Quantum Computing Hardware

A qubit in the state |0⟩ has a Hadamard (H) gate applied once and is then measured. What is the probability of getting 1?

  1. A0%
  2. B50%
  3. C100%
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Answer: B. 50%

The H gate turns |0⟩ into (|0⟩ + |1⟩)/√2. The amplitude of |1⟩ is 1/√2, so the probability of getting 1 is (1/√2)² = 1/2 = 50%. The H gate is often used at the start of quantum algorithms to create an equal superposition of all basis states. Applying the H gate twice in a row returns the qubit to |0⟩, and a measurement then always gives 0.

  • A. This is the value when the H gate is applied twice (returning to |0⟩).
  • B. Correct. (1/√2)² = 1/2.
  • C. This is the value for the X gate, which turns |0⟩ into |1⟩.
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More Quantum Computing Hardware questions

  1. Starting from the two-qubit state |00⟩ (the left digit is the first qubit), an H gate is applied to the first qubit, followed by a CNOT gate with the first qubit as control and the second as target. What is the resulting state?
  2. Gates H, Z and H are applied in that order to a qubit in |0⟩, which is then measured. What is the probability of getting 1?
  3. Why can't quantum error correction directly use the classical approach of making three copies of the data and taking a majority vote?
  4. Which correctly describes the energy relaxation time T1 of a qubit?
  5. Which correctly describes the relationship between the dephasing time T2 and the energy relaxation time T1 of a qubit?