A proof of RH using quantum physics

  1. A proof of RH using quantum physics...:)

    Here you are a proof of Riemann hypothesis, using quantum physics,first we associate the roots of the Riemann function [tex]\zeta(1/2+is) [/tex] with the energies of a certain Hamiltonian with potential V and we prove v is real so all the energies will be real and Rh is true..you can take a look at my theorem attached to this message...

    also i prove that the roots of the function [tex]\zeta(a+is)[/tex] with a different from 1/2 can not be real....
     

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  2. jcsd
  3. mathman

    mathman 6,569
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    If your proof is valid, there is a prize of $1000000 for you. Try submitting it to a math journal.
     
  4. Gokul43201

    Gokul43201 11,141
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    Reading just the OP (not the file), it appears that you are merely using the result that a hermitian matrix has real eigenvalues, and not anything based on the postulates of QM.

    I may be mistaken, though.
     
  5. matt grime

    matt grime 9,395
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    Often when Eljose says he has a proof he means he has an idea about how someone might start a proof.

    There are methods of attack using this (well, not necessarily eljose's idea) QM idea, see eg Berry, Keating et al.

    Of course, there is a problem that eljose claims to have proven something that is false: there are real zeroes with a different from a half. And I'm not sure how proving that there are complex roots with a not equal to half would be of any use since that is the antithesis of RH,
     
    Last edited: Sep 8, 2005
  6. No,no i have proved that all the zeroes with of the function [tex]\zeta(1/2+is) [/tex] are real..and that the zeroes of the functions [tex]\zeta(a+is) [/tex] with a different from 1/2 are all complex, so there can not be a root of the zeta function in the form [tex]\zeta(a+is)[/tex] with s real.

    this last fact comes from the fact that for a different from 1/2 are "energies" of the form
    s+(2a-1)i so the potential for this a must be complex,but this implies that all energies would be complex as the mean value of b <b> (the complex part of the potential) is non-zero.

    The other part of the proof is that the roots of [tex]\zeta(1/2+is)[/tex] are all real (this is precisely the RH).
     
    Last edited: Sep 8, 2005
  7. matt grime

    matt grime 9,395
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    But the roots of zeta(1/2+is) are not all real (that isn't RH, which states that the nontrivial zeroes are in the critical region), s=(-n-1/2)/i is a zero of zeta (1/2+is) for any n in N.

    As you write in doc format I am unable to tell you where your mistake is.
     
    Last edited: Sep 8, 2005
  8. yes ,when i am refering to roots i am refering only to the non-trivial ones...in fact i have used that if [tex]\zeta(1/2+is) [/tex] is a root also [tex]\zeta(1/2+is*) [/tex] is also a root...but with -2,-4,-6,-8,... and so on (i think you are referring to these non-trivial zeroes) -2 is a zero but 3 is not.... throughout the paper i am refering only to the non-trivial zeroes these that lie on the critical line 0<Re(s)<1 if you are interested i can send you a .pdf file of my manuscript....
     
  9. Zurtex

    Zurtex 1,123
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    But the equation on line 25 only is true when Re(s) > 1, you however are incorrectly saying, and misquoting Riemann that it is true [itex]\forall s \in \mathbb{C}[/itex]

    Furthermore, assuming it was true, your next line goes on to say that "Have real part ½" which would also be untrue.
     
  10. Yes but you could continue the Riemann zeta function Analitically for Re(s)>0 by using Dirichlet Etha function....

    In fact Rh didn,t mean that all the NON TRIVIAL Zeros (sorry for this) had real part 1/2?...

    Ok i will correct the fact in line 25 by adding Dirichlet etha function
     
  11. matt grime

    matt grime 9,395
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    you are really making several mistakes at once here.

    given that you are using zeta(1/2+is) then the nontrivial zeroes are now not those on the critical line Re(s)=1/2, but in your translated setting on the line Im(s)=0

    i'm not surprised that you are assuming the series expansoin n^{-s} for all s, you've done it repeatedly in the past as well.
     
  12. i Really don,t understand math teachers,math referees and math journals..they keep saying "solve Riemann Hypothesis and th eworld will lend a path to your door..." but i have solved it,provided proof for its solution but no journal wants to publish it..they say "you don,t prove any theorem" or "your work is only especulative".

    I have provided proof for RH,the existence theorem says y```=F(x,y)=Ay+V(x)y with F and its partial derivative respect to y being continuous so the Differential equation exists, after that i prove that the potential V is real...(so far in this forum anybody has found a counterexample of my method) for a=1/2 and V is complex for a<>1/2 so the RH is proved.

