Can a quantum gate send $|a,b\rangle$ to $|0,a\rangle+|1,b\rangle$?

Question

I am looking for a quantum gate (or a circuit) that operates on two quantum registers of equal size and in states $|a \rangle$ and $| b \rangle$, respectively, and prepares the state: $\frac{1}{\sqrt{2}} |0, a \rangle + \frac{1}{\sqrt{2}} | 1, b \rangle$.

Let's say that $a$ and $b$ are $N$ qubit registers. This means that we are going from an $2N$ qubit system to an $N + 1$ qubit system. So it may be best to use an output register like so, where $U$ is a quantum circuit:

$$ U |a, b \rangle |0 \rangle | 0 ^ {\otimes N} \rangle = \frac{1}{\sqrt{2}} |a, b \rangle (|0, a \rangle + |1, b \rangle). $$

But I think that is impossible by the no-cloning theorem; there exists no $V$ such that $V |x, 0 \rangle = | x, x \rangle$. We would need to clone both the $a$ and $b$ register in my $U$ circuit.

So is this possible some other way?

Answer 1

TL;DR: This is impossible for arbitrary $|a\rangle$ and $|b\rangle$. Remember how in the proof of the no-cloning theorem we pick two distinct and non-orthogonal states $|\phi\rangle$ and $|\psi\rangle$, define $x=\langle\phi|\psi\rangle$ and use the quantum copier to derive the equation $x=x^2$ for contradiction? Well, the same strategy works here.

Assume that $U$ is the circuit described in the question. Let $|a\rangle$ and $|a'\rangle$ be two distinct, but not orthogonal states of the first register and likewise for $|b\rangle$ and $|b'\rangle$. Define $x:=\langle a|a'\rangle$ and $y:=\langle b|b'\rangle$ and calculate $$ \begin{align} xy&=\langle a|a'\rangle\langle b|b'\rangle\tag1\\ &=\langle a|a'\rangle\langle b|b'\rangle\langle 0|0\rangle\langle 0^{\otimes N}|0^{\otimes N}\rangle\tag2\\ &=\langle a|\langle b|\langle 0|\langle 0^{\otimes N}|U^\dagger U|a'\rangle|b'\rangle|0\rangle|0^{\otimes N}\rangle\tag3\\ &=\langle a|a'\rangle\langle b|b'\rangle\frac{\langle 0|\langle a|+\langle 1|\langle b|}{\sqrt2}\frac{|0\rangle|a'\rangle+|1\rangle|b'\rangle}{\sqrt2}\tag4\\ &=\langle a|a'\rangle\langle b|b'\rangle\frac{\langle 0|0\rangle\langle a|a'\rangle+\langle 1|1\rangle\langle b|b'\rangle}{2}\tag5\\ &=\langle a|a'\rangle\langle b|b'\rangle\frac{\langle a|a'\rangle+\langle b|b'\rangle}{2}\tag6\\ &=xy\frac{x+y}{2}.\tag7 \end{align} $$ As in the proof of no-cloning theorem, the solutions $x,y\in\mathbb{C}$ of $2xy=xy(x+y)$ with $|x|\leqslant 1$ and $|y|\leqslant 1$ are quite restricted. Namely, there are three: either $x=0$, or $y=0$ or $x=y=1$. But that contradicts the choice of the four states. Therefore, no unitary $U$ can satisfy the desired property for all $|a\rangle$ and $|b\rangle$.