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RE: [nanotech] Quantum dot wire less logic

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  • Jean ROCH
    Beats me... :) but I don t think it s appropriate to use the term lower energy state : in a computer system you need two different states to implement binary
    Message 1 of 23 , Jan 31, 2004
      Beats me... :) but I don't think it's appropriate to use the term "lower
      energy state" : in a computer system you need two different states to
      implement binary logic, but it is vital these states are equally stable, and
      that means they must have the same energy level.

      Really, I'm not the most knowledgeable guy, there :))

      Jean

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    • Eugen Leitl
      ... Not exactly, just the energetic bareer between them must be high enough so you won t get too frequent crossovers between two states through system noise.
      Message 2 of 23 , Feb 1, 2004
        On Sun, Feb 01, 2004 at 07:12:14AM +0000, Jean ROCH wrote:
        > Beats me... :) but I don't think it's appropriate to use the term "lower
        > energy state" : in a computer system you need two different states to
        > implement binary logic, but it is vital these states are equally stable, and
        > that means they must have the same energy level.

        Not exactly, just the energetic bareer between them must be high enough so
        you won't get too frequent crossovers between two states through system noise.

        Many molecular systems fit the bill, the difficulty is to build a stable
        operational device addressable at single molecule level -- and make it
        mass-producable. By now we have several candidates which fit the criteria but
        the last one.

        Individual switches are easy, orthogonal 2d and 3d arrays of individual
        switches are doable (albeit not done -- buckytube crossbars do not count as
        such), once you want to grow a Pentium by way
        of autoassembly -- forget it.

        -- Eugen* Leitl <a href="http://leitl.org">leitl</a>
        ______________________________________________________________
        ICBM: 48.07078, 11.61144 http://www.leitl.org
        8B29F6BE: 099D 78BA 2FD3 B014 B08A 7779 75B0 2443 8B29 F6BE
        http://moleculardevices.org http://nanomachines.net


        [Non-text portions of this message have been removed]
      • Mark Gubrud
        At least in the case of classical computing, where the physical state of the computer (continuous physical degrees of freedom, as opposed to a logical state)
        Message 3 of 23 , Feb 1, 2004
          At least in the case of classical computing, where the physical state of
          the computer (continuous physical degrees of freedom, as opposed to a
          logical state) always has definite values, including during a state
          transition, the only requirement is that the energy barrier (energy of the
          transition state) be high enough to be inacessible by thermal or other
          noise, i.e. that there are no thermal or other sources of excitation large
          enough to supply the energy needed to reach the energy of the transition
          state from the bottoms of the energy wells defining each of the two
          logical states (1 and 0). Not only is there no particular reason why the
          1 and 0 states should be equal in energy, but obviously they almost never
          really are.

          > Beats me... :) but I don't think it's appropriate to use the term "lower
          > energy state" : in a computer system you need two different states to
          > implement binary logic, but it is vital these states are equally stable, and
          > that means they must have the same energy level.

          --
          Mark Avrum Gubrud | "The Farce?"
          Center for Superconductivity Research | "Well, the Farce is what
          Physics Dept., University of Maryland | gives a Jolli his power.
          College Park, MD 20742-4111 USA | It's a comedy field created
          ph 301-405-7673 fx 301-314-9541 | by all suffering things..."
        • E M
          Not only is there no particular reason why the 1 and 0 states should be equal in energy, but obviously they almost never really are. OK Mark,,the question is
          Message 4 of 23 , Feb 4, 2004
            Not only is there no particular reason why the
            1 and 0 states should be equal in energy, but obviously they almost never
            really are.

            OK Mark,,the question is why these quantum cells tend to have the same energy when they are aligned ,,is that because they want to reach the stable point?

            by the way ,,,I got the these ideas from http://www.mitre.org/tech/nanotech/quantum_dot_cell3.html



            Mark Gubrud <mgubrud@...> wrote:
            At least in the case of classical computing, where the physical state of
            the computer (continuous physical degrees of freedom, as opposed to a
            logical state) always has definite values, including during a state
            transition, the only requirement is that the energy barrier (energy of the
            transition state) be high enough to be inacessible by thermal or other
            noise, i.e. that there are no thermal or other sources of excitation large
            enough to supply the energy needed to reach the energy of the transition
            state from the bottoms of the energy wells defining each of the two
            logical states (1 and 0). Not only is there no particular reason why the
            1 and 0 states should be equal in energy, but obviously they almost never
            really are.