    I really hate referees and math teachers,you solve something send it to them and get no response of them as if they hadn,t read your manuscript at all,i don,t know what more else to do,they are simply very snobish (for example Louis de Branges famous math teacher claimed to have a proof and they listened to him and publish his paper although Li and others found a counterexample providing it was not real) but to me that i,m simply an unknown math-physicist asking for an opportunity i got no response from them,i don,t know what else to do perhaps i will e-mail it to several teachers in my country (Spain) hoping they give me a chance to publish it under their name.

    You will think i,m greedy that i,m only interested in the prize,well if they give me an amount of money (sharing it with others that have provided a valid proof to RH) wouldn, be bad,but i prefer to get a bit of fame to be able to dedicate to mathematical-physics (String theory,NOn-renormalizable theories,Quantum Gravity and so on..) the question is what must a man do to be heard by math community?..
     
  13. matt grime

    matt grime 9,395
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    we (well, I) think you are not capable of judging the quality of your own work. you have repeatedly demonstrated that you do not understand many parts of mathematics. For instance not knowing that the common series expansion of zeta is not valid for re(s)<1, and in this case confusing s and 1/2+is. You have made many mistakes and admitted them; we do not think you are any different this time. Email me the damn latest thing in pdf if you must or put it somewhere and I'll tell you where it is wrong, though given your previous writings there is little chance of anyone being able to understand what you say. Not because you are using something earth shatteringly clever but simply because of the standard of writing and lack of references.

    to get any credit from the mathematics community the least you could do would be to bother to learn about other poeples' work and not claim originality for things that are not original, or that you method is the best for calculating pi(x) without bothering to quantify 'best' or prove that it indeed satsified this quantification.

    you might also wish to bother reading the reasons for your rejections of submissions (to what journals?) and try to learn how to not make this mistake again. You could try going to a conference and giving a talk; there are many conferences and you would not be rejected by all of them. try a post graduate conference in spain in your own language with your peers.

    i do not recommend this course of action, but if it appeases you and is the only way to get you to listen to people who might know something about the subject and with whom you can communicate then so be it. you will probably not like what you hear, and i advise against it and would suggest you learn some mathematics first.
     
    Last edited: Sep 10, 2005
  14. First of all you are saying i made the mistake of supposing the expansion of the zeta function in the form 1+2^{-s}+3^{-s}+.... is valid for Re(s)<1,you are right but using Dirichlet Eta function you can extend the domain of convergence to Re(s)>0 (i have corrected this fact).


    So far you and none of this forum has provided with a counterexample to my manuscript, and the journals (London math society journal,American Math society journal,Australian math society journal,Journal of number theroy of Burdeaux,Israel math journal,and so on...)also didn,t provide a counterexample so there is no reason to think that my results aren,t valid at all,i have checked similar works to mine (these using QM to solve RH ) but they only postulate the existence of an operator (not Hermitian) whose roots are the roots of the Zeta function [tex]\zeta(s)[/tex] but they don,t proof anything (tell me a webpage including a work on QM that proves that all the roots of the Riemann Zeta function s=1/2+it with t always real....

    When it comes to the fact that i dind,t get any response i consider it a lack of respect and a "snobish" actitude..what are they afraid of to listen to me?.
    I know i have made mistakes on the past (the pi function,but i would like to remember that i provide a formula for [tex]\pi(x^{a})[/tex] so the error would go like O(x^{d/a}) but there are no reasons to think why my formula would be incorrect...for the fact i don,t know mathematics i would like to remember you that i,m a physicist so i think i know enough QM to solve this problem.
     
  15. Zurtex

    Zurtex 1,123
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    Ask yourself eljose, if people actually thought after looking over this that it was close to a proof of the Riemann Hypothesis then why on earth would they just dismiss it so quickly?

    This proof of yours, to me, appears to be riddled with mistakes. I understand very little of the kind of mathematics here, but I can point out another mistake after line 25 is line 33, with your representation of the logarithmic integral.

    As for the journals acting snobbish, even if this was a proof of the Riemann Hypothesis it is neither rigorous enough nor of a good enough format to publish in a journal.
     
  16. shmoe

    shmoe 1,994
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    My understanding of de Branges work is that he proved some theorems that are actually correct and would imply RH if the zeta function (and other Dirichlet L-functions) satisfied certain properties, which it turns out they don't. That is, his work was correct, but doesn't apply where we want it to. I could be wrong though. You'll notice that his most recent announcement was pretty much ignored by mathematicians, as his method is now known to not work.

