Coherent state of second quantized hamiltonian

Question

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$.)

1. 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 $b$.

2. 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.

Answer 1

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.