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

2017-06-14 07:43:28

Suppose you have $N$ particles, each of which can occupy any of $s$ states. In general, you can write the $N$ particle Hilbert space $\mathcal{H}^N$ as a product of $1$ particle Hilbert spaces $\mathcal{H}^1$:

$$\mathcal{H}^N = \mathcal{H}^1 \otimes \mathcal{H}^1 \otimes \dots \otimes \mathcal{H}^1,$$

with $\mathrm{dim}[\mathcal{H}^1]=s$.

This means that the $N$ particle space will have $\mathrm{dim}[\mathcal{H}^N]=s^N$.

Now we look at the usual subspaces: $\mathcal{F}^N$ for fermions and $\mathcal{B}^N$ for bosons. For their dimensions, we have

$$\mathrm{dim}[\mathcal{F}^N] = \binom{s}{N}$$

and

$$\mathrm{dim}[\mathcal{B}^N] = \binom{s+N-1}{N}.$$

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 $\mathcal{H}^N$ into a direct sum

$$\mathcal{H}^N = \mathcal{F}^N\oplus\mathcal{B}^N.$$

However, as one can easily check (e.