There are many ways to go around this. You can start from the coherent states and apply the unitary $\hat{O}$ directly on them. That will not be that simple because you will get a term $\hat{O}e^{\alpha\hat{a}^\dagger}e^{\beta\hat{b}^\dagger}$. Now, the typical approach would be to exchange the order of the operators to get something like $e^{\alpha\hat{a}^\dagger}e^{\beta\hat{b}^\dagger}\hat{O}$ (up to some extra term due to the commutator). This not really a simple task but once you are done, you can Taylor expand the operator $\hat{O}$ and keep only the zeroth order (all other terms contain annihilation operators acting on vacuum). I am not going to dig into the calculation in more detail, there are many ways to do it and none of them is really pleasant.

But there is a better way. You can define a displacement operator by the action $\hat{D}(\alpha)|0\rangle = |\alpha\rangle$ and then you have $\hat{D}(\alpha)\hat{a}\hat{D}^\dagger(\alpha) = \hat{a}+\alpha$. You can combine this with the formulas for $\hat{O}\hat{a}\hat{O}^\dagger$, $\hat{O}\hat{b}\hat{O}^\dagger$ to see how the annihilation operators are transformed. What you should get is a beam-splitting of the two coherent states, i.e.,
$$|\alpha,\beta\rangle\to|t\alpha+r\beta,t\beta-r\alpha\rangle$$,
where $t = \cos\theta$, $r = \sin\theta$.

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

$$\chi^{A}(M_{1},M_{2})=\frac{1}{\sqrt{2}}(\uparrow_{1}\downarrow_{2}-\uparrow_{2}\downarrow_{2}) \hspace{5mm} M_{s=}0,\hspace{5mm} S=0$$

$$\chi^{S}(M_{1},M_{2})=\left\{\begin{array}{ll}
\uparrow_{1}\uparrow_{2},\hspace{29mm} M_{s}=1\\
\frac{1}{\sqrt{2}}(\uparrow_{1}\downarrow_{2}+\uparrow_{2}\downarrow_{2}) \hspace{5mm} M_{s=0}\\
\downarrow_{1}\downarrow_{2} \hspace{29mm} M_{s}=-1
\end{array}\right \} , S=1$$

## Best Answer

## Mathematical answer #1

Mathematically, you can use the definition $a|z\rangle \equiv z|z\rangle$ of coherent states to explicitly construct them in terms of energy eigenkets. Up to a normalization constant, the result is: $$\left|z\right\rangle = \sum_{n=0}^\infty \frac{z^n}{\sqrt{n!}}\left|n\right\rangle$$ Now, what happens if you let $z \rightarrow 0$ in the expression above? The only surviving term in the right-hand sum will be the one with $n=0$, and we get the result: $$\left|z=0\right\rangle = \left|n=0\right\rangle$$ So the ground state of the system is a coherent state with parameter $z=0$.

## Mathematical answer #2

You could also see this directly from definitions. Coherent states are defined as $a|z\rangle \equiv z\left|z\right\rangle$. Per definition, the coherent state with $z=0$ will then have the following property: $$a\left|z=0\right\rangle = 0$$ The definition of the ground state of the harmonic oscillator is: $$a\left|n=0\right\rangle = 0$$ Putting the two definitions together, we get the same result as earlier: $$\left|z=0\right\rangle = \left|n=0\right\rangle$$

## Physical answer

One of the interesting properties of the coherent states $\left|z\right\rangle$, is that the expectation values of position and momentum follow the equations of a classical harmonic oscillator: $$\langle x \rangle = \sqrt{\frac{2\hbar}{m\omega}}\Re\left\{z \exp(-i\omega t)\right\}$$ $$\langle p \rangle = \sqrt{2\hbar m\omega}\Im\left\{z \exp(-i\omega t)\right\}$$ It might look a bit more familiar if we choose a real value of $z$: $$\langle x \rangle = \sqrt{\frac{2\hbar}{m\omega}} \; z \cos(\omega t)$$ $$\langle p \rangle = -\sqrt{2\hbar m\omega} \; z \sin(\omega t)$$ Now, what would happen if you let $z\rightarrow 0$? Then you basically get a state with $\langle x \rangle = \langle p \rangle = 0$.

So if you think of the coherent states as the states where expectation values behave like a classical harmonic oscillator, then the ground state is the state with

zero energy, i.e. the state with expectation values thatdoesn't oscillate at all.