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It is said in some texts, that quantum computer undergoes only 2 types of transofrmations:

1) unitary evolution while computing

and

2) non-unitary transformation while reading result (output).

What about input?

0) initializing qubits with values

Is it unitary or non-unitary?

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  • $\begingroup$ That's the Copenhagen interpretation of quantum mechanics and extends beyond quantum computing. 0) and 2) are identical processes. So far we haven't found counterexamples. Interaction with the environment throws a couple of wrenches into this simplified picture, but we know how to work with those, too. $\endgroup$ – CuriousOne Jan 4 '15 at 16:01
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    $\begingroup$ Since initialization destroys the previous value, it is not reversible and hence not unitary. The fact that unitary erasure of information is not possible (or the "no-erasure theorem") is the time-reversed counterpart of the famous no-cloning theorem. See nature.com/nature/journal/v404/n6774/abs/404164a0.html and sciencedirect.com/science/article/pii/S0375960103010478 $\endgroup$ – Māris Ozols Jan 4 '15 at 23:47
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Initialization is non-unitary (at least if you don't know the state your qubits are in before that). Let's say you want to initialize your qubits in the state $\lvert0\rangle$. First, you need to measure each qubit in some basis (say, $\{\lvert0\rangle,\lvert1\rangle\}$) -- this is the non-unitary part -- and now, if the result was $\lvert1\rangle$, you need to rotate $\lvert1\rangle$ to $\lvert0\rangle$ (which is unitary).

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