Timeline for States and Probability distributions that the 5-qubits IBM computer can produce
Current License: CC BY-SA 3.0
23 events
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| Apr 4, 2018 at 0:45 | comment | added | Rob | You can ask this question at: quantumcomputing.stackexchange.com . | |
| Jan 13, 2018 at 19:48 | history | edited | Gil Kalai | CC BY-SA 3.0 |
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| Sep 15, 2017 at 13:22 | comment | added | Gil Kalai | @JoeFitzsimons Joe, does IBM provides an opportunity to experiment on the cloud for they larger than 5-qubits quantum computers (perhaps 16 or 17 qubits)? Is it possible to tell what kind of quantum states can be reached by this computers? Also, there were news that IBM used the quantum computer for simulation certain molecules. What kind of states were involved in thes simulations? | |
| Jul 10, 2017 at 20:38 | history | edited | Gil Kalai | CC BY-SA 3.0 |
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| Jun 19, 2016 at 16:06 | comment | added | Joe Fitzsimons | @GilKalai: It is simply enough to run the stabilizer state, but the IBM system uses an approximately universal gate set where only T is non-Clifford. Also, programming is done via circuit diagrams. The result is that producing the W states you mention is extremely difficult in practice for normal users due to the need to approximate the required non-Clifford group gates. | |
| Jun 19, 2016 at 15:51 | answer | added | Joe Fitzsimons | timeline score: 7 | |
| S Jun 19, 2016 at 15:11 | history | suggested | Omar Shehab | CC BY-SA 3.0 |
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| Jun 19, 2016 at 14:34 | review | Suggested edits | |||
| S Jun 19, 2016 at 15:11 | |||||
| Jun 19, 2016 at 14:13 | comment | added | Peter Shor | In terms of density matrices, this is a 1023-dimensional space. I don't think stating it in terms of probability distributions makes it substantially easier. For arbitrarily many qubits, a generalization of this problem is probably unsolvable. This is not an easy question. | |
| Jun 19, 2016 at 10:29 | history | edited | Gil Kalai | CC BY-SA 3.0 |
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| Jun 13, 2016 at 19:22 | comment | added | Gil Kalai | Peter, Thanks for the clarification. It will be interesting to see how well can the IBM 5-qubit machine achieve this state then. My theory of decoherence is based entirely on the standard noise model (in fact I need to worry only about depolarizing noise). The non standard modeling of noise for larger systems emerges as a consequence of failure to lower the rate of noise to the level needed for quantum fault-tolerance for small systems. (There is nothing non-standard to start with.) | |
| Jun 13, 2016 at 18:16 | comment | added | Peter Shor | Gil: ctually, in standard quantum theory, that is a relatively noise-sensitive state. However, maybe you think there are states which are much more noise-sensitive in your theory of decoherence. I'd like to see what they are and why they're more noise-sensitive. | |
| Jun 13, 2016 at 5:22 | history | edited | Gil Kalai | CC BY-SA 3.0 |
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| Jun 13, 2016 at 5:19 | comment | added | Gil Kalai | Dear Peter, right! I suppose I have to think more carefully about good test cases for "noise sensitive" states for this case. | |
| Jun 11, 2016 at 19:36 | comment | added | Peter Shor | I hope you realize that the state you wrote down, in a different basis, is $(|+++++\rangle - |-----\rangle)/\sqrt{2}$, where $|+\rangle = (|0\rangle +|1\rangle)/\sqrt{2}$ and $|-\rangle = (|0\rangle -|1\rangle)/\sqrt{2}$. | |
| Jun 10, 2016 at 15:35 | comment | added | Frédéric Grosshans | The state you chose is a stabilizer state (stabilized by $-ZZZZZ$, $XXIII$, $IXXII$, $IIXXI$, $IIIXX$). Did you chose it on purpose ? Even if a quantum machine could produce and sample all stabilizer states, it could not prove any supremacy, because of the Gottesmann-Knill theorem. | |
| Jun 10, 2016 at 15:18 | history | tweeted | twitter.com/StackCSTheory/status/741288441091088385 | ||
| Jun 10, 2016 at 7:02 | comment | added | Andrew Morgan | I think I get where I was coming from and why it was confusing: I was generalizing the problem to one where "the IBM computer" was replaced by "some fixed, idealized quantum computer with 5 qubits". I didn't realize that this could (ostensibly) drastically influence the answer, while your question is particular to the IBM computer. | |
| Jun 9, 2016 at 20:53 | history | edited | Gil Kalai | CC BY-SA 3.0 |
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| Jun 8, 2016 at 22:04 | comment | added | Andrew Morgan | Sorry, I'm just trying to get my head around what all is going on. I'm assuming that you're trying to understand when, by starting from a simple proscribed initial state, applying some local unitary operations, and then sampling the state, a given distribution over bit strings can be reached. Qubit-efficient universal quantum computation is relevant since that helps understand what intermediate steps we can take. I just meant to give an example where (if I'm understanding your question and conjecture right) the output distribution might be more complicated than you suspect. | |
| Jun 8, 2016 at 21:50 | comment | added | Gil Kalai | I am not sure I understand you. (Anyway this condition may well be naive.) Also, I am not sure what kind of computations the IBM machine allows. Essentially I want to consider only probability distributions obtained by measuring a pure (as possible) state on the 5 qubits. (but maybe I miss something.) | |
| Jun 8, 2016 at 21:42 | comment | added | Andrew Morgan | In what sense might qubit-efficient universal computation fit into this? I'm thinking of the following: Imagine for a moment that that any boolean function on $n$ bits can be computed using only the output qubit and no ancillary bits as $|x b\rangle \mapsto |x(b\oplus f(x))\rangle$. Then one might compute the uniform distribution over all even-weight strings by preparing a uniform superposition in the first $n-1$ bits and $|0\rangle$ as the last qubit, and then computing parity of the first $n-1$ bits into the last bit. Am I understanding right that this would disagree with your conjecture? | |
| Jun 8, 2016 at 18:56 | history | asked | Gil Kalai | CC BY-SA 3.0 |