A proof of RH using quantum physics

[eljose]

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...

 

[matt grime]

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?

 

[matt grime]

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.

 

[shmoe]

...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.

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.

 

[eljose]

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?..

 

[matt grime]

i don't have to. you have to prove it exists if you are claiming a proof of the riemann hypothesis. making assumptions you can't prove and then deducing a theorem is mostly useless since usually you have now got to show something harder than your original claim. of course that isn't always the case. if you can show the converse is true (ie the RH implies that there is a potential) then you have created an "equivalent" problem. you hve not either explained why your assumption is easier to prove (ineed, you've not proven it), or shown that it is an equivalenet proble. Looks like the million dollars is safe after all...

when will you get tired of claiming a proof for something you barely comprehend?

and when will you shut up about this conspiravcy by snobbish mathematicians? If I, someone who knows no nubmer theory or quantum physics, can spot the errors in your idea whcih is not written in a manner that conforms with any of the basic requirements of publication, what makes you think that anyone else should bother to read it and even then to reply to your submission? the errors are easy to spot and they will dismiss such a submission out of hand as they ought to.

incidentally, can you disprove that there are infinitely many arithmetic progressions of primes with any common difference? If not then I claim I've just solved the twin prime conjecture. see how easy it is? I'm sure i can write down a couple of "trivial" observations whcih, if true, would net me about $6,000,000 (US) from the clay instutite. of course i could never prove any of them is true, but according to you that isn't important, is it?

 

[eljose]

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...

 

[TenaliRaman]

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, doesn't and shouldn't 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 that's the source of their creativity in most cases. Infact, most abstractions in mathematics were realized in an inductive way nonetheless, they were pinned down with axioms and rigorous arguments. If it weren't 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 wouldn't 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

 

[matt grime]

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.

 

[eljose]

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...

 

[matt grime]

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.

 

[eljose]

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.

$$y=exp(\int[b(x)]^{1/2})$$ $$(y`/y)^{2}=b(x)$$

$$\ey``+b(x)y=0$$ $$e(y``+(y`)^{2}/y)=0$$

 

[matt grime]

and what has all that got to do with creatin anythgin with the spectrum the nontrivial zeroes of the 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.

 

[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)

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 continuous (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 $$W=\int(2m(E_{n}-V))^{1/2}$$

 

[matt grime]

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.

 

[matt grime]

to illustrate the circularity:

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 continuous (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 because 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.

 

[eljose]

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

 

[matt grime]

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)

 

[arildno]

Are you talking about WKB-approximations, eljose?

 

[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)

 

[matt grime]

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.

 

[eljose]

why is not my method acceptable?..from the integral equation above you could construct the potential V(x) by solving a non-linear equation for numerical values V(xj) j=1,2,3...n or for example we could use an existence theorem for integral equations so we can deduce that the potential exists...wouldn,t that be acceptable?. we have the system of equations for the Potential in the form:

$$\sum_{j}K(n,V(x_{j}))=g(n)$$ now set n=1,2,3...k so we have the roots E1,E2,E3,...Ek upto afinite number k (we know the roots of Riemann function upto 10^{12} or even more.

I have a proof of RH the only "inconsistency" according to you is that i have not proved that the potential exist...but here i am giving an example of how to obtain its numerical value,now with those numerical value we could "modelize" our potential...

The other chance is to use the existence theorem of integral equations for the Kernel K(n,V(x)) and show that this integral exist..a question on rigour,if you prove that can calculate numerical values of the potential wouldn,t you have proved this potential exists?...

 

[matt grime]

your method isn't acceptable because it doesn't prove anything. i can tell that by the way you haven't proven anything; bit of a give away. all i see is a load of:

if i do this, then i might be able to do something, that might up to an error term do something else, but i can't prove any of it.

that isn't a proof.


given the zeroes, I'm sure you can construct some differential operator as required, however even if that were the case then you've not shown how one may determine that it is hermitian.

to do that one would need to make V(x) a function of zeta. zeta has not appeared at any point in your calculations.

since you are not assuming that the non trivial zeroes lie on the critical line you are somehow creating a hermitian operator from these values. since zeta has not been mentioned at all in your working then the same logic applies to any complex numbers, and indeed one can conclude, accoridng to your logic, that all complex numbers are in fact real.

you see, I don't need to show that something cannot be done (since it might be doable) to show that a method given is nonsense. as yours is as best we can tell.

example: let S be the set of zeroes of the function

sin(iz)

create a potential according to our method, and from this i conclude that all zeroes of this must satisfy z=z* ie are real since the operator so created is hermitian. however, I know where the zeroes are, thank you, and they are certainly not at real numbers, they are at purely imaginary numbers.

so, why doesn't your method apply here? i see no reason for it not to.

 

[EL]

Credit to matt grime for his patient arguing with a wall!

