Question

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

Answer 1

Hey mate ^_^

The symmetrization postulate indeed says that only either completely symmetric or completely antisymmetric states are realized in nature, depending on whether the spin is integral or non Integral...

#Be Brainly❤️

Answer 2

Suppose you have N particles, each of which can occupy any of s states. In general, you can write the N particle Hilbert space HN as a product of 1 particle Hilbert spaces H1: HN=H1⊗H1⊗⋯⊗H1, with dim[H1]=s.  This means that the N particle space will have dim[HN]=sN.  Now we look at the usual subspaces: FN for fermions and BN for bosons. For their dimensions, we have dim[FN]=(sN) and 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 HN into a direct sum HN=FN⊕BN. However, as one can easily check (e.g. for N=3), this cannot be true since the dimensions don't add up, dim[H3]=g3≠dim[BN]+dim[FN]≈13g3.  What happens with the "missing" dimensions? Can something be said about a decomposition of HN as a consequence of the symmetrization postulate?