# A question on the no-cloning theorem

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.)

• 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). – Sebastian Riese Jun 14 '18 at 13:20
• @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." – THC Jun 14 '18 at 13:38
• 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 $\left|e\right>$. – Sebastian Riese Jun 14 '18 at 13:55

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.