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

Physics Asked by Wojciech Morawiec on December 13, 2020

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
$$
mathcal{H}^N = mathcal{F}^Noplusmathcal{B}^N.
$$
However, as one can easily check (e.g. for $N=3$), this cannot be true since the dimensions don’t add up, $mathrm{dim}[mathcal{H}^3] = g^3 neq mathrm{dim}[mathcal{B}^N] + mathrm{dim}[mathcal{F}^N] approx frac13 g^3$.

What happens with the “missing” dimensions? Can something be said about a decomposition of $mathcal{H}^N$ as a consequence of the symmetrization postulate?

One Answer

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 nonintegral (and assuming that all particles are identical - otherwise the symmetrization is to be done only over each group of identical particles). Thus the used part of $H^N$ is either $B^N$ or $F^N$ (no superpositions as in your direct sum). Nonsymmetrized states are simply not eligible for realistic systems, although there are plenty of them in $H^N$.

However, from a purely mathematical perspective, one can decompose $H^N$ into a direct sum of subspaces each carrying an irreducible representation of the symmetric group of the $N$ particles. Two of these subspaces have a physical meaning - precisely those where the representation is 1-dimensional. The remaining subspaces would correspond to hypothetical ''parafermionic'' particles not found in nature.

Answered by Arnold Neumaier on December 13, 2020

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