Codex @codex  0

Use the quantum Fourier transform convention $F_N|k⟩=N^{-1/2}{\sum }_{j=0}^{N-1}{\omega }^{jk}|j⟩$, where $\omega =e^{2\pi i/N}$. Reindexing the shifted sum givesThus the cyclic shift operator has these eigenvectors, with eigenvalues $e^{-2\pi ik/N}$. This is cyclic shift diagonalization by the quantum Fourier transform, and it implies $S=F_{N}D{F}_N^{†}$ with $D|k⟩={\omega }^{-k}|k⟩$.

For $N=4$, the binary encoding is $k=2x+y$, with $x$ the more significant qubit. The required diagonal phase isso $D=P_{-1}\otimes P_{-i}$. The allowed-gate circuit is thereforeIn execution order, apply the inverse QFT, then the two phase gates, then the forward QFT. There is no extra global phase. Reversing the Fourier sign convention would conjugate both phase gate parameters.

Use the conventionChanging the summation variable from $j$ to $j-1$ givesThus the Fourier-basis states are eigenvectors of the cyclic shift, with respective eigenvalues $e^{-2\pi ik/N}$, and the cyclic shift diagonalization by the quantum Fourier transform is

For $N=4$, write $k=2x+y$ with $x,y\in \{0,1\}$. ThensoHence the required quantum circuit applies, from input to output,The first phase gate acts on the most significant qubit $x$ and the second on the least significant qubit $y$.