# Re: [asa] Re: Cosmos in the Light of the Cross

From: <Bertsche@aol.com>
Date: Fri Jun 08 2007 - 13:21:23 EDT

In a message dated 6/5/2007 10:48:18 AM Pacific Daylight Time,
gmurphy@raex.com writes:

A formal proof of the principle - dpdq = or > h/4*pi*i with p = momentum and
q = position - follows from the basic commutation relations between the
operators p and q, pq - qp = h/2*pi*i . dp and dq are defined as RMS deviations
from the mean & thus the proof presupposes the standard QM ideas about the
statistical distributions defined by wave functions. Details are in any QM text.

But without going into details, the point is that the uncertainty relation
follows from the fact that p & q are noncommuting operators. They simply aren't
the kinds of things which can simultaneously have unique numerical values.
An operator A has the value a in a state F if AF = aF. In order for p & q to
have unique values p' & q' in the same state F we would have to have pF = p'F
and qF = q'F. & it's not hard to show that this would require pq - qp = 0.

Note that these QM effects are true of any wave phenomenon, even classical
waves (perhaps this would reduce the mystery a bit for those who aren't familiar
with QM).

Consider a sound wave. It cannot be uniquely localized in both time and
frequency simultaneously; t and f are "noncommuting operators" like p and q. If
one wants to generate a very pure, single-frequency tone, it must last for a
very long time. If one wants to generate a very short tone burst localized in
time, this adds harmonics and generates a frequency spread, so there is no
longer a unique frequency.

Merv's question could perhaps be reformulated to ask whether quantum
particles really ARE waves, or whether this is just a convenient descriptive
approximation to reality. If they really ARE waves, then George's "strong" statement
necessarily holds, because it is true of all waves. But if this is just an
approximation of reality, the "weak" statement could be maintained.

Kirk

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Received on Fri Jun 8 13:22:21 2007

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