I'm reading Lectures on Quantum Mechanics by Gordon Baym (1969). In his discussion of 3-identical fermions Baym writes: "One way to make $\Psi(1,2,3)$ [the total wave-function] antisymmetric is to take a symmetric $\chi\left(s_{1}, s_{2}, s_{3}\right)$ times an antisymmetric $\psi\left(\mathbf{r}_{1}, \mathbf{r}_{2}, \mathbf{r}_{3}\right) .$ The other way around won't work, since it isn't possible to construct a completely antisymmetric spin wave function $\chi\left(\mathrm{s}_{1}, \mathrm{s}_{2}, \mathrm{s}_{3}\right)$ from just the two choices, up or down, for each spin. There is another possibility though. Suppose that we take a $\chi\left(\mathrm{s}_{1}, \mathrm{s}_{2}, \mathrm{s}_{3}\right)$ that is antisymmetric in $\mathrm{s}_{2}$ and $\mathrm{s}_{3},$ for example," $$ \chi(s_1,s_2,s_3)=\chi_{\uparrow}\left(\mathrm{s}_{1}\right)\left[\chi_{\uparrow}\left(\mathrm{s}_{2}\right) \chi_{\downarrow}\left(\mathrm{s}_{3}\right)-\chi_{\downarrow}\left(\mathrm{s}_{2}\right) \chi_{\uparrow}\left(\mathrm{s}_{3}\right)\right] $$ Baym goes on to construct the totally anti-symmetric wave-function: $$ \Psi(1,2,3)=\chi\left(\mathrm{s}_{1}, \mathrm{s}_{2}, \mathrm{s}_{3}\right) \psi\left(\mathrm{r}_{1}, \mathrm{r}_{2}, \mathrm{r}_{3}\right)+\chi\left(\mathrm{s}_{2}, \mathrm{s}_{3}, \mathrm{s}_{1}\right) \psi\left(\mathrm{r}_{2}, \mathrm{r}_{3}, \mathrm{r}_{1}\right) + \chi\left(\mathrm{s}_{3}, \mathrm{s}_{1}, \mathrm{s}_{2}\right) \psi\left(\mathrm{r}_{3}, \mathrm{r}_{1}, \mathrm{r}_{2}\right) $$ My question is what exactly Baym means when he says "it isn't possible to construct a completely antisymmetric spin wave function $χ(s_1,s_2,s_3)$ from just the two choices, up or down, for each spin.", and how his latter construction is different from that.
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The only possible completely antisymmetric wave function for three spins $1/2$ is identical zero. From three spin variables $s_1, s_2, s_3$, each being equal $1/2$ or $-1/2$, at least two have same value. The antisymmetry of wave function leads to its zero value in this case.
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$\begingroup$ Does this imply Bayms construction is the zero vector? $\endgroup$– CraigCommented Jan 11, 2020 at 19:09
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1$\begingroup$ @Craig. This implies there is no completely antisymmetric function of the form $\chi(s_1,s_2,s_3)\psi(r_1,r_2,r_3)$ where $\chi$ is antisymmetric and $\psi$ is symmetric. $\endgroup$– GecCommented Jan 11, 2020 at 21:04
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$\begingroup$ Ah thank you so much! Do you know how one would translate this spin cooridnate description to the typical Bra-Ket notation? Thanks $\endgroup$– CraigCommented Jan 11, 2020 at 22:24
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1$\begingroup$ As I understand, the wave function is a particular representation of a ket-vector $\Psi(1,2,3) = \langle 1,2,3|\Psi\rangle$. $\endgroup$– GecCommented Jan 12, 2020 at 20:03