# Phase Kickback and Controlled Rotation - Problem in proving symmetry

I am reading through the Phase Kickback chapter of the IBM online textbook about quantum computation. It is stated that, when applying any controlled Z-rotation, the concept of test and control qubit is lost. To show this phenomenon, the effect of a Controlled-T gate (Z-rotation of $$\pi/4$$) on the state $$|1+ \rangle$$ is demonstrated mathematically. The formal control qubit would be $$|+\rangle$$ while the target qubit would be $$|1\rangle$$. As I show in the image, the "target" qubit stays unchanged, while the "control" qubit has got rotated (it has now a relative phase).

I then wanted to prove myself that the same result would be delivered when applying the Controlled-T gate on the state $$|+1 \rangle$$, which physically would mean to swap the two qubits with respect to the former situation. I would then expect the same result as above, but with the two qubits "flipped" on opposite sides of the tensor product. In this case, however, I get a different result (a global phase). I think that I cannot commutate the tensor product, because this would mean physically swapping the two qubits, right? Can somebody explain to me the conceptual mistake that I am doing?

• In your penultimate line, why have you put $e^{i\pi/4}$ in the $|01\rangle$ term? Sep 21 '20 at 14:59
• Also - would you be able to transcribe your notes to mathjax? Lovely handwriting, we just tend to prefer LaTeX to images of math Sep 21 '20 at 15:11
• @DaftWullie because I thought that, for the controlled-T operation, the only condition needed to be fulfilled is that the "control" qubit should be in the $|1\rangle$ state, but not necessary also for the "target" qubit. The answer of Davit shows it correctly. Sep 22 '20 at 10:41
• Yes, exactly - if the control is in $|1\rangle$, apply the $T$ gate to the target. That's different from just applying a phase. Sep 22 '20 at 11:17

When we apply a phase gate, a relative phase is added (this is the definition that I will use in this answer for the phase gate). In the Qiskit's textbook (and in the textbook by M. Nielsen and I. Chuang) $$T$$ is defined as a phase gate $$P(\frac{\pi}{4})$$:

$$P |\psi \rangle = P (\alpha |0\rangle + \beta |1\rangle) = \alpha |0\rangle + e^{i\varphi}\beta |1\rangle \\ P(\varphi) = \begin{pmatrix} 1&0 \\ 0&e^{i \phi} \end{pmatrix} \qquad T = P(\frac{\pi}{4}) = \begin{pmatrix} 1&0 \\ 0&e^{i \frac{\pi}{4}} \end{pmatrix}$$

where $$P$$ is the phase gate, $$\alpha$$ and $$\beta$$ are some initial amplitudes, $$\varphi$$ is the phase defined by the $$P$$ gate. Note that only $$|1\rangle$$ in the superposition state has obtained the phase. The same works for the controlled phase gate: only $$|11\rangle$$ obtains a phase because the control qubit should be $$|1\rangle$$ and the target qubit also should be $$|1\rangle$$:

$$CP_{2 \rightarrow 1} |+1 \rangle = CP_{2 \rightarrow 1} \frac{1}{\sqrt{2}} (|01\rangle + |11\rangle) = \\ = \frac{1}{\sqrt{2}} (|01\rangle + e^{i \varphi}|11\rangle) = \frac{1}{\sqrt{2}} (|0\rangle + e^{i \varphi}|1\rangle) \otimes |1\rangle$$

where $$CP$$ is the controlled phase gate, $$2 \rightarrow 1$$ subscript denotes that the $$CP$$ gate is controlled by the second qubit. More general proof for $$CP_{1 \rightarrow 2} = CP_{2 \rightarrow 1}$$ can be derived by using matrix representation of the $$CP$$ gate. This proof is similar to the proof for $$CZ_{1 \rightarrow 2} = CZ_{2 \rightarrow 1}$$ that can be found in this answer.

$$CP_{1 \rightarrow 2} = |0\rangle \langle 0| \otimes I + |1\rangle \langle 1| \otimes P = \\ = \begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&e^{i \varphi} \\ \end{pmatrix} = \\ =I \otimes |0\rangle \langle 0| + P \otimes |1\rangle \langle 1| = CP_{2 \rightarrow 1}$$

Side note about why the "symmetry" is not true for controlled $$R_z$$ gate in contrast to controlled $$P$$ gate:

If for the general case instead of $$P(\varphi)$$ we will use $$R_z(\varphi)$$ gate then we will have a different result:

$$CRZ_{1 \rightarrow 2} = |0\rangle \langle 0| \otimes I + |1\rangle \langle 1| \otimes R_z = \begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&e^{-i \frac{\varphi}{2}}&0 \\ 0&0&0&e^{i \frac{\varphi}{2}} \\ \end{pmatrix}\\ CRZ_{2 \rightarrow 1} = I \otimes |0\rangle \langle 0| + R_z \otimes |1\rangle \langle 1| = \begin{pmatrix} 1&0&0&0 \\ 0&e^{-i \frac{\varphi}{2}}&0&0 \\ 0&0&1&0 \\ 0&0&0&e^{i \frac{\varphi}{2}} \\ \end{pmatrix}$$

where $$R_z(\varphi) = \begin{pmatrix} e^{-i \frac{\varphi}{2}}&0 \\ 0&e^{i \frac{\varphi}{2}} \end{pmatrix}$$. So $$CRZ_{1 \rightarrow 2} \ne CRZ_{2 \rightarrow 1}$$. This answer also might be relevant where the difference between the controlled versions of $$R_z$$ and $$U1 = P$$ is discussed.