How to prove there's no quantum channel that clones all classical states?

Considering a qubit $$\scr H =\Bbb C^2$$ I have seen a proof of the no-cloning theorem for pure states. I wonder how do you prove it for a classical state?

1)That is, how do I prove that there is no quantum channel that clones all classical states?

2) For a state that is both pure and classical my books says that there exists a channel that clones all of them. How do I find it?

Definition of quantum channel: We say that Is that it is a superoperator $$\Phi$$ (a map in $$L(L(\scr {H_A}),L(\scr {H_B})$$) that is trace preserving and completely positive( For all $$\scr H_R$$ and $$M_{AR}\ge 0$$, it holds that $$(\Phi\otimes I_R)[M_{AR}]\ge 0$$ )

Definition of clonning:

A quantum channel $$\Phi \in C(\scr H,\scr H\otimes \scr H)$$ (C is just the set of all quantum channels between the specified spaces)clones a state $$\rho \in D(\scr H)$$ if $$\Phi[\rho]=\rho\otimes \rho$$

Definition of classical state

A quantum state $$\rho$$ on $$C(\scr H^{\Sigma})$$ is classical if it is of the form $$\sum_{x \in \Sigma}p_x|x\rangle\langle x|$$ where $$(p_x)_{x \in \Sigma}\in P(\Sigma)$$ is an arbitrary probability distribution.

Edit: My try: For the classical case:

I have for the more simple case of the question which is $$\scr H =\Bbb C^2$$.

$$\rho= \begin{pmatrix} p_0 & 0 \\ 0 & p_1 \end{pmatrix}$$

and if by the sake of contradiction I assume that it can be cloned, there exists $$\Phi$$ such that $$\Phi(\rho)=\begin{pmatrix} p_0 & 0 & 0 & 0\\ 0 &p_1 & 0 & 0 \\ 0 & 0 & p_0 & 0 \\ 0 & 0 & 0 &p_1\end{pmatrix}$$. Now I am supposed to use the lineary of the channel to arrive at a contradiction. OR I can observe that the trace is 2 and not 1 as it should, so it is not trace-preserving. A contradiction. Is this approach correct?

• What do you mean by a classical state''? By the standard definition, it should be clonable. Commented Mar 8 at 10:19
• @Rammus It means diagonal with respect to the standard basis Commented Mar 8 at 10:19
• @Rammus Who are the pure and classical states? Commented Mar 8 at 10:27
• @darkside maybe start by considering a single classical bit. Say you have a pure state $|b\rangle$ where $b \in \{0, 1\}$, what is a map that takes $|b\rangle|0\rangle$ to $|b\rangle|b\rangle$? Is there a map that works for single qubit mixed states? Commented Mar 8 at 19:31

Consider a classical state $$\rho = \sum_x p(x) |x\rangle \langle x|\,.$$ Now take a quantum channel $$\Phi(M) = \sum_x K_x M K_x^\dagger$$ defined by the Kraus operators $$\{K_x\}_x$$ where $$K_x = |xx\rangle\langle x|$$, then $$\Phi(\rho) = \sum_x K_x \rho K_x^{\dagger} = \sum_x p(x) |xx\rangle \langle xx|\,.$$ Now this is not your definition of cloning, but I would argue it is the correct definition. The state represents two random variables $$XX'$$ which both have the same marginal distribution $$p(x)$$ and $$X=X'$$, i.e. $$X'$$ is an exact copy of $$X$$.
Your definition of cloning You are asking for a state representing two random variables $$XX'$$ that have the same marginal distribution $$p(x)$$ but are independent, that is you want $$\Phi\left(\sum_x p(x) |x\rangle \langle x|\right) = \left(\sum_x p(x) |x\rangle \langle x|\right) \otimes \left(\sum_x p(x) |x\rangle \langle x|\right).$$ But you have the immediate problem that this mapping is not linear, you should have $$\Phi(\lambda \rho) = \lambda \Phi(\rho)$$ but with the above map you get $$\Phi(\lambda \rho) = \lambda^2 \Phi(\rho)$$. For the case of pure states, $$p(x)=1$$ for some $$x$$ and then this is covered by first map I described.
• If it is linear $\lambda \in \mathbb{C}$. Commented Mar 8 at 14:03
• No your approach is not correct, you have written $\rho \oplus \rho$ and not $\rho \otimes \rho$. Commented Mar 8 at 14:05