Question

Which specific gate must be applied to a qubit in the state |0⟩ to flip its phase without changing its computational basis probability amplitudes?

Accepted Answer

The specific gate required to flip the phase of a qubit without changing its computational basis probability amplitudes is the Z gate, also known as the Pauli-Z gate. A qubit in the state |0⟩ is represented as the vector [1, 0], meaning it has a probability amplitude of 1 for the |0⟩ state and 0 for the |1⟩ state. The Z gate is represented by the matrix [[1, 0], [0, -1]]. When applied to a state vector, the Z gate leaves the amplitude of the |0⟩ state unchanged because it multiplies it by 1, but it multiplies the amplitude of the |1⟩ state by -1. Because the amplitude of the |1⟩ state in |0⟩ is 0, multiplying it by -1 results in 0, leaving the probability amplitudes for measuring |0⟩ or |1⟩ identical to the original state. However, the phase of the state is effectively flipped if there is any component in the |1⟩ position. In the specific case of |0⟩, the Z gate acts as an identity operator because it leaves the state |0⟩ unchanged, as the phase flip only manifests on the |1⟩ component of a superposition. If the goal is to flip the phase of a state that is in a superposition, such as the plus state |+⟩ which equals (1/√2)|0⟩ + (1/√2)|1⟩, the Z gate transforms it into the minus state |-⟩, which is (1/√2)|0⟩ - (1/√2)|1⟩. The probability amplitudes, defined as the square of the absolute value of the complex coefficients, remain (1/√2)² = 0.5 for both |0⟩ and |1⟩, confirming that the measurement probabilities are unchanged while the phase relationship between the basis states is inverted.

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Which specific gate must be applied to a qubit in the state |0⟩ to flip its phase without changing its computational basis probability amplitudes?

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