Entanglement
H then CX turns |00⟩ into the Bell state (|00⟩+|11⟩)/√2. Each bit on its own is a fair coin. Yet the two bits agree on every shot. Simple arithmetic shows that no pair of separate qubit descriptions can make this state. The link lives in the joint state, and it can't be used to send a message.
Why care? Entanglement is the resource that separates many quantum methods from ordinary ones. It is also the most over-hyped word in the field. Here you will build it and check it with plain arithmetic.
Build the state by hand first. Start at 00 with amplitude 1. (An amplitude is a number you square to get a chance.)
- Apply
Hto q0. The amplitudes become 0.7071 on 00 and 0.7071 on 01. - Apply
CXwith control q0 and target q1. The 00 part has control 0, so it stays. The 01 part has control 1, so q1 flips and it becomes 11. - Final amplitudes: 0.7071 on 00, 0.7071 on 11, and zero everywhere else.
The squares are 0.5 and 0.5. So the prediction is: half the shots read 00, half read 11, and 01 and 10 never appear.
This state, (|00⟩ + |11⟩)/√2, is called a Bell state. It is entangled. That means it is a joint state that can't be split into a separate description for each qubit. The next section proves that with arithmetic, not just a claim.
- The Bell circuitINTERACTIVE
- Why can't this state be split into two qubit descriptions?
- Where did the correlation come from?INTERACTIVE
- Can entanglement send a message?
- How close should the counts be?
- Worked exampleINTERACTIVE
- How do Bell pairs test real hardware?
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