Timeline for Is a probabilistic implementation of unitaries invertible?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 26 at 19:03 | answer | added | Michael Hardy | timeline score: 0 | |
| Aug 26 at 18:58 | comment | added | Michael Hardy | Is this the same as asking whether the expectation of a random unitary matrix invertible? (Where "random" does not mean uniformly distributed, but just means having some probability distribution.) | |
| Aug 26 at 4:32 | history | became hot network question | |||
| Aug 25 at 22:08 | vote | accept | Hans Schmuber | ||
| Aug 25 at 21:28 | answer | added | user196574 | timeline score: 4 | |
| Aug 25 at 21:24 | comment | added | Hans Schmuber | If I restrict to the invariant subspace of Hermitian operators, then it looks like this has a trivial kernel, right? | |
| Aug 25 at 21:23 | comment | added | Hans Schmuber | From your answer, I can see that if $n=2$, $A=((0,1)(0,0))$ then it is possible to find unitaries with 1 and -1 eigenvalues, then it looks like it has a non-trivial kernel. | |
| Aug 25 at 21:07 | history | edited | Hans Schmuber | CC BY-SA 4.0 |
added 173 characters in body
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| Aug 25 at 20:59 | comment | added | Hans Schmuber | Thanks for your answer! I'm not sure I understand your counter example. If $A\in \mathbb{R}$, then surely the linear mapping in your example is E(A) = (1/2)*(-1)A(-1) + (1/2)*(1)A(1) = A? But I feel like I've misunderstood you here? | |
| Aug 25 at 19:34 | review | Close votes | |||
| Sep 3 at 3:08 | |||||
| S Aug 25 at 17:53 | review | First questions | |||
| Aug 25 at 18:08 | |||||
| S Aug 25 at 17:53 | history | asked | Hans Schmuber | CC BY-SA 4.0 |