Physics Asked on March 27, 2021
If you asked me what physical result would be naturally referred to as "the no-deleting theorem", then I would probably guess something like this:
Given a designated "blank" state $|0rangle$ in a system’s Hilbert space and two fixed states $|arangle$ and $|a’rangle$ in an ancilla Hilbert
space, there is no single linear map that takes $|psirangle|arangle$ to $|0rangle |a’rangle$ for all system states $|psirangle$.
But that’s not what the actual result known as the "no-deleting theorem" says. Instead, it talks about deleting only one of two identical qubits: it says that there’s no single linear map that takes $|psirangle |psirangle|arangle$ to $|psirangle|0rangle|a’rangle$ for all $|psirangle$.
This seems to me like a really weird and artificial way to formalize the concept of "deleting". Why consider only deleting one of two copies of the state? Why not one of three, or two of five, (most naturally, in my mind) one of one? Is deleting possible if you start with more than two copies of the state?
I will try to explain from perspective of gate-based quantum computing. Please anyone feel free to add general comments for any other quantum entity other than qubits.
It seems to me threre should be "...no unitary map that takes state...to...". Since in practical QC there is a reset gate which switches any state to $|0rangle$. But this gate is of course not unitary because it is not reversible. So you cannot have a unitary map from $|psirangle|arangle$ to $|0rangle|arangle$.
Concerning the part
...it says that there's no single linear map that takes $|psiranglepsirangle|arangle$ to $|psirangle|0rangle|a'rangle$ for all $|psirangle$.
I would say that this is a consequence of non-clonning theorem. If state $|psirangle$ is prepared via fan-out gate both "copies" are entangled. So deleting one state should influence other state. Because of non-clonning theorem you cannot prepare independent copies of $|psirangle$ and hence delete only one.
Answered by Martin Vesely on March 27, 2021
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