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Re: A question about the spin 2 particle.



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



On Fri, 22 Dec 2000, Robert B Zannelli wrote:

OK Let me see if I get this. The number of a particles' eigenstates is given
by the equation n=2s+1. Any particle will be not be in any defined state
until a measurement is made of that particles spin. An unmeasured particle
will be in a superposition of all it's possible eigenstates. For a spin 2
particle there are 5 possible states which are-2,-1,0,1,2 . Now if we take
the vector in the Hilbert space which defines the mix of these states and our
2 spin particle superposition states consist of only amplitudes in even
number states then a 180 degree rotation will be a complete cycle. Should
there be any probability amplitude for any odd numbered state we would need a
360 degree rotation to complete a full cycle. In any case a 360 degree
rotation always returns the same amplitude.


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