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EDIT: In the following I am using the Feynman notation for controlled operations - e.g. $\land_{ab}(X)$ is equivalent to a $CNOT$ with control qubit $q_a$ and target $q_b$. Ultimately, for any single-qubit unitary $U$ applied to some qubit $q_a$, it has the compact notation $U_a$.

Consider a circuit scenario $\land_{ab}(X)U_b$. Is there any generalized "push" rule such that $\land_{ab}(X)U_b\equiv (U'_a\otimes U''_b)\land_{ab}(U''')$?

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    $\begingroup$ It would probably be helpful to give more details on the notation you used: what are $a$ and $b$, the $\land$ notation might not be understood by everyone, and what are $U_a'$, $U_b''$ and $U'''$? $\endgroup$ Jul 6 at 9:33
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Certainly, it is sufficient to look at a two-qubit system (the relevant subsystem given by $a$ and $b$).

Any controlled unitary $\land_{12}(V)$ is of the form $$ \land_{12}(V) = |0\rangle\langle 0|\otimes\mathbb{I} + |1\rangle\langle 1|\otimes V \quad ( = \mathbb{I}\oplus V). $$ Hence, $$ \land_{12}(V) (\mathbb{I}\otimes U) = (|0\rangle\langle 0|)\otimes\mathbb{I}) (\mathbb{I}\otimes U) + (|1\rangle\langle 1|\otimes V)(\mathbb{I}\otimes U) = (\mathbb{I}\otimes U) \land_{12}(U^\dagger V U). $$

Does this answer your question?

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  • $\begingroup$ Thanks a lot! Yes it does. $\endgroup$ Jul 6 at 10:11

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