Square of the squeezing operator

Question

What is the square of the squeezing operator $S(z)=\exp[\frac12\left((z(a^{\dagger})^{ 2}−z^\ast a^2\right)]$?

I mean, with $z \in \mathbb{R}$, what is $S(Z)S(Z)$? Is there any formula ?

Answer 1

Indeed, it is a double squeeze, with $z\to 2z$, since the two operators in the exponents commute with each other (!), so you can simply CBH-add them, $$ S(z)S(z)= S(2z). $$

You may also see this (in overkill, but good to know for more elaborate combinations with the displacement operator) by squaring the middle line matrix of (3.5) in Fisher, R. A., Nieto, M. M., & Sandberg, V. D. (1984). "Impossibility of naively generalizing squeezed coherent states", Physical Review D29 (6), 1107.

The idea is to map the oscillator operators $a^\dagger a^\dagger$, $-aa$ and $a^\dagger a +1/2$ to the generators of SU(1,1), and then work out the combination of the exponentials as 2x2 matrix multiplications, in the easier-to-handle faithful 2x2 matrix representation of SU(1,1). Any group relation holding for this matrix rep will then hold for the oscillator realization!

(See a simple application in this.)