A proof of RH using quantum physics

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  • #1
eljose
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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
mathman
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If your proof is valid, there is a prize of $1000000 for you. Try submitting it to a math journal.
 
  • #3
Gokul43201
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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.
 
  • #4
matt grime
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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,
 
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  • #5
eljose
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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).
 
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  • #6
matt grime
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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.
 
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  • #7
eljose
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yes ,when i am referring to roots i am referring 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 referring 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...
 
  • #8
Zurtex
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eljose said:
yes ,when i am referring to roots i am referring 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 referring 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...
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.
 
  • #9
eljose
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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
 
  • #10
matt grime
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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.
 
  • #11
eljose
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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?..
 
  • #12
matt grime
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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.
 
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  • #13
eljose
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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.
 
  • #14
Zurtex
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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.
 
  • #15
shmoe
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eljose said:
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.

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.
 
  • #16
Hurkyl
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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.
 
  • #17
eljose
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-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 continuous so the differential equation exists..(prove a counterexample that V(x) is continuous everywhere)...

Is always the same thing if i were famous the hypothesis would have been published long ago..
 
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  • #18
shmoe
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eljose said:
-tell me where i have mispelled words or i have used bad english grammar..

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.
 
  • #19
matt grime
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eljose said:
-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)


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?
 
  • #20
matt grime
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eljose said:
-The existence theorem provides that F(x,y)=Ay+V(x)y say that the potential must be continuous so the differential equation exists..(prove a counterexample that V(x) is continuous everywhere)...


"The" existence theorem? Whcih one? Have you proven that your functions and equations etc satisfy the hyptheses of the theorem?
 
  • #21
matt grime
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eljose said:
]-tell me where i have mispelled words or i have used bad english grammar..


dunno in the .doc. can't read it at work. i keep telling you .doc is not a valid file format for mathematical journals. did you even check the journals' publishing standards? But you cannot use apostrophes at all if your posting here is anything to go by.

Is always the same thing if i were famous the hypothesis would have been published long ago..

to use a vernacular of english, that is complete bollocks.
 
  • #22
matt grime
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Ok, am borrowing a windows machine so here goes:

the abstract:

"i" shuld be capitalized in line 1, quantum physics shouldn't be, nor should eigenvalues. the word "first" shold be preceded by a fullstop and then capitalized.

now, the assertion that "all roots have real part equal to 1/2" is false given there are trivial zeroes.


section titled Riemann hypthesis

"Have" after the first displayed equation should not be capitalized.

"which is similar to prove" makes no gramatical sense, or mathematical. I don't care what it is "similar to prove" but "which is equivalent to proving"

you are still ignoring the roots at {-1,-2,...}

the comma before "Riemann" after the word "real" needs to be a fullstop and there ought to be a space there.

"if *IT*is true", not "if is true"

lastly for spelling etc that is not how you spell schrodinger.

onto maths. you declare zeta in the dirichlet series but then talk about it in the region where the dirchlet series is not valid without mentioning its analyitc continuation to other values.

now, it appears that you must have that V(x) is continuous, what is the V(x)? I ask because at no point do you spcify what it is. you keep saying "if V" and so on but never show that any such V exists, and no you can't appeal to the existecne theorem since that requires V to be given, so simply and clearly state what V is. so how do you know there is a potential that has exactly the specifed eigen values?

further, there is nothing in your reasoning that appears to explain why it is your method fails to detect the trivial zeroes.


it is very hard to make anymore detailed matematical observations than those since it is almost unintelligible, I'm afraid.
 
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  • #23
eljose
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I use the property that if [tex]\zeta(1/2+is)=0[/tex] then [tex]\zeta(1/2+is*)=0[/tex] this only happens for the non-trivial zeroes.

The potential V(x) is continuous but on a number of points...this perhaps is the weaker point of my argument so the existence theorem works (proof a counterexample).

The assumption that the eigenfunctions are square-integrable comes from the postulates of QM , the probability P of finding the particle in all R must be 1,you can impose to your equation as a boundary condition.

i always send a .pdf version of my work to journals...

