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From: rimey@ucbmiro.ARPA (Ken Rimey)
Newsgroups: net.physics
Subject: Re: Heisenberg Uncertainty Principle
Message-ID: <9761@ucbvax.ARPA>
Date: Thu, 8-Aug-85 09:34:23 EDT
Article-I.D.: ucbvax.9761
Posted: Thu Aug  8 09:34:23 1985
Date-Received: Mon, 12-Aug-85 00:42:05 EDT
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Reply-To: rimey@ucbmiro.UUCP (Ken rimey)
Organization: U.C. Berkeley
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>In over-simplified terms Heisenberg's Uncertainty Principle says that we
>cannot know the simultaneous position and momentum of an individual
>elementary particle with unlimited accuracy.  Yet, we are able to
>determine the simultaneous position and momentum of conglomerations of
>these elementary particles.

No, you cannot simultaneously determine the exact position and momentum of a
conglomeration either.

>What is different about the individual particles and groups of particles?

Your intuition about groups of particles is really an intuition about
relatively massive objects in the relatively large-scale everyday world.
The uncertainty principle is

	(unc. in pos) (unc. in momentum) >= h = 6.6E-34 Js

If you want to insure that some object is in its proper place to an
accuracy of 1 micrometer, the uncertainty in its momentum will be
6.6E-28 kg m / s.  To get the uncertainty in velocity, we divide by
the mass.  Let's say the thing weighs a milligram.  Then it will
likely have a velocity of the order of 6.6E-22 m/s - very small.
The problem is that everyday distances are large, and that everyday
masses are huge, by atomic standards.  The mass of an electron is
9.1E-31 kg; experiment with that number.

>Is it strictly a case of the measurement process itself disturbing the
>individual particle, or is something else going on here?
>
>	Mike Augeri, DEC, Maynard MA USA

Something else is going on here.  In classical mechanics, a particle
has a position and a momentum.  You can be uncertain about the
position and momentum, but so what?  Lack of knowledge doesn't
account for the discrete energy levels of a hydrogen atom.  In
quantum mechanics, what you don't know is also irrelevant.  However,
the state of particle simply isn't described by a position and
a momentum as in classical mechanics.

It makes sense to ask, given a particle in some state what might I
get if I measure its position or its momentum?  Given the state, I
could calculate probability distributions describing what I might
get.  If the distribution for the position measurement is very
tightly peaked around a single position, I say that the state has a
well-defined position.  If I measure the position, I almost always
get values very close to that particular one.  Similarly, a state
might have a well-defined momentum.  Can I have a state that has both
a well-defined position and a well-defined momentum?  No, there is no
such thing.

Say you have a single particle.  It is described by wavefunction, a
complex number for every point in space.  The Schrodinger equation
determines how this wavefunction changes with time.  If at some time
the wavefunction is zero someplace, looking for the particle there
then is guaranteed fruitless.  The probability of finding the
particle in a given place is given by the magnitude of the complex
number there squared.  (I should say "The probability of finding it
in a given region ..." but you know what I mean.)

Take the fourier transform of this wavefunction.  This gives you the
momentum space wavefunction, a complex number for each momentum.  This
determines the probability of finding the particle with a given momentum
in the same way as the position wavefunction determines the probability
of finding the particle with a given position.  Note that the momentum
wavefunction cannot be chosen independently of the position wavefunction -
it is the fourier transform of the latter.

Is it possible to have a function that is nonzero only in a very small
region, such that its fourier transform is nonzero only in a very small
region?  No, that is not possible.  The electrical engineers among you
may recall that the fourier transform of a narrow gaussian is a wide
gaussian and vice versa, that the fourier transform of a spike has
infinite extent, and so on.  This is the uncertainty principle.

I'll mention two questions that my explanation above doesn't address.
First, does this momentum wavefunction that I have defined really
relate to the p=mv we all know and love?  Sure.  Take a wavefunction
with a moderately peaked position distribution, and a moderately peaked
momentum distribution.  If you calculate how this wavefunction evolves
in time according to the Schrodinger equation, you will find that the
peak in the position wavefunction moves with a velocity corresponding
to the momentum at which the peak in the momentum wavefunction lies.

Second, if it is impossible to determine exact position and momentum
simultaneously, then what happens if I try?  Does lightning strike me
or what?  The only way I can think of to measure the position of a
free particle is to bounce another particle off of it.  If that other
particle is to have a well-defined position, there will be a large
uncertainty in its momentum, and it will transfer some of that
momentum to the particle whose position is being determined.  Some
people's reaction to this thought experiment is "Aha! So that's
what's really going on."  They are fooling themselves if they think
this is a non-mathematical explanation of why the uncertainty
principle is true.  The argument is circular.  The probe particle is
a large disturbance only because it itself must satisfy the
uncertainty principle.

						Ken Rimey
						rimey@berkeley