Timeline for Why do the controlled unitary operations in quantum phase estimation have $2^n$ in their exponents?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 8, 2021 at 18:02 | comment | added | Condo | @mikanim if you apply $U$ just once you will only learn one bit of the phase... not so useful. | |
| Feb 7, 2021 at 9:27 | vote | accept | mikanim | ||
| Feb 6, 2021 at 21:25 | answer | added | sheesymcdeezy | timeline score: 1 | |
| Feb 6, 2021 at 14:40 | history | edited | Mark Spinelli | CC BY-SA 4.0 |
Expand question based on comments.
|
| Feb 6, 2021 at 8:18 | comment | added | mikanim | Why a power at all? Why not just U with no power for all unitary gates? | |
| S Feb 6, 2021 at 8:07 | history | edited | Martin Vesely | CC BY-SA 4.0 |
Added relevant tags
|
| S Feb 6, 2021 at 8:07 | history | suggested | Gokul Alex |
Added relevant tags
|
|
| Feb 6, 2021 at 6:53 | review | Suggested edits | |||
| S Feb 6, 2021 at 8:07 | |||||
| Feb 6, 2021 at 1:07 | comment | added | Mark Spinelli | Is your question why $2^n$ as opposed to some other power? The QFT factors nicely to determine a binary representation of the phase. Remember we also like to pad the (classical) FFT to a power of $2$. Cooley and Tukey (rediscovering the work of Gauss) . | |
| Feb 6, 2021 at 0:07 | history | asked | mikanim | CC BY-SA 4.0 |