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

i can,t prove it :( :( :( :( :( i can not prove the existence of the potential even with the variational principle for Schroedinguer equation:

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

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?..

 an "approximate" solution to the potential is putting $$\psi=exp(iW/\hbar)$$ (1) as its WKB solution (this solution works fine for "big" masses for example m=1pound) $$W=\int[2m(E_{n}-V)]^{1/2}$$ or differentiating (1) we would get the potential: $$(-\hbar^{2})\frac{d\psi}{\psi}=2m(E_{n}-V)dt$$ from this we can get the potential and substitute in our second order differential equation so we get a new "approximate" equation in the form: $$a\frac{d^{2}\psi}{dt}+b(1/\psi)(\frac{d\psi}{dt})^{2}+c\psi=0$$ then you now can apply existence theorem to prove that the $$\psi$$ exist...but if the Psi functions exist by the expression (1) the potential also exists... I really hate mathematicians they help physicist of course but are always with the nasty rigour,rigour and rigour....as Fourier once said (sorry for the quote if is not exact) "they prefer a good building with a good entrance and a good appearance rather than the building is useful", the extreme rigour is bad it doesn,t help to the development of mathematics...

 Quote by eljose I really hate mathematicians they help physicist of course but are always with the nasty rigour,rigour and rigour....as Fourier once said (sorry for the quote if is not exact) "they prefer a good building with a good entrance and a good appearance rather than the building is useful", the extreme rigour is bad it doesn,t help to the development of mathematics...
LOL!! I am really not sure, how a mathematician would react to that particular statement (IANAMathematician), but you know there are reasons why physics works they way it works and why mathematics works the way it works. In terms of logic, physics works using inductive logic (infact most science works in inductive logic), which makes sense since we are observers of certain facts and try to "induce" generalisations from those facts and make further predictions. In other words, its a bottom up approach, we start with the leaf and try to reach the roots (the roots in case of physics being convergence towards a theory of everything). Mathematics, doesnt and shouldnt work using inductive logic. This is because any non-sense in mathematics only generates more non-sense and there is no way to cross check things until some good amount of time has been lost. This is not certainly the case with physics, where predictions can be verified and a hypothesis can be overthrown or accepted based on results. However, this does not stop mathematicians from thinking in an inductive way and thats the source of their creativity in most cases. Infact, most abstractions in mathematics were realised in an inductive way nonetheless, they were pinned down with axioms and rigorous arguments. If it werent so, we would have never seen the face of applied mathematics at any point of time in history.

As for fourier's statement, i am not sure whether he said it or not, but really if he did say it and in the context in which we are talking, then he, i am afraid, was certainly wrong. I wouldnt mind saying this to him, even if he were standing in front of me right now. As an engineer, i respect him, but this statement only shows his short sightedness towards the world of mathematics.

The point of the whole post is "leave mathematics alone, its a great tool for many real developments, but how mathematics itself develops is well left to those who know it".

-- AI

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 Quote by eljose I really hate mathematicians they help physicist of course but are always with the nasty rigour,rigour and rigour....as Fourier once said (sorry for the quote if is not exact) "they prefer a good building with a good entrance and a good appearance rather than the building is useful", the extreme rigour is bad it doesn,t help to the development of mathematics...

Extreme rigour? My God, all we're asking you to do is justify your claim. Since you want to be a physicist in this go on give me some evidence that your claim is true (this actually is not impossible since the zeroes of the zeta function are related to spectra in QM). However, since you want to use the word proof you must accept that you must meet th current burden of proof.

Quoting Fourier out of context doesn't help you. There are times in physics when certain mathematical requirements are ignored, eg convergence of series, stating that a function equals its fourier series, arbitrarily truncating infinite divergent sequences, using intuition for what ought to be true and so on. Indeed mathematicians even do this to give simplified explanations of phenomena. HOwever, just as a physicist mustmake sure his loose reasoning fits with any evidence, so must a mathematician makes sure he can do it rigourously as well. I would hazard that this is what Fourier was talking about. Not that wild speculation and false claims were better than rigour but that sometimes it is better to play around with things and ignore the minor details that will generally take care of themselves.

