Coherent state of second quantized hamiltonian In my preparation for the exam I tried to solve the exercise 2.4 in Coleman's Introduction to Many-Body Physics. I like diagonalizing Hamiltonian, so I picked this problem. Also to learn more about coherent states, which we just accidentally discussed. 
In this problem one has the bosonic Hamiltonian
\begin{align}
H = \omega \left( a ^ { \dagger } a + \frac { 1 } { 2 } \right) + \frac { 1 } { 2 } \Delta \left( a ^ { \dagger } a ^ { \dagger } + a a \right)
\end{align}
and transforms it with the Bogoliubov transformation
\begin{align}
\begin{aligned} b & = u a + v a ^ { \dagger } \\ b ^ { \dagger } & = u a ^ { \dagger } + v a \end{aligned}
\end{align}
 to \begin{align}
H = \tilde { \omega } \left( b ^ { \dagger } b + \frac { 1 } { 2 } \right)
\end{align}
This is done by $\tilde \omega = \frac{1}{2uv} \Delta$. Now the coherent state comes into play: 

The Hamiltonian has a boson pairing term. Show that the ground state of $H$ can be written as a coherent condensate of paired bosons, given by
  \begin{align}
| \tilde { 0 } \rangle = e ^ { - \alpha \left( a ^ { \dagger } a ^ { \dagger } \right) } | 0 \rangle
\end{align}
  Calculate the value of $\alpha$ in terms of $u$ and $v$. (Hint: $ | \tilde { 0 } \rangle $ is the vacuum for $b$, i.e. $ b | \tilde { 0 } \rangle = \left( u a + v a ^ { \dagger } \right) | \tilde { 0 } \rangle = 0 $. Calculate the commutator of $ \left[ a , e ^ { - \alpha a ^ { \dagger } a ^ { \dagger } } \right] $ by expanding the exponential as a power series. Find a value of $\alpha$ that guarantees that $b$ annihilates the vacuum $ | \tilde { 0 } \rangle$.)



*

*Why is this a coherent state? - I don't know much about coherent states, but basically it means that $ \hat { a } | \alpha \rangle = \alpha | \alpha \rangle $. So I don't see this condition being fulfilled. They state it's the ground state because $b$ annihilates the vacuum and simultaneously it should be the coherent state. But Wikipedia says: 



Physically, this formula ($ \hat { a } | \alpha \rangle = \alpha | \alpha \rangle $)  means that a coherent state remains unchanged by the annihilation of field excitation or, say, a particle. 

As far as I understood this, the coherent state shouldn't be zero after application of $b$. 


*How to calculate the commutator? - I tried the following: $$ \left[ a , \mathrm { e } ^ { - \alpha a ^ { \dagger } a ^ { \dagger } } \right] = \left[ a , \sum _ { k = 0 } ^ { \infty } \frac { \left( - \alpha a ^ { \dagger } a ^ { \dagger k } \right) } { k ! } \right] = \left[ a , - \alpha a ^ { \dagger } a ^ { \dagger } + \frac { \alpha ^ { 2 } } { 2 } \left( a ^ { \dagger } a ^ { \dagger } \right) ^ { 2 } + \ldots \right] = - 2 \alpha a ^ { \dagger } - 2 \alpha ^ { 2 } \left( a ^ { \dagger } \right) ^ { 3 } - \ldots $$
After this I cannot compute the next element in the sum and didn't know how to continue.  

 A: A coherent state (in the Perelomov sense) is a displaced ground state.  Here, the operators 
$$
K_0=\frac{1}{2}\left(a^\dagger a+a a^\dagger\right)\, ,
K_+=a^\dagger a^\dagger\, ,\qquad K_-=a\,a
$$
span an $\mathfrak{su}(1,1)$ algebra.  The displacement operator for this is an $SU(1,1)$ transformation, which can be normal-ordered to give the (unnormalized) states
$$
\vert\xi\rangle = e^{\xi a^\dagger a^\dagger}\vert 0\rangle\, ,\qquad \hbox{or}\qquad 
\vert\xi\rangle = e^{\xi a^\dagger a^\dagger}\vert 1\rangle\, .
$$
In the unitary form one can write
$$
e^{-i\alpha K_0}e^{-i\beta K_y}\vert 0\rangle
$$
where $2iK_y=K_+-K_-$.  The overlap
$$
\langle n\vert e^{-i\alpha K_0}e^{-i\beta K_y}\vert 0\rangle
$$
is actually an $SU(1,1)$ group function; such functions have a closed form expression closely related the usual Wigner $D$-functions for $SU(2)$.  For $SU(1,1)$ they can be found (along with other details) in

Ui, Haruo. "Clebsch-Gordan formulas of the SU (1, 1) group." Progress of Theoretical Physics 44.3 (1970): 689-702.

Searching for "su(1,1) coherent states" in Google will produce multiple helpful hits.
The key hint is that, if your write your $H$ in terms of $K_0$ and $2K_x=K_++K_-$, then the $SU(1,1)$ transformation $e^{-i\beta K_y}$ will digonalize $H$ for some $\beta\in \mathbb{R}$.  This is similar to the way a Hamiltonian $J_0+bJ_x$ is diagonalised by a rotation about $\hat y$.  So the key is to work through commutators like $[K_\pm,K_y]$ to unwrap the effect of the exponential on $K_0$ and $K_x$; if done correctly some magic will happen.
