A5: Generate state $\frac{1}{\sqrt{3}} (\ket{0} + \ket{1} + \ket{2}))$ II

Editorial

In problem A4, we prepared the superposition state $\frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2}(\ket{10} + \ket{01})$, composed of the three computational basis states $\ket{00}$, $\ket{10}$, and $\ket{01}$.

In this problem, we establish a uniform superposition of these states, changing their probability amplitudes to $1/\sqrt{3}$.

This can be achieved by replacing the Hadamard gate in Equation $(1)$ of editorial A4 with a rotation gate. If we can find a rotation gate $R$ which acts on the left qubit to transform the initial $\ket{00}$ state to $\frac{1}{\sqrt{3}} \ket{00} + \sqrt{\frac{2}{3}} \ket{10}$, then the subsequent controlled-Hadamard and controlled-$X$ gates would produce our desired uniform superposition:

We will now find the axis of rotation and the angle to perform such a transformation.

An arbitrary state $\ket{\psi}$ of a single qubit can be expressed by two angle parameters, $\theta$ and $\phi$, as follows:

The method of representing the quantum state of one qubit on the surface of a sphere with radius 1 is called the Bloch sphere, and by using phase gates and rotation gates based on the angular parameters $\theta$ and $\phi$, it is possible to transition any one-qubit state to any other arbitrary state.

Bloch sphere

Let us find the Bloch angular parameters $\theta$ and $\phi$ for the state $\frac{1}{\sqrt{3}} \ket{00} + \sqrt{\frac{2}{3}} \ket{10}$.

Since the imaginary component is $0$, we can set $\phi = 0$, and establish the following system of equations:

By solving the above equation within the range $0 \leq \theta < 2 \pi$, we obtain the following solution:

Looking at the Bloch sphere, we see that the angle parameter $\theta$ represents the rotation angle about the $y$-axis.

Thus, using the $R_y$ gate, which is a rotation gate about the $y$-axis, we can implement the rotation by $\theta_0$ to achieve the transition

As a result, the problem can be solved through the following transitions:

Sample Code

Below is a sample program:

from qiskit import QuantumCircuit
import math
 
def solve() -> QuantumCircuit:
    qc = QuantumCircuit(2)
 
    theta = 4 * math.atan(math.sqrt(6)/ (3 + math.sqrt(3)))
 
    qc.ry(theta, 0)
    qc.ch(0, 1)
    qc.cx(1, 0)
 
    return qc