Crypto and new computing strategies

[hughes@ah.com (Eric Hughes)]

>> The Bekenstein Bound gives limits both on the expected maximum number
>> of quantum states encodable in a given volume of space and on the
>> expected maximum number os transitions between these states.  If this
>> bound holds (and it certainly seems to hold for EM fields), then a
>> probabilistic Turing machine will be able to simulate it.

>First off, EM fields are NOT QM.

The "EM fields" I was referring to mean electromagnetic interactions,
that's all.  The argument on the Bekenstein bound does not depend on
the nature of the particles mediating the field, but on the existence
of non-zero commutators for position and momentum, i.e. Heisenberg
uncertainty.  Bekenstein uses his argument to try to constrain the
possibilities of interaction inside the proton, for example.  I'm not
sure it works for that, but the argument is pretty clear about states
mediated by electromagnetic interaction.

>As to infinite precision and its non-presence....Beeep....wrong answer...

You must not understand what the Bekenstein bound says.  It says, very
clearly, infinite precision does not exist.  If you disagree with the
applicability of the result, then say so, but you'd better know what
the result is before you go haplessly denying it.

>Electrons change state in zero time, this implies at least some form o f
>infinite precision 

The second half of the Bekenstein bound says that infinitely fast
state changes do not occur.  Again, no infinite precision.

"Zero time" is a different statement than "almost zero time" or "so
small that we can't measure how small."  What may be reasonably taken
to be instantaneous in one model, with it's own characteristic
approximations, need not be instantaneous in another.  

Eric