# Are these two different forms of the squeeze operator equivalent?

As far as I know, the squeeze operator can be presented as: $$S(z)=exp(\frac{1}{2}z a^\dagger a^\dagger-\frac{1}{2}z^* a a),$$ where $$z=re^{i\theta}$$.

When I tried to use Baker–Campbell–Hausdorff formula to expand $$S(z)$$, I found a paper "Impossibility of naively generalizing squeezed coherent states" PRD 29, 1107(1984), where

$$S'(z)=exp(\frac{1}{2}e^{i\theta}\tanh{r} a^\dagger a^\dagger - \frac{1}{2}e^{-i\theta}\tanh{r} a a + (\text{sech} r-1)a^\dagger a - \frac{1}{2}\ln{(\cosh{r})}).$$

I couldn't see they are equivalent to each other. I know

$$\lim_{r->0} S'(z) = S(z),$$

but I don't think there is such assumption when we deal with most cases. Did I misunderstand something here?

• Does this result from the action of the displacement operator $D(\alpha)~=~exp(\alpha a^\dagger - \alpha^*a)$ on the squeezed state operator $S(z)$? The BCH formula is usually applied to the multiplication of exponentiated operators. – Lawrence B. Crowell Jan 18 '17 at 17:58
• @LawrenceB.Crowell I don't think so, there is no $\alpha$ in the second formula, and it is for squeezed vacuum state I think. There is a Zassenhaus formula for an expression like $exp(a+b)$, as I see from the BCH formula wikipedia page: en.wikipedia.org/wiki/…. – Lu Zhang Jan 18 '17 at 18:06

• @LuZhang That's not what I meant - the emphasis is on summary. You don't need to (and shouldn't) paste the full paper, but a paragraph detailing the key points would vastly improve this answer. – Emilio Pisanty Feb 2 '17 at 14:28
The usual decomposition is $$S(z)= \exp\left\{{\textstyle \frac12}(z {a^\dagger}^2 -z^* a^2)\right\}\\ =\exp\left\{e^{i\theta}{\textstyle \frac12}\tanh |z|\, {a^\dagger}^2\right\}\exp\left\{ -\ln\cosh |z| (a^\dagger a+{\textstyle \frac12})\right\} \exp\left\{-e^{-i\theta}{\textstyle \frac12}\tanh |z| \, a^2\right\},\\ =\exp\left\{-e^{-i\theta}{\textstyle \frac12}\tanh |z| \,a^2\right\}\exp\left\{ +\ln\cosh |z| (a^\dagger a+\textstyle \frac12)\right\} \exp\left\{e^{i\theta}{\textstyle \frac12}\tanh |z|\, {a^\dagger}^2\right\}.$$ This comes very quickly if you use from a Gaussian decomposition and the faithful, but non-unitary representation of the $$\mathfrak{su}(1,1)$$ algebra: $$a^2\mapsto 2i\sigma_-,\\ {a^\dagger}^2 \mapsto 2i\sigma_+,\\ (a^\dagger a+\textstyle \frac12)\mapsto \sigma_3.$$ Your original paper has the key steps over the papge from the auors eq 3.1. I can't see how the authors of your paper put all the exponentials together in their eq 3.1 though.