From the expression of the Hamiltonian you can see that $\hat{H}=\hat{H}_{1}+\hat{H}_{2}$. Also you have the following commutation relation $[\hat{H}_{1},\hat{H}_{2}]=0$. Thus, you can label the eigenstates as $|n_{1}n_{2},SM_{s}\rangle$. Where $|SM_{s}\rangle$ is the two particle total spin state.
Now, the ket corresponding to the ground state is $|n_{1}n_{2},SM_{s}\rangle=|00,00\rangle$. To see where the spin comes into play, we look at the first excited state. The first excited energy is $E_{1}=E(10)=E(01)$ with four possible kets:
For the singlet state $\left(\frac{1}{\sqrt{2}}|10\rangle+\frac{1}{\sqrt{2}}|01\rangle\right)|00\rangle$
For the triplet state $\left(\frac{1}{\sqrt{2}}|10\rangle+\frac{1}{\sqrt{2}}|01\rangle\right)|S=1,M_{s}=0,\pm1\rangle$
Having fermions, the antisymmetric wave function is
$$\psi=\frac{1}{\sqrt{2}}(\psi_{1}(x_{1})\psi_{2}(x_{2})-\psi_{1}(x_{2})\psi_{2}(x_{1}))$$
(there's a plus in your wave function and that is for integer spin particles). This wave function can be split into a spatial and spin part. Being an antisymmetric wave function, when the spatial part is symmetric the spin part is antisymmetric and vice versa.
$$\psi(x_{1},x_{2},M_{1},M_{2})=\left\{
\begin{array}{ll}
\psi^{S}(x_{1},x_{2})\chi^{A}(M_{1},M_{2})\\
\psi^{A}(x_{1},x_{2})\chi^{S}(M_{1},M_{2})
\end{array}\right.$$
where the spatial part is
$$\psi^{S}(x_{1},x_{2})=\frac{1}{\sqrt{2}}(\psi_{1}(x_{1})\psi_{2}(x_{2})+\psi_{1}(x_{2})\psi_{2}(x_{1}))$$
$$\psi^{A}(x_{1},x_{2})=\frac{1}{\sqrt{2}}(\psi_{1}(x_{1})\psi_{2}(x_{2})-\psi_{1}(x_{2})\psi_{2}(x_{1}))$$
and the spin part
$$\chi^{A}(M_{1},M_{2})=\frac{1}{\sqrt{2}}(\uparrow_{1}\downarrow_{2}-\uparrow_{2}\downarrow_{2}) \hspace{5mm} M_{s=}0,\hspace{5mm} S=0$$
$$\chi^{S}(M_{1},M_{2})=\left\{\begin{array}{ll}
\uparrow_{1}\uparrow_{2},\hspace{29mm} M_{s}=1\\
\frac{1}{\sqrt{2}}(\uparrow_{1}\downarrow_{2}+\uparrow_{2}\downarrow_{2}) \hspace{5mm} M_{s=0}\\
\downarrow_{1}\downarrow_{2} \hspace{29mm} M_{s}=-1
\end{array}\right \} , S=1$$
Your "snap of the fingers" really means that your Hamiltonian is
$$H(t) = \frac{p^2}{2m} + \frac{1}{2}m \omega^2 - \theta(t)C$$
where $\theta(t)$ is the Heaviside step function.
The eigenstates will only be equivalent asymptotically for $t \rightarrow\pm\infty$ since you are now dealing with a time-dependent Hamiltonian.
Best Answer
The Hamiltonian of your problem is given by $$\begin{align} H &= \frac{P^2}{2m}+ V_0\left(\frac{X}{a} +1\right)\left(\frac{X}{a}-1\right) \theta(X+a) \theta(a-X) \\ &=\frac{P^2}{2m}+\frac{V_0}{a^2}\left(X^2 -a^2\right)\theta(X+a)\theta(a-X),\end{align}$$ describing neither a harmonic oscillator nor being identical to the (obviously wrong) expression for $\hat{H}$ given in your second equation.
If you wish to establish a relation between your system and a harmonic oscillator, you could proceed as follows: as the inequality $x^2-a^2 \ge 0$ holds for $|x|\ge a$, the Hamiltonian $H$ and the operator $$H_{\rm HO} =\frac{P^2}{2m} + \frac{V_0}{a^2}(X^2-a^2)= \frac{P^2}{2m}+\frac{V_0}{a^2} X^2-V_0$$ obey the operator inequality $H \le H_{\rm HO}$. $H_{\rm HO}$ describes a harmonic oscillator with angular frequency $\omega =\frac{1}{a}\sqrt{2 V_0/m}$ and the pure point spectrum $\hbar \omega (n+1/2)-V_0$ (with $n=0, 1, 2, \ldots$). In addition, $H$ fulfills the (trivial) inequality $-V_0 \le H$ and we conclude that the two operator inequalities $$-V_0 \le H \le H_{\rm HO}$$ imply$^\ast$ that the ground-state energy $E_0$ of the Hamiltonian $H$ is bounded from below and above by $$-V_0 \le E_0 \le \frac{\hbar \omega}{2}-V_0, \qquad \omega=\frac{1}{a}\sqrt{\frac{2 V_0}{m}}.$$ In the case of large values of $\omega$ with $\hbar \omega /2 \ge V_0$, the upper bound can be improved to $E_0 \le 0$.
The spectrum of $H$ consists of a point spectrum with eigenvalues $-V_0 \le E_r \le 0$ and a continuous spectrum $[0, \infty)$. There is no reason why the well-known theorem about the existence of at least one bound state in the case of an attractive potential in one spatial dimension should be violated in this specific case.
$^\ast$ $H \le H_{\rm HO}$ means that $\langle \psi | H \psi \rangle \le \langle \psi | H_{\rm HO} \psi \rangle$ for all $\psi \in D(H) \cap D(H_{\rm HO})= D(H_{\rm HO})$. $\inf\limits_{||\psi ||=1} \langle \psi | H \psi \rangle \le \inf\limits_{||\psi||=1} \langle \psi |H_{\rm HO} \psi \rangle$ implies the inequality $E_0 \le \hbar \omega/2 -V_0$ for the ground-state energies of $H$ and $H_{\rm HO}$. Analogously, $ -V_0 \langle \psi |\psi \rangle \le \langle \psi | H \psi \rangle$ for all $\psi \in D(H)$, implies $-V_0 \le E_0$.