Referees of journals are very smart but can not give a counterexample to my affirmations (that the potential for this case is always discontinous and that the eigenfunction are not square integrable and so on)...they only (as you do) keep only saying "you did not do it and that" if they had been so rigorous with Euler,Gauss and Ramanujan (at last to this one a teacher helped him) they would never have become famous..i really hate you instead of supporting a proof to RH based on QM (and QM is true) you keep as these nasty referees do saying fails and fails,math and physics are made of supositions, the only condition to V(x) to fulfill is that is continuous and not even everywhere,it can have some discontinuities ( i would be gratefull if someone proved that the potential that generates all the roots of the Zeta function is discontinous everywhere instead of keep saying only "this is not possible" and so on)..

And another thing to say to famous mathematician L de Branges they found a counterexample,because the Zeta function according to him should fulfill a serie of condition that did not, i have used the existence theorem and that the potential is continuous but on a number of point (a supposition perhaps..but could you prove is untrue?) and a thing more to say,even if it were discontinous everywhere but was integrable (in the sense of Riemann or Lebesgue) you could still calculate the roots (energies) in the form: [tex]E_{n}=<\phi|H|\phi>[/tex]
 
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  • #24
matt grime
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Can you ever accept that someone else's criticism is relevant? From the article as you write it it is impossible to

1. decide what V is

2. why a V must exist that has its engeries at exactly the "non trivial roots of zeta(1/2+is)".

There is no proof of the second and no explanation of the first.

Give the full statement of the existence theorem.

They cannot affirm or deny your statements sicne they are incomprehensible.

And that doesn't correct any of the other mistakes in it either, such as using the wrong definition of the zeta function..
 
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  • #25
eljose
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-The existence theorem (as far as they taught me) is that given a second order differential equation y``=F(x,y) then F and its partial derivative of f respect to y dF/dy must be continous,in my case we have a constants (complex in general ) A and B so F(x,y)=AV(x)y+By where the constant B includes the energies E_{n},if we assume potential V is continuous (is this affirmation impossible) then the existence theorem would hold and no matter what E_{n} would be the differential equation y``=AV(x)y+By would exist so there would be a Hamiltonian whose energies would be precisely the root of Z(1/2+is), in fact i have used the famous HIlbert-Polya conjecture to obtain this Hamiltonian...

i say the potential must be chosen so the Eigenvalues of the Hamiltonian are the roots of Z(1/2+is),perhaps a quantum mechanics teacher would say to you that my argument is right,critics hurts but at least you answer me and not these snobbish referees that give me no reason for rejection or those teachers i send my work to that don,t even answer me..perhaps i should try a math physics or a physics journal to send my manuscript.

The main key in my manuscript is to show that V is real ,the existence theorem (assuming V is continous) holds for this V so the differential equation has (infinite) solution so there is a potential continuous (almost everywhere) that has the roots of Z(1/2+is) as its eigenvalues,the only possibility (and it would be an exceptional case of bad luck) would be that the differential equation has solution nowhere in R.
 
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  • #26
matt grime
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the differential equation is, up to constants k_1 and k_2:

y''+k_1V(x)y=k_2y

and this has a solution IF k_2y-k_1V(x) satisfies some conditions. That does not tell us that suich a V exists, it tells us if a V exists then there are eigen functions. So, what is the V(x) that has thecorrect eigenf****ions? How do I know that given some discrete set of points (the zeroes to zeta(1/2+is) that I can create a potential that has these and exactly these as its spectrum?

YOu have given a series of conditions that allegedly if met will prove RH. YOu have not shown any of these can be met. That is not a proof.
 
  • #27
eljose
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A V(x) that is continuous everywhere but a number of point so the differential equation..with that we would have enough proof,the only supposition we make is that V(x) is continuous if this happens the differential equation:

y``+K_1V(x)y=K_2y exists with K1 and K2...

in fact i have put another post in QM forum abouth this problem (the problem of existence)

All makes clear assuming V(x) is continous... :biggrin: :biggrin:

Another question let be the classical problem similar to our quantum one with the potential V(x) and let,s put ourselves in the worst case (the potential is nowhere continous) then we would have the Hamilton Jacobi equation:

[tex]S=-Et+W[/tex] with W satisfying (being V(x) our potential)...

