What is the final state of the second qubit after applying the Hadamard gate and the CNOT gate to the initial state |0⟩|1⟩?
The final state of the second qubit after applying the Hadamard gate and the CNOT gate to the initial state |0⟩|1⟩ can be determined by applying the gates sequentially and calculating the resulting state vector.
Let's start with the initial state |0⟩|1⟩. The first qubit is in the state |0⟩ and the second qubit is in the state |1⟩. The Hadamard gate (H) acts on the first qubit, transforming it into the superposition state (|0⟩ + |1⟩)/√2. The state of the second qubit remains unchanged at this point.
Next, we apply the CNOT gate, which is a controlled-X gate. It flips the second qubit (target qubit) if and only if the first qubit (control qubit) is in the state |1⟩. In our case, the control qubit is in the state (|0⟩ + |1⟩)/√2 and the target qubit is in the state |1⟩.
To determine the final state, we need to consider all possible combinations of the control and target qubit states and apply the gate accordingly. Let's denote the control qubit as C and the target qubit as T. We have four possible combinations: |C⟩|T⟩ = |0⟩|0⟩, |0⟩|1⟩, |1⟩|0⟩, and |1⟩|1⟩.
For the combination |0⟩|0⟩, the CNOT gate does not flip the target qubit since the control qubit is in the state |0⟩. The resulting state is still |0⟩|0⟩.
For the combination |0⟩|1⟩, the CNOT gate also does not flip the target qubit since the control qubit is still in the state |0⟩. The resulting state remains |0⟩|1⟩.
For the combination |1⟩|0⟩, the CNOT gate flips the target qubit since the control qubit is in the state |1⟩. The resulting state becomes |1⟩|1⟩.
For the combination |1⟩|1⟩, the CNOT gate flips the target qubit as the control qubit is in the state |1⟩. The resulting state is |1⟩|0⟩.
Combining all these results, we obtain the final state of the two-qubit system after applying the Hadamard gate and the CNOT gate as follows:
(1/√2) * |0⟩ * |0⟩ + (1/√2) * |0⟩ * |1⟩ + (1/√2) * |1⟩ * |1⟩ + (1/√2) * |1⟩ * |0⟩
Simplifying this expression, we get:
(1/√2) * (|0⟩ * |0⟩ + |0⟩ * |1⟩ + |1⟩ * |1⟩ + |1⟩ * |0⟩)
This is the final state of the second qubit after applying the Hadamard gate and the CNOT gate to the initial state |0⟩|1⟩.
Other recent questions and answers regarding Examination review:
- What are the four Bell basis states and why are they important in quantum information processing and quantum teleportation?
- What is the final state of the first qubit after applying the Hadamard gate and the CNOT gate to the initial state |0⟩|0⟩?