximally entangled two-qubit states, and small edits to the same circuit reach them all:

• Φ⁺ = (|00⟩ + |11⟩)/√2 — H, then CX. Outcomes always match.
• Φ⁻ = (|00⟩ − |11⟩)/√2 — insert a Z on q0 after the H. Outcomes still always match.
• Ψ⁺ = (|01⟩ + |10⟩)/√2 — insert an X on q1 before the CX. Outcomes always differ.
• Ψ⁻ = (|01⟩ − |10⟩)/√2 — both edits at once.

Note what the computational basis can and can't see: Φ⁺ and Φ⁻ produce identical histograms here, and so do Ψ⁺ and Ψ⁻. The sign between the branches is a phase, and phases only become visible when you measure in another basis — the same lesson as the H sandwich.

No — and it's worth being precise about why, because this is one of the most-hyped claims around quantum computing. Suppose you keep q0 and send q1 far away. Whatever you do to your qubit — measure it, rotate it first, choose a clever basis — the statistics of your partner's measurements stay exactly 50/50. Nothing they can observe locally changes. The correlation exists only in the joint record, and the only way to see it is to bring both records together over an ordinary classical channel, at ordinary speed.

This is the no-communication theorem: a consequence of how measurement works, not an engineering limitation someone might one day work around. Entanglement is a resource for correlation, not a channel for signalling.

Run the Bell circuit on real hardware and 01 and 10 do appear — typically a few per cent of shots. That isn't the physics changing its mind; it's noise: an imperfect two-qubit gate, readout errors that misreport a bit, and decoherence during the circuit. The ideal state still assigns those outcomes amplitude zero.

Because the circuit is tiny and the ideal answer is sharp, Bell-pair quality is a standard first benchmark for any two-qubit system — one two-qubit gate and both readouts exercised in a single shot. Try it in the lab, then see how the devices you'd run it on compare on the QPU index.