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Uniform Quantum Superposition States
Uniform quantum superposition states are a fundamental concept in quantum mechanics, representing a state where a quantum system exists in a linear combination of multiple basis states, with each basis state contributing equally to the overall superposition.

Definition
In the context of an $$ n $$-qubit system, a uniform quantum superposition state is defined as:

Here, $$ |j\rangle $$ represents the computational basis states of the $$ n $$-qubit system, and  $$ N $$ is the total number of distinct states in the superposition. The normalization factor $$ \frac{1}{\sqrt{N}}$$ ensures that the total probability of finding the system in one of the basis states is equal to 1.

Importance in Quantum Computation
Uniform superposition states play a crucial role in quantum computation algorithms. They are often utilized as initial states or intermediate states during quantum computations. The ability to efficiently prepare uniform superposition states is essential for the implementation of various quantum algorithms (e.g., Grover's algorithm, Quantum Fourier Transform), as it impacts the overall efficiency and success of quantum computations.

Preparation of uniform quantum superposition states when $$N=2^n$$
For an $$n$$-qubit system, Hadamard gates acting on each of the $$n$$ qubits (each initialized to the $$|0\rangle$$) can be used to prepare uniform quantum superposition states when $$N$$ is of the form $$N = 2^n$$. In this case case with $$n$$ qubits, the combined Hadamard gate $$H_n$$ is expressed as the tensor product of $$n$$ Hadamard gates: $$H_n = \underbrace{H \otimes H \otimes \ldots \otimes H}_{n \text{ times}}$$

The resulting uniform quantum superposition state is then: $$ H_{n} |0\rangle^{\otimes n} = \frac{1}{\sqrt{2^n}} \sum_{j=0}^{2^n-1} |j\rangle$$ This generalizes the preparation of uniform quantum states using Hadamard gates for any $$N = 2^n$$ .

Measurement of this uniform quantum state results in a random random state between $$|0\rangle$$ and $

Example 1: $$N = 2 $$
For a system with $$n = 1$$ qubit, the Hadamard gate is applied to the single qubit:

$$H_1 = H$$

Applying $$H_1$$ to $$|0\rangle$$ yields the uniform quantum superposition state: $$\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$

Example 2: $$N = 4 $$
For a system with $$n = 2$$ qubits, the combined Hadamard gate $$H_2$$ is the tensor product of two Hadamard gates:

$$H_2 = H \otimes H$$

Mathematically, this is expressed as:

$$H_2 = \frac{1}{2} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix}$$ Applying $$H_2$$ to $$|00\rangle$$, yields the superposition states with equal weights.

Preparation of uniform quantum superposition states in the general case, $$N $$ ≠ $$2^n$$
An efficient and deterministic approach for preparing the superposition state

with a gate complexity and circuit depth of only $$ O(\log_2 N)$$ for all $$N$$ was presented in. This approach requires only $$ n = \lceil \log_2 N \rceil$$ qubits. Importantly, neither ancilla qubits nor any quantum gates with multiple controls are needed in this approach for creating the uniform superposition state $$ |\Psi\rangle $$.