Exact eigenfunctions of two interacting identical particles
While I was reading about quantum states of $N$ interacting identical particles, I realized that I don't understand some fundamental things. So In order to clear my confusion, I decided to consider a toy model and work everything out for it: ground state wave-function, energy and first excited state wave-function and energy.
Consider two interacting identical particles (let's say bosons with spin $0$ or fermions with spin $\frac{1}{2}$) with Hamiltonian: $$ H=\frac{p_1^2+p_2^2}{2m}+\frac{m\omega^2}{2}(x_1^2+x_2^2)+\frac{m\Omega^2}{2}(x_1-x_2)^2. $$ This Hamiltonian can be easily diagonalized using the coordinates $\xi=\frac{x_1+x_2}{\sqrt{2}}$, $\eta=\frac{x_1-x_2}{\sqrt{2}}$:
eigenfunctions $$ \psi_{n_1,n_2}(x_1,x_2)=A\exp\left{-\frac{m\omega}{2}\xi^2-\frac{m\sqrt{\omega^2+2\Omega^2}}{2}\eta^2\right}H_{n_1}\left(\sqrt{m\omega}\xi\right)H_{n_2}\left(\sqrt{m(\omega^2+2\Omega^2)}\eta\right), $$ and spectrum $$ E_{n_1,n_2}=\omega\left(n_1+\frac{1}{2}\right)+\sqrt{\omega^2+2\Omega^2}\left(n_2+\frac{1}{2}\right),\quad n_1,n_2=0,1,2... . $$ [here $H_n$ - are Hermite polynomials.]
1) two identical bosons with spin $0$.
To get the ground state wave function, which is symmetric in $x_1, x_2$, we take the eigenfunction with $n_1=n_2=0$ $$ \psi_0(x_1,x_2)=\psi_{0,0}(x_1,x_2)=A\exp\left{-\frac{m\omega}{2}\xi^2-\frac{m\sqrt{\omega^2+2\Omega^2}}{2}\eta^2\right} $$ since it is automatically symmetric in $x_1$ and $x_2$. 281
Ground state energy is $E_{0,0}$. First exited state energy will be $E_{1,0}$ with eigenfunction $$ \frac{1}{\sqrt{2}}\left(\psi_{1,0}(x_1,x_2)+\psi_{1,0}(x_2,x_1)\right). $$ We also get that this system can not have energy $E_{0,1}$ because in that case $$ \psi_{0,1}(x_1,x_2)+\psi_{0,1}(x_2,x_1)\equiv 0 $$ since $H_1(x)$ is an odd function.
2) fermions with spin $\frac{1}{2}$.
Ground state is given by $\psi_0(x_1,x_2)=\psi_{0,0}(x_1,x_2)$ and corresponds to the singlet state (total spin $S=0$) of the system of two spin $\frac{1}{2}$ fermions. Since the singlet state spin wave-function is antisymmetric, this means spatial wave-function is symmetric. First exited state energy will be $E_{1,0}$ with symmetric spatial eigenfunction $$ \frac{1}{\sqrt{2}}\left(\psi_{1,0}(x_1,x_2)+\psi_{1,0}(x_2,x_1)\right). $$
However, unlike the boson case above, the energy $E_{0,1}$ is attainable for fermions of spin $\frac{1}{2}$ in the triplet state (with total spin $S=1$). 66