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

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?

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 $$|b\rangle$$.
• Does this reasoning generalize to prohibit any process that sends $|a,b\rangle \rightarrow (|0,a\rangle + |1,b\rangle)/\sqrt{2}$, or just the unitary OP proposed? May 8, 2023 at 13:45
• No quantum gate or channel sends $|a\rangle|b\rangle$ to $(|0\rangle|a\rangle+|1\rangle|b\rangle)/\sqrt2$. The proposed mapping isn't even a function when we consider it as defined on rays in the Hilbert space (any map on quantum states implicitly acts on rays due to global phase equivalence). To see this note that the state $|a\rangle\otimes|b\rangle\equiv|a\rangle\otimes i|b\rangle$ is sent to both $(|0\rangle|a\rangle+|1\rangle|b\rangle)/\sqrt2$ and $(|0\rangle|a\rangle+i|1\rangle|b\rangle)/\sqrt2$ (and infinitely many other states) depending on the vector representative we choose. May 10, 2023 at 5:53