From: Richard.Mathews@West.Sun.COM (Richard M. Mathews)
Newsgroups: sci.physics
Subject: Re: the nature of exclusion
Date: 31 Aug 1992 19:04:24 -0700
Organization: Sunsoft Inc., Los Angeles, CA.
NNTP-Posting-Host: smaug.West.Sun.COM

barry@arnold.math.ucla.edu (Barry Merriman) writes:
>Ok, but one would like to understand this in physical, causal 
>terms if possible.

I don't know if the following helps or not.  I think it does at least
help to show how the "exclusion principle" implies "statistical repulsion".

Take two "different" particles (different either in being fundamentally
different particles or just identical particles which happen to be in
different initial states).  Try to put them into the same state by
scattering one or both of them into new states.

Consider an arbitrary scattering center.  For particle 1 it has some
amplitude to scatter the particles into state theta.  Call this
	A1(theta)

(For visualization, you might want to think of theta as just the angle
of the path of the outgoing particle in a 2 dimensional scattering problem,
but theta can really represent whatever complete set of observables is
necessary to describe the state of any particle leaving the scattering
center.  In the following, I will even refer to scattering off at "the
angle theta", but remember that that interpretation of theta is just an
example.)

Now particle 2 is a different particle, so it may scatter differently.
Let the amplitude for it to scatter at angle theta to be
	A2(theta)

The amplitude for particle 1 to scatter at angle theta1 and particle 2
to scatter at angle theta2 is
	A1(theta1) * A2(theta2)

Now let's say that the two particles really are fundamentally different,
but our detector doesn't bother to distinguishing them.  In this case,
the probability of getting a detection at both theta1 and theta2 (the
probability of particle 1 scattering at angle theta1 and particle 2
scattering at angle theta2 *OR* particle 1 scattering at angle theta2
and particle 2 scattering at angle theta1)
	Pdiff(theta1,theta2) = [A1(theta1) * A2(theta2)]^2
			     + [A1(theta2) * A2(theta1)]^2
(all squares in this article really mean complex magnitude squared).

On the other hand if the particles are identical bosons, the states must
be symmetric at all times.  When we symmetrize the state and *then*
calculate the probability, we get
	Pboson(theta1,theta2) = [A1(theta1) * A2(theta2)
			       + A1(theta2) * A2(theta1)]^2
And if the particles are identical fermions, we get the same thing with
a different sign:
	Pfermion(theta1,theta2) = [A1(theta1) * A2(theta2)
			         - A1(theta2) * A2(theta1)]^2

Now take these in the limit that theta1 and theta2 approach theta (where
in the general case, this means that all of the parameters in each set
approach a common set of values).  Then
	Pdiff(theta,theta)  -> 2 * [A1(theta) * A2(theta)]^2
	Pboson(theta,theta) -> 4 * [A1(theta) * A2(theta)]^2
	Pfermion(theta,theta) -> 0

Now, I think this result is interesting for several reasons.  First,
it doesn't depend on the nature of the scattering center.  Bosons are
always twice as likely to end up in identical states as are different
particles, and fermions will never be able to scatter into the same
state.  No "force" is introduced to make this happen.  It is purely a
result of the symmetry rules.  Second, it shows that these probabilities
must continuously approach these values as theta1 and theta2 approach a
common value as long as A1 and A2 are continuous (of course, when any
of the variables included in the set, theta, are confined to discrete
values, you can't call A1 and A2 continuous).  In other words, not only
is it impossible for 2 fermions to scatter into identical states, it
is highly unlikely for them to scatter into even very similar states.
How similar the states have to be (how narrow the peaks are for bosons
and how narrow the nodes are for fermions, so to speak), however, does
depend on the nature of the scattering function.  Finally, it shows that
there is no magical force involved.  You might try to hypothesize a force
which maintains the symmetry rules, but this force would have to act to
maintain such symmetry even between particles which are separated by light
years and across space-like intervals.  It is easier to imagine that this
symmetry is just a property of each of the forces of nature, and so it
is maintained regardless.  Since all forces respect this symmetry, all
forces are in essence partially responsible for maintaining the exclusion
principle.

     Richard M. Mathews			D efend
					 E stonian-Latvian-Lithuanian
Richard.Mathews@West.Sun.COM		  I ndependence
Newsgroups: sci.physics