# What does Baym mean here in his Lecture on Identical Particles?

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.

## 1 Answer

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.

• Does this imply Bayms construction is the zero vector? Jan 11, 2020 at 19:09
• @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.
– Gec
Jan 11, 2020 at 21:04
• Ah thank you so much! Do you know how one would translate this spin cooridnate description to the typical Bra-Ket notation? Thanks Jan 11, 2020 at 22:24
• As I understand, the wave function is a particular representation of a ket-vector $\Psi(1,2,3) = \langle 1,2,3|\Psi\rangle$.
– Gec
Jan 12, 2020 at 20:03