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

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.

  • I'm not very familiar with Young diagrams in this context, but I think you are answering me assuming that: the boson and antiboson are just states. Thus, a pair boson antiboson is just a system of two identical particle with two orthogonal states: boson and antiboson. Is that right? – MBolin Nov 2 '17 at 17:26
  • I was saying: if you combine any 2 same spin bosons and consider states of total spin, $J$, the they're even for even $J$ and odd for odd $J$, which I thought was the question. The reason is "2" can only be partitioned into $2=2$ (symmetric) and $2=1+1$ (antisymmetric)--and then after all the representation theory, your question is answered. – JEB Nov 3 '17 at 23:15

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