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Suppose there are two arbitrary $n$-dimensional input states $|x_1\rangle$ and $|x_2\rangle$. Let $R_{x_1} = 2|x_1\rangle\langle x_1|-1$, which is a unitary reflection operator, with $1$ being identity operator. Now I want to design a quantum circuit that takes $|x_1\rangle$ and $|x_2\rangle$ and outputs $|x_1\rangle \left(R_{x_1}|x_2\rangle\right)$, either probabilistically or deterministically.

Is such a quantum circuit possible?

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2 Answers 2

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Deterministic case

The map sends $$ |+\rangle|+\rangle\mapsto|+\rangle|+\rangle\tag1 $$ and $$ \begin{align} |0\rangle|+\rangle\mapsto&|0\rangle|-\rangle\tag2\\ |1\rangle|+\rangle\mapsto&-|1\rangle|-\rangle\tag3 \end{align} $$ so it fails to be linear (let alone unitary) and therefore cannot be realized as a quantum circuit.

Probabilistic case

Let $\mathcal{E}$ be a completely positive, but not necessarily trace-preserving linear map that implements the desired probabilistic protocol. Then, similarly to $(1)$-$(3)$, we have $$ \begin{align} \mathcal{E}(|+\rangle\langle+|\otimes|+\rangle\langle+|)&=p_1|+\rangle\langle+|\otimes|+\rangle\langle+|\tag4\\ \mathcal{E}(|-\rangle\langle-|\otimes|+\rangle\langle+|)&=p_2|-\rangle\langle-|\otimes|+\rangle\langle+|\tag5 \end{align} $$ and $$ \begin{align} \mathcal{E}(|0\rangle\langle 0|\otimes|+\rangle\langle+|)&=p_3|0\rangle\langle 0|\otimes|-\rangle\langle-|\tag6\\ \mathcal{E}(|1\rangle\langle 1|\otimes|+\rangle\langle+|)&=p_4|1\rangle\langle 1|\otimes|-\rangle\langle-|\tag7 \end{align} $$ for some $p_1,p_2,p_3,p_4\in[0,1]$. But the maximally mixed state $$ I=\frac12|0\rangle\langle 0|+\frac12|1\rangle\langle 1|=\frac12|+\rangle\langle +|+\frac12|-\rangle\langle -|\tag8 $$ so $\mathcal{E}(\frac{I}{2}\otimes|+\rangle\langle+|)$ is simultaneously proportional to $\rho_1\otimes|+\rangle\langle+|$ and to $\rho_2\otimes|-\rangle\langle-|$ for some states $\rho_1$ and $\rho_2$. This is possible only if $\mathcal{E}\left(\frac{I}{2}\otimes|+\rangle\langle+|\right)=0$. Similarly, $\mathcal{E}\left(\frac{I}{2}\otimes|-\rangle\langle-|\right)=0$. Consequently, $\mathcal{E}\left(\frac{I}{4}\right)=0$. But $\mathcal{E}$ is completely positive, so $\mathcal{E}(\rho)=0$ for all $\rho$.

Given an input state $\rho$, the probability that the process described by $\mathcal{E}$ occurs is $\mathrm{tr}(\mathcal{E}(\rho))$. But $\mathcal{E}(\rho)=0$, so this probability is zero.

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  • $\begingroup$ How do eqs 1 2 3 imply nonlinearity? $\endgroup$
    – MonteNero
    Feb 5, 2023 at 6:05
  • $\begingroup$ $$\begin{align}f\left(\frac{1}{\sqrt2}|0\rangle|+\rangle\right)+f\left(\frac{1}{\sqrt2}|1\rangle|+\rangle\right)&=\frac{1}{\sqrt2}|0\rangle|-\rangle-\frac{1}{\sqrt2}|1\rangle|-\rangle\\&=|-\rangle|-\rangle\\&\ne|+\rangle|+\rangle\\&=f(|+\rangle|+\rangle)\\&=f\left(\frac{1}{\sqrt2}|0\rangle|+\rangle+\frac{1}{\sqrt2}|1\rangle|+\rangle\right)\end{align}$$. $\endgroup$ Feb 5, 2023 at 7:11
  • $\begingroup$ BTW: This assumes $f$ is homogeneous $f(a|x\rangle|y\rangle)=af(|x\rangle|y\rangle)$ which is another condition that must be satisfied for linearity. The function actually fails that, too, but this is easy to remedy by explicit normalization in the definition of $R_x$. The failure of additivity is the real issue. $\endgroup$ Feb 5, 2023 at 8:02
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This could be used for cloning and is therefore impossible.

Given an unknown single-qubit state $|\psi\rangle$, you can repeatedly prepare and sample $|R_\psi 0\rangle$ in order to estimate the Z coordinate of the Bloch vector of $|\psi\rangle$ to arbitrary accuracy without harming $|\psi\rangle$. Then $|R_\psi +\rangle$ and $|R_\psi i\rangle$ give you the X and Y coordinates. Those three coordinates tell you $|\psi\rangle$, so now you can start printing out copies.

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