    Another big difference between de Branges and you is he is capable of writing clearly and formatting his paper properly for submission to the appropriate journal. The basics like spelling, grammer, punctuation, and appropriate formatting (using equation editor in word is rarely appropriate) are not to be ignored. English as a second language is no excuse if you try to submit to an English journal- they have standards that can only be bent so far.

    matt sounds willing to read your latest attempt but you don't seem to have supplied him with a format he can read? Here you have someone who is willing to actually read your work and yet you still aren't willing to meet this humble formatting request? There may be other ways to convert a .doc to a .pdf, but I know OpenOffice will work (and it's free).

    I'm personally tired of your attitude and your repeated claims to have proven RH (and other things). I'm also tired of reading poorly formatted documents along with your rants about why journals are ignoring them as submissions, I mean it's really not that suprising. You keep saying "i don,t know mathematics," well then learn! If you want mathematicians to pay attention there are basic standards that you're proving time and time again to be incapable of meeting. So learn how and do it.
     
  17. Hurkyl

    Hurkyl 16,089
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    First off, in your abstract, you don't even claim to have a proof, just an idea for a proof.

    Secondly, the thing you call the Schrödinger equation is not. The equation you wrote is merely the equation stating that En is an eigenvalue of the Hamiltonian.

    Thirdly, you claim to write "a easier formula for the potential V(x)", but in reality you have infinitely many formulae, one for each desired eigenvalue.

    Fourthly, you made a sign mistake in your "easier formula for the potential V(x)".

    Fifthly, for the partial derivative of F w.r.t. y to be continuous, then V(x) must be a continuous function, it cannot even have a single point of discontinuity.

    Sixthly, you've made several of these mistakes (and probably all):

    (1) You've not noticed that the differential equation:
    [tex]\frac{-\hbar^2}{m} \frac{d^2 \phi}{dx^2} + V(x) \phi = K \phi[/tex]
    can be solved for φ no matter what value is used for K.

    (2) You've not shown that the φn are square-integrable, which is a requirement for writing things like <φnn>.

    (3) You've not shown that when φ is restricted to be a square integrable function that the only eigenvalues of the Hamiltonian are the En.

    (4) You've assumed it's possible for a Hamiltonian to have all of the En as its eigenvalues.

    (5) You've assumed that V(x) is a real function.


    You've made so many silly mistakes that it's easily understandable why most people would not bother even responding: whether or not you are a crackpot, you sound like one.
     
  18. -I have proved that the potential is real....V*(x)=V(x) by proving that if E_{n} is an energy also E*_{n}=E_{k} will be also an energy..see it more carefully, another detail for a<>1/2 the potential is complex (you have not mentioned it)

    -In QM a potential can have discontinuities..(see the QM examples in the books for the function [tex]\delta(x)[/tex] or a potential V(x) that is equal to the Window function (tell a physicist that according to the existence theorem the solution of Schroedinguer Eigenvalue equation can not be obtained for these potentials).

    -I have considered the physical problem that the roots of the Riemann Zeta function are the eigenvalues of a Hamiltonian with a certain potential V(x) the eigenvalues E_{n} then would have all the same potential V(x) for a fixed a.

    -The condition of the [tex]\phi[/tex] to be square integrable functions is a physical condition or a boundary condition imposed to the differential equation (is an axiom of QM)

    -tell me where i have mispelled words or i have used bad english grammar..

    -The existence theorem provides that F(x,y)=Ay+V(x)y say that the potential must be continous so the differential equation exists..(prove a counterexample that V(x) is continous everywhere)...

    Is always the same thing if i were famous the hypothesis would have been published long ago..
     
    Last edited: Sep 11, 2005
  19. shmoe

    shmoe 1,994
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    The obvious things that leap out: "let,s", "i", missing articles (like "the") in many places, sentences run into one another, not capitalizing "Riemann" in some places, and so on. None of this is a huge deal in an informal setting like a forum such as this, but if this is an example of what you've been sending off to journals in hopes of being published you'll be lucky if they even bother to respond to you.
     
  20. matt grime

    matt grime 9,395
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    Hurkyl said that you have not shown that these phi's satisfy this axiom. It may be an axiom but you have not shown it is satisfied in this case. You do understand the requirement here, right?
     
  21. matt grime

    matt grime 9,395
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    "The" existence theorem? Whcih one? Have you proven that your functions and equations etc satisfy the hyptheses of the theorem?
     
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