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Let the state $|\Psi\rangle\equiv a|0\rangle\otimes|\psi_0\rangle + b|1\rangle\otimes|\psi_1\rangle$, where $|\psi_0\rangle$ and $\psi_1\rangle$ belong to a multi-qubit register $R$ and the coefficients $a,b$ are not known a priori. I would like to extract the state $|\psi_0\rangle$. Essentially, I would like to apply the projector $\hat P_0\equiv|0\rangle\langle0|$ on $|\Psi\rangle$.

The standard approach seems to be to measure the solitary qubit to collapse $R$ into either $|\psi_0\rangle$ if you measured $0$ or $|\psi_1\rangle$ if you measured $1$. In the latter case, simply discard the state and try again from scratch. This is a probabilistic method and obviously undesirable if preparing $\Psi$ is difficult or if $|a|$ is very small.

Is there some quantum circuit to implement $\hat P_0$ and deterministically extract $|\psi_0\rangle$? It is clearly not unitary, so such a circuit clearly is not simple. But my intuition tells me that there ought to be a strategy involving an ancilla register and/or a second copy of $|\Psi\rangle$. I am however self-taught and my intuition is not terribly reliable; I would happily accept a proof that it can't be done. ^_^

The closest I've gotten so far is recognizing that it's easy to do with two entangled copies of $|\Psi\rangle$: let $|\Phi\rangle\equiv \frac{a+b}{\sqrt{2}}|01\rangle\otimes|\psi_0\rangle\otimes|\psi_1\rangle + \frac{a-b}{\sqrt{2}}|10\rangle\otimes|\psi_1\rangle\otimes|\psi_0\rangle$, where the two solitary qubits are lumped into one register. Then measure either solitary qubit. If it's $0/1$, $|\psi_0\rangle$ is the state in the first register $R_1$. If it's $1/0$, take the second register $R_2$. But I don't know how to generate the entangled state $|\Phi\rangle$, or whether it is possible to do so from two initially independent copies of $|\Psi\rangle$.

Bonus points if the procedure extends naturally to arbitrary multi-qubit projection operators without a tedious basis transformation!

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    $\begingroup$ Did you try to apply amplitude amplification? $\endgroup$
    – M. Stern
    Jul 25 at 7:29
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    $\begingroup$ I need to read a bit more to be sure, but it seems like amplitude amplification will remain probabilistic if $a$ and $b$ aren't known..? $\endgroup$
    – jecado
    Jul 26 at 14:04
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    $\begingroup$ depends on the state, but in general yes. However, you might get a success probability that suffices in practice. $\endgroup$
    – M. Stern
    Jul 26 at 14:39
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    $\begingroup$ I suppose the circuit should also not depend on $|\psi_0\rangle$? Otherwise a trivial answer would be a circuit that completely disregard the inputs and always outputs $|\psi_0\rangle$ $\endgroup$
    – glS
    Aug 2 at 22:44
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    $\begingroup$ Hah. Yes, I forgot to explicitly state that $|\psi_0\rangle$ is not known a priori. I would probably not accept your solution as an answer. :P $\endgroup$
    – jecado
    Aug 3 at 20:42
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Consider the states $|\psi\rangle = a|0\rangle|\psi_0\rangle + b|1\rangle|\psi_1\rangle$, and $|\psi'\rangle = a|0\rangle|\psi_0'\rangle + b|1\rangle|\psi_1\rangle$, where $a$ and $b$ are non-zero and $\psi_0$ and $\psi_0'$ are known and orthogonal. By applying your proposed circuit and measuring the circuit output (the orthogonal $\psi_0$ or $\psi_0'$) you could perfectly distinguish between the non-orthogonal $\psi$ and $\psi'$, which is impossible. Likewise for any finite-copy case.

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    $\begingroup$ ...I find myself far less happy to accept it now that it has been presented to me... $\endgroup$
    – jecado
    Aug 3 at 20:46
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    $\begingroup$ Do you by chance have any insights on the entangled state solution? Evidently the entangled state $|\Phi\rangle$ cannot be prepared given two copies of $|\Psi\rangle$, but suppose I have a circuit $V$ which prepares $|\Psi\rangle$ on $n$ qubits. Is there a modified procedure to generate the entangled version $|\Phi\rangle$ on $2n$ qubits? I'll probably pose this as an independent question in a day or two but I may as well ask here first. $\endgroup$
    – jecado
    Aug 3 at 20:55
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    $\begingroup$ Not totally clear to me how many copies of $\Phi$ you're obtaining from how many copies of $\Psi$ in your question, but I think the answer to whether it can be done reliably will be no, by the same reasoning: you can reliably obtain $\psi_0$ from $\Phi$, so you can use a circuit which creates $\Phi$ to reliably distinguish between non-orthogonal states $\psi$ and $\psi'$ in my example. It doesn't change the reasoning much if the input is now $\Psi^{\otimes n}$, because for finite $n$, $\psi^{\otimes n}$ and $\psi'^{\otimes n}$ are still non-orthogonal. $\endgroup$
    – GotCarter
    Aug 3 at 22:47
  • $\begingroup$ The input wouldn't be $|\Psi\rangle^{\otimes n}$ but rather the operator $V$ such that $V|0\rangle=|\Psi\rangle$. As you've proven, the procedure to generate $|\Phi\rangle$ wouldn't begin by preparing copies of $|\Psi\rangle$ independently, but I am imagining some sort of complex procedure in which $V$ is applied as a controlled operation, or perhaps it is applied to pre-entangled qubits, or something like that. But I admit I haven't thought very carefully about it yet. $\endgroup$
    – jecado
    Aug 4 at 15:51

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