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Is there any possible quantum circuit, containing gates like CNOT and single qubit rotations, that can take (let's say) 100 qubits in unknown states and drive one of them towards $|0\rangle$ without invoking the magic of any "measurements"?

After the self-contained process has completed, we should be able to take the output qubit, measure it in the computational basis and see that it's set to $|0\rangle$ with a high probability.

The other qubits can be left in any required state.

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No, there is no such circuit.

Quantum circuits are unitary (except for measurement). They are reversible. Forcing one of the qubits to be zero is not reversible because it would cut the state space in half. You can't squeeze $2^n$ possible input states into $2^{n-1}$ possible output states. You wouldn't be able to run the circuit in reverse.

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  • $\begingroup$ I accept your entropy-based argument. I think that the question I'm really driving at here is summed up by the phrase "preferred-basis problem", which I'll try and read more about. $\endgroup$ – David B May 15 '17 at 9:18

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