In the recent paper "Riemann Hypothesis as an Uncertainty Relation" (http://arxiv.org/abs/1304.2435) the author claims that the presence of zeros out of the critical line may lead to the violation of a Heisenberg-type uncertainty relation. Is that work a proof of Riemann's hypothesis?
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1$\begingroup$ That paper seems making little sense to me, because in equation $(1)$, $\sum_{n=1}^\infty(-\log n)^k$ does not converge. $\endgroup$– 23rdCommented Apr 14, 2013 at 18:46
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1$\begingroup$ I have seen mathematics abused many times to "explain" physical phenomena, but this is an interesting take on the other direction! $\endgroup$– Asaf Karagila ♦Commented May 25, 2013 at 11:48
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3 Answers
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The author is transforming the Riemann hypothesis into an equivalent mathematical problem that satisfies the same math as the uncertainty principle. Even if you could show that this leads to a violation of the uncertainty principle, it has nothing to say about the physical uncertainty principle. Math similarity does not imply causality (in the sense that the uncertainty principle (HUP) of physics is due to something related to the Rieman hypothesis) ,nor the other way around (use the validity of the HUP to prove the Riemann Hypothesis
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$\begingroup$ Then you understand that the HUB cannot be used, in essence, as tool to help solve the Riemann Hypothesis? $\endgroup$ Commented Apr 14, 2013 at 18:33
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1$\begingroup$ Even if you could transform the RH into a mathematical problem that is equivalent to the violation of an uncertainty principle. Then to prove the RH you should also mathematically show that the principle is violated (but prove it mathematically, under certain assumptions). The fact that QM violates the HUP doesn't say anything about the RH because, because the violation could have multiple other causes, or be not exact, or whatever else, that are not exactly reflected in the initial mathematical analogy $\endgroup$– noeliaCommented Apr 14, 2013 at 18:41
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No. He says his result may lead to a violation of the uncertainty relationship, not that it actually does.
answered Apr 14, 2013 at 17:11
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1$\begingroup$ Could you explain better? What is missing? $\endgroup$ Commented Apr 14, 2013 at 17:15
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$\begingroup$ @user901366 What is missing from the explanation that
may$\ne$does? $\endgroup$– DidCommented May 25, 2013 at 12:58
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I developed an algorithm that is based on uncertainty principle, the result is consistent with RH. The basic idea of the algorithm is representation of objects movement in space and time using two operators. I still can not answer the question whether this could prove RH.
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$\begingroup$ How I can best know your work? $\endgroup$ Commented Sep 6, 2013 at 18:42