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2

I think it helps here to write things explicitly. Suppose $\mathcal E(\rho)=\operatorname{Tr}_E[U(\rho\otimes|\mathbf e_0\rangle\!\langle\mathbf e_0|)U^\dagger]$. Pick a basis for the environment in which $|\mathbf e_0\rangle$ is the first element. Note that here $U$ is a unitary matrix in a bipartite system. The operator before taking the partial trace ...

2

$|e_k\rangle$ is the basis of the environment. Taking the sum of projections onto an orthonormal basis of one subsystem is the definition of the partial trace over that subsystem.

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We can proceed by applying the parity check matrix $H_2$ for $C_2^\perp$ on the ancilla, which would become $\vert H_2(z'+e_2)\rangle =\vert H_2(z'+u + e_2 -u)\rangle =\vert H_2(e_2-u)\rangle$. Measuring the ancilla gives us $H_2(e_2-u)$. As $u$ is known. we then know $H_2 e_2$, and in turn $e_2$. After correcting the now bit flip $e_2$, we have: \dfrac{1}{...

1

Thanks Peter for the clarification about information vs. outcomes. I accept his answer to acknowledge that, and want to add the possible construction of such measurement. In the same book section 2.2.8, a general method is described. In this case, one can add two qubits prepared as $|00\rangle$, apply a unitary on the three qubits and measure the two ...

4

Three outcomes amounts to more than one bit if the outcomes are all deterministic, and give you information about the original qubit. But suppose I have a coin (that is either heads or tails). I roll a dice, and if it comes 1 through 5, I tell you "H" or "T", depending on what the coin is. If it comes up 6, I tell you "6". There are three outcomes, but ...

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The real missing keyword in the stating the theorem is "arbitrary unknown state"! If you have some information about $|\psi\rangle$, i.e. specific state, then perhaps you can reconstruct that state!

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Yes. It is outdated, however, to explain something without sounding too abstract the old stationary orbiting solutions come handy.

7

The proof does not seem to rule out the case that there exists a specific U that can clone only the specific state |ψ⟩. That's because you can clone specific states. Cloning is only impossible if the set of possible input states includes a pair of states that are not orthogonal. For example, here is a circuit that performs $|\psi⟩ \to |\psi⟩|\psi⟩$ as long ...

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