Entanglement

The circuit is just two gates. H puts q0 into (|0⟩+|1⟩)/√2, while q1 stays at |0⟩. Then CX (a controlled NOT) flips q1 only when q0 is 1. It acts on each part of the state separately. In the part where q0 = 0, it does nothing. In the part where q0 = 1, it flips q1. The result is (|00⟩ + |11⟩)/√2. That means "both qubits zero" plus "both qubits one," and nothing else.

Read the amplitudes as they are. The parts |01⟩ and |10⟩ have amplitude exactly zero. Not small: zero. Mismatched answers aren't just unlikely. In the ideal state they can't happen. Yet each qubit on its own acts like a fair coin. Measure either one alone and you get 50/50.

This state can't be written as "q0's state" times "q1's state." The qubits no longer have their own states, only a shared one. That is what entangled means.

That is a fair question. Imagine two coins glued so they always land the same way. They would make the same chart. If you could only ever measure 0 versus 1, a Bell pair would look just like a bag of pre-agreed bit pairs.

The difference shows up when you change the measurement basis, meaning the axis you measure along. Add an H to each qubit just before measuring. This measures along a turned axis. The Bell state (|00⟩+|11⟩)/√2 is left exactly the same by that pair of H gates. So the answers are still only 00 and 11.

Now try it with the glued coins. Each pre-agreed pair, like 00, turns into all four answers equally after the two H gates. So the glued coins would spread out evenly: about 25% each of 00, 01, 10 and 11. Here the coin picture stops working. A match that holds along more than one axis is the truly quantum part. It is what Bell-test experiments measure.

There are four Bell states. Each is as entangled as two qubits can be. Small edits to the same circuit reach all of them:

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

Notice what a plain 0-versus-1 measurement can and can't see. Φ⁺ and Φ⁻ give the same chart here. So do Ψ⁺ and Ψ⁻. The sign between the two parts is a phase. Phases only show up when you measure along another axis. It is the same idea as the H sandwich.

No. It is worth being clear about why, because this is one of the most hyped claims about quantum computing.

Say you keep q0 and send q1 far away to a friend. Whatever you do to your qubit, your friend's results stay exactly 50/50. You can measure it, turn it first, or pick a clever axis. Nothing your friend can see on their own changes. The link only shows up when you put both lists of results side by side. To do that, you have to send your list over a normal channel, like a phone call, at normal speed.

This rule is called the no-communication theorem. It comes from how measurement works. It is not an engineering limit that someone might one day get around. Entanglement gives you linked results, not a way to send signals.

Run the Bell circuit on real hardware and 01 and 10 do show up, usually a few percent of shots. The physics hasn't changed. It's noise. The two-qubit gate is not perfect. Readout sometimes reports the wrong bit. And the qubits slowly lose their state during the circuit, which is called decoherence. The ideal state still gives those answers amplitude zero.

The circuit is tiny and the ideal answer is sharp. That makes Bell-pair quality a standard first test for any two-qubit system. It checks one two-qubit gate and both readouts in a single shot. Try it in the lab. Then see how the devices you'd run it on compare on the QPU index.