Excusion principle for particle and antiparticles

[Paul Colby]

Hi,

Lately I've been looking at how multi-particle wave functions appear
in quantum field theory. What I wanted to understand was how,
say in two electron systems, one would handle spin degrees of
freedom. The answer appears to be just the Kronecker product of the
usual gamma matrices as one would expect for independent
particle degrees of freedom. In addition the anti-symmetry of the
wave function follows from the creation operators anti-commuting.
What _bothers_ me is that electron-positron wave functions are also
required to be anti-symmetric under interchange.

The explanation of the exclusion principle I grew up with has that
indistinguishable particle wave functions must be either symmetric
or anti-symmetric under interchange. One can certainly distinguish
a positron from an electron so it's surprising symmetric wave functions
for particle/anti particles aren't allowed. Clearly the spin statistics
theorem of quantum field theory is the final say on this but is there a
simpler explanation?

Regards
Paul Colby

[Jos Bergervoet]

On 7/26/2012 10:19 PM, Paul Colby wrote:
> Hi,
>
> Lately I've been looking at how multi-particle wave functions appear
> in quantum field theory. What I wanted to understand was how,
> say in two electron systems, one would handle spin degrees of
> freedom. The answer appears to be just the Kronecker product of the
> usual gamma matrices as one would expect for independent
> particle degrees of freedom. In addition the anti-symmetry of the
> wave function follows from the creation operators anti-commuting.

Actually the wave function doesn't
have to be anti-symmetric. The total
state is anti-symmetric. This can
mean that the wave function is
antisymmetric and the spin state is
symmetric, but also that the wave
function is symmetric and the spin
state is antisymmetric. It can even
mean that the wave function has mixed
symmetry (not symmetric and also not
antisymmetric) and the spin state has
mixed symmetry as well.

In the latter case, the two mixed-
symmetry parts must be matched in
such a way that their tensor product
becomes exactly antisymmetric. This
can be done using Young tableaux.

--
Jos

[Paul Colby]

Thanks,

Turns out there is a 3S1(para) and a 1S0 (ortho) state of positronium.
Both states are observed in nature so my thoughts on the matter are just
confused. The only way for the 3S1 state to be antisymmetric is to be
anti-symmetric in charge state? This would be consistent with my
understanding of how two particle states are constructed in field
theory.

Paul C

[Jos Bergervoet]

I should add that with only 2 particles there
isn't the 'mixed-symmetry' case I mentioned..
(Since the permutation group S2 only has two
irreducible representations, the ones for the
symmetric and asymmetric combination. With S3
it is much more fun!)

> Both states are observed in nature so my thoughts on the matter are just
> confused. The only way for the 3S1 state to be antisymmetric is to be
> anti-symmetric in charge state? This would be consistent with my
> understanding of how two particle states are constructed in field
> theory.

This is the same in proton-neutron states, i.e.
the 3S1 state of the deuteron! There you have
anti-symmetry in isospin (which describes the
particles as one species). The new problem now
is, that this isospin is not an exact symmetry.

The same holds for 'flavour' if we have quarks,
and you could always make up more symmetries
that may or may not exist at some GUT scale!
So if we do not know exactly how many quantum
numbers there are, how can we ever choose the
correct symmetry for the wave function? (Isn't
that a nice paradox?!)

Or, stated differently: If we would *not* use
the theory of isospin and just treat P and N as
non-identical particles (no symmetry requirement
attached!) then what could go wrong? And if
nothing does go wrong when neglecting symmetry,
then why should we ever bother about it?!

--
Jos

[dimit...@yahoo.gr]

I had the same problem.. but I think I solved it. Lets call wave functions
"states" (this makes it more official :) )
The problem is that you think that antisymmetry implies the exclusion
principle (this was my problem). This is not true because the
electron and the positron
have different charges and thus they are not identical. think 2 electrons
at the same state (same momentum, spin etc)
|p,s...;p,s...>=a^+_{p,s...}a^+_{p,s...}|0>=-a^+_{p,s...}a^+_{p,s...}|0>=-|p,s...;p,s...> => |p,s...;p,s...>=0.

for an electron-positron pair this is:
|p,s,q=+1...;p,s,q=-1...>a^+_{p,s,q=+1...}b^+_{p,s,q=-1...}|0>
-b^+_{p,s,q=-1...}a^+_{p,s,q=+1...}|0>=|p,s,q=-1...;p,s,q=+1...>.
That is, the state is anti symmetric but does not vanish, because the two
particle have different charge.

I used p for momentum, q for charge, s for spin, "..." for other quantum
numbers and "+" for dagger (hermitian conjugation). If I am wrong please
correct me!

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