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 $|0\rangle$ in a system's Hilbert space and two fixed states $|a\rangle$ and $|a'\rangle$ in an ancilla Hilbert space, there is no single linear map that takes $|\psi\rangle|a\rangle$ to $|0\rangle |a'\rangle$ for all system states $|\psi\rangle$.

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 $|\psi\rangle |\psi\rangle|a\rangle$ to $|\psi\rangle|0\rangle|a'\rangle$ for all $|\psi\rangle$.

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?

  • $\begingroup$ Maybe it’s just meant to be the opposite of the no cloning theorem? $\endgroup$
    – knzhou
    Nov 21 '20 at 19:44
  • $\begingroup$ @knzhou Yeah, that's probably it, but that seems like ... not a great motivation for a named theorem with its own Wikipedia page. $\endgroup$
    – tparker
    Nov 21 '20 at 20:09
  • $\begingroup$ @NiharKarve Could you elaborate on what that means and why it's an interesting/useful formulation? $\endgroup$
    – tparker
    Nov 22 '20 at 4:35
  • $\begingroup$ According to arxiv.org/abs/quant-ph/0306044, even orthogonal states cannot be deleted if there is no copy (with the unitarity assumption). And you might find arxiv.org/abs/quant-ph/0007121 useful for $N\rightarrow M$ deleting $\endgroup$ Nov 22 '20 at 15:20
  • 1
    $\begingroup$ @NiharKarve Thanks. Honestly, I find the no-deleting theorem to be pretty useless in practice, because the unitarity assumption and the requirement that the final ancilla state can't depend on the initial state effectively rule out the possibility of making measurements. (Depending on your interpretation of QM, the measurement process violates one or the other of those assumptions.) I get the interest from a fundamental physics perspective, but for (e.g.) practical quantum computing, it's trivially easy to delete a qubit state by just consecutively measuring it in two mutually unbiased bases. $\endgroup$
    – tparker
    Nov 22 '20 at 19:46

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 $|0\rangle$. But this gate is of course not unitary because it is not reversible. So you cannot have a unitary map from $|\psi\rangle|a\rangle$ to $|0\rangle|a\rangle$.

Concerning the part

...it says that there's no single linear map that takes $|\psi\rangle\psi\rangle|a\rangle$ to $|\psi\rangle|0\rangle|a'\rangle$ for all $|\psi\rangle$.

I would say that this is a consequence of non-clonning theorem. If state $|\psi\rangle$ 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 $|\psi\rangle$ and hence delete only one.


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