# Two-Electron System

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:

1. What's the role of the Hamiltonian in that section?
2. Why can we decouple position and spin, i.e. find independent function $\phi$ and $\chi$ yielding $\psi$ at all?
3. 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.

If $S_\text{tot}^2$ is a symmetry i.e., commutes with the Hamiltonian, then one can choose a basis in where it is diagonal. This corresponds to going to the coupled basis, the triplet (total spin=1) and singlet (total spin=0) combinations of the two spins. Those are the four choices that you have listed for the `spin functions' -- the first three are the triplet $m=+1,0,-1$ states and the last one is the singlet.