A2: Generate State $\frac{1}{\sqrt{2}}\lparen\ket{0}-\ket{3}\rparen$

Editorial

In problem A1, we addressed the inversion of the global phase.

Now, how can we prepare a state with a different phase between two computational basis states?

First, prepare $\frac{1}{\sqrt{2}} \lparen \ket{00} + \ket{11} \rparen$ without considering the phase. Apply the Hadamard ($H$) gate to the $1$st qubit.

By applying the Hadamard gate, we can create a superposition state of qubits.

Next, we need to transform $\ket{10}$ to $\ket{11}$. To do this, we use the controlled $X$ ($CX$) gate, with the $1$st qubit as the control qubit and the $2$nd qubit as the target qubit. This controlled $X$ gate applies the $X$ gate to the $2$nd qubit only in $\ket{10}$, where the $1$st qubit is $\ket{1}$.

Finally, apply a $Z$ gate to the $1$st qubit of $\frac{1}{\sqrt{2}} \lparen \ket{00} + \ket{11} \rparen$ to transform $\ket{11}$ into $-\ket{11}$.

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.z(0)
 
    return qc

Supplementary Information

• Measuring either qubit in the state $\frac{1}{\sqrt{2}} (\ket{00} - \ket{11})$ 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 $\ket{1}$.