Re: A question about the spin 2 particle.

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding  Bob Zannelli's question of 16 DEC 2000:

> Dear List Members.
>        As is well known in QM, a rotation in space for a fermion requires
> 720 degrees to restore the sign of the De Broglie amplitude. For a vector
> particle the sign is restored after 360 degree rotation. Is it correct to say
> that for a spin 2 particle (Graviton) the sign is restored after 180 degrees
> and then again at 360 degrees?

In general the answer is *no*, a generic spin-2 state is not brought back
to itself under a 180 degree rotation.  But there are *some* 'special
case' spin-2 states that *do* have this property.  The way to find these
states is as follows:  First choose an axis about which we are going to
do our 180 degree rotation.  Consider this axis as the quantization axis
which defines a convenient basis for our relevant spin-2 subspace of our
Hilbert space.  Now expand the spin-2 state of interest as a
superposition of the 5 different m-value eigenstates of the diagonal
rotation generator (i.e. component of spin along our quantization axis)
that will be a convenient basis for us.  *If* our spin-2 state is made up
of only even-m value amplitudes (i.e. m = -2, 0, and/or +2) and has *no*
projection along the odd-m basis states (m = +/- 1), *then* a rotation of
this state about the chosen axis by 180 degrees will return the original
state to itself with a restored sign.  If the projection onto this
quantization basis has any nonzero amplitude for any of the odd-m values
then our state must be rotated by a full 360 degrees to restore it to its
original value.

>        Any help on this would be appreciated.

I hope this helps.

David Bowman
David_Bowman@georgetowncollege.edu