# Does a noiseless quantum channel violate no-cloning theorem?

A quantum channel is defined as CPTP map from Alice to Bob $$\mathcal{N}: \mathcal{H}_A \to \mathcal{H}_B$$. In particular, a noiseless quantum channel is such that $$\mathcal{N}(\rho_A) = \rho_A$$. My question is: why does this not violate no-cloning? In other words, what is the initial composite system state $$\rho^{initial}_{AB}$$ and what is the final composite system state $$\rho^{final}_{AB}$$ after the transmission takes place?

To clarify, I'm thinking about a scenario where Alice is at one end of an optical fiber. Alice transmits her state over the channel to Bob. At the end of the transmission, Bob has the state.

• How do you clone your state if your channel does nothing? Commented Oct 31, 2022 at 17:44
• I edited the question. Commented Oct 31, 2022 at 17:48
• If Alice transmits her state she no longer has her state. Nothing has been cloned because there is still only one system. Physically she has sent her photon to Bob, so she no longer has it. Commented Oct 31, 2022 at 19:14
• Physically, I understand this. However, within the formalism of quantum information theory, suppose Bob was in a state $|0\rangle_{B_1}$ so that the initial state of the system is $\rho_A \otimes |0\rangle_{B_1}$. After the transmission, would the state be $\rho_{B_2} \otimes |0\rangle_{B_1}$ and the system A would be dropped? Commented Oct 31, 2022 at 19:53
• Yes, exactly. As you have defined the problem your channel transforms a system $A$ into a system $B$, afterwards there is no system $A$. Commented Oct 31, 2022 at 20:07

No. Quantum channels typically represent evolutions of open quantum systems. The input is a state at a time $$t_1$$ and the output is a state at a latter time $$t_2>t_1$$. Having a state $$\rho$$ at time $$t_1$$ and then still having it at time $$t_2>t_1$$ is not cloning, it's just quantum memory. A cloner would send $$\rho$$ at time $$t_1$$ to $$\rho\otimes\rho$$ at time $$t_2$$.