Re: Pauli Exclusion

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding Eric Scerri's question:

> A few days ago I was teaching the Pauli Exclusion Principle and
> stated that in its most general form this is stated as;
>
> On the interchange of any two particles the wavefunction for a
> system of fermions is anti-symmetric.

I don't think this is the most general formulation.  The above is
only for states with a definite number of indistinguishable fermions.
The most general formulation is the requirement (based on the Spin-
Statistics Theorem's need for consistency) that fermion fields be
quantized using anticommutator bracket relations rather than using
commutator bracket relations as is done for boson fields.  The
antisymmetry of fermion states under particle exchange comes from
the anticommutivity of the single fermion creation and destruction
operators referring to different fermion modes.

> This followed a discussion of electrons being indistinguishable
> particles.
>
> So up pops a student and asks "if electrons are really
> indistinguishable why does it make ANY difference to the
> wavefunction if two electrons are interchanged.  Why should even
> just the sign change on performing an interchange between two
> supposedly identical particles?

The student is correct in the following sense.  After the particles
are exchanged the state *is* the same *physical* state as before the
exchange.  In fact, if the state was represented by a pure-state
density matrix (i.e. a projection operator onto a 1-dimensional
subspace of the quantum Hilbert space) rather than by a wave
function (i.e. a unit vector along the 1-dim subspace of that
Hilbert space) then there is no change whatsoever in the physical
system's state.  The physically observable properties of the system
depend on bilinear products of fermion fields (or wave functions if
you prefer).  What is required for physical indistinguishability is
that this product be unchanged when the particles are exchanged.
This identity can be a due to *either* a complete identity of
both factors separately, *or* by a sign change of both factors.
This is ultimately because the 2 square roots of 1 are both +1 and
-1.  When the +1 root is relevant we have bosons.  When the -1 root
is relevant we have bosons.

> Any thoughts?  I am looking for a conceptual response that meets
> the question at the appropriate level if possible.  But I am also
> interested in any  more fundamental explanation.
>
> But I am sorry to say that I am not interested in responses from
> John Denker.

You should be sorry to say it--and, apparently, you are.

> eric scerri

David Bowman