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I'm trying to figure out what is the computational complexity of initializing a quantum register of N qubits.

For my research, I have used the initialize method of qiskit, in which you set the amplitudes of the 2^N basis states to get the desired quantum state. The point is that I can't find a reference that discusses the cost of such an operation. Moreover, if we consider a standard Grover's algorithm, why the initialization of a uniform superposition of basis states is not counted in the complexity?

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  • $\begingroup$ Beware that the initialize method probably simply initializes the simulator with the statevector you give it. I doubt it implements the requisite circuit at all, and you shouldn't rely upon it when developing an algorithm for use in quantum hardware. $\endgroup$
    – jecado
    Oct 8 at 17:47
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    $\begingroup$ @jecado the initialize method can be decomposed, although it is a non-unitary Instruction since it contains resets at the beginning, but these can be removed when sending the circuit to a real device since anyways the initial state of the hardware is the ground state. But yes, there are definitely more efficient methods depending on the exact algorithm they are developing. $\endgroup$
    – epelaaez
    Oct 9 at 7:23
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The running time of creating any fully separable pure initial state is $O(1)$. This includes the standard $|0\rangle^{\otimes n}$ or something like $|+\rangle^{\otimes n}$, which is the uniform superposition of all basis states. That's because this is just single-qubit preparation performed in parallel.

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For general initial states, i.e., not fully separable, the paper Synthesis of Quantum Logic Circuits gives the Quantum Shannon Decomposition (QSD) algorithm that has a gate count of $\frac{23}{48}4^n-\frac{3}{2}2^n+\frac{4}{3}$ where $n$ is the number of qubits (look at table 1 of the linked paper).

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