 i have justified my claim two post above yours....take the solution $$\psi=exp(iW/\hbar)$$ then you have that the potential is proportional to: $$(a/\psi^{2})(\frac{d\psi}{dt})^{2}+b$$with a and b constants including the energies E_{n} add this quadratic term to the SE and apply existence theorem...then you see that the $$\psi$$ exists so the potential function will also exist...unless you are again my approximate deduction...well if you think my approach is wrong due again to the nasty rigour i will say that WKB approach is valid for m=1 (although not exact) and that SE is exact for whatever real mass.. I have also send my manuscript to several physics journals to see what they,ve got to answer me $$V=(a/\psi^{2})(\frac{d\psi}{dt})^{2}+b$$ (this comes from WKB approach).. $$W=[2m(E_{n}-V]^{1/2}$$ is related to the Zeta function as the Energies E_{n} are the zeroes of the Z(1/2+is) as you can see the W function is related to the eigenfunctions $$\psi$$ and the potential V so you can form in this case an "approximate" differential equation (non-linear) in the form: $$c1\frac{d^{2}\psi}{dx}+c2(a/\psi^{2})(\frac{d\psi}{dt})^{2}+c3\psi=0$$(3) where here i have used that WKB approach and SE equation then apply existence theorem to this approximate equation to get that the $$\psi$$ will exist and as the potential is realted by WKB approach to the $$\psi$$ functions you get that also the potential MUST exist.....
 Recognitions: Homework Help Science Advisor And W is what? remember we aren't all physicists. Since your post doesn't even mention zeta at all we are left wondering what the hell W has to do with the zeroes of the zeta function.
 As an example we can always say that for a WKB approach the function always exist: let be the WKB idfferential equation: $$\ey+b(x)y=0$$ with e<<<<<<<<1 for example e=10^{-34} then its WKB solution is: $$y=exp(\int[b(x)]^{1/2})$$ $$(y/y)^{2}=b(x)$$ then substitute this solution into the original equation for the b(x) function we would get $$\epsilon(y+(y)^{2}/y)=0$$ from this last equation and the existence theorem you would get that the y function exist for whatever b(x) in the approach WKB but if y exist then b(x) also exists. $$y=exp(\int[b(x)]^{1/2})$$ $$(y/y)^{2}=b(x)$$ $$\ey+b(x)y=0$$ $$e(y+(y)^{2}/y)=0$$
 Recognitions: Homework Help Science Advisor and what has all that got to do with creatin anythgin with the spectrum the nontrivial zeroes of teh zeta function? still can't see any mention of zeta in that at all. not that any of your latex works. If you can't simply and clearly starting from the basics explain why you can create a potential V(x) such that the eigenvalues of some differential operator are exactly the non-trivial zeroes of the zeta function, and that this operator is hermitian, and that thus the zeroes do indeed al have real part 1/2 you should not bother posting. That is all that is asked of you: to explain what you claim to have shown.
 for any potential V this includes the one that provides the roots of the zeta function as its "energies" we can use the WKB approach so $$\psi=Exp(iW/\hbar)$$ (1) then we take a look to our existence equation $$F(x,\psi)=AV(x)\psi+B\psi$$ from equation (1) we would get that $$V(x)=c+d/(\psi)^{2}(\frac{d\psi}{dx})^{2}$$ then we know substitute our expresssion for the potential in F(x,y)... $$F(x,\psi,d\psi/dx)=A(c\psi+d/(\psi)(\frac{d\psi}{dx})^{2})+B\psi$$ now if we apply the existence theorem we would get that F and its partial derivative respect to $$\psi$$ are continous (the WKB solution is never 0) so for any potential (including this that gives the zeros of Riemann zeta function) we get that the eigenfunctions exists.. and if $$\psi$$ exists also $$V(x)=(\hbar^{2}/2m)\frac{d^{2}\psi}{dx^{2}}+E_{n}$$ from SE equation...we have just proved that the functions exist.. and of course the potential is unique as also the Psi functions depends on n and Energies are a functions of n E_{n}=f(n) as you can see matt i have given an expression for the potential in terms of the second derivative of the Psi function and the roots of the Riemann zeta functions E_{n},usgin the existence theorem we prove the Psi functions exists and using SE equation we prove that the potential can be written in terms of these functions... The L^2(R)) functions are $$\psi=Sen(W/\hbar)$$ where we have called $$W=\int(2m(E_{n}-V))^{1/2}$$
 Recognitions: Homework Help Science Advisor that is meaningless. what is psi? what is W? for god's sake are you incapable of explaining anything? what existence theorem (state it in full) what makes you think the relevant functions satisfy the htpothesis of the existence theorem that you've delcined to state properly in this thread. HOw can an existence theorem whcih surely can only tell you if psi exists GIVEN a V tell you that V exists? what now is f as ub E_n=f(n)? what you've said is given psi, then V exists, given V exists then psi exists. that is circular logic and is complete bollocks.

