Loading ...
Sorry, an error occurred while loading the content.

Re: [nanotech] Quantum dot wire less logic

Expand Messages
  • 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 1 of 23 , Feb 1 2:15 AM
    • 0 Attachment
      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 2 of 23 , Feb 1 9:01 AM
      • 0 Attachment
        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 3 of 23 , Feb 4 1:14 PM
        • 0 Attachment
          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..."



          The Nanotechnology Industries mailing list.
          "Nanotechnology: solutions for the future."
          www.nanoindustries.com
        • 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 4 of 23 , Feb 4 7:49 PM
          • 0 Attachment
            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..."
          Your message has been successfully submitted and would be delivered to recipients shortly.