 

[matt grime]

And I further see no reason why your method picks out exactly the non-trivial zeroes and nothing else. but then the dependence on zeta has never been explained.

so, if you want to make this metho work:

take g(z) some function frmo C to C.

explain how we may take a restricted set of its zeroes, call them T, and from T create a differential operator

$$L=\partial^2_x + V_g$$

such that the spectrum is exactly T

now specify the conditions on g and T that allow us to conclude L is hermitian, and hence T must be a set of real numbers.

You have not in any decipherable way done any of those things.

I am perfectly willing to believe that it can be done. but I have not seen you do it in a way that convinces me at all.

 

[eljose]

let be perturbation theory:(at first order of the energies)...

$$E_{n}-E^{0}_{n}=\delta{E(n)}=<\phi|V|\phi>$$ (1)

where E_{n} are the "energies" roots of the Riemann Z(1/2+is) and E^0_{n} are the Eneriges of the Hamiltonian H0=P^{2}/2m then (1) provides an integral equation for V we could find a resolvent kernel R for this so:

$$V(x)=\int_{-\infty}^{\infty}dnR(n,x)\delta{E(n)}$$

with the resolvent Kernel: $$R=\sum_{m=0}^{\infty}b_{m}(K-cI)^{m}$$ is the Taylor expansion of the operator K^{-1} and K^{m} is the m-th iterated kernel,I is the identity operator and c is a real constant so the series converge for ||K||<c

$$\delta{E(n)}=K[V(x)]$$

c is a constant so the series converges,the radius of convergence is given by the norm of K operator ||K||<c

we have proved that V exist at first order in perturbation theory...

with K the kernel given by $$K(x,n)=|\phi(x,n)|^{2}$$
phi are the eigenfunctions of the Hamiltonian H0 epending on n i have applied Neumann series to the operator to obtain the resolvent Kernel...

 

[David]

I'll let Matt continue to point out the circularity of the argument being presented here adn tackle it from a technical viewpoint but will add one other little piece of information to the discussion, just to demonstrate the futility of this in a completely different way.

Supposing this were a valid proof, then credit for it should go to other people who have discovered it long ago. The connection between certain quantum systems and the RH is even well-known enough to be discussed at length in some of the recent popular books on the Reimann Hypothesis. So supposing this proof were correct, then it has already been demonstrated by numerous other people.

But seeing as it's not a valid proof (which all these other people have recognized), there's not much to worry about. None of those people are going to claim you "stole their proof".

 

[eljose]

if they discovered the proof long ago...why did they not publish it?...in a post above i have proof that the potential exist and even calculated it to first order in perturbation theory depending on $$\delta{E(n)}$$ and the E_{n} and E^0_{n} are known even more if we call Z the inverse of the function $$\zeta(s)$$ then we could write: $$E_{n}=i(1/2-Z(0))$$ (i have oly inverted the function, and i have used simple integral equation theory to prove the existence of the potential...

The problem with Matt (as happen with most of math teachers) is that they have assumed certain conceptions in math and if you are out of these,you are nothing,i don,t know what argument will now matt grime have to say my maths are wrong,but is only an "approach to the potential" (is would be only correct to first order in perturbation theory, the whole serie of values of energy is perhaps even divergent) and the WKB is also an approach,to say that we can choose some functions that are on L^{2}(R) function space...$$\psi=Asen(S(x)/\hbar)$$ for example.

another question as we are dealing with rigour an other things...can anyone of you brilliant,smart intelligent mathematician to prove the existence of infinitesimals,i,ll put even more easier,write (with numerical value) an infinitesimal...

 

[Zurtex]

Hyper real numbers have infinitesimals in them:

http://mathforum.org/dr.math/faq/analysis_hyperreals.html

http://en.wikipedia.org/wiki/Hyperreal_number

But what has that got to do with this?

Anyway, eljose, I am unsure if the mathematics in your post makes any sense or not, it could well do, but I can clearly see it does not show that all s in the critical line of $\zeta (s) = 0$ satisfy $\Re (s) = 1/2$.

And with all due respect, my 12 year old brother could see sentences in your proof that logically made no sense.

 

[eljose]

hyperreal numbers including infinitesimal..but could you write "an" infinitesimal?..

i have shown that the potential (and calculated it too) for $$\zeta(1/2+is)$$ is real,(the proof is that for any existing E_{n} also E*{n}=E_{k} is also an energy from this we deduce using the expected value of the Hamiltonian that V=V* so V is real,as you can see this only happens with a=1/2 the other cases there are complex energies in the form E*{n}+(2a-1)i, but a complex energies will come from a complex potential so $$\zeta(a+is)=0$$ can not have any real root except a=1/2 that have all the roots real as the potential is real) then i have also calculated (given an integral expression for) the potential upto first order in perturbation theory.

I am checking my grammar and spelling to make it the most clearer of possible zurtex,i am not ofended .