QCoder

Editorial

In problem A1, we prepared the "uniform superposition state" on $1$ qubit.

Now, how can we express the state of $n$ qubits when each of the $n$ qubits is in a uniform superposition state?

When measuring one qubit, we observe either of the computational basis states $\ket{0}$ or $\ket{1}$ with equal probability $1/2$ . Therefore, when measuring $n$ qubits in sequence, we can observe any one of the $2^n$ possible computational basis states — from $\ket{0...0}$ to $\ket{1...1}$ — with equal probabilities of $1/2^{n}$ .

Consequently, the state $\ket{A}$ of the uniform superposition state of all $2^n$ computational basis states can be expressed as

\begin{align} \ket{A} & = \frac{1}{\sqrt{2^n}} (\ket{0...0}_n + ... + \ket{1...1}_n) \nonumber\\ &= \frac{1}{\sqrt{2^n}} \sum_{i=0}^{2^n-1} \ket{i} \end{align}

As a result, we can prepare a uniform superposition state by applying a Hadamard gate to each of the $n$ qubits:

\begin{equation} \ket{0...0}_n \xrightarrow{H^{\otimes n}} \frac{1}{\sqrt{2^n}} (\ket{0...0}_n + ... + \ket{1...1}_n) \end{equation}

Sample Code

Below is a sample program:

from qiskit import QuantumCircuit
 
 
def solve(n: int) -> QuantumCircuit:
    qc = QuantumCircuit(n)
 
    qc.h(range(n))
 
    return qc

Alternatively, you can use a for-loop to apply the Hadamard gates in turn.

from qiskit import QuantumCircuit
 
 
def solve(n: int) -> QuantumCircuit:
    qc = QuantumCircuit(n)
 
    for i in range(n):
      qc.h(i)
 
    return qc

A2: Generate Uniform Superposition State

Editorial

Sample Code