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When I build a quantum circuit and my initial state is the one composed only by zeros ($|000\ldots 0\rangle$), I have a final state $|\psi\rangle$ that is the result of the application of the quantum circuit to $|000\ldots 0\rangle$.

My question is: if I have the final state $|\psi\rangle$ (or something similar, like the probability related to each element of the computational basis) and I know that my initial state is $|000\ldots 0\rangle$, is there a way (exact, variational, etc.) to find one of the quantum circuit that applied to $|000\ldots 0\rangle$ give me $|\psi\rangle$?

Mark Spinelli
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stopper
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1 Answers1

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If you initial state is $| 0\dots0\rangle$ and your final state is $|\psi\rangle$, than it is trivial to come up with some unitary matrix $U$, which preforms such transformation (it's first column will be equal to the coeficients of $|\psi\rangle$ in computational basis).

Later $U$ can be decomposed in no more than $n$ unitary matrices $U'_i$, each of which acts non-trivially only on two basis vectors.

Each $U'_i$ can be decomposed in no more than $n$ matrices $U''_i$, each of which acts non-trivially only on two qubits. (see the Gray's encoding)

$U''_i$ is a tensor product of $n-2$ identity matrices and an arbitrary two-qubit gate.

It can be proven that an arbitrary two qubit gate can be constructed with CNOT and arbitrary one-qubit gates.

Any arbitrary one-qubit gate is rotation on the Bloch sphere. It can be approximated with only H, S and T gates.

Disclaimer: I should mention that it is not an optimal solution, for arbitrary $U$ this can lead to $O(2^n)$ number of gates.

totikom
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