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Entanglement and the Bell states
Build the four Bell states, test them with correlations, and see why they cannot be factored.
- 01Derive |Φ⁺⟩ from H then CNOT.
- 02Show that |Φ⁺⟩ is not a product state.
- 03Make all four Bell states with one extra X or Z.
H on qubit 0 gives (|0⟩+|1⟩)|0⟩/√2. CNOT with control 0 flips qubit 1 in the second branch only: (|00⟩+|11⟩)/√2. Try to write this as (a|0⟩+b|1⟩)(c|0⟩+d|1⟩): you need ac ≠ 0 and bd ≠ 0 but ad = bc = 0, which is impossible. The state has no description qubit by qubit.
Perfect correlation
Simulate: half the shots read 00, half 11, never 01 or 10. Then add z q[0] before the CNOT to make |Φ⁻⟩: the histogram is identical. Only the State tab's phases tell the two apart.
OPENQASM 3.0;
include "stdgates.inc";
qubit[2] q;
bit[2] c;
h q[0];
cx q[0], q[1];
c = measure q;
| State | Before the CNOT, add |
|---|---|
| |Φ⁺⟩ | nothing |
| |Φ⁻⟩ | z q[0] |
| |Ψ⁺⟩ | x q[1] |
| |Ψ⁻⟩ | z q[0] and x q[1] |
01Entanglement you can measure
Correlation in one basis is not enough: two classical coins glued together also always agree. Entanglement shows when the correlation survives a change of basis. Measure both qubits of |Φ⁺⟩ in the X basis (H on each before measuring) and they still agree every time. Classical coins cannot do that.
Agreement in the X basis
Simulate: again only 00 and 11. Replace the Bell pair by two independent H gates (remove the CNOT) and the X-basis results become uniform.
OPENQASM 3.0;
include "stdgates.inc";
qubit[2] q;
bit[2] c;
h q[0];
cx q[0], q[1];
h q[0];
h q[1];
c = measure q;
2 questions · one try each counts
Which circuit prepares |Ψ⁺⟩ = (|01⟩+|10⟩)/√2?
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Two qubits always give the same bit in the Z basis. Are they necessarily entangled?
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