Scalar product of squeezed coherent states

Consider two states of the type $$|\alpha,\xi \rangle = \hat{D}(\alpha) \hat{S}(\xi) |0\rangle$$, where $$D$$ and $$S$$ are the displacement and squeeze operators, respectively, and $$|0\rangle$$ is a 1D harmonic oscillator vacuum state.

My question is: Is there a closed formula for $$\langle \alpha, \xi | \beta, \eta \rangle$$?

I know how to calculate this for two coherent states ($$\xi = \eta = 0$$), but since the commutator of $$[a^2,a] \neq I$$ (which comes from $$S$$) the same strategy I use in that case does not work (i.e., using the Zassenhaus formula).

I saw that there is a way to express the wave function of this state in position representation, so I could calculate this as $$\int dx \langle \alpha, \xi | x \rangle \langle x \beta, \eta \rangle$$, but this seems really unwieldly. Is there a simpler way analogous to the coherent case?

• Please show your attempt, i.e. whatever you tried with the Zassenhaus formula. – DanielSank Jan 9 at 19:42
• Can inserting $\int_{z \in \mathbb{C}}^{}\frac{dz_{}^{*}dz_{}^{}}{2\pi i}| z\rangle\langle z|$ help? Atleast it will reduce the problem to overlap of coherent state and squeezed vaccum state and perhaps a gaussian integral. – Sunyam Jan 9 at 20:19
• @DanielSank I tried using it to move all the $a$'s in the exponentials to the right side (to act over the vacuum), but since the commutator I posted was not central I could not find a simple way to relate $e^A e^B$ to $e^B e^A$. Using the Zassenhaus formula didn't get me very far. – Gabriel Cozzella Jan 9 at 21:07
• @Sunyam Didn't try it, but maybe it helps. Thanks for the suggestion. – Gabriel Cozzella Jan 9 at 21:13

For the pure squeezing case where $$S(z) \equiv e^{{\textstyle \frac12}( z {a^\dagger}^2- z^*a^2)}, \quad z= e^{\theta} |z|$$ $$= e^{ {\textstyle \frac12} e^{i\theta} \tanh |z| {a^\dagger}^2}e^{ - \ln \cosh |z| \left(a^\dagger a+\textstyle \frac12\right)} e^{- {\textstyle \frac12} e^{-i\theta} \tanh |z|{a}^2}$$ and defining $$\alpha= e^{i\theta} \tanh |z|$$ we can use the formula $$\hat S(\alpha_2) \hat S(\alpha_1)= \hat S(\alpha_3) \exp\{i\chi(\alpha_1,\alpha_2) (a^\dagger a +{\textstyle \frac 12})\}$$ where $$\alpha_3= \frac{\alpha_1+\alpha_2}{1+\alpha_1\alpha_2^*}, \quad \exp\{2i\chi\}= \frac{1+\alpha_1^*\alpha_2}{1+\alpha_1\alpha_2^* }.$$ to compute the overlap. I have not tried to add in the displacement operators, but it should not be too hard.
• Thanks! I guess this solves my problem, since it is just a question of commuting $S$'s and $D$'s now in the right way. Do you have a reference for the formula you posted? – Gabriel Cozzella Jan 9 at 21:03
• They are bit complicated as the use the failthful but non-unitary 2d representation of the $\mathfrak{sp}(2)$ lie algebra. I think that I learned of them from Prelemov's book on generalized coherent states. – mike stone Jan 9 at 21:05