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

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  • 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 1 of 12 , Aug 1, 2001
      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 2 of 12 , Aug 1, 2001
        > 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 3 of 12 , Aug 1, 2001
          > 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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