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RE: [PrimeNumbers] Re: Cracking RSA: Relationship between prime numbers and quantum theory

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  • Paul Leyland
    ... Not true in general. It is true for fermions (particles with spin (2i+1)/2) and is known as the Pauli exclusion principle. For bosons (particles with
    Message 1 of 12 , Aug 1, 2001
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      > One of the postulate in quantum theory states that no two
      > particle can occupy the same place at the same time.

      Not true in general. It is true for fermions (particles with spin
      (2i+1)/2) and is known as the Pauli exclusion principle. For bosons
      (particles with integral spin) particles can, and do, occupy precisely
      the same state at the same time. It's why lasers, superfluids and
      superconductors have such interesting properties.

      I fail to see what any of this has to do with factorization, but it will
      be interesting to see if anything comes from it.


      Paul
    • Aleksey D. Tetyorko
      ... Hi, Kent! I took a look at your paper. As I see it there *is* function C1-- (P1, P2), but existence does not mean fast computability . If C1=P1*P2 and
      Message 2 of 12 , Aug 1, 2001
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        > > > I'm actively trying to crack the RSA code. My paper:
        > > >
        > > > http://www.mslinux.com/research/cracking_pki/cracking_pki.html
        Hi, Kent!
        I took a look at your paper. As I see it there *is* function
        C1-->(P1, P2), but 'existence' does not mean 'fast
        computability'. If C1=P1*P2 and Pi are odd primes, then we
        can use the following (where sigma(n) is the sum of all the
        divisors of n).

        {sigma(C1)=1+P1+P2+C1
        {C1=P1*P2

        And we have the equation:

        x^2-(sigma(C1)-C1-1)*x+C1=0
        {P1,P2}=
        {((sigma(C1)-C1-1)+-sqrt((sigma(C1)-C1-1)^2-4*C1))/2}

        Then sigma(x) can be computed from the following
        equations:

        sigma(1)=1
        sum(((x-1)-5*k*(k+1))/2*sigma(x-k*(k+1)),
        for k=0 to FLOOR((SQRT(1+4*x)-1)/2))=0

        These equations make the infinite matrix equation for
        sigma(x). Maybe you can use it as density matrix equation.
        But I don't know how it can be done, alas.

        Aleksey
      • Kent Nguyen
        ... occupy precisely ... can you elaborate? Why don t you use the word occupy exactly ? ... If there is a wavefunction to describe prime number, and I
        Message 3 of 12 , Aug 1, 2001
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          On Wednesday 01 August 2001 08:10, Paul Leyland wrote:
          > > One of the postulate in quantum theory states that no two
          > > particle can occupy the same place at the same time.
          >
          > Not true in general. It is true for fermions (particles with spin
          > (2i+1)/2) and is known as the Pauli exclusion principle. For bosons
          > (particles with integral spin) particles can, and do, occupy precisely
          > the same state at the same time. It's why lasers, superfluids and
          > superconductors have such interesting properties.

          "occupy precisely" ... can you elaborate? Why don't you use the word "occupy
          exactly"?

          >
          > I fail to see what any of this has to do with factorization, but it will
          > be interesting to see if anything comes from it.

          If there is a wavefunction to describe prime number, and I believe there is.
          A few scientists have report a spectral in nature that resemble prime number
          sequence.

          If this wavefunction exist we can use the "position" of the prime numbers to
          figure out primes to a composite number.

          (1, 2) (2, 3) (3, 5) (4, 7) ... (n, p)

          Rather than finding p we find n for the wavefunction.

          Such that sigma(C1) = { n1, n2 }.

          --kent
        • Kent Nguyen
          ... Hi Aleksey! ... If I find the wavefunction, I will use Newton s method of approximation to find out the primes. In order to use Newton s method, I need to
          Message 4 of 12 , Aug 1, 2001
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            On Wednesday 01 August 2001 11:41, Aleksey D. Tetyorko wrote:
            > > > > I'm actively trying to crack the RSA code. My paper:
            > > > >
            > > > > http://www.mslinux.com/research/cracking_pki/cracking_pki.html
            >
            > Hi, Kent!

            Hi Aleksey!

