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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?

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  • $\begingroup$ What do you mean by a ``classical state''? By the standard definition, it should be clonable. $\endgroup$
    – Rammus
    Commented Mar 8 at 10:19
  • $\begingroup$ @Rammus It means diagonal with respect to the standard basis $\endgroup$
    – darkside
    Commented Mar 8 at 10:19
  • $\begingroup$ @Rammus Who are the pure and classical states? $\endgroup$
    – darkside
    Commented Mar 8 at 10:27
  • $\begingroup$ @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? $\endgroup$
    – xzkxyz
    Commented Mar 8 at 19:31

1 Answer 1

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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.

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  • $\begingroup$ what is the range of values of lambda here? $\endgroup$
    – darkside
    Commented Mar 8 at 10:54
  • $\begingroup$ is it real in [0,1] only? $\endgroup$
    – darkside
    Commented Mar 8 at 10:55
  • $\begingroup$ Would my trace argument work in my try in the edit? $\endgroup$
    – darkside
    Commented Mar 8 at 11:07
  • $\begingroup$ If it is linear $\lambda \in \mathbb{C}$. $\endgroup$
    – Rammus
    Commented Mar 8 at 14:03
  • $\begingroup$ No your approach is not correct, you have written $\rho \oplus \rho$ and not $\rho \otimes \rho$. $\endgroup$
    – Rammus
    Commented Mar 8 at 14:05

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