# Probabilistic quantum cloning

I have a question with regard to probabilistic quantum cloning - see for example http://prl.aps.org/abstract/PRL/v80/i22/p4999_1.

It does seems like I can use the proof for no-cloning theorem to disproof the existence of equation (7) in the paper quant-ph/9704020. Do you think this is true?

Suppose probabilistic quantum cloning is actually possible, now according to the current papers, if the cloning process fails, the state that we intend to clone would be destroyed. Looking at the equation in the original paper, however, it does seem that a repair or a reverse operation might be possible in the event of a failure. More specifically, in the paper quant-ph/9704020, if I replace the $\left|\phi\right>$ state in Equation (7) with say, $\left|\psi_T\right>\left|0\right>$ where $\left|\psi_T\right>$ is a state orthogonal to the input state $\left|\psi\right>$, the proof would work just as fine. Do you think this is true? If this is true, a repair would be possible in case of cloning failure. The process would still be probabilistic, but I can repeat till it is successful.

I do hope for some feedback from experts out there. Thanks!

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What to you mean exactly? In eq (7) of quant-ph/9704020, you vant to replace the $|\Phi_{AB}\rangle$ by $|\Psi_0^\perp\rangle$ if the input is $|\Psi_0\rangle$ and by $|\Psi_1^\perp\rangle$ if the input is $|\Psi_1\rangle$ ? –  Frédéric Grosshans Feb 25 '11 at 11:18
Yes, I meant that. From answers below, I know now however, that is not possible. –  Nordin Feb 28 '11 at 7:54

The act of measurement of a system which is not an eigenstate of the measurement must destroy information. That is, you cannot return to the initial states after making the measurement.

The only information coming into the problem are the states that are being cloned. So by conservation of information it's not possible to clone two states with probability 1 using a sequence of measurements and unitary operations. Maybe you can improve the probabilities this way.

I would think that the easiest way to attack the problem would be to take a look at the qubit example they do at the end of the paper. From this you can see just how much information is remaining.

I think this is an attractive problem and when I have more time, Sunday, I'll work on it, unless someone solves it before then.

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Thanks. Please see my comment to Peter Shor's reply. –  Nordin Feb 28 '11 at 7:53
What is unambiguous state discrimination? Suppose you have a set of $k$ linearly independent pure quantum states. There is a POVM measurement whose outputs are either "state $j$" or "I don't know." If the input is state $j$, then the output is "state $j$" with some probability $p_j$ and "I don't know" with probability $1-p_j$. On the "I don't know" outputs, the input state is destroyed.
Unambiguous state discrimination can be used to build a probabilistic cloner: if the output is "state $j$," prepare an additional copy of state $j$. Postselected on the output not being "I don't know," you have cloned the input state. However, this is presumably not optimal; the optimal cloner should only be able to produce two copies of a state (the intuition for this is that otherwise it is extracting more quantum information from the state than it really needs to clone). This process can produce arbitrarily many copies.
As for your question about equation 7 in arxiv quant.ph/9704020 (the PRL version has a different numbering), if you replace $|\Phi_{AB}\rangle$ with a state that will let you restore $|\Psi_0\rangle$ and $|\Psi_1\rangle$, as suggested in Frederic's comment, the operator $U$ is no longer unitary. If it were, you would have a unitary operator that doesn't preserve inner products, which is impossible.