# The control phase gate in Quantum fourier transform and the question it brings up regarding control gates in general

I have a question that's arose from reading "Quantum computing explained" by David McMahon. On page 212 there's an aspect of his description of the quantum Fourier transform which I don't understand .

He writes that after the Hadamard gate has been applied to the zero'th qubit and then we apply the control phase gate where the zero'th qubit is the target and the first is the control that the state becomes :

$$I \otimes R_2|x_1>\otimes \tfrac{1}{\sqrt{2}}(|0\rangle +|1\rangle) = |x_1\rangle \otimes \tfrac{1}{\sqrt{2}}(|0\rangle +e^{2\pi i (0.j_0)}|1\rangle)$$.

I have two questions regarding this :

1) The $$R_2$$ gate is a control gate which changes the phase of the second qubit if the if the first is 1 and does nothing if it's zero. However here he rights that the $$|0\rangle$$ qubit is never affected by the control gate regardless of the state of the control qubit ? How can this be ?

2) I was under the impression that all control gates were of the form $$P_0\otimes I+ P_1 \otimes \hat{A}$$ ( where $$\hat{A}$$ is an observable ) if the first qubit was the control and the second was the target and they were of the form $$I\otimes P_0+ \hat{A} \otimes P_1$$ , if the second was the control and the first was the target , yet here he writes $$I \otimes R_2$$ Which I thought simply meant do nothing to the first but act on the second without their being any controls involved ? What's going on here ?

• Quick remark about your point (2): $\hat A$ should be unitary, not necessarily an observable (which would be Hermitian). That is, $\hat A$ should be not just "a linear transformation" on the mathematical level, but more specifically a way that the state of a system could transform. – Niel de Beaudrap Aug 15 '19 at 8:04

You are correct that a controlled-gate should be written in the form $$P_0\otimes I+P_1\otimes U$$ for a unitary $$U$$. It certainly should not be of the form $$I\otimes U$$. I don't have a copy of your textbook to hand, but could you perhaps have misread it? Perhaps the author meant $$I\oplus U$$, which is an alternative way of describing the matrix structure?
It is, however, true that if either of the qubits is in the $$|0\rangle$$ state, controlled-$$R_2$$ does nothing. For example, if the target qubit is in $$|0\rangle$$, then either (i) the control is in $$|0\rangle$$, and so $$I$$ was applied to the target. This leaves $$|0\rangle$$ as $$|0\rangle$$, or (ii) the control is in $$|1\rangle$$, so $$R_2$$ is applied to the target. But $$R_2|0\rangle=|0\rangle$$ so, again, nothing happens.