X-Message-Number: 8556
From: Andre Robatino <>
Subject: Re: CryoNet #8546 - #8551
Date: Sun, 7 Sep 97 20:04:04 EDT

> 
> Message #8550
> Date: Sat, 6 Sep 1997 20:57:12 -0700 (PDT)
> From: John K Clark <>
> Subject: Digital Shakespeare
> 
<snip>

>                 
> In #8545  Andre Robatino <> On Fri, 5 Sep 97 Wrote:
>                
<snip>


>         >Seth Lloyd's quantum simulators do indeed have an infinite number of

>         >possible internal states.  Don't let the fact that the input and
>         >output are discrete fool you. 
>                
> 
> If "different" internal states  produces the same output for the same input 

  They don't.  Like I said, because of nonzero decoherence, the quantum state
ends up being changed slightly which means that for a given input there are
more than one possible output with nonzero probability.  Even a quantum
computer which is adequate for factoring, say, will not function determinis-
tically.  In this type of problem, it doesn't have to.  Since it's easy to
use a regular digital computer to check the QC's factorization (just do a
division!) the QC only has to have a significant probability of getting the
correct answer on each run.  So suppose it has a 1/5000 probability of
putting out the correct factorization of a 10,000-digit number which is the
product of two primes.  No problem.  Just keep running it over and over, and
check the output each time by doing a division with a conventional digital
computer until the correct answer comes out.  This will take an average of
5000 tries, which is a lot faster than a digital computer would take doing
the job all by itself.  What I said before about analog always being useless
for discrete problems is sometimes false for this reason, if one can get
partial information from analog and then use digital to finish the job then
it can be useful.  In fact, this is how people function in doing things like
programming and theorem-proving, the creative process is largely analog but
the result isn't.

> then exactly what's different about those states? 
>                
> 

>         >Thus even though the input and output are discrete (when measured),

>         >the output has a nonzero probability of being wrong if one attempts

>         >to use it for any form of digital computation.  So instead of the

>         >uncertainty being manifested by a small error in the output, it shows
>         >up as a nonzero probability of getting the wrong discrete answer. 
>                
> 
> The output is either correct or it's not, there are no other possibilities. 

> Are you saying the output is correct if the machine was in one internal state
> but the exact same answer is wrong if the machine is in another internal 
> state???       
> 
  See above...
> 

>         >Though an electron's measured spin has just two possible values,

>         >+1/2 and -1/2, its spin state is described by a superposition of

>         >pure spin states of the form a|+1/2> + b|-1/2> where a and b are
>         >complex numbers. Thus it lies in a continuum. 
>                
> 

> That's true, by measurement I can tell if an electron is spin up or spin down,
> but that still doesn't completely define the axis, it can be anywhere on the 
> surface of a cone. I will go further, the quantum wave function itself is 

  After you do the measurement, you know the wave function up to a global
phase factor (a complex multiplicative factor with unit length) which has
no observable effect.  The above expression has 4 apparent degrees of freedom
(2 for the real and imaginary parts of each of a and b), but one of them
goes away because the expression only matters up to a phase factor, and
another goes away due to the normalization condition |a|^2 + |b|^2 = 1.
In this case, there are still 2 left.

> continuos, but that does us no good because experimentally we can't measure 
> the quantum wave function F(x) of a particle, we can only measure the 
> intensity of the wave function [F(x)]^2 because that's a probability and  
> probability we can measure. 
> 
  True if you're doing a measurement of position, but a measurement of
momentum involves taking partial derivatives of the wave function, so any
change in the wave function other than a global phase factor leads to changes
in the probabilities of measurements of various quantities.  Anyway, the
change in the wave function due to decoherence _does_ in general affect the
intensity, along with the phase.

> You complain that I'm only talking about observable qualities, but we're   

  I'm not sure what you're talking about here, I'm a firm believer that only
observable quantities are meaningful and I don't think I ever suggested
otherwise.

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