Is there a unique way to construct the overall spatial wavefunction for identical particles?

Question

While studying the quantum mechanics of $N$ identical particles, I stumbled upon formulas for generalizing the spatial wavefunction for bosons:

$$\psi(x_1,...,x_N)=\frac{1}{\sqrt{N!\prod_\alpha N_\alpha!}}\sum_p \psi_1(x_1)...\psi_N(x_N)$$

and for fermions, using the Slater Determinant: $$ \psi(x_1,...,x_N)=\frac{1}{\sqrt{N!}} \begin{vmatrix} \psi_1(x_1) & \psi_1(x_2) & ... & \psi_1(x_N)\\ \psi_2(x_1) & \psi_2(x_2) & ... & \psi_2(x_N)\\ \vdots & \vdots & \ddots & \vdots \\ \psi_N(x_1) & \psi_N(x_2) & ... & \psi_N(x_N)\\ \end{vmatrix} $$

My question is this: Are these the only ways of constructing wavefunctions for bosons and fermions? If no, how else do we do it? If yes, does the uniqueness of the formulas above point toward some deep fundamental fact about particles?

The way I understand it, the only requirement here is that the spatial wavefunction for the system of bosons be symmetric, and for the system of fermions, anti-symmetric. That said, I have not come across other schemes for constructing these wavefunctions, and I am curious if there are such schemes at all.

Answer 1

Determinants and permanents are not the only way.

There is a general result for the permutation group that, if you want to construct a fully symmetric wavefunction by combining the spatial and spin parts, then both parts must have the same permutation symmetry (i.e. must transform by identical irreps). This applies also for multidimensional representation of the permutation group.

If you want to construct a fully antisymmetric wavefunction, you must combine a spatial and spin part of conjugate symmetries.

As an example, it is possible to construct a fully symmetric 4-particle state by combining spatial states and spin states in the 3-dimensional irrep $\{3,1\}$, and a fully antisymmetric state by combining a spatial part in $\{2,1,1\}$ and a spin part in $\{3,1\}$ (both of which of course are $3$-dimensional).

The explicit sums of products of the various parts are given by the Clebsch-Gordan coefficients for the permutation group (in my example, those would be CG’s for $S_4$).

This method does have the advantage of providing you with state of specific total spin but algebraic cost increases with $n$, and tables of CG’s for $S_n$ aren’t so common (much less pre-packaged codes).

On balance, determinants or permanents are more common, and chemists in particular have developed various clever methods to extract angular momentum or spin information when needed.

Answer 2

The fact is that all states of an N-particle system can be written in this way. This is because the Hilbert space of such a system is a subspace of the $n$-fold tensor product of the one-particle Hilbert space. However, there are certainly other methods of describing states in this Hilbert space. A particularly useful one is in terms of occupation number. Since the particles are indistinguishable, the state is completely determined by saying how many particles are in a given state. In the case of bosons this number is unrestricted. In the case of fermions this number is restricted to be $0$ or $1$ due to antisymmetry.

To be specific, consider a one particle Hilbert space $\mathfrak{H}$ spanned by states $\{|\phi_1,\phi_2,\cdots\rangle\}$. The correspondic bosonic/fermionic $N$ particle space is spanned by kets of the form $|n_1,n_2,\cdots\rangle$ with $n_1+n_2+\cdots=N$. It corresponds to $n_1$ particles being in the state $|\phi_1\rangle$, $n_2$ particles being in the state $|\phi_2\rangle$, etc... In the bosonic case $n_1,n_2,\dots \in\{0,1,2,\dots\}$. In the fermionic case $n_1,n_2,\dots \in\{0,1\}$.