Symmetry of spin states with three electrons

Question

The exchange symmetry of the spin wave function is under consideration. To simplify things I start with two fermionic particles with spin 1/2. One possibility for basis vectors for two spins is

$uu, ud, du ,dd$,

where $u$ and $d$ are the up and down spins, respectively. $uu$ and $dd$ are symmetric if the two particles are exchanged, but $ud$ and $du$ are neither symmetric nor antisymmetric. However, if we create the linear combinations

$ud+du$ and $ud-du$

the first one is symmetric, while the second is antisymmetric with respect to particle exchange.

What if there are three particles? The basis vectors corresponding to the total momentum eigenstates are given here. For example, the state

$(uud) + x (udu) + y (duu)$

is neither symmetric nor antisymmetric in general. Here, $x,y$ are some complex multipliers. If I exchange particle 1 and 2, the result is

$(uud) + y (udu) + x (duu)$,

which is not proportional to the original vector unless $x=y$, which is the trivial choice. If $x=y=1$, that is the full symmetric wave function. My questions are:

1. How to understand symmetry if there are more than two particles?
2. What to do with the statement that the wave function should be antisymmetric if two particles are exchanged?
3. Is it possible to separate the wave function into the product of the coordinate and spin parts for three and more particles?

Answer 1

The Pauli principle says that the wave function of identical fermions must be antisymmetric under the simultaneous interchange of spatial and spin coordinates between any two fermions. This is why in the two-particle case the total wave function is written as either "symmetric spatial part $\times$ antisymmetric spin part" (singlet state) or "antisymmetric spatial part $\times$ symmetric spin part" (triplet state). This requirement of total antisymmetry remains the same for more than two particles, but the permutational properties of the spatial and the spin parts are less trivial.

The case of two particles is simple because the two-element permutation group $S_2$ is abelian, so it has only one-dimensional irreducible representations (a totally symmetric and a totally antisymmetric one). This means that you can simply construct the $\lvert S,M_S\rangle$ eigenfunctions of $\hat{S}^2$ and $\hat{S}_z$ as symmetric/antisymmetric combinations of primitive spin functions: $$ \begin{aligned} |0,0\rangle&=\frac{1}{\sqrt{2}}\Big[\lvert\uparrow\downarrow\rangle-\lvert\downarrow\uparrow\rangle\Big] \ , \\ |1,0\rangle&=\frac{1}{\sqrt{2}}\Big[\lvert\uparrow\downarrow\rangle+\lvert\downarrow\uparrow\rangle\Big] \ , \\ \lvert1,+1\rangle&=\lvert\uparrow\uparrow\rangle \ , \\ \lvert1,-1\rangle&=\lvert\downarrow\downarrow\rangle \ , \end{aligned} $$ and then build the two-fermion states as simple products $\lvert\Psi_{S,M_S}\rangle=\lvert\Phi_{S}\rangle\otimes\lvert S,M_S\rangle$, with the permutation property $\Phi_S(\vec{r}_1,\vec{r}_2)=(-1)^S\Phi_S(\vec{r}_2,\vec{r}_1)$ of the spatial part enforcing the Pauli principle.

