A2: Generate State $\frac{1}{\sqrt{2}}\lparen\ket{10}+\ket{01}\rparen$

Editorial

To begin with, we demonstrate the necessary gate operations in the following quantum circuit.

First, we apply a Hadamard ($H$) gate to the first qubit.

By applying the Hadamard gate in this way, we can bring the qubit into a superposition state.

Next, we need an operation that transforms $\ket{10}$ to $\ket{11}$. For this, we can use a controlled-$X$ (controlled-X; $CX$) gate with the first qubit as the control and the second qubit as the target. The controlled-$X$ gate applies an $X$ gate to the second qubit only when the first qubit is in state $\ket{1}$, acting on $\ket{10}$ in particular.

Finally, by applying an $X$ gate to the first qubit, we arrive at the solution.

Sample Code

Below is a sample program:

from qiskit import QuantumCircuit
 
 
def solve() -> QuantumCircuit:
    qc = QuantumCircuit(2)
 
    qc.h(0)
    qc.cx(0, 1)
    qc.x(0)
 
    return qc

Supplementary Information

• Measuring either qubit in the state $\frac{1}{\sqrt{2}} (\ket{10} + \ket{01})$ will determine the outcome of the other qubit. This phenomenon is known as quantum entanglement.
• The controlled gate can be generalized to the multi-controlled gate which operates upon multiple control qubits. This modifies computational basis states wherein all control qubits are in the state $\ket{1}$.