            > I took a look at your paper. As I see it there *is* function
            > C1-->(P1, P2), but 'existence' does not mean 'fast
            > computability'. If C1=P1*P2 and Pi are odd primes, then we
            > can use the following (where sigma(n) is the sum of all the
            > divisors of n).
            >
            > {sigma(C1)=1+P1+P2+C1
            > {C1=P1*P2
            >
            > And we have the equation:
            >
            > x^2-(sigma(C1)-C1-1)*x+C1=0
            > {P1,P2}=
            > {((sigma(C1)-C1-1)+-sqrt((sigma(C1)-C1-1)^2-4*C1))/2}
            >
            > Then sigma(x) can be computed from the following
            > equations:
            >
            > sigma(1)=1
            > sum(((x-1)-5*k*(k+1))/2*sigma(x-k*(k+1)),
            > for k=0 to FLOOR((SQRT(1+4*x)-1)/2))=0
            >
            > These equations make the infinite matrix equation for
            > sigma(x). Maybe you can use it as density matrix equation.
            > But I don't know how it can be done, alas.

            If I find the wavefunction, I will use Newton's method of approximation to
            find out the primes. In order to use Newton's method, I need to make a good
            guess, a good guess is sqrt(C1)/2. Newton's method of approximation is very
            fast, it is the method used in calculator to find square root, cubic root,
            and inverses.

            For example to find sqrt of 2, using Netwon's method, it only takes 5 adding,
            dividing, and multiplying iterations to come close to the answer you see in
            your calculator.

            A good reference of Newton's method:
            http://gamba.ugrad.math.ubc.ca/coursedoc/math100/notes/approx/newton.html

            --kent
          • Paul Leyland
            ... precisely ... You re quibbling. If you prefer it, I m equally happy with occupy exactly . If you really want to get pedantic, I d be even happier with
            Message 5 of 12 , Aug 1, 2001
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              > > Not true in general. It is true for fermions (particles with spin
              > > (2i+1)/2) and is known as the Pauli exclusion principle. For bosons
              > > (particles with integral spin) particles can, and do, occupy
              precisely
              > > the same state at the same time. It's why lasers, superfluids and
              > > superconductors have such interesting properties.
              >
              > "occupy precisely" ... can you elaborate? Why don't you use
              > the word "occupy exactly"?

              You're quibbling. If you prefer it, I'm equally happy with "occupy
              exactly". If you really want to get pedantic, I'd be even happier with
              phrasing along the lines of "there is no constraint on the occupancy
              number of an eigenstate of a system of bosons" but that seems unduly
              wordy.

              If we're being pedantic, your original statement is unequivocably false.
              It's not even strictly true for fermions in that the Pauli exclusion
              principle is not really a postulate of quantum mechanics (in the sense
              of a presupposed truth which is not amenable to question) but rather a
              consequence of the anticommutativity of operators acting on fermion
              quantum fields. I was being generous and assumed you meant
              "consequence" or "feature" where you wrote "postulate".

              Even being that generous, your statement "no two particle (sic) can
              occupy the same place at the same time" is false, if by "place" you mean
              spatial location. For a start, two fermions differing only in spin can
              occupy the same energy state. Further, from Heisenberg's uncertainty
              principle, the spatial location of each of two particles can only be
              precisely determined if their momenta are completely undetermined. If
              you know anything about the momenta of the particles, their wave
              functions *will* overlap in space. Trying to nail down "the same time"
              is equally difficult: the particle's energy is then the conjugate
              quantity. But I'll be generous again and assume that by "particle" you
              meant fermion and that by "place" you meant eigenstate.

              If you really want to make progress, I suggest that you consult an
              introductory text or two on quantum field theory. It's 19 years since I
              last studied QFT so the references I can quote from memory are now
              outdated and possibly unavailable, but I'm sure there must be
              contemporary works available.

              (Just checked on Amazon: a search on Quantum Field Theory yields 596
              hits, so you ought to be able to find something. Further, the book I
              own, Elements of Advanced Quantum Theory written by John M Ziman in
              1975, is still in print.)


              Paul
            • Paul Leyland
              ... precisely ... You re quibbling. If you prefer it, I m equally happy with occupy exactly . If you really want to get pedantic, I d be even happier with
              Message 6 of 12 , Aug 1, 2001
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                > > Not true in general. It is true for fermions (particles with spin
                > > (2i+1)/2) and is known as the Pauli exclusion principle. For bosons
                > > (particles with integral spin) particles can, and do, occupy
                precisely
                > > the same state at the same time. It's why lasers, superfluids and
                > > superconductors have such interesting properties.
                >
                > "occupy precisely" ... can you elaborate? Why don't you use
                > the word "occupy exactly"?

                You're quibbling. If you prefer it, I'm equally happy with "occupy
                exactly". If you really want to get pedantic, I'd be even happier with
                phrasing along the lines of "there is no constraint on the occupancy
                number of an eigenstate of a system of bosons" but that seems unduly
                wordy.

