How to show that BB84 is safe against imperfect cloning?

Question

One of the reasons why BB84 is usually considered safe is due to the no-cloning principle. However, as far as I know imperfect cloning of quantum states is possible. I've been thinking about a potential attack on the BB84 where Eve makes imperfect copies of the states of qubits sent by Alice, and then measures the imperfect copies.

I see imperfect cloning as an operation described with a unitary operator $U$ which acts in the following way:
$U \vert \psi \rangle \vert 0 \rangle \rightarrow \vert \psi \rangle \vert \psi^{\prime} \rangle$ where $\psi$ and $\psi^{\prime}$ are at least somewhat similar, i.e. $\langle \psi \vert \psi^{\prime} \rangle$ > 0.5.
Cannot think of a reason why such operator should not exist.

Obviously, Eve cannot retrieve the entire key in this way but I guess even if she gains knowledge about a significant percentage of the key this can considerably facilitate other attacks (even if just some classical brute force attack). And since she measures the copies, she will be able to remain undetected. Moreover - the more similar the copy is to the original state, the more information Eve should be able to gain but I suppose as long as there is virtually any correlation between the copy and the original state, Eve should be able to retrieve some information about the qubits sent by Alice, therefore weakening the protocol's security.

I am aware that a proof of BB84's security exists and hence I reason that such an attack cannot be possible. However, I am not sure why it is the case. I am wondering whether there is some kind of limitation that you cannot get any information about a state at all without influencing it, but this seems like a very strong statement and I don't know how to show it. Or is there some other reason?

I've read all results of the searches no-cloning, no-cloning theorem and BB84 attack but, surprisingly, to no avail. I will be grateful for any insights.

Answer 1

If you impose that $U|\psi\rangle|0\rangle=|\psi\rangle|\psi'\rangle$ for all $|\psi\rangle$ (or at least for a pair of non-orthogonal states), then you'll be able to prove that $|\psi'\rangle$ must be independent of $\psi$, and hence is useless from Eve's perspective.

To see this, let's start by assuming that $\psi_0'=0$ in the case of $\psi=0$. It really doesn't matter what we set it to because we can always apply an extra single-qubit unitary at the end. What could $\psi_+'$ be when the input is $\psi=+$? By linearity, we know that $$ U(|0\rangle+|1\rangle)=|0\rangle|0\rangle+|1\rangle|\psi_1'\rangle $$ BUT you also specified that the output state has to be separable (so that we still have $|+\rangle$ on the first qubit). The only way you can do that is if $|\psi_1'\rangle=|0\rangle$ (up to a phase). It follows that for every possible input state, $\psi'=0$.