Question

What does the symmetrization postulate mean for the decomposition of the $N$ particle Hilbert space $\mathcal{H}^N$?

Answer 1

Suppose you have NN particles, each of which can occupy any of ss states. In general, you can write the NN particle Hilbert space HNHN as a product of 11particle Hilbert spaces H1H1:

HN=H1⊗H1⊗⋯⊗H1,HN=H1⊗H1⊗⋯⊗H1,

with dim[H1]=sdim[H1]=s.

This means that the NN particle space will have dim[HN]=sNdim[HN]=sN.

Now we look at the usual subspaces: FNFN for fermions and BNBN for bosons. For their dimensions, we have

dim[FN]=(sN)dim[FN]=(sN)

and

dim[BN]=(s+N−1N).dim[BN]=(s+N−1N).

Now, I was under the impression that the symmetrization postulate, saying that there are only either completely symmetric or completely antisymmetric states, means that there is a decomposition of HNHN into a direct sum

HN=FN⊕BN.HN=FN⊕BN.

However, as one can easily check (e.g. for N=3N=3), this cannot be true since the dimensions don't add up, dim[H3]=g3≠dim[BN]+dim[FN]≈13

Answer 2

The symmetrization postulate indeed says that only either ... is to be done only over each group of identical particles).