Timeline for Find the conditions under which the state $|\phi\rangle = \sum_{y=0}^{2^n -1} e^{\frac{2 \pi i a y}{2^n}} |y\rangle$ is unentangled
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 10, 2021 at 15:43 | vote | accept | vivek kumar | ||
| Jul 5, 2021 at 21:21 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
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| Jul 5, 2021 at 21:19 | answer | added | Adam Zalcman | timeline score: 2 | |
| Jul 5, 2021 at 16:10 | comment | added | Condo | It's kind of easy to what happens if you take @DaftWullie's advice and believe the combinatorial formula $\prod_{v\in V}(1+x_v)=\sum_{S\subseteq V}\prod_{v\in S}x_v$, where $\{x_v\}_{v\in V}$ are a finite set of indeterminates. | |
| Jul 5, 2021 at 7:01 | comment | added | DaftWullie | Try expressing $y$ in terms of binary. | |
| Jul 4, 2021 at 20:54 | history | edited | glS♦ | CC BY-SA 4.0 |
edited title; edited tags
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| Jul 4, 2021 at 19:08 | comment | added | keisuke.akira | Any state of the form $| a_{1} \rangle \otimes | a_{2} \rangle \otimes \cdots | a_{n} \rangle$ is unentangled. In your question if $| \phi \rangle$ is of the form $\otimes_{i=1}^{n} \left( | 0 \rangle + \alpha_{i} | 0 \rangle \right)$ then, by definition, it is unentangled. | |
| Jul 4, 2021 at 11:49 | comment | added | Harshit Gupta | I think this is very similar to the Quantum Fourier Transform of a state. If you refer to the Circuit Implementation heading of this wiki article you would be able to find your answer. Also, I think there needs to be an overall amplitude factor of $\frac{1}{\sqrt{2^{n}}}$ multiplied with the state $|\phi \rangle$ to make it normalized. | |
| Jul 4, 2021 at 5:49 | history | asked | vivek kumar | CC BY-SA 4.0 |