The case of three particles is not that simple, since $S_3$ is non-abelian with one- and two-dimensional irreducible representations. A brute force way to obtain states with the correct symmetry is to first construct the spin eigenfunctions: $$ \begin{aligned} \left|\frac{1}{2},+\frac{1}{2}\right>&=\frac{1}{\sqrt{2}}\Big[\lvert\uparrow\uparrow\downarrow\rangle-\lvert\uparrow\downarrow\uparrow\rangle\Big] \ , \\ \left|\frac{1}{2},-\frac{1}{2}\right>&=\frac{1}{\sqrt{2}}\Big[\lvert\downarrow\downarrow\uparrow\rangle-\lvert\downarrow\uparrow\downarrow\rangle\Big] \ , \\ \left|\frac{1}{2},+\frac{1}{2}\right>'&=\frac{1}{\sqrt{6}}\Big[\lvert\uparrow\uparrow\downarrow\rangle+\lvert\uparrow\downarrow\uparrow\rangle-2\lvert\downarrow\uparrow\uparrow\rangle\Big] \ , \\ \left|\frac{1}{2},-\frac{1}{2}\right>'&=\frac{1}{\sqrt{6}}\Big[\lvert\downarrow\downarrow\uparrow\rangle+\lvert\downarrow\uparrow\downarrow\rangle-2\lvert\uparrow\downarrow\downarrow\rangle\Big] \ , \\ \left|\frac{3}{2},+\frac{1}{2}\right>&=\frac{1}{\sqrt{3}}\Big[\lvert\uparrow\uparrow\downarrow\rangle+\lvert\uparrow\downarrow\uparrow\rangle+\lvert\downarrow\uparrow\uparrow\rangle\Big] \ , \\ \left|\frac{3}{2},-\frac{1}{2}\right>&=\frac{1}{\sqrt{3}}\Big[\lvert\downarrow\downarrow\uparrow\rangle+\lvert\downarrow\uparrow\downarrow\rangle+\lvert\uparrow\downarrow\downarrow\rangle\Big] \ , \\ \left|\frac{3}{2},+\frac{3}{2}\right>&=\lvert\uparrow\uparrow\uparrow\rangle \ , \\ \left|\frac{3}{2},-\frac{3}{2}\right>&=\lvert\downarrow\downarrow\downarrow\rangle \ , \\ \end{aligned} $$ and then antisymmetrize the product of spatial and spin functions. For $S=3/2$, this is easy, because the spin function is totally symmetric, so $\Phi_{3/2}(\vec{r}_1,\vec{r}_2,\vec{r}_3)$ must be totally antisymmetric to satisfy the Pauli principle, and then we can write $$ \begin{aligned} \lvert\Psi_{3/2,M_S}\rangle&=\lvert\Phi_{3/2}\rangle\otimes\left|\frac{3}{2},M_S\right>\ . \end{aligned} $$ For $S=1/2$, there are two sets of spin states, and the general spin function can be written as an arbitrary linear combination: $$ \lvert\Theta_{1/2,M}\rangle=c\left|\frac{1}{2},M\right>+c'\,\left|\frac{1}{2},M\right>' \ , $$ as long as $|c|^2+|c'|^2=1$. This spin state is "permutationally invariant" in the sense that any permutation of the spins just corresponds to a different choice for $c$, $c'$. You can see this by building the matrix representation of $\hat{S}^2$ in e.g. the $M_S=1/2$ subspace: you will find that any state of the form $$ \alpha_1\lvert\uparrow\uparrow\downarrow\rangle +\alpha_2\lvert\uparrow\downarrow\uparrow\rangle +\alpha_3\lvert\downarrow\uparrow\uparrow\rangle $$ is a $\lvert1/2,+1/2\rangle$ state as long as $\alpha_1+\alpha_2+\alpha_3=0$, so you can choose any two orthonormal vectors.

