Re: Question: about relation between particle exchange symmetry and angular momentum

Thank you for your reply. But I still do not understand it. I
understand that the charge conjugation operator in this case is
equivalent to swapping the position of the particles, and their
relative position vector turns negative of itself. But I cannot see why
this "turning negative" means a equivalent application of a parity
operator, because a parity operation results in the position vectors of
the TWO particle respective to the origin of the coordinate turning
negative of theirselves, NOT their relative position vector. The
parity operator, which turns x-->-x for ALL particles, do result in
(-1)^L factor in the spatial wavefunction, however, Is an operation
which turns (x_1-x_2)-->-(x_1-x_2), also result in (-1)^L factor in
the spatial wavefunction of the total system. Thanx.

.