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Q8.Hadamard Self-Inverse Identity (H · H = I)Easy

Q8.Hadamard Self-Inverse Identity (H · H = I)

Single-Qubit OperationsQubits: 1Q
Demonstrate that the Hadamard transformation is unitary and its own inverse (H†=HH^\dagger = H).
Target Objective:
Apply two successive Hadamard gates on Qubit 0 initialized in ∣0⟩|0\rangle to recover the initial ∣0⟩|0\rangle state with 100% fidelity.
Judge Acceptance Criteria:

Statevector fidelity must reach ≥ 99.9% with the target unitary solution.

Gates:
q[0]|0⟩
S0
S1
S2
S3
S4
S5
S6
S7
Live Python / Qiskit Code (Edit to Update Circuit)
 
Type Qiskit Python code above to update gates on wires automatically0 gates · 1Q
0 gates placed1 Qubits · 8 Steps
State Measurement Distribution:2 non-zero basis states
|0⟩100.0%
|1⟩0.0%