I am very new to Quantum Information. I had the following question on two no-go theorems:

The No cloning theorem states there is no unitary operator $U$ such that $U|\psi\rangle|0\rangle=|\psi\rangle|\psi\rangle$ for all state $|\psi\rangle$. The No deletion states there is no unitary operator $U$ such that $U|\psi\rangle|\psi\rangle=|\psi\rangle|0\rangle$ for all state $|\psi\rangle$. Doesn't the last statement follows from the first after taking $U^*$ on both sides. Also the corresponding wikipedia article mentions it is a time-reversed dual to the no-cloning theorem, so is there some categorical implication?

  • 1
    $\begingroup$ By categorical implications are you thinking of the distinction between affine and linear logic? $\endgroup$
    – AHusain
    Apr 13 at 0:58

2 Answers 2


Yes! Your reasoning is correct about applying $U^*$ to both sides, since $U^*U=I$. You can derive $U|\psi\rangle|0\rangle = |\psi\rangle|\psi\rangle$ from $U|\psi\rangle|\psi\rangle = |\psi\rangle|0\rangle$ and vice-versa. If one is forbidden for all unitary $U$ then so is the other.

The Schrodinger equation tells us that $U^*$ is the time-reversed $U$ operator, so this is how they are related. If a system has a Hamiltonian $H$ and evolves under this Hamiltonian for a time $t$, then the time-evolution operator is $U = e^{-iHt}$. $U^* = e^{iHt}$, which is the same as taking $t$ to $-t$ i.e. time-reversal.


The no cloning and no-deletion theorems are a much stronger claim than you've stated. For example, the no-cloning theorem states that there does not exist any quantum operation that takes $|\psi\rangle|0\rangle$ (plus any state of any other quantum systems you might choose to include) to $|\psi\rangle|\psi\rangle$ (plus some state of the other systems, that may depend on $\psi$) for some non-orthogonal set of states $|\psi\rangle$. So, this allows you measurements and operations across ancillary systems. In other words, something that, when restricted just to the two qubits of primary interest, is not unitary.

Thus, while your reasoning works for unitaries, you're really interested in a broader class of operations which may not be invertible. Put another way, the symmetry is broken when considering ancilla systems because I would write the two cases as \begin{align*} |\psi\rangle|0\rangle|0\rangle&\mapsto|\psi\rangle|\psi\rangle|\phi(\psi)\rangle \\ |\psi\rangle|\psi\rangle|0\rangle&\mapsto|\psi\rangle|0\rangle|\phi'(\psi)\rangle \end{align*} It is now not so clear that one is the inverse of the other.

  • $\begingroup$ If you allow non-unitary operations deletion is not really a problem, is it? Measure in some basis, then apply some unitary conditioned on the measurement outcome. I haven't heard about no-deletion before and now struggle to understand its significance. $\endgroup$ Apr 14 at 11:39
  • $\begingroup$ @NikitaNemkov I'm not saying that this breaks the statement of no cloning. But the whole operation is not straightforwardly invertible, so the OP's equivalence proof cannot be applied directly. $\endgroup$
    – DaftWullie
    Apr 14 at 11:46

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