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 $|\psi\rangle$?

  • 1
    $\begingroup$ Do you want an analytical algorithm or do you want to use some software e.g. Qiskit? $\endgroup$
    – Mauricio
    Nov 24, 2021 at 17:30
  • $\begingroup$ @Mauricio if there is some documentation about the algorithm used by the software (e.g. qiskit libraries) for me is ok. If it's a black box it would not be really helpful since I would like to know if it is efficient with respect to the number of qubit or not. $\endgroup$
    – stopper
    Nov 24, 2021 at 18:10
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    $\begingroup$ The paper Synthesis of Quantum Logic Circuits may be what you are looking for $\endgroup$
    – epelaaez
    Nov 24, 2021 at 18:42

1 Answer 1


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.

  • $\begingroup$ Thank you, what's the state of the art for decomposition of unitaries? Are there algorithms that in some cases do It efficiently? $\endgroup$
    – stopper
    Nov 26, 2021 at 17:38
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    $\begingroup$ For arbitrary unitaries this is (AFAIK) the state of the art. It is a interesting question, what constrains should we apply to unitary matrix, so it is possible to decompose it effectively. $\endgroup$
    – totikom
    Nov 27, 2021 at 8:44

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