Sticking to real coefficients, I choose $c=\cos(\theta)$, $c'=\sin(\theta)$. Then $$ |\Theta_{1/2,+1/2}\rangle=c_1(\theta)\lvert\downarrow\uparrow\uparrow\rangle +c_2(\theta)\lvert\uparrow\downarrow\uparrow\rangle +c_3(\theta)\lvert\uparrow\uparrow\downarrow\rangle \ , $$ where $$ \begin{aligned} c_1(\theta)&=-\frac{2}{\sqrt{6}}\sin(\theta) \ , \\ c_2(\theta)&=-\frac{1}{\sqrt{2}}\cos(\theta)+\frac{1}{\sqrt{6}}\sin(\theta) \ , \\ c_3(\theta)&=+\frac{1}{\sqrt{2}}\cos(\theta)+\frac{1}{\sqrt{6}}\sin(\theta) \ . \\ \end{aligned} $$ Multiplying with some spatial function and antisymmetrizing the product yield $$ \begin{aligned} \lvert\Psi_{1/2,+1/2}\rangle &= {\cal{A}}\Big[\lvert\Phi_{1/2}\rangle\otimes\lvert\Theta_{1/2,+1/2}\rangle\Big] \\ &= \frac{1}{\sqrt{6}}\Big[ \lvert\chi_{1}\rangle\otimes\lvert\downarrow\uparrow\uparrow\rangle+ \lvert\chi_{2}\rangle\otimes\lvert\uparrow\downarrow\uparrow\rangle+ \lvert\chi_{3}\rangle\otimes\lvert\uparrow\uparrow\downarrow\rangle \Big] \ , \end{aligned} $$ where $$ \begin{aligned} \chi_{1}(\vec{r}_1,\vec{r}_2,\vec{r}_3)=&{\color{white}+}c_1(\theta)\Big[\Phi_{1/2}(\vec{r}_1,\vec{r}_2,\vec{r}_3)-\Phi_{1/2}(\vec{r}_1,\vec{r}_3,\vec{r}_2)\Big] \\ &{\color{black}+}c_2(\theta)\Big[\Phi_{1/2}(\vec{r}_2,\vec{r}_3,\vec{r}_1)-\Phi_{1/2}(\vec{r}_2,\vec{r}_1,\vec{r}_3)\Big] \\ &{\color{black}+}c_3(\theta)\Big[\Phi_{1/2}(\vec{r}_3,\vec{r}_1,\vec{r}_2)-\Phi_{1/2}(\vec{r}_3,\vec{r}_2,\vec{r}_1)\Big] \ , \end{aligned} $$ $$ \begin{aligned} \chi_{2}(\vec{r}_1,\vec{r}_2,\vec{r}_3)=&{\color{white}+}c_1(\theta)\Big[\Phi_{1/2}(\vec{r}_3,\vec{r}_1,\vec{r}_2)-\Phi_{1/2}(\vec{r}_2,\vec{r}_1,\vec{r}_3)\Big] \\ &{\color{black}+}c_2(\theta)\Big[\Phi_{1/2}(\vec{r}_1,\vec{r}_2,\vec{r}_3)-\Phi_{1/2}(\vec{r}_3,\vec{r}_2,\vec{r}_1)\Big] \\ &{\color{black}+}c_3(\theta)\Big[\Phi_{1/2}(\vec{r}_2,\vec{r}_3,\vec{r}_1)-\Phi_{1/2}(\vec{r}_1,\vec{r}_3,\vec{r}_2)\Big] \ , \end{aligned} $$ $$ \begin{aligned} \chi_{3}(\vec{r}_1,\vec{r}_2,\vec{r}_3)=&{\color{white}+}c_1(\theta)\Big[\Phi_{1/2}(\vec{r}_2,\vec{r}_3,\vec{r}_1)-\Phi_{1/2}(\vec{r}_3,\vec{r}_2,\vec{r}_1)\Big] \\ &{\color{black}+}c_2(\theta)\Big[\Phi_{1/2}(\vec{r}_3,\vec{r}_1,\vec{r}_2)-\Phi_{1/2}(\vec{r}_1,\vec{r}_3,\vec{r}_2)\Big] \\ &{\color{black}+}c_3(\theta)\Big[\Phi_{1/2}(\vec{r}_1,\vec{r}_2,\vec{r}_3)-\Phi_{1/2}(\vec{r}_2,\vec{r}_1,\vec{r}_3)\Big] \ . \end{aligned} $$ The above symmetry adaptation of the spatial wave function will satisfy the Pauli principle, while the spin eigenfunction property is built into the $c_1,c_2,c_3$ coefficients; I considered $M_S=+1/2$, but the same will be true for $M_S=-1/2$ with arrows "flipped" in the spin functions. Note that this wave function is yet to be normalized (in the general case, not much can be said about the overlap of the original and the permuted $\Phi$).

So, the short answer to your third question is that for more than two fermions, the spin part is not simply multiplied by a spatial function, but by a non-trivial linear combination enforcing the overall antisymmetry.

Addendum

Up to this point, we made no reference to the structure of the spatial function that we subjected to symmetry adaptation. It might be instructive to show the connection with Slater determinants as special cases of the above.

