Entanglement and the Bell states

Build the four Bell states, test them with correlations, and see why they cannot be factored.

You will learn to

Derive |Φ⁺⟩ from H then CNOT.
Show that |Φ⁺⟩ is not a product state.
Make all four Bell states with one extra X or Z.

∣ Φ^{\pm} ⟩ = \frac{∣00 ⟩ \pm ∣11 ⟩}{2}, ∣ Ψ^{\pm} ⟩ = \frac{∣01 ⟩ \pm ∣10 ⟩}{2}

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.

Try it

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;

Open in the IDE Open in the Playground

State	Before the CNOT, add
\|Φ⁺⟩	nothing
\|Φ⁻⟩	z q[0]
\|Ψ⁺⟩	x q[1]
\|Ψ⁻⟩	z q[0] and x q[1]

Entanglement 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.

Try it

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;

Open in the IDE Open in the Playground

Quiz 1 · 10 XP

Which circuit prepares |Ψ⁺⟩ = (|01⟩+|10⟩)/√2?

Quiz 2 · 10 XP

Two qubits always give the same bit in the Z basis. Are they necessarily entangled?

Key takeaway

H then CNOT makes a Bell state: a two-qubit state with no description qubit by qubit, whose correlations survive a change of basis. One X or Z before the CNOT selects which of the four.

Docs: entanglement chapter

Up next: Reading a circuit: depth, width and cost

State

Before the CNOT, add

|Φ⁺⟩

nothing

|Φ⁻⟩

z q[0]

|Ψ⁺⟩

x q[1]

|Ψ⁻⟩

z q[0] and x q[1]