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


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – MBolin
    Nov 2, 2017 at 17:26
  • $\begingroup$ 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. $\endgroup$
    – JEB
    Nov 3, 2017 at 23:15

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