Let us choose the product form $\lvert\Phi_{1/2}\rangle=\lvert\phi_a\phi_b\phi_c\rangle$, where the $\phi$-s are spatial orbitals of particles 1, 2 and 3, in this order. Then (in bra-ket notation) $$ \begin{aligned} \lvert\chi_{1}\rangle=&{\color{white}+}c_1(\theta)\Big[\lvert\phi_a\phi_b\phi_c\rangle-\lvert\phi_a\phi_c\phi_b\rangle\Big] \\ &{\color{black}+}c_2(\theta)\Big[\lvert\phi_b\phi_c\phi_a\rangle-\lvert\phi_b\phi_a\phi_c\rangle\Big] \\ &{\color{black}+}c_3(\theta)\Big[\lvert\phi_c\phi_a\phi_b\rangle-\lvert\phi_c\phi_b\phi_a\rangle\Big] \ , \end{aligned} $$ $$ \begin{aligned} \lvert\chi_{2}\rangle=&{\color{white}+}c_1(\theta)\Big[\lvert\phi_c\phi_a\phi_b\rangle-\lvert\phi_b\phi_a\phi_c\rangle\Big] \\ &{\color{black}+}c_2(\theta)\Big[\lvert\phi_a\phi_b\phi_c\rangle-\lvert\phi_c\phi_b\phi_a\rangle\Big] \\ &{\color{black}+}c_3(\theta)\Big[\lvert\phi_b\phi_c\phi_a\rangle-\lvert\phi_a\phi_c\phi_b\rangle\Big] \ , \end{aligned} $$ $$ \begin{aligned} \lvert\chi_{3}\rangle=&{\color{white}+}c_1(\theta)\Big[\lvert\phi_b\phi_c\phi_a\rangle-\lvert\phi_c\phi_b\phi_a\rangle\Big] \\ &{\color{black}+}c_2(\theta)\Big[\lvert\phi_c\phi_a\phi_b\rangle-\lvert\phi_a\phi_c\phi_b\rangle\Big] \\ &{\color{black}+}c_3(\theta)\Big[\lvert\phi_a\phi_b\phi_c\rangle-\lvert\phi_b\phi_a\phi_c\rangle\Big] \ . \end{aligned} $$ Let us introduce spin orbitals as $\lvert a\rangle=\lvert\phi_a\rangle\otimes\lvert\uparrow\rangle$, $\lvert\overline{a}\rangle=\lvert\phi_a\rangle\otimes\lvert\downarrow\rangle$, so that $$ \begin{aligned} \lvert\Psi_{1/2,+1/2}\rangle =&{\color{white}+}\frac{c_1(\theta)}{\sqrt{6}}\Big[\lvert \overline{a}bc\rangle+\lvert c\overline{a}b\rangle+\lvert bc\overline{a}\rangle-\lvert b\overline{a}c\rangle-\lvert \overline{a}cb\rangle-\lvert cb\overline{a}\rangle\Big] \\ &{\color{black}+}\frac{c_2(\theta)}{\sqrt{6}}\Big[\lvert a\overline{b}c\rangle+\lvert ca\overline{b}\rangle+\lvert \overline{b}ca\rangle-\lvert \overline{b}ac\rangle-\lvert ac\overline{b}\rangle-\lvert c\overline{b}a\rangle\Big] \\ &{\color{black}+}\frac{c_3(\theta)}{\sqrt{6}}\Big[\lvert ab\overline{c}\rangle+\lvert \overline{c}ab\rangle+\lvert b\overline{c}a\rangle-\lvert ba\overline{c}\rangle-\lvert a\overline{c}b\rangle-\lvert \overline{c}ba\rangle\Big] \ . \end{aligned} $$ All three terms contain an antisymmetrized product of spin orbitals (Slater determinants). In special cases, we can further reduce this. For example, if you wanted to do a symmetry-adapted Hartree-Fock calculation for the Li atom, then two electrons would be loaded in the same $S$ orbital (with opposite spins), and the third one in another $S$ orbital. Let us set two of the spatial orbital indices equal (say $a=b=1$ and $c=2$). Then the third term vanishes (since two electrons with the same spin would occupy the same orbital), and we are left with $$ \begin{aligned} \lvert\Psi_{1/2,+1/2}\rangle =&{\color{white}+}\frac{c_1(\theta)-c_2(\theta)}{\sqrt{6}}\Big[\lvert \overline{1}12\rangle+\lvert 2\overline{1}1\rangle+\lvert 12\overline{1}\rangle-\lvert 1\overline{1}2\rangle-\lvert \overline{1}21\rangle-\lvert 21\overline{1}\rangle\Big] \ . \end{aligned} $$ Assuming orthonormal orbitals, we can simply normalize the wave function to 1: $$ \begin{aligned} \lvert\Psi_{1/2,+1/2}\rangle =&{\color{white}+}\frac{1}{\sqrt{6}}\Big[\lvert 1\overline{1}2\rangle+\lvert \overline{1}21\rangle+\lvert 21\overline{1}\rangle-\lvert \overline{1}12\rangle-\lvert 2\overline{1}1\rangle-\lvert 12\overline{1}\rangle\Big] \ . \end{aligned} $$ So indeed, the symmetry adapted HF wave function of Li is a single Slater determinant. By again separating spatial and spin degrees of freedom, we can check that we did nothing illegal, this is still a spin eigenfunction with $S=1/2$: $$ \begin{aligned} \lvert\Psi_{1/2,+1/2}\rangle =\frac{1}{\sqrt{3}}\Big[ {\color{white}+}&\lvert \phi_1\phi_1\phi_2\rangle\otimes\frac{1}{\sqrt{2}}\left[\lvert\uparrow\downarrow\uparrow\rangle-\lvert\downarrow\uparrow\uparrow\rangle\right] \\ {\color{black}+}&\lvert \phi_2\phi_1\phi_1\rangle\otimes\frac{1}{\sqrt{2}}\left[\lvert\uparrow\uparrow\downarrow\rangle-\lvert\uparrow\downarrow\uparrow\rangle\right] \\ {\color{black}+}&\lvert \phi_1\phi_2\phi_1\rangle\otimes\frac{1}{\sqrt{2}}\left[\lvert\downarrow\uparrow\uparrow\rangle-\lvert\uparrow\uparrow\downarrow\rangle\right] \Big] \ . \end{aligned} $$ It is important to remember that antisymmetry is always understood in the combined coordinate+spin space. Sometimes it is easier to think about this by introducing projections with formal coordinate+spin eigenstates; see this earlier answer of mine.

