Explore Bell-state phase
Change a Bell state's relative phase without changing its computational-basis probabilities.
Intuition
A Bell state is a precise quantum state, not just a histogram. A local Z changes Φ+ into Φ−; a local X can change matching outcomes into opposite outcomes.
Build H(q0), CX(0,1), then Z on either qubit. In the Statevector tab, the |11⟩ amplitude changes sign. The probabilities remain unchanged. This is why fidelity, not histogram matching, judges state preparation.
The mathematics
Look a little deeper
The four Bell states Φ± and Ψ± form an orthonormal basis of the two-qubit space. A global minus sign does not change a state, but a minus sign on only one of the two nonzero amplitudes changes it to an orthogonal Bell state.
Use it
Prepare Φ− in at most three gates, using only H, CX, and Z.
This exercise uses the same server-side state and circuit checks as its linked practice problem.
Build your circuit
Click a gate to append it, or drag it onto a wire. CX uses the selected control; SWAP uses it as the second operand. Basis order: |qₙ … q₀⟩.
Returned logical circuit · before transpilation · q0 on top
Connect to code
qc.h(0) qc.cx(0, 1) qc.z(0)
These operations go inside a circuit-building program. Practice problems provide a complete solve() template.
Check your understanding
First complete the circuit exercise successfully.