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!