Answer 2

(2) What to do is: you can't put 3 electrons in the same state (disregarding spin). This is how atoms work: H, He get all required electron(s) in the $^1S$ orbital, but when we get to Li, that shell is FULL, and you need to introduce the $^2S$ orbital, and that orbital is (spatially) orthogonal to the $^1S$ orbital.

Note: the reason we're topped out at 2 is because the underlying symmetry group, $SU(2)$, has a 2 dimensional fundamental representation $(u, d)$. Had we been discussing color symmetry, $SU(3)$, then the fundamental rep is $(r, g, b)$, and we can put 3 particles in an antisymmetric state, e.g. the $\Omega^-$ baryon.

(1) The symmetry can be calculated. The question: "How to understand it" is somewhat subjective. I will outline a very mathematical method that starts with the Robinson–Schensted correspondence (https://en.wikipedia.org/wiki/Robinson–Schensted_correspondence) and then uses Schur–Weyl duality (https://en.wikipedia.org/wiki/Schur–Weyl_duality) to get the answer. [note: those wiki links should be suitably abstract as to be useless]. The technique works for any number of particles, in any number of spatial or spinor dimensions, over any field ($\mathbb R$, $\mathbb C$, and I think some more mathy ones).

Suppose we have $N$ particles (I will use $N=3$). Then, take the integer $N$ and enumerate its partitions (https://en.wikipedia.org/wiki/Integer_partition), e.g.:

$$ 3 = 1 + 1 + 1 $$ $$ 3 = 1 + 2$$ $$ 3 = 3 $$

Those can be expressed as Young diagrams (via the sum of the number of boxes in a row):

Each diagram shows the symmetry: rows are symmetric and columns are antisymmetric, so for $N=2$, the two choices are 1 row ($uu, ud+du, dd$), or 1 column ($uu-du$).

When you set $N=3$, as you pointed out, there are now sets with mixed symmetry. That is the one diagram that is upside-down "L" shaped. It's symmetric in one pair of indices and antisymmetric in the other.

The questions are then: which ones, and how do I enumerate them?

You need to find all the standard Young tableux that fill the diagram. In this case, there are 2, as show here:

Each of those represent a representation of the permutation group on $N=3$ symbols, and the numbers in the boxes represent particle label (or index-count, if you're doing cartesian tensors). From there, there is a procedure to find a permutation called "The Young Symmetrizer" that is applied to particle labels, and that gives you an irreducible basis of states (irrep) to use.

There is also "The Remarkable Hook Length Formula" (https://en.wikipedia.org/wiki/Hook_length_formula) that allows you to compute the dimension of the irrep. What is amazing is that it works for any initial number of dimensions. So for spin (d=2), (aka doublets):

$$ {\bf 2} \otimes {\bf 2} \otimes {\bf 2} = {\bf 4_S} \oplus {\bf 2} \oplus{\bf 2}\oplus{\bf 0_A} $$

which means there is a 4D symmetric quartet $(J=\frac 3 2$), two mixed symmetry doublets ($J=\frac 1 2$), and zero antisymmetric combinations.

