Timeline for Is it true that if $U$ sends computational basis states to product states, then it sends product states to product states?
Current License: CC BY-SA 4.0
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| Apr 16, 2023 at 17:23 | comment | added | Adam Zalcman | I don't know if this is true for all such unitaries, but it is true for many more than just the QFT. We can certainly say the following: if $$U|x\rangle=|f_1(x_1,\dots,x_n)\rangle\otimes|f_2(x_2,\dots,x_n)\rangle\otimes\dots\otimes|f_n(x_n)\rangle$$ and if each factor $|f_k(x_k,\dots,x_n)\rangle$ can be implemented efficiently by a circuit which uses $x_{k+1}\dots x_n$ as controls (and hence doesn't disturb computational basis states on qubits $k+1$ to $n$) then we can implement $U$ efficiently, too (by essentially following the recursive construction of the standard QFT circuit). | |
| Apr 16, 2023 at 15:06 | comment | added | glS♦ | "On one hand, the fact that QFT sends computational basis states to product states is the key observation behind the construction of an efficient quantum circuit for it" our of curiosity: are you saying that any unitary with this property has an efficient circuit decomposition? Or just that the decomposition of the QFT you're referring to is a standard step in the derivation of the decomposition for the QFT? | |
| Apr 16, 2023 at 2:04 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Apr 16, 2023 at 1:28 | vote | accept | trillianhaze | ||
| Apr 15, 2023 at 23:26 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Apr 15, 2023 at 23:20 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Apr 15, 2023 at 23:02 | history | answered | Adam Zalcman | CC BY-SA 4.0 |