> At 09:23 AM 1/22/97 0600, you wrote:
> >On Tue, 21 Jan 1997, Dan Plante wrote:
>
> snip (sorry to whoever; the recursion was 5 deep at this point, and my reader
> doesn't preserve poster identity..)
>
> >> >ManyWorlds represents the arbitrariness of "freewill" as which
> >> >timeline one is in.
> >> >
> >> >Copenhagen represents the arbitrariness of "freewill" as an "agent" that
> >> >is explicitly NOT described by any axioms, except that it causes wave
> >> >function collapses.
>
> snip (references to Many Worlds and Copenhagen interpretations re: free will)
>
> >> >There's some interesting psychophysics in the most recent Scientific
> >> >American theme issue on Consciousness. "Researcher tries to disbelieve
> >> >his own experiments, and *fails*."
> >>
> >> My understanding of the peerreviewed literature on this subject (if I
> >> remember correctly), represents quantum interactions as completely
> >> predictable through computation, and therefore deterministic. Assuming
> >> any and all aspects of (our own particular instance of?) existence are
> >> determined by these interactions, then there is no such thing as "real"
> >> free will.
> >
> >That is *not* my understanding of the literature.
> >
> >What is completely predictable, in principle [to my best knowledge] is a
> >probabilistic tree. *That's* deterministic.
>
> I'm not sure what you mean by 'probabilistic tree'. I'll have to claim
> ignorance, being one who aspires to 'rank amateur' status :)
> but I /am/ pretty sure that 'probabilistic' and 'deterministic' are mutually
> exclusive terms.
Oh, no.
The *outcomes* of *measurements* are random. However, the probability
distribution of measurement results is quite deterministic. And if you
consider a series of measurements, you get the tree I'm talking about. I.e.:
[grossly oversimplified]


    [measurement 1]
   
/ \ / \ / \ / \ [measurement 2]
While the result of these two measurements is random, the probability
distribution is deterministic. [if some of the resulting endconditions
are indistiguishable, we have to consider interference as wellbut even
that leaves the distribution of results deterministic.]
The evolution of an unmeasured state function *is* deterministic. Perhaps
that is what you were referring to?
>
> >The computation isn't exactly numerically stable; some work is required.
>
> Sorry, you lost me again. Numerically 'stable'?
The finite precision at all stages of a floatingpoint calculation
creates roundoff error. When computing eigenvalues [required for
computing the probablistic tree, above], this can easily destroy *ALL*
significant digits, if proper care is not taken in the algorithms used.
In extremely bad cases, they're destroyed completely no matter what
algorithm is used.
>
> >I clipped your further comments. You do see the relative uselessness of
> >taking 1 year to emulate a few nanoseconds?
> >
> >/ Kenneth Boyd
>
> I think I see where the problem is. A crucial point was made in the clipped
> comments regarding initial states. I guess I could have put it at the
> beginning.
> To reiterate:
>
> The state vector describes the evelotution of a quantum system in a way that
> is wholly deterministic /once the initial conditions have been established/.
> With this caveat, the future behaviour of a quantum particle(s) is predictable
> through quantum mechanical laws. (Ref: J. Baggot, The Meaning of Quantum
> Theory,
> Oxford Science Publications, 1992, ch 2, sec 6).
I agree with that for a "free" system*no* interactions.
[CLIP]
//////////////////////////////////////////////////////////////////////////
/ Towards the conversion of data into information....
/
/ Kenneth Boyd
//////////////////////////////////////////////////////////////////////////