Question

Implement the Hadamard gate on a qubit in the state ∣1⟩. What is the resulting state? Implement the T gate on a qubit in the state ∣+⟩. What is the resulting state?

Answer 1

The Hadamard gate applied to |1\rangle results in the state (|0\rangle - |1\rangle)/\sqrt{2}, and the T gate applied to |+\rangle results in the state (|0\rangle + e^{i\pi/4}|1\rangle)/\sqrt{2}.

Step-by-step explanation:

When implementing the Hadamard gate on a qubit in the state |1\rangle, the resulting state is a superposition of the |0\rangle and |1\rangle states, with equal probability amplitude but opposite phases. The resulting state is given by (|0\rangle - |1\rangle)/\sqrt{2}. On the other hand, when implementing the T gate (which is a pi/8 phase shift) on a qubit in the state |+\rangle, which is already a superposition of the |0\rangle and |1\rangle states (|0\rangle + |1\rangle)/\sqrt{2}, the |1\rangle component of the state acquires an additional phase of e^{i\pi/4}, resulting in the state (|0\rangle + e^{i\pi/4}|1\rangle)/\sqrt{2}.