Spin state of two distinguishable bosons I was reading about the $C$-parity of a particle-antiparticle pair. Since charge conjugation has the effect of swapping the particle and antiparticle, the $C$-parity can be found from the symmetry of their overall wave function. 
For finding the contribution of the spin state, it argued:

The spin state of two bosons is even if their total spin is even, and
  odd if the total is odd.

Noting they are not identical particles, since one is a boson and the other an antiboson, where does this property of the spin state come from? Thank you 
 A: The complete answer to your question requires understanding representations of the permutation group(s), and the Robinson-Schensted correspondence--which associates them with Young Tableaux. Schur-Weyl duality then associates those with rotational symmetries (closed subspaces) of tensor-spaces of any rank. This is a deep subject--that was core part of at least 3 Nobel prizes (Wigner, Pauli, Gell-Mann).
In terms of young diagrams and representation theory:
[] + [][] = [] x []
[]
[can we do this in LaTex?]
which says that combining two "basic" representations gives a totally symmetric
combination and a totally antisymmetric combinations.
In practice, that means:
So for spin 1/2:
$ 2 \otimes 2 = 3_S \oplus 1_A $
which is a symmetric vector (triplet) and antisymmetric scalar (singlet).
For spin 1:
$ 3 \otimes 3 = 6_S \oplus 3_A $
If you break out the Clebsch-Gordon coefficients, you'll find that $6_S$ is a symmetric spin-2 (5) and a symmetric spin-0 (1). Likewise, the 3 is an antisymmetric spin-1. Note that even spin is symmetric and odd spin is antisymmetric.
For spin 2 the multiplicities get hard to follow-but the amazing Hook-Length Formula will show you that:
$ 5 \otimes 5 = 15_S \oplus 10_A $
where the symmetric part breaks down into:
$ 15 = 9 + 5 + 1$ which is spin 4, 2, and 0,
and the antisymmetric part is:
$10 = 7 + 3$ which is spin 3 and spin 1.
The amazing part is that tensor product expansion of the Young Diagrams can enumerate all these SU(2) combinations; moreover, jumping to SU(3), it can be used to describe the meson spectrum.
