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?
