Pure State No Cloning

For each unit representative $\psi$ of a ray in $S$, the reduced state of $V\psi$ on the two output registers is the pure projector $P_{\psi }\otimes P_{\psi }$. By the Schmidt decomposition [L3], there is a unit environment vector $e_{\psi }$ and a phase absorbed into it such that $V\psi =\psi \otimes \psi \otimes e_{\psi }$.

For any two unit representatives $\psi ,\varphi \in S$, the isometry preserves their inner product. By step 1.1, $⟨\psi ,\varphi ⟩=⟨V\psi ,V\varphi ⟩=⟨\psi ,\varphi {⟩}^2⟨e_{\psi },e_{\varphi }⟩\mathrm{.}$ Taking absolute values and applying Cauchy–Schwarz to the unit environment vectors gives $c=c^2\mathrm{\mid }⟨e_{\psi },e_{\varphi }⟩\mathrm{\mid }\le c^2$, where $c=\mathrm{\mid }⟨\psi ,\varphi ⟩\mathrm{\mid }\in [0,1]$. Hence $c(1-c)\le 0$ with both factors nonnegative, so $c=0$ or $c=1$. The first case is orthogonality; in the second, equality in Cauchy–Schwarz makes the two vectors scalar multiples and their rays identical (Cauchy–Schwarz: $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, with equality exactly for dependent pairs).

If $\dim H\ge 2$, choose orthonormal $e_1,e_2$; the rays of $e_1$ and $(e_1+e_2)\mathrm{/}\sqrt{2}$ are neither equal nor orthogonal, contradicting step 2.1 for a universal exact cloner. If $\dim H=1$, fix a unit vector $e$ and take the one-dimensional environment $E=\mathbb{C}$ with unit vector $f$. Define $V(e)=e\otimes e\otimes f$ and extend linearly; because this sends the unit basis vector to a unit product vector, it is an isometry. By [L4], $X\mapsto {\mathrm{Tr}}_E(VX{V}^{\ast })$ is a channel; on the sole pure-state projector it outputs $P_e\otimes P_e$. Hence the unique ray is cloned by a channel.