I'm reading the section "Two Electron System" in Sakurai's textbook and I'm stuck on the following reasoning:
Let us now consider specifically a two-electron system. The eigenvalue of the permutation operator is necessarily $-1$. Suppose the base kets we use may be specified by $\mathbf x_1, \mathbf x_2, m_{s1}$, and $m_{s2}$, where $m_{s1}$ and $m_{s2}$ stand for the spin-magnetic quantum numbers of electron 1 and electron 2, respectively.
We can express the wave function for a two-electron system as a linear combination of the state ket with eigenbras of $\mathbf x_1, \mathbf x_2, m_{s1}$, and $m_{s2}$ as follows:
$$ \psi = \sum_{m_{s1}}\sum_{m_{s2}} C(m_{s1}, m_{s2})\cdot \langle \mathbf x_1, m_{s1}; ~\mathbf x_2, m_{s2} \,\lvert \, \alpha\rangle. $$
If the Hamiltonian commutes with $S_{\text{tot}}^2$,
$$ \left[ S_{\text{tot}}^2, H \right] = 0,$$
then the energy eigenfunction is expected to be an eigenfunction of $S_{\text{tot}}^2$, and if $\psi$ is written as
$$ \psi = \phi(\mathbf x_1, \mathbf x_2)\chi(m_{s1},m_{s2}), $$
then the spin function $\chi$ is expected to be one of the following:
$$ \chi(m_{s1},m_{s2}) = \begin{cases} \chi_{++} & \\ \frac{1}{\sqrt{2}}\cdot (\chi_{+-} + \chi_{-+}) & \\ \chi_{--} & \\ \frac{1}{\sqrt{2}}\cdot (\chi_{+-} - \chi_{-+}) &\end{cases} $$
I have the following questions:
- What's the role of the Hamiltonian in that section?
- Why can we decouple position and spin, i.e. find independent function $\phi$ and $\chi$ yielding $\psi$ at all?
- Why, in principle, couldn't $\chi$ be of form $a\cdot (\chi_{++} + \chi_{+-} + \chi_{-+})$? I understand that this has to do something with the symmetrization postulate, but I can't see why exactly this needs to be the case: If $P$ is the permutation operator, then $$ P(\chi_{++} + \chi_{+-} + \chi_{-+}) = (\chi_{++} + \chi_{-+} + \chi_{+-}),$$ which is exactly the same and thus should be symmetrical.