                If we're being pedantic, your original statement is unequivocably false.
                It's not even strictly true for fermions in that the Pauli exclusion
                principle is not really a postulate of quantum mechanics (in the sense
                of a presupposed truth which is not amenable to question) but rather a
                consequence of the anticommutativity of operators acting on fermion
                quantum fields. I was being generous and assumed you meant
                "consequence" or "feature" where you wrote "postulate".

                Even being that generous, your statement "no two particle (sic) can
                occupy the same place at the same time" is false, if by "place" you mean
                spatial location. For a start, two fermions differing only in spin can
                occupy the same energy state. Further, from Heisenberg's uncertainty
                principle, the spatial location of each of two particles can only be
                precisely determined if their momenta are completely undetermined. If
                you know anything about the momenta of the particles, their wave
                functions *will* overlap in space. Trying to nail down "the same time"
                is equally difficult: the particle's energy is then the conjugate
                quantity. But I'll be generous again and assume that by "particle" you
                meant fermion and that by "place" you meant eigenstate.

                If you really want to make progress, I suggest that you consult an
                introductory text or two on quantum field theory. It's 19 years since I
                last studied QFT so the references I can quote from memory are now
                outdated and possibly unavailable, but I'm sure there must be
                contemporary works available.

                (Just checked on Amazon: a search on Quantum Field Theory yields 596
                hits, so you ought to be able to find something. Further, the book I
                own, Elements of Advanced Quantum Theory written by John M Ziman in
                1975, is still in print.)


                Paul
              • Kent Nguyen
                ... First Pauli exclusion principle states: In a closed system, no two electrons can occupy the same state. http://theory.uwinnipeg.ca/mod_tech/node168.html
                Message 7 of 12 , Aug 1, 2001
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                  On Wednesday 01 August 2001 13:01, Paul Leyland wrote:
                  > > > Not true in general. It is true for fermions (particles with spin
                  > > > (2i+1)/2) and is known as the Pauli exclusion principle. For bosons
                  > > > (particles with integral spin) particles can, and do, occupy
                  >
                  > precisely
                  >
                  > > > the same state at the same time. It's why lasers, superfluids and
                  > > > superconductors have such interesting properties.
                  > >
                  > > "occupy precisely" ... can you elaborate? Why don't you use
                  > > the word "occupy exactly"?
                  >
                  > You're quibbling. If you prefer it, I'm equally happy with "occupy
                  > exactly". If you really want to get pedantic, I'd be even happier with
                  > phrasing along the lines of "there is no constraint on the occupancy
                  > number of an eigenstate of a system of bosons" but that seems unduly
                  > wordy.

                  First Pauli exclusion principle states:
                  "In a closed system, no two electrons can occupy the same state."
                  http://theory.uwinnipeg.ca/mod_tech/node168.html

                  Note Pauli only states for occupance of same state not same time.

                  This isn't "exactly" what I'm saying. Go back and read "exactly" what I
                  said. Because when you say "occupy exactly" ... I'm very suspicious whether
                  you've figured a way to violate Heinsberg uncertainity principle.
                  http://www.srikant.org/core/node12.html

                  >
                  > If we're being pedantic, your original statement is unequivocably false.
                  > It's not even strictly true for fermions in that the Pauli exclusion
                  > principle is not really a postulate of quantum mechanics (in the sense
                  > of a presupposed truth which is not amenable to question) but rather a
                  > consequence of the anticommutativity of operators acting on fermion
                  > quantum fields. I was being generous and assumed you meant
                  > "consequence" or "feature" where you wrote "postulate".

                  You don't have to be generous. You need to understand what I wrote. You are
                  assuming what I wrote is "Pauli exclusion principle". Having bad assumption
                  leads to bad argument.

                  >
                  > If you really want to make progress, I suggest that you consult an
                  > introductory text or two on quantum field theory. It's 19 years since I
                  > last studied QFT so the references I can quote from memory are now
                  > outdated and possibly unavailable, but I'm sure there must be
                  > contemporary works available.

                  Working for your employer really makes you *think* you are making progress.
                  :)

                  >
                  > (Just checked on Amazon: a search on Quantum Field Theory yields 596
                  > hits, so you ought to be able to find something. Further, the book I
                  > own, Elements of Advanced Quantum Theory written by John M Ziman in
                  > 1975, is still in print.)