Had I started with a 3D basis (QCD color, $uds$-flavor, or Euclidean $(x, y, z)$, the formula says:

$$ {\bf 3} \otimes {\bf 3} \otimes {\bf 3} = {\bf 10_S} \oplus {\bf 8} \oplus{\bf 8}\oplus{\bf 1_A} $$

where you may recognize some of the dimensions from say Gell-Mann's Eight-Fold Way and the famed Baryon decuplet:

and octet:

or for the one dimensional antisymmetric Cartesian rank-3 tensor, aka $\epsilon_{ijk}$:

So it's a giant field, with a monster answer.

On question (3): I'm not sure what you mean. I think yes, I mean that's what atomic physics does for atoms bigger than lithium.

Answer 3

1. How to understand symmetry if there are more than two particles?
2. What to do with the statement that the wave function should be antisymmetric if two particles are exchanged?
3. Is it possible to separate the wave function into the product of the coordinate and spin parts for three and more particles?

For more than two particles, the permutation group contains sets of states that are neither symmetric nor antisymmetric under permutation. For $3$ spin-1/2 particles for instance, there are $2$ possible states with $S=1/2$ and $m_s=1/2$. You construct them by first combining spins 1 and 2 into $S_{12}=1$ and $S_{12}=0$ respectively, and then couple these to the last spin. You get: \begin{align} \vert \psi_1\rangle &= \sqrt{\frac23}\vert ++-\rangle-\frac{1}{\sqrt{6}}\vert +-+\rangle -\frac{1}{\sqrt{6}}\vert -++\rangle\\ \vert \psi_2\rangle&=\frac{1}{\sqrt{2}}\vert +-+\rangle -\frac{1}{\sqrt{2}}\vert -++\rangle \, , \end{align} where $\vert abc\rangle=\vert\frac12,\frac{a}{2}\rangle_1 \vert\frac12,\frac{b}{2}\rangle_2\vert\frac12,\frac{c}{2}\rangle_3$. Notice how $\vert\psi_1\rangle$ is symmetric under interchange of particles $1$ and $2$. However, if you permute $2$ and $3$ then $\vert \psi_1\rangle$ will become a linear combination of $\vert \psi_1\rangle$ and $\vert \psi_2\rangle$.

All is not lost because you also need to permute the spatial indices, so it is in fact possible combinations of spatial kets $\vert\phi_1\rangle$ and $\vert \phi_2\rangle$, each product of 3 spatial states, so that some specific linear combination of the form $$ \alpha_{11} \vert\phi_1\rangle\vert\psi_1\rangle +\alpha_{12}\vert\phi_1\rangle\vert\psi_2\rangle +\alpha_{21}\vert\phi_2\rangle\vert\psi_1\rangle +\alpha_{22}\vert\phi_2\rangle\vert\psi_2\rangle $$ is actually fully antisymmetric. The resulting states do not cleanly separate the spatial and spin degrees of freedom although it guarantees the resulting states have total $S=1/2$

You can also use determinants. For instance $$ \text{Det}\left\vert \begin{array}{ccc} \xi(x_1)\vert +\rangle_1 & \xi(x_1)\vert -\rangle_1 & \zeta(x_1)\vert +\rangle_1 \\ \xi(x_2)\vert +\rangle_2 & \xi(x_2)\vert -\rangle_2 & \zeta(x_2)\vert +\rangle_2 \\ \xi(x_3)\vert +\rangle_3 & \xi(x_3)\vert -\rangle_3 & \zeta(x_3)\vert +\rangle_3 \end{array} \right\vert $$ where $\xi(x)$, and $\zeta(x)$ are spatial states. In this case, the resulting 3-particle state does not necessarily have good total $S$ but permuting particles $i$ and $j$ amounts to permuting rows $i$ and $j$ in the determinant, so the result automatically picks up a sign.

It is only possible to cleanly separate space and spin degrees of freedom for symmetric spin states. For say $n=3$ spins, the state $\vert +++\rangle$ is clearly symmetric. To go with this you need to generate an antisymmetric spatial part, which is done again using a determinant, this time of 3 separate spatial states: $$ \text{Det}\left\vert \begin{array}{ccc} \xi(x_1) & \eta(x_1) & \zeta(x_1) \\ \xi(x_2) & \eta(x_2) & \zeta(x_2) \\ \xi(x_3) & \eta(x_3) & \zeta(x_3) \end{array} \right\vert $$