Assume one is given two oracle circuits providing access to matrices $A_{ij}$ and $B_{ij}$ as follows (see eq. (6.2) here): \begin{equation} O_A |0\rangle|i\rangle|j\rangle=\left(A_{ij}|0\rangle+\sqrt{1-|A_{ij}|^2}|1\rangle\right)|i\rangle|j\rangle \, ,\\ O_B |0\rangle|i\rangle|j\rangle=\left(B_{ij}|0\rangle+\sqrt{1-|B_{ij}|^2}|1\rangle\right)|i\rangle|j\rangle \, , \end{equation} where indices $i$ and $j$ are encoded with qubits in a the binary form. The said representation will be referred to as block encoding.

I am wondering if one could construct a circuit $O_{A+B}$ implementing \begin{equation} O_{A+B} |0\rangle|i\rangle|j\rangle=\Bigl((A_{ij}+B_{ij})|0\rangle+\ldots|1\rangle\Bigr)|i\rangle|j\rangle \, . \end{equation}

I placed $\ldots$ instead of $\sqrt{1-|A_{ij}+B_{ij}|^2}$ just to indicate that using ancillas and measurements would be OK.


For completeness I add a straightforward implementation based on the conversion to query oracles.

  1. Use the controlled rotation circuit (see Proposition 4.7 here) enter image description here to convert the block-encoded oracles to query-encoded oracles $\widetilde{O}_A$ and $\widetilde{O}_B$: $$ \widetilde{O}_A |0\rangle|i\rangle|j\rangle = |\widetilde{A}_{ij}\rangle|i\rangle|j\rangle \, ,\\ \widetilde{O}_B |0\rangle|i\rangle|j\rangle = |\widetilde{B}_{ij}\rangle|i\rangle|j\rangle \, . $$

  2. The consecutive application of oracles renders then $$ \widetilde{O}_A \widetilde{O}_B |0\rangle|i\rangle|j\rangle = \widetilde{O}_A |\widetilde{B}_{ij}\rangle|i\rangle|j\rangle = |\widetilde{A}_{ij} + \widetilde{B}_{ij}\rangle|i\rangle|j\rangle \, . $$

Would be nice to avoid converting to query oracles.

  • $\begingroup$ Given that $A_{ij}+B_{ij}$ is not necessarily in the unit ball anymore, what other restrictions are you envisioning for $A$ and $B$? Should they be normalized? And since generally $A$ and $B$ would be known classically, why not add them classically and then query the result? $\endgroup$
    – Sam Jaques
    Feb 14 at 9:45

1 Answer 1


You can do this using linear combination of unitaries (LCU) technique. It is described in the paper you mentioned (section 7.3). More specifically, see example 7.12

  • $\begingroup$ Seems like an overkill. I added some edits to the question. $\endgroup$
    – mavzolej
    May 19, 2022 at 1:57

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