QCoder

Editorial

By applying an $X$ gate to a qubit followed by a Hadamard ( $H$ ) gate, the zero state is transformed into the minus state:

\begin{align} \ket{0} &\xrightarrow{X(0)} \ket{1} \nonumber\\ &\xrightarrow{H(0)} \frac{1}{\sqrt{2}} (\ket{0} - \ket{1}) = \ket{-} \end{align}

Since the observation probability is given by the squared absolute value of its probability amplitudes, measuring the qubit results in observing the computational basis states $\ket{0}$ and $\ket{1}$ with equal probabilities $1/2$ each.

Such states in a probabilistic superposition are one of the unique properties of quantum computers that do not exist in classical computers.

Sample Code

Below is a sample code:

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

Supplements

There are many types of quantum gates besides the $X$ $X$ gate and the Hadamard gate that we use in this contest. Understanding the meaning and function of each quantum gate can be helpful in solving future problems:
- List of quantum logic gates (Wikipedia)

B: Generate Minus state

Editorial

Sample Code

Supplements