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When reading papers on the no-cloning theorem, I understand that one is searching for a unitary matrix $U$, such that for any state $\vert \phi_A \rangle$ in a Hilbert state $H_A$, we have that $U\Big(\vert \phi \rangle_A \otimes \vert e \rangle_B\Big) = \vert \phi \rangle_A \otimes \vert \phi \rangle_B$, where $e_B$ is a blank state in the Hilbert space $H_B$, and $\vert \phi \rangle_B$ is a copy of $\vert \phi \rangle_A$ in $H_B$. Such $U$ cannot exist.

Still, I wonder what "blank state'' really means -- is one allowed to choose $\vert e \rangle_B$ as a function of $\vert \phi \rangle_A$ when building cloning machines, or generalizations ? (So that the cloning machine not only comes with the matrix $U$, but also with a choice function.)

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  • $\begingroup$ To choose $\left|e\right>$ in dependence of $\left|\phi\right>$ you already would have to know $\left|\phi\right>$. The point of the no cloning theorem is, that you cannot clone an unknown quantum state. So $\left|e\right>$ must be some arbitrary fixed state (or some arbitrary random state, but that would make it not less impossible). $\endgroup$
    – Sebastian Riese
    Jun 14, 2018 at 13:20
  • $\begingroup$ @SebastianRiese : ok, but isn't it so that if we consider two unknown states $\vert \phi \rangle_A$ and $\vert \phi' \rangle_A$ to be cloned, the blank states that will be used also differ ? (If one, say $\vert e \rangle_B$, is used, we cannot re-use it. As in the case of a real photocopier.) In that case, it couldn't be "fixed." $\endgroup$
    – THC
    Jun 14, 2018 at 13:38
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    $\begingroup$ Well, of course we need yet another qubit prepared in state $\left|e\right>$ if we want to try to clone the state of yet another qubit. Also, if we know the state of a qubit (and it is not entangled with other qubits) we can always find a unitary operation that changes it to our fixed state $\left|e\right>$. $\endgroup$
    – Sebastian Riese
    Jun 14, 2018 at 13:55

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As was already pointed out in the comments, the "blank state" cannot be chosen as a function of $|\phi\rangle$.

The reason is that a cloning operation is a unitary $U$ such that for any state $|\phi\rangle$, one has $U|\phi\rangle\mapsto|\phi\rangle|\phi\rangle$. Put in a different way, $U$ represents a procedure that would be able to clone arbitrary input states without loss of information.

The "blank state" is just part of this protocol, and as such has to be chosen beforehand, and cannot depend on the input being cloned.

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