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In the quantum circuit model we prepare some inital state $|\psi\rangle$ and throw our algorithm in form of some unitary transformation(s) $U$ on it to get our result $|\Psi\rangle$:

$|\Psi\rangle$=$U$ $|\psi\rangle$.

Why does $U$ need to be unitary in this quantum computation model? I think traditional (classical) computer stuff is not time-reversible so why do we want/demand it here?

The time-reversal thing is all I have in mind when I think about unitarity or why it could be nice but it does not seem necessary. In the end we just want the output/result $|\Psi\rangle$ so I don't see any reason why we should restrict our transformations in that way.

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You can also define a quantum computation model based on completely positive trace-preserving (CPTP) maps, which are the most general physically allowed maps. These are generally irreversible. Since CPTP maps contain all unitaries, and conversely, any CPTP map can be simulated using unitaries and ancilla qubits, this model is just as powerful as a model based on unitaries.

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  • $\begingroup$ Ok, I read know about some quantum computation models but have not encountered that one yet. But I still would like to know why in the quantum circuit model "we" make this demand for unitary transformations. $\endgroup$ – Thomas Elliot Jun 24 '15 at 11:43
  • $\begingroup$ @ThomasElliot The other model is not commonly used, because the unitary model is easier. But as I said: You don't need to assume unitarity, but any circuit model defined with physical gates is equivalent to it, as it can be simulated using unitaries (known as the Stinespring dilation). $\endgroup$ – Norbert Schuch Jun 24 '15 at 11:46
  • $\begingroup$ @ThomasElliot Also note that irreversibility means that information is lost to the environment, i.e., the system becomes less quantum, which is not what we want for a quantum computer. $\endgroup$ – Norbert Schuch Jun 24 '15 at 11:52

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