Question

Implement the CNOT gate on a pair of qubits in the Bell state 1/√2(|100) + |11)). What is the resulting state?

Implement the SWAP gate on a pair of qubits in the state |01). What is the resulting state?

Answer 1

To implement the CNOT gate on the given pair of qubits in the Bell state 1/√2(|100) + |11)), apply the X gate on the second qubit. The resulting state would be 1/√2(|100) - |11)). To apply the SWAP gate on the state |01), the resulting state would be |10).

Step-by-step explanation:

The CNOT (Controlled-NOT) gate is a two-qubit gate in quantum computing that applies a NOT gate (X gate) on the second qubit if and only if the first qubit is in the state |1>. To implement the CNOT gate on the given pair of qubits in the Bell state 1/√2(|100) + |11)), we can apply the X gate on the second qubit. The resulting state would be 1/√2(|100) - |11)).

The SWAP gate, as the name suggests, swaps the states of two qubits. In the given state |01), applying the SWAP gate would result in the state |10).