# Phase kickback without controlled operator

I have some doubts about the phase kickback trick when used in some algorithm.

Consider a quantum system with two qubits, and suppose we have a boolean oracle $$U_f$$. We know that if we put the second qubit in the state $$|-\rangle$$, the boolean oracle acts as a phase oracle because of the phase kickback trick. However, it is also known that the phase kickback requires:

1. the target qubit must be an eigenstate of the operator used in the circuit;
2. the operator must be a controlled one, (e.g., CNOT);
3. the first qubit (control) must be in superposition.

Now, I understand how the phase kickback works and the reason why. Let's move onto the next step: consider the Deutsch algorithm. The circuit for a generic function implemented by an oracle is the following:

However, the first $$H$$ gate in the second qubit change it from $$|1\rangle$$ to $$|-\rangle$$; this operation triggers the phase kickback, so the boolean oracle $$U_f$$ acts as a phase oracle instead; we further measure the first qubit after re-applying an $$H$$ gate in order to find if the function is constant or balanced.

So, basically, the second qubit acts as an "ancilla" qubit and we need it just for the phase kickback operation.

My question is the following: if we want to evaluat the function $$f(x) = 1 \;\;\forall x \in \{0,1\}$$, the oracle is simply an $$X$$ gate in the second qubit, between the steps $$|a_1\rangle$$ and $$|a_2\rangle$$; how exactly is the phase kickback involved, if the operator (the oracle) is not controlled by the first qubit?

• Point 3 in your list is not required to get a phase kick-back. (Of course, it is not useful in that case, but that's a different story.) Feb 20 at 19:19

In the case where $$f(0)=f(1)=1$$, $$U_f$$ is just the controlled-Identity gate. This is still a controlled gate, though a trivial one. And, as one should expect, the phase we get is the eigenvalue of $$\lvert-\rangle$$ in the identity operator.
Note that "controlled gate" should be read in a generalized way: It just means that there are gates $$G_0$$ and $$G_1$$ which are applied if the control is zero or one, respectively. (E.g., for $$f(0)=f(1)=1$$, $$G_0=G_1=X$$.) Or, more generally, for a function with $$n$$ input bits, $$G_0,\dots G_{2^n-1}$$.
• @aghin00 The best way to analyze the system is to not put the control qubit(s) in superposition, but in a classical state |x>. (Then you can use linearity.) With this, it is easy to see that given the control is |x>, you apply $G_x$ to the target in the |-> state (or whatever state you choose), and thus, you get the eigenvalue corresponding to the eigenvector $|x>$ of $G_x$ as a phase. --- Note that in the Deutsch algorithm, the four functions correspond to Identity, CNOT, X only on the target (no control, if you wish), and CNOT + X. Feb 21 at 12:21
• Actually, I think I found the cause of my misunderstanding. The thing is that I was looking at the phase kickback example itself vs. the relationship between the boolean oracle and the phase oracle in Deutsch algorithm; the former comes almost always with examples of CNOT gate, in order to explain the controlling gate; in the latter, actually the operation is "controlled" by the function f(x) which is based on the state of the qubit $|x\rangle$, thus acting as a control for our gate even if it is not a CNOT (which explicitly has the control shown). Feb 21 at 15:40
• In your second comment, though, the eigenvalue that we get, shouldn't it be corresponding to the eigenvector $|-\rangle$ of the gate? I mean, in order for the phase kickback to work, we need to have the state of the qubit where the gate operator is applied as an eigenvector of the gate, not the control qubit's state. Feb 21 at 15:41