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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?

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  • $\begingroup$ 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. $\endgroup$ – Lawrence B. Crowell Jan 18 '17 at 17:58
  • $\begingroup$ @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/…. $\endgroup$ – Lu Zhang Jan 18 '17 at 18:06
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The eq. (23a) in paper "Baker-Campbell-Hausdorff relations and unitarity of SU(2) and SU(1,1) squeeze operators" (PRD 31, 8) solves my question. Detailed proof are shown in this paper.

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    $\begingroup$ This is a bit short for this site, and it will be of limited usefulness to future visitors with similar questions. Please include a summary of the resolution to your problem here. $\endgroup$ – Emilio Pisanty Jan 31 '17 at 23:59
  • $\begingroup$ @EmilioPisanty Sorry, I'm not going to paste the whole paper here, since the paper is basically about the derivation of my question. I'm answering my own question because it might be helpful to others too. I can delete the question if it doesn't satisfy the rule for this site. $\endgroup$ – Lu Zhang Feb 2 '17 at 14:26
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    $\begingroup$ @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. $\endgroup$ – Emilio Pisanty Feb 2 '17 at 14:28
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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.

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