Can indistinguishable particle wavefunctions be written as a product of total observable eigenstates?

Question

Consider the wavefunction of say two electrons in an external potential, associated with two possible states $\phi_a$ and $\phi_b$. Furthermore, each electron can have two spin states $\chi_1$ and $\chi_2$. A convenient basis for the anti-symmetric Hilbert space of the two electrons is (omitting the normalizations) \begin{equation} (\phi_a\phi_b + \phi_b\phi_a)(\chi_1\chi_2 - \chi_2\chi_1),\quad (\phi_a\phi_a)(\chi_1\chi_2 - \chi_2\chi_1),\quad (\phi_b\phi_b)(\chi_1\chi_2 - \chi_2\chi_1)\\ (\phi_a\phi_b - \phi_b\phi_a)(\chi_1\chi_2 + \chi_2\chi_1),\quad (\phi_a\phi_b - \phi_b\phi_a)(\chi_1\chi_1),\quad (\phi_a\phi_b - \phi_b\phi_a)(\chi_2\chi_2). \end{equation} This is convenient since the total wavefunction factors out in a spatial- and spin-wavefunction with even or odd exchange symmetry and, moreover, the spin part are just the different eigenstates of the total spin of the system. The fact that the above system is indeed a basis of the anti-symmetric Hilbert space is a speciality of a two particle system with two "available" quantum numbers (here the "spatial quantum number" ($a$ and $b$) and the spin).

Ever so often, however, one comes across a situation where people write down the total wavefunction of a larger system. E.g. of the wavefunction of a baryon. The document states that the total wavefunction is given as a product of properly symmetrized wavefunctions of the different quantum numbers, i.e. \begin{equation} \Psi_{total} = \psi(\text{space})\psi(\text{spin})\psi(\text{colour})\psi(\text{flavour}). \end{equation}

Why does this make sense? One would obviously miss a potentially large part of the accessible Hilbert space. Also the above wavefunction is used as evidence for the colour quantum number in the first place, as otherwise certain (observed) states would violate the exchange-symmetry condition (precisely when the other products together yield a symmetric wavefunction). This argument seems to be wrong since more states than the ones captured by the above "basis" are present.

Answer 1

Let us formulate your question a bit more abstractly:

You are starting from a one-particle space of the form $H = H_1\otimes\dots \otimes H_k$, where the $H_i$ are spaces like the "position space", "spin space", "flavour space", etc.

You are then forming the $n$-particle space $H^{\otimes n} = (H_1\otimes \dots H_n)^{\otimes n} = H^{\otimes n}_1 \otimes\dots\otimes H^{\otimes n}_k $ and you are applying either the symmetrization operator $\mathrm{Sym}$ or the anti-symmetrization operator $\Lambda$ to it.

You are asking why a basis for e.g. $\mathrm{Sym}^n(H)$ is given by tensors of the form $\psi = \psi_1 \otimes \dots\otimes \psi_k$ where each of the $\psi_i$ is in $\mathrm{Sym}^n(H_i)$ or $\Lambda^n(H_i)$ such that the total tensor has the correct symmetry properties.

The answer is: It's not, this is indeed just one piece of the total Hilbert space. According to this MO answer and the book by Fulton/Harris on representation theory of the symmetric group one can use the Littlewood-Richardson rule for Schur polynomials to argue that $$ \mathrm{Sym}^n(V\otimes W) = \bigoplus_{\alpha|n} S^\alpha(V)\otimes S^\alpha(W),$$ where $S^\alpha$ is a Schur factor corresponding to a Young diagram $\alpha$ with $n$ boxes and a similar decomposition for the anti-symmetric product. The spaces $\mathrm{Sym}^n(V)\otimes \mathrm{Sym}^n(W)$ and $\Lambda^n(V)\otimes \Lambda^n(W)$ are just summands in this decomposition, but for $n> 2$ there are indeed more.

Answer 2

Basis does not mean each unit vector is a physically valid state. For instance look at the states you wrote:

\begin{equation} (\phi_a\phi_b + \phi_b\phi_a)(\chi_1\chi_2 - \chi_2\chi_1),\quad (\phi_a\phi_a)(\chi_1\chi_2 - \chi_2\chi_1),\quad (\phi_b\phi_b)(\chi_1\chi_2 - \chi_2\chi_1)\\ (\phi_a\phi_b - \phi_b\phi_a)(\chi_1\chi_2 + \chi_2\chi_1),\quad (\phi_a\phi_b - \phi_b\phi_a)(\chi_1\chi_1),\quad (\phi_a\phi_b - \phi_b\phi_a)(\chi_2\chi_2). \end{equation}

These can all be decomposed and written into the basis

\begin{equation} \rvert \phi_i \rangle \rvert \phi_j \rangle\rvert \chi_m \rangle\rvert \chi_n \rangle \end{equation}

for all valid $i,j,m,n$, e.g. \begin{equation} (\phi_a\phi_b + \phi_b\phi_a)(\chi_1\chi_2 - \chi_2\chi_1) = \rvert \phi_a \rangle \rvert \phi_b \rangle\rvert \chi_1 \rangle\rvert \chi_2 \rangle + \rvert \phi_b \rangle \rvert \phi_a \rangle\rvert \chi_1 \rangle\rvert \chi_2 \rangle - \rvert \phi_a \rangle \rvert \phi_b \rangle\rvert \chi_2 \rangle\rvert \chi_1 \rangle - \rvert \phi_b \rangle \rvert \phi_a \rangle\rvert \chi_2 \rangle\rvert \chi_1 \rangle. \end{equation}

Why is this the case? Because the basis has a state that represents all possible measurement outcomes on the system.