Phase Kickback and Controlled Operations

Question

I am reading the chapter about phase kickback from "An Introduction to Quantum Computing" by Kaye, Laflamme, Mosca. I understand why $U_f :|x \rangle |-\rangle \rightarrow (-1)^{f(x)} |x\rangle |-\rangle$ is correct. But then it is written that $U_f$ can be thought as a 1-qubit operator $\hat{U}_{f(x)}$ acting on the second qubit controlled by the state $|x\rangle$. I don't get why this is true and why the circuits below are equivalent.

For instance if $f(0)=1$, then $U_f:|0\rangle|-\rangle \rightarrow - |0\rangle |-\rangle$, but this is not the case for the circuit given on the right since if $|x\rangle=0$, no operation is applied on the second register.

Can someone show me what I am missing?

Answer 1

Without a bit more context of how they use the notation in the rest of the book, I'm not certain, but the way I would interpret that is saying "if the control qubit is 0, apply unitary $\hat U_{f(0)}$ on the target. if the control qubit is 1, apply $\hat U_{f(1)}$ on the target".

If this is to be the case, let's see what the action is supposed to be. As you say, $$ |x\rangle|-\rangle\rightarrow (-1)^{f(x)}|x\rangle|-\rangle. $$ You also have to check the action on the other basis state, $$ |x\rangle|+\rangle\rightarrow|x\rangle|+\rangle. $$ So, to summarise, if $x$ is 0, the unitary on the second qubit is $$ |+\rangle\langle +|+(-1)^{f(0)}|-\rangle\langle -|. $$ If $x$ is 1, the unitary on the second qubit is $$ |+\rangle\langle +|+(-1)^{f(1)}|-\rangle\langle -|. $$ So, if we defined $U_y$ as $$ \hat U_y=|+\rangle\langle +|+(-1)^{y}|-\rangle\langle -|, $$ then this would be consistent with my statement.