Quantum Computing Hardware
Which correctly describes dispersive readout, which is widely used to read out superconducting qubits?
- AA DC current is passed through the qubit, and the different resistances for 0 and 1 are read as a voltage
- BThe small shift in the resonance frequency of a resonator coupled to the qubit, which depends on the qubit state, is read as a change in the phase or amplitude of a microwave signal
- CVisible-light photons emitted by the qubit when it is 1 are counted with a photodiode
Show answer and explanation
Answer: B. The small shift in the resonance frequency of a resonator coupled to the qubit, which depends on the qubit state, is read as a change in the phase or amplitude of a microwave signal
In dispersive readout, a readout resonator whose frequency is far from the qubit's (for example, around 7 GHz) is coupled to the qubit. The resonator frequency shifts slightly depending on whether the qubit is in |0⟩ or |1⟩, so a weak microwave signal is sent to the resonator and the state is determined from differences in the phase and amplitude of the reflected or transmitted signal. The signal is extremely weak, so it is amplified with as little added noise as possible by cryogenic Josephson parametric amplifiers, HEMT amplifiers at the 4 K stage and so on, and processed at room temperature. Because the readout does not directly absorb the qubit's energy, the qubit is left in the state corresponding to the measurement result.
- A. A superconducting qubit has no resistance, and its state is not read from a DC resistance.
- B. Correct. The shift of the resonator frequency is detected as a difference in microwave phase or amplitude.
- C. The energy of a superconducting qubit lies in the microwave range, and it does not emit visible light. Readout by counting light is used in trapped-ion and neutral-atom platforms.