A qubit is rotated with microwave pulses tuned to its resonance frequency, and the Rabi frequency (the frequency of one round trip between |0⟩ and |1⟩) is 10 MHz. What is the length of the π/2 pulse that takes |0⟩ to a superposition in which the amplitudes of |0⟩ and |1⟩ have equal magnitude?

Answer: A. 25 ns

Under continuous microwave driving, the probability of being in |1⟩ oscillates as sin²(Ωt/2) with Ω = 2π × 10 MHz, and the period of one round trip is 1 ÷ 10 MHz = 100 ns. The rotation angle Ωt reaches π/2 at t = 25 ns, when the probability of being in |1⟩ is 50%. A rotation angle of π (50 ns) is an X gate (π pulse) that swaps |0⟩ and |1⟩, and 2π (100 ns) returns the qubit to |0⟩. The rotation angle is set by the pulse amplitude and length, so errors in the microwave amplitude or timing translate directly into gate errors.

• A. Correct. A quarter of the 100 ns period gives a rotation angle of π/2.
• B. This is the length for a rotation angle of π, the π pulse that swaps |0⟩ and |1⟩.
• C. This is one full Rabi period; the rotation angle is 2π and the qubit returns to |0⟩.