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Let's say that I know the decomposition of a unitary operator $\hat{A}$ in terms of other unitary operators $U_{k=0, \dots, M}$, i.e:

$$ \hat{A} = \sum_k \alpha_k U_k$$

I know how to implement in circuit form $U_k$ but not $\hat{A}$. Is there a way to obtain a decomposition:

$$ \hat{A} = \sum_k \alpha_k U_k = B_1 B_2 \dots B_M$$

where $B_{i=1,\dots, M}$ are unitary operators and are obtained from $\alpha_k$ and $U_k$ ? I'd like the $B_i$ to use the same number of qubits as $U_k$, so no ancillary register (no LCU method).

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    $\begingroup$ Maybe this question is too broad. I mean the clearly the optimal answer is to write $\sum \alpha_kU_k = AI\cdots I$ or $\sqrt[M]{A}$ (multiplied by itself $M$ times), maybe you should restrict the $B_k$s a little bit more. $\endgroup$
    – Mauricio
    Sep 9, 2022 at 17:27
  • $\begingroup$ Mmm ok. Being more precise. Let's say I know to implement in a circuit the unitary gates $U_k$. I do not know how to implement the operator $\hat{A}$ in the circuit. There is a way that, from knowing $\alpha_k$ and $U_k$ I can implement the operator $\hat{A}$ on the circuit? PS: Thanks for your answer. I'd change the word 'optimal' to 'trivial'. I was not precise enough, but I didn't mention these solutions since they were trivial. $\endgroup$
    – Andrés Ruiz
    Sep 12, 2022 at 12:33
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    $\begingroup$ Maybe you should indicate that in your post (you can edit it). If I understand correctly you want $B_{i}$ just in terms of $U_k$ and $\alpha_k$ (unless you allow for a series of other gates then you should indicate all of them). $\endgroup$
    – Mauricio
    Sep 12, 2022 at 12:45
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    $\begingroup$ I think something is still missing. Imagine that you want to write $H=(X+Z)/\sqrt{2}$. Clearly it is not possible to write $H$ as just products of $X$ and $Z$. So if there is a solution to your answer it is very restricted to some particular cases. $\endgroup$
    – Mauricio
    Sep 12, 2022 at 13:45
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    $\begingroup$ Here is a proof of no-go for single qubits: every unitary matrix $U$ can be written as a linear combination of Pauli matrices. Pauli matrices form a group. Products of Pauli matrices stay in the Pauli group. So you are only able to decompose $U$ if $U$ is in the Pauli group (which is incredibly restrictive). $\endgroup$
    – Mauricio
    Sep 12, 2022 at 14:35

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