In a message dated 12/22/00 7:17:31 PM Eastern Standard Time, jlu@HEP.ANL.GOV
writes:
<< Hi all-
This is unnecessarily complicated and confusing. The answer is
that a 1-particle bosonic (integer spin) state is invariant under a
360 degree rotation. A 1-particle fermionic state changes sign under such
a rotation.
End of story. For more detail see Weinberg, Vol. I, p. 89, where
you will see that the result is not quantum-mechanical but has to do
with representations of the Poincare' group.
Regards,
Jack
>>
Hi Jack
This has been an ongoing discussion on another Physics list where, after
a few days with no answer from this list, I submitted this question. The very
same division occurred among list members there between those who held there
was no difference between a spin 2 boson and a spin 1 boson and those who
said the De Broglie amplitude would be the same after 180 degree rotation.
The problem that also developed was that according to Feynman a
particle exchange was equivalent to a 360 rotation. Of course no one disputes
that for any 360 degree rotation of a boson or any boson exchange that the De
Broglie amplitude is returned. The dispute is over if it occurs also at 180
degree for a spin 2 particle.
Now David Bowmen has posted and has said as long as the probability
amplitude is zero for all odd numbered eigenstates the De Broglie amplitude
would be returned after a 180 degree rotation. So I am not sure if the
problem is so simple but I don't really know, which is why I posted my
question. I never expected to split the opinion on two physics list.