[tex]\frac{1}{2m}(\frac{dW}{dx})^{2}+V(x)=E [/tex]

then we can clearly find a solution to W in the form:

[tex]W=\int[2m(E-V)]^{1/2}dx [/tex] so we (in spite of what the existence theorem says) have found a solution to the Classical problem.

another example for the Quantum Schroedinguer equation we can always obtain an "approximate" solution in the form: [tex]\psi=exp(iW/\hbar)[/tex]
 
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  • #28
matt grime
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Eh? You have still not shown that any such V(x) exist continuous or otherwise that has these the correct eigenfunctions. You keep saying there is one but have given no prrof that any such thing exists.
 
  • #29
eljose
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If you assume "A priori" that the potential is continuous all the doubts disappear...why can not we assume that potential V(x) is continuous but on a number of points?...

as for the eigenfunctions you can write them as i pointed you in the form [tex]\Psi=exp(iW/\hbar) [/tex] where W is this one given above.
 
  • #30
matt grime
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You really don't get it do you?


Right, let's try again.

PROVE that there exists a potential V(x) with the correct properties of continuity (whatever you want them to be) that has exactly the values of s as its eigenvalues as you require. YOu haev said suich a V exists but not justified this step. The existence theorem you keep invoking only applies to the phi (or y) once the V has been given. What is the V?

I mean, if I assume that given any d there are infinitely many arithemetic progression of primes of common difference d then I have just proved the twin primes conjecture. So, prove that your "if" statements are true.
 
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  • #31
eljose
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V(x) could be a given continuous function...i don,t know how to calculate it but the existence theorem exiges it to be continuous 8but on a finite number of points..I DON,T Know what the potential is if i knew i wouldn,t have needed to prove is real...i don,t know how to calculate it...all i know is that it must be continous...
 
  • #32
matt grime
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exiges? what does that mean? so you're claiming that this V exists but you can't prove it? Come one, prove it exists then. Don't tell me what it is if that is too hard, which is frequently the case in maths (I know that the quantity pi(x)-li(x) changes sign infinitely many times but I don't know when it happens first). So, prove that there is a V by some existence theorem then. The only theorem of this kind you've stated is that given

y''=F(x,y)


with F satisfying certain properties, then there is a solution in y. But that doesn't prove that a V exists for the situation we are in. Indeed V is part of the definition of F, and the y is the phi.

So, again, why does there exist a V with the propeties you require?
 
  • #33
matt grime
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one more thing, on journal submission, is that maths journals require either camera ready print or latex sources with the correctly formatted preamble and correct usepackage arguments. pdf is not admissable in any journal that I know of. But then I don't create pdfs.
 
  • #34
shmoe
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eljose said:
...they only (as you do) keep only saying "you did not do it and that" if they had been so rigorous with Euler,Gauss and Ramanujan (at last to this one a teacher helped him) they would never have become famous..

This is nonsense. Euler lived in a time with a very different standard of rigorous. I'm sure if he were alive he'd be up to the task of meeting these standards. Gauss only published very polished work, do you have an example of something he's published that you wouldn't consider rigorous? I can't think of any. Ramanujan is in a league of his own, essentially untrained in mathmematics yet totally brilliant. To attempt a comparison with his notes, which contained many profound things (yet lacking in proof), to what you've been trying to pass off is delusional and makes you sound even more like a crackpot.

eljose said:
And another thing to say to famous mathematician L de Branges they found a counterexample,because the Zeta function according to him should fulfill a serie of condition that did not,...

Yes and? You think it's up to referees (and random mathematicians) to produce counter examples to every piece of junk that comes in the mail? Time was spent on de Branges method because it showed some promise and it was original, I assure you it wasn't his name alone that caused people to look up and show it didn't satisfy the conditions he thought it would (also note that initially he didn't claim to have proved RH in his '94 paper and mentions that the conditions he requires were still in doubt). The burden of proof is on the author, not the reader.
 
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eljose
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i can,t prove it :( :( :( :( :( i can not prove the existence of the potential even with the variational principle for Schroedinguer equation:

[tex]J[\phi]=\int_{-\infty}^{\infty}dx(V(x)\phi^{2}/2-(\hbar^{2}/4m)(D\phi)^{2})[/tex] with the condition [tex]\int_{-\infty}^{\infty}dx|\phi|^{2}=Constant [/tex]

we don,t prove anything,in fact i think that in math you sometimes MUST make assumptions and after that prove or disprove them..at least in physics is what we do.

I can,t prove if the potential V exists or not only that if exist must be continous.:( but i have a question for you how would you "disprove" that the potential does not exist?..
 
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