I was reading about Flexible Representation of Quantum Images (FRQI) encoding in Qiskit textbook. It says that, given $$\{\theta_0, \theta_1, ..., \theta_{4^{n}-1}\} \quad (\theta_i \in [0,\pi/2])$$ we can transform $\mid{0}\rangle^{\otimes2n+1}$ into FRQI state $\mid{I(\theta)}\rangle$ by applying unitary operator $\mathcal{R}\mathcal{H}$, with $\mathcal{H}= I \otimes H^{\otimes 2n}$ and $\mathcal{R}= \prod_{i=0}^{4^n-1}R_i$. where $$\mid{I(\theta)}\rangle= \frac{1}{2^n}\sum_{i=0}^{4^n-1}(cos \theta_i|{0}\rangle+sin \theta_i|{1}\rangle)\otimes |{i}\rangle$$

And $R_i$ is a unitary matrix named controlled rotation, given by $$R_i= I \otimes \sum_{j=0, j \neq i}^{4^n-1} |{j}\rangle \langle {j}| + R_y(2\theta_i) \otimes |{i}\rangle \langle {i}\mid$$ I would like to implement the controlled rotation $R_i$ or $\mathcal{R}$ using $C^{2n}R_y$ (controlled $R_y$) and $X$ gate.

Edit: I think I have made some progress. We can write, $$ R_i= I^{\otimes 2n+1}+(R_y(2\theta_i)-I) \otimes |{i}\rangle \langle {i}\mid$$ $$ C^{2n}R_y(2\theta_i)= I^{\otimes 2n+1}+(R_y(2\theta_i)-I) \otimes |{4^n-1}\rangle \langle {4^n-1}\mid $$ Please review my results and suggest me how to approach further.


1 Answer 1


If you already have controlled$^{2n}-R_Y$, this is fairly straightforward. This means that what you already have is $$ C^{2n} R_y(2\theta_i)=I^{\otimes(2n+1)}+(R_Y(2\theta_i)-I)\otimes |1\rangle\langle 1|^{\otimes 2n} $$ and you are aiming to make $$ R_i=I^{\otimes(2n+1)}+(R_Y(2\theta_i)-I)\otimes |i\rangle\langle i|. $$ Now, let $X_x$ be a tensor product of Pauli $X$ matrices that have the action $$ X_x|i\rangle=|1\rangle^{\otimes 2n}\rangle. $$ In other words, if $y$ is the binary representation of $i$, then $$ x=y\oplus 11\ldots 1=\bar y. $$ Then \begin{align*} X_x C^{2n} R_y(2\theta_i)X_x&=X_x(I^{\otimes(2n+1)}+(R_Y(2\theta_i)-I)\otimes |1\rangle\langle 1|^{\otimes 2n})X_x\\ &=I^{\otimes(2n+1)}+(R_Y(2\theta_i)-I)\otimes |i\rangle\langle i|\\ &=R_i. \end{align*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.