# Is there a circuit that takes two single-qubit states and reflects the second around the first?

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?

## 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.

• How do eqs 1 2 3 imply nonlinearity? Feb 5 at 6:05
• \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}. Feb 5 at 7:11
• 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. Feb 5 at 8:02

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.