                  Thanks for using amazon.com, it's better than bn.com don't you think? :)

                  --kent
                • Paul Leyland
                  ... No it does not! Just because that web page makes that claim that doesn t mean that the PEP is as stated. The PEP states that no two fermions can occupy
                  Message 8 of 12 , Aug 1, 2001
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                    > First Pauli exclusion principle states:
                    > "In a closed system, no two electrons can occupy the same state."
                    > http://theory.uwinnipeg.ca/mod_tech/node168.html

                    No it does not! Just because that web page makes that claim that
                    doesn't mean that the PEP is as stated. The PEP states that no two
                    fermions can occupy the same quantum state. Electrons are fermions,
                    indeed, but electrons can pair up to form "Cooper pairs" which
                    themselves are bosons. These bosons can indeed occupy the same quantum
                    state and, when they do, give rise to the phenomenon of
                    supercoductivity.

                    The web page itself goes on to state "actually, protons and neutrons
                    obey the same principle, while photons do not)" something you seem to
                    have missed. Lasers function precisely because photons do not obey the
                    same principle. Protons and neutrons are spin-half particles and thus
                    fermions; photons are spin-zero bosons. Photons, as far as we know,
                    have no sub-structure but both protons and neutrons are composite
                    particles (as are Cooper pairs and helium nuclei). The helium-4 nucleus
                    is a spin-zero boson and so can violate the PEP. When it does, bulk
                    helium-4 becomes superfluid. The helium-3 nucleus is a spin-half
                    fermion and so liquid helium-3 doesn't become superfluid until the
                    temperature is low enough for pairs of nuclei to form spin-zero bosons,
                    whereupon it too shows superfluidity.

                    > This isn't "exactly" what I'm saying. Go back and read
                    > "exactly" what I said.

                    Very well, I quote: "One of the postulate in quantum theory states that
                    no two particle can occupy the same place at the same time."

                    This statement is just plain wrong, for the reasons I went into
                    previously.

                    > Because when you say "occupy exactly" ... I'm very
                    > suspicious whether
                    > you've figured a way to violate Heinsberg uncertainity principle.
                    > http://www.srikant.org/core/node12.html

                    For a start, Heisenberg's uncertainty principle only applies to
                    conjugate quantities, such as energy/time and linear momentum/position
                    (these two quantities are, of course, special cases of the more general
                    4-momentum / spacetime coordinates). It does *not* apply to
                    non-conjugate measurements, such as the x-component of momentum and the
                    y coordinate, which can be simultaneously measured to arbitrary
                    accuracy.

                    In general, if the operators corresponding to observables anti-commute,
                    HUP applies. If they commute, they do not.

                    Please read some real books on quantum theory.


                    > assuming what I wrote is "Pauli exclusion principle". Having
                    > bad assumption leads to bad argument.

                    But that is precisely what you did write!

                    > Working for your employer really makes you *think* you are
                    > making progress. :)

                    I don't think I understand that comment. Don't bother elucidating, as
                    the smiley suggests that it's probably not that important.


                    I'm becoming ever more convinced that this thread has very little, if
                    anything, to do with prime numbers. I've probably already bored the
                    majority of readers, so I'll drop out of it here.


                    Paul
                  • Kent Nguyen
                    ... You miss the point of the relation to prime number. Cracking the RSA code is a linear problem, thus a one-dimensional problem. You come and talk about the
                    Message 9 of 12 , Aug 1, 2001
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                      > I'm becoming ever more convinced that this thread has very little, if
                      > anything, to do with prime numbers. I've probably already bored the
                      > majority of readers, so I'll drop out of it here.

                      You miss the point of the relation to prime number.

                      Cracking the RSA code is a linear problem, thus a one-dimensional problem.
                      You come and talk about the 4th dimension, which to me doesn't seem relevant.
                      So you ya, you convince yourself.

                      As I've said before, there exist a very close spectra that resemble prime
                      number sequence.
                      http://www.maths.ex.ac.uk/~mwatkins/zeta/physics1.htm

                      My equation with two variables:

                      Assume = C1 = P1*P2
                      f(x) = x^2 - (P1 + P2)*x + C1 = 0

                      I only have one equation with two variables. I need another equation to
                      solve for P1 and P2. That's what lead me to quantum mechanic in trying to
                      find the wavefunction that describes prime number sequence.

                      If P1 = P2, I can use the quadraic formula to solve for x. Resulting in
                      sqrt(C1).

                      If P1 < P2 or P1 > P2, it's a more difficult situation.

                      --kent
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