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# Re: A fermion question from a Chemistry teacher

• From: John Denker <jsd@AV8N.COM>
• Date: Sun, 15 Feb 2004 11:27:45 -0800

Quoting SSHS KPHOX <kphox@CHERRYCREEKSCHOOLS.ORG>:

a) Can bosons break Pauli's exclusion principle??? How do you explain=

Since the Pauli exclusion principle never applied to bosons
in the first place, nothing is "broken". It's like asking
whether a squirrel breaks the law by walking across the street.
I don't think jaywalking laws apply to squirrels.

The explanation goes like this: The familiar law
"one and one makes two"
applies only in the classical limit. In the real (i.e. quantum)
world, the laws of wave mechanics apply, and one and one could
make zero (destructive interference) or one and one could make
four (constructive interference) or anything in between.

Suppose we have two identical particles (a and b) in a box, and
we do an experiment in hopes of observing one of them. In the
high-temperature limit, we can assume there is only one process
by which one of the particles can leave the box and enter the
detector, and we can assign this process the probability
P = |Phi(a,b)|^2
where Phi() is the wavefunction. We don't at the moment know
much about Phi(), but for what follows we don't need to know

At lower temperatures, we must account for other ways a particle
could reach out detector.
In particular, there is some chance that the two particles can
exchange places while we aren't looking, so the particle we detect
isn't the one we thought we would detect. This process adds
wave-mechanically to the other process, so we have
P = |Phi(a,b) + X Phi(a,b)|^2
where X describes the exchange process.

More generally, the particles can exchange multiple times, so we
should write
P = |Phi(a,b) + X Phi(a,b) + X^2 Phi(a,b) + X^3 Phi(a,b) ... |^2
but that gets pretty uninteresting after a while, because in the
low-temperature limit we know X^2 = 1. That's because if you have
two particles and exchange them twice, you get back exactly what
you started with.

The crux of the argument is that we have X^2=1, as opposed to |X|^2=1.
That leaves us with only two possibilities: either X=1 or X=-1. For
some particles, X=1 and we call them bosons. For other particles, X=-1
and we call them fermions. For fermions, the probability of observing
two of them in the same state is
P = |Phi(a,b) + (-1) Phi(a,b)|^2
= 0
That is, you will never see two fermions in the same state. For
bosons, the same calculation gives a non-zero probability.

b) Can somebody briefly explain how is it that paired fermions may be=
have as bosons??

Very briefly: If you exchange TWO fermions with TWO other fermions,
you pick up two factors of X simultaneously (one for each member of
the pair) so the pair acts like a boson.

Is it because of the spin number???

It is true that all fermions have half-integral spin, while bosons
have integral spin. This is called the spin/statistics theorem.
For now you will have to take it as an experimentally-observed
fact ... nobody has ever given an elementary proof of the theorem.