# Generalized push for $\land_{ab}(X)$ gate

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 $$\land_{ab}(X)U_b\equiv (U'_a\otimes U''_b)\land_{ab}(U''')$$?

• 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 $U'''$? Jul 6, 2021 at 9:33

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, $$\land_{12}(V) (\mathbb{I}\otimes U) = (|0\rangle\langle 0|)\otimes\mathbb{I}) (\mathbb{I}\otimes U) + (|1\rangle\langle 1|\otimes V)(\mathbb{I}\otimes U) = (\mathbb{I}\otimes U) \land_{12}(U^\dagger V U).$$