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to illustrate the circularity:

[QUOTE=eljose]for any potential V this includes the one that provides the roots of the zeta function as its "energies" we can use the WKB approach so $$\psi=Exp(iW/\hbar)$$ (1){quote]

so we're assuming that V exists, why does V exists? anyway, V apparently exists so we can deduce the psi exists

 then we take a look to our existence equation $$F(x,\psi)=AV(x)\psi+B\psi$$ from equation (1) we would get that $$V(x)=c+d/(\psi)^{2}(\frac{d\psi}{dx})^{2}$$ then we know substitute our expresssion for the potential in F(x,y)... $$F(x,\psi,d\psi/dx)=A(c\psi+d/(\psi)(\frac{d\psi}{dx})^{2})+B\psi$$ now if we apply the existence theorem we would get that F and its partial derivative respect to $$\psi$$ are continous (the WKB solution is never 0) so for any potential (including this that gives the zeros of Riemann zeta function) we get that the eigenfunctions exists.. and if $$\psi$$ exists also $$V(x)=(\hbar^{2}/2m)\frac{d^{2}\psi}{dx^{2}}+E_{n}$$ from SE equation...we have just proved that the functions exist..
so now youre concluding that since psi exists then V exists. but the only reaosn you thought V existed was becuase the psi exists.

see, that is circular.

so

1. why is there a V such that the spectrum of the differential operator is exactly the zeroes of the zeta function? you have not explained this. AGAIN.

 NO,with the existence theorem i have proved that the Psi function exist....and from the existence of Psi we deduce the existence of the potential... you will agree that you can express $$F(x,\psi,d\psi/dx)=A(c\psi+d/(\psi)(\frac{d\psi}{dx})^{2})+B\psi$$ by using the WKB function [/tex]\psi=exp(iW/\hbar)[/tex] and another quote you have the integral equation for the potential... $$\Int_C[2m(E_{n}-V(x)]^{1/2}=2\pi(n+1/2)\hbar$$ from this you could deduce the existence of the potential couldn,t you?...as we know that the "energies" or roots of the Riemann function exists where C is a line between two points a and b where E=V..now apply the eixstence theorem for integral equations and you get that the potential will exist in fact our integral equation is of the form: $$\int_{a}^{b}dxK(n,f(x))=g(n)$$ where g(n) is (n+1/2)hbar
 Recognitions: Homework Help Science Advisor eljose, I have no idea what the WKB function is, as i said before. psi is a function that satisfies a differential operator. the V(x) is part of the definition of the differential operator. thus the V must exist and satisfy certain properties before you can conclude psi exists. hencve you cannot use the existence of psi to deduce the existence of V(x) by rearranging a differential equation. that is what you have said you are doing. you might not intend that but that is because you comminucation skills are not sufficioent to write mathematics in english and expect people to understand you. that is an observation, not a criticism. either prove V exists or show why does psi exist in a way that is independent of V(x)
 Recognitions: Gold Member Homework Help Science Advisor Are you talking about WKB-approximations, eljose?
 Yes Arildno i,m talking about WKB approach as it can be done for mass m=1 so $$\hbar^{2}$$=10^{-68} (small enough isn,t it).... then by using Bohr.Sommerfeld quantization formula (valid for WKB) you get that the potential must satisfy an integral equation of the form: $$\int_{a}^{b}dxK(n,V(x))=g(n)$$ from this integral equation we could obtain numerical valours for the potential V(x) and "construct" it (unles Mr. Matt Grime the super-mathematician defender of the extreme rigour in math disagrees)
 Recognitions: Homework Help Science Advisor I do not deny that given a set of complex numbers satisfying certain properties that one can construct a potential with these and only these as the as its spectrum. HOwever, I have yet to see any stated reason why this will work for the zeta function, and that from it one can clconlude the Riemann hypothesis. approximate solutions are not accepetable. opinion of what is and isn't rigorous is completely irrelevant, jose. since you are claiming to do maths you must accpet the standards of mathematics. if we were taking "suggestive" reasons then of course RH is true since we know it is emprically true for such a ragne of s that it cannot help but be true for all s. but that is different from proving it. you are the person who keeps claiming to have a proof. if you dont' like what that really entails tehn stop abusign the word.