Is a ccH, ccX and ccH equivalent to a cH, ccX and cH sequence?

Question

It appears to me that a ccH, ccX and ccH sequence is exactly equivalent to a cH, ccX and cH gate sequence. Is there any quick way to see/verify this?

Answer 1

Start

No control equals each control $\forall U : U = C(U) \cdot \bar{C}(U)$

Opposite controls commute $\forall U, V : [C(U), \bar{C}(V)] = 0$

No control equals each control $\forall U : U = C(U) \cdot \bar{C}(U)$

Self-inverse operations self-cancel

Done

More generally, for any "V conjugated by U" operation of the form $U_a \cdot C_b(V_a) \cdot U_a^{-1}$, the $U$ operation can gain or lose any controls that $V$ has without changing the circuit's effect.

Answer 2

There is a way to simplify it down slightly - for some controlled unitary $C\left(U\right)=I\oplus U$ and some arbitrary unitary $V$ (of the same dimension as $U$), $$\left(I\otimes V\right).C\left(U\right).\left(I\otimes V^\dagger\right) = \begin{pmatrix}V&0\\0&V\end{pmatrix}\begin{pmatrix}I&0\\0&U\end{pmatrix}\begin{pmatrix}V^\dagger&0\\0&V^\dagger\end{pmatrix} = C\left(VUV^\dagger\right),$$ so it's now clear that $\left[I\otimes C\left(H\right)\right].C\left(C\left(X\right)\right).\left[I\otimes C\left(H\right)\right] = C\left(C\left(H\right).C\left(X\right).C\left(H\right)\right)$.

Funnily enough, if instead of using $I\otimes V$ in the above and instead used $C\left(V\right)$, we'd find that $$C\left(V\right).C\left(U\right).C\left(V^\dagger\right) = \begin{pmatrix}I&0\\0&V\end{pmatrix}\begin{pmatrix}I&0\\0&U\end{pmatrix}\begin{pmatrix}I&0\\0&V^\dagger\end{pmatrix} = C\left(VUV^\dagger\right)$$ and so it's not just clear that \begin{align*}C\left(C\left(H\right)\right).C\left(C\left(X\right)\right).C\left(C\left(H\right)\right) &= C\left(C\left(H\right).C\left(X\right).C\left(H\right)\right) \\ &=\left[I\otimes C\left(H\right)\right].C\left(C\left(X\right)\right).\left[I\otimes C\left(H\right)\right]\end{align*} but the more general version $$\left(I\otimes V\right).C\left(U\right).\left(I\otimes V^\dagger\right) = C\left(V\right).C\left(U\right).C\left(V^\dagger\right)$$ is also true, for any unitaries $U$ and $V$

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^{My use of brackets in the control operation is maybe slightly more unusual, but hopefully it makes it more obvious what the operation is}