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Consider the following problem from Nielsen and Chuang's Quantum Computation and Quantum Information:

Explain how a device which, upon input of one of two non-orthogonal quantum states $\left|\psi\right>$ or $\left|\phi\right>$ correctly identified the state, could be used to build a device which cloned the states $\left|\psi\right>$ and $\left|\phi\right>$, in violation of the no-cloning theorem. Conversely, explain how a device for cloning could be used to distinguish non-orthogonal quantum states.

I know that this problem has already been posted at least two times here and here, but I am not satisfied with the answers to the first part given there. As you can see, all answers to the first part basically tell you to use the device to distinguish between $\left|\psi\right>$ and $\left|\phi\right>$ and then just prepare an identical state. However, nowhere does this exercise say that we know how to prepare states $\left|\psi\right>$ and $\left|\phi\right>$. Moreover, preparing an identical state is not the same as cloning. A cloning device has to be able to replicate a state without having any prior knowledge about its "internal" structure. With this in mind, I am posting this problem once again.

Here is my attempt to address this problem. Let us say that we have two non-orthogonal states $\left|\psi\right>$ and $\left|\phi\right>$. We can measure $\left|\phi\right>$ in the basis of $\left|\psi\right>$, and feed the result to our magical device. Since $\left|\phi\right>$ is non-orthogonal to $\left|\psi\right>$, there is non-zero probability that measurement will give us another $\left|\psi\right>$. In this case the device will identify it as $\left|\psi\right>$, which means we have created a clone. If the measurement projects $\left|\phi\right>$ to some other state $\left|\psi'\right>$, orthogonal to $\left|\psi\right>$, then the device will not identify it as $\left|\psi\right>$, and we can repeat the process by measuring $\left|\psi'\right>$ again in the basis of $\left|\phi\right>$, which will create another state non-orthogonal to $\left|\psi\right>$.

This solution is also quite flimsy for the following reasons. First, I am not sure if an arbitrary unknown state can be used as a basis for measurement. Second, this solution never actually used the device to distinguish between $\left|\phi\right>$ and $\left|\psi\right>$. It only used it to distinguish between $\left|\psi\right>$ and not $\left|\psi\right>$, but it is unclear what happens if we try to feed a state other than $\left|\phi\right>$ and $\left|\psi\right>$ to the device, so this way was probably not implied by the authors of the exercise.

Any ideas?

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I think you've maybe missed some of the reasoning in the premise.

If you are trying to clone an unknown qubit, but you know it to be one of two states $|\psi\rangle$ or $|\phi\rangle$, then the point is that you do know what $|\psi\rangle$ and $|\phi\rangle$ are. And, because we're assuming you can perfectly perform arbitrary unitaries (we don't care in this context about how long it might take), we can easily specify two unitaries which act as $$ U_1|0\rangle=|\psi\rangle,\qquad U_2|0\rangle=|\phi\rangle. $$
In fact, for this argument, I don't even have to talk quite so generically. I could be specific and say $|\psi\rangle=|0\rangle$ and $|\phi\rangle=|+\rangle$. So you definitely know how to make them.

Cloning is absolutely the case where you start from your unknown state $|\gamma\rangle$, and produce one (or more) copies $|\gamma\rangle|\gamma\rangle$. In this particular setting, we can produce as many copies as we want. The "internal" structure to which you are referring in this case is only the question of which of the two states it is, and not what those two possible states are.

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  • $\begingroup$ In what sense do we know what $\left|\psi\right>$ and $\left|\phi\right>$ are? Let me try to rephrase the problem one more time the way I see it. In our lab, we have an information source which randomly sends us one of the two unknown states: either $\left|\psi\right>$ or $\left|\phi\right>$. For the sake or simplicity let us assume that $\left|\psi\right>$ and $\left|\phi\right>$ are single qubit states, i.e. $\left|\psi\right>=a_{1}\left|0\right>+b_{1}\left|1\right>$ and $\left|\phi\right>=a_{2}\left|0\right>+b_{2}\left|1\right>$. $\endgroup$
    – DartLenin
    Oct 4, 2021 at 15:32
  • $\begingroup$ The only thing we know about these states is that they are non-orthogonal. We also have a magical device, which tells us whether it's state #1 ($\left|\psi\right>$) or #2 ($\left|\phi\right>$). It does not tell us the values of a and b (what I referred to as "internal" structure), and we have no idea where these states came from or how to prepare more copies. However, the information source sends us these states rather infrequently, so we would like to build a cloning device to get more copies to experiment with. Is this possible? $\endgroup$
    – DartLenin
    Oct 4, 2021 at 15:32
  • $\begingroup$ In the context that N&C are discussing, you do know the $a_i$ and $b_i$s. Yes, you could ask the other case (but why would you be talking about two unknown states? It would be entirely equivalent to being given a single entirely unknown state) but that is not what they're doing. The text just before Exercise 1.2 talks about cryptographic protocols, which are fixed protocols that used known bases, for example. $\endgroup$
    – DaftWullie
    Oct 5, 2021 at 6:47
  • $\begingroup$ First, the exercise does not explicitly tell us whether the states are known or unknown, so they could be unknown. Second, the exercise asks us to build a cloning device "in violation of the no-cloning theorem". This part makes me believe the states are unknown, because the no-cloning theorem talks specifically about inability to clone an unknown state. We can always prepare more copies of known states, that does not violate the no-cloning theorem and we don't need no magical devices for that. $\endgroup$
    – DartLenin
    Oct 5, 2021 at 7:08
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    $\begingroup$ A formal statement of the no-cloning theorem would be that "if you are provided with a state which is an unknown choice from a known set of states, you cannot perfectly clone that state if the set contains a pair of non-orthogonal states". A bit more colloquially, people often talk of cloning an unknown state where they mean the special case where the unknown set is all possible pure states of a given dimension. Here the case really is that the known set is just two states. $\endgroup$
    – DaftWullie
    Oct 6, 2021 at 6:48

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