Timeline for How does Landauer's Principle apply in quantum (and generally reversible) computing
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 23, 2019 at 9:14 | answer | added | Rob | timeline score: 2 | |
| Jul 17, 2015 at 8:04 | comment | added | Selene Routley | @ArnoldNeumaier Thanks Arnold. I think Ron is right - I need to think on it a bit. | |
| Jul 17, 2015 at 7:55 | comment | added | Arnold Neumaier | There is an answer at physicsoverflow.org/10366 | |
| Jan 24, 2015 at 18:10 | history | edited | glS |
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| Oct 3, 2014 at 18:51 | history | tweeted | twitter.com/#!/StackPhysics/status/518111121249939456 | ||
| Aug 21, 2014 at 6:42 | answer | added | Incnis Mrsi | timeline score: 3 | |
| Aug 21, 2014 at 6:18 | comment | added | Incnis Mrsi | @Norbert Schuch: A system composed of N qubits is described by $2^N$-dimensional space, in other words, it has $2^N$ distinct/independent quantum states. Exactly the same thing as N classical bits that have $2^N$ states. I suppose a confusion between number of states and amount of information (that grows logarithmically). | |
| Nov 2, 2013 at 19:54 | comment | added | Norbert Schuch | ... you might also want to check arxiv.org/abs/1306.4352. | |
| Nov 2, 2013 at 12:55 | comment | added | Norbert Schuch | Yes, no, and no. Initializing one qubit dissipates $kT$ of energy, and thus, initializing N qubits dissipates an energy of $NkT$. (Note that if the energy would not scale linearly with the number of qubits, this would likely give rise to all kind of contradictions!) This is closely related to the question whether $N$ qubits "contain" $N$ bits or $2^N$ bits of information (and typically $N$ is the more appropriate answer) -- e.g., arxiv.org/abs/quant-ph/0507242 contains some arguments about that. | |
| Nov 2, 2013 at 11:57 | comment | added | Selene Routley | @Trimok Ca marche bien pour moi! How very interesting indeed, thanks heaps. | |
| Nov 2, 2013 at 10:04 | comment | added | Trimok | Interestingly, I found a paper ("unfortunately" in French), here, which states , that in the case of a hybrid optomechanic system (See fig $2$ page $7$, fig $5$ page $10$), the work necessary to initialize a qbit is proportionnal to the Rabi frequency, which plays the role of a temperature (formula $23$ page $12$) | |
| Nov 2, 2013 at 3:24 | history | asked | Selene Routley | CC BY-SA 3.0 |