            > Beats me... :) but I don't think it's appropriate to use the term "lower
            > energy state" : in a computer system you need two different states to
            > implement binary logic, but it is vital these states are equally stable, and
            > that means they must have the same energy level.

            --
            Mark Avrum Gubrud | "The Farce?"
            Center for Superconductivity Research | "Well, the Farce is what
            Physics Dept., University of Maryland | gives a Jolli his power.
            College Park, MD 20742-4111 USA | It's a comedy field created
            ph 301-405-7673 fx 301-314-9541 | by all suffering things..."



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          • Mark Gubrud
            ... You have two charges on each molecule (cell) and four sites per molecule, arranged in a square. Porphyrins come to mind. The two charges can t get off
            Message 5 of 23 , Feb 4, 2004
              E M wrote:

              > OK Mark,,the question is why these quantum cells tend to have the same energy when they are aligned ,,is that because they want to reach the stable point?
              >
              > by the way ,,,I got the these ideas from http://www.mitre.org/tech/nanotech/quantum_dot_cell3.html
              >

              You have two charges on each molecule (cell) and four sites per
              molecule, arranged in a square. Porphyrins come to mind. The two
              charges can't get off the molecule because they are counterbalanced by
              the nuclear charges. So they will go to opposite corners because that
              way they are furthest apart.

              At least, that's the argument; the quantum mechanics of this gets more
              complicated. The two states / and \ can't actually have one charge on
              one site and the other on the other. They must be spin pairs of
              two-electron states, with each spin equally likely to be found on each
              site.

              Then, neighboring cells align because otherwise you have two charges
              that are quite close to each other, a high energy configuration. But
              now we really have 4-electron states, although the tunneling rates from
              one cell to another may be very small.

              Actually an isolated molecule must have equal charge density on each of
              the equivalent sites. It is only the proximity of two neighboring cells
              which breaks the symmetry and creates four states, two high energy
              states \/ and /\, and two low energy states // and \\. (This can't be
              exactly correct; for the high energy states, the charges would
              distribute unequally between the two corners of each molecule. And I'm
              not counting spin states here.) Only then is it possible for a single
              cell to be held in one state or the other. Even then // and \\ must
              tunnel into one another eventually.

              Now you build some big maze of QCA logic, and all these electrons are
              interacting so to be completely correct you have to talk about the
              collective many-electron state of the entire system. The argument is
              that you can force the free ends of the cell strings, using
              externally-supplied electric fields. So, you apply some configuration
              of voltages to forcing electrodes, which you might call setting boundary
              conditions, and now the entire system has a ground (lowest energy)
              state, and that state is the desired result of the computation.

              It's hard to believe, though, that the forcing doesn't peter out after a
              few cells downline, leaving "solitons" or walls between domains of QCA
              cellstring which are polarized oppositely. The idea is that these
              solitons should propagate rapidly from the point where the polarity was
              flipped by the external forcing, because the energy cost of their
              existence is higher than the temperature. But intuitively, one expects
              the solitons to relax somehow - e.g. delocalized collective states over
              many cells - and stall after some distance from the cell that is
              directly forced. Lent et al. claim this won't happen, but I don't know
              if they've ever used high-level simulation for a molecular realization.

              Many years ago (mid 1970s), Forrest Carter proposed using solitons on
              polyacetylene and similar linear pi-bonded systems where you have an
              alternation of single and double bonds, and you can flip which bonds are
              single and which double. So you force one end to have a 12121... bond
              alternation pattern instead of 21212.... Somewhere down the line you've
              got a soliton, 1211212, which carries a positive charge, or 1212212,
              which carries a negative charge. Except what actually happens is you
              find that the solitons relax, the extra or missing electron delocalizing
              over several units, but still localized within that region, due to
              scattering. The solitons will have no particular reason to move left or
              right, because there is no change in the energy when they move. So
              they'll just sit there unless there is a voltage applied to drive them,
              in which case the polyacetlyene chain is acting as a resistive conductor
              ("organic metal") rather than a QCA.

              I haven't heard a lot of buzz about these schemes in recent years.

              --
              Mark Avrum Gubrud | "The Farce?"
              Center for Superconductivity Research | "Well, the Farce is what
              Physics Dept., University of Maryland | gives a Jolli his power.
              College Park, MD 20742-4111 USA | It's a comedy field created
              ph 301-405-7673 fx 301-314-9541 | by all suffering things..."
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