I'm wondering if it exists a composition law for the squeezing operation ? I guess so for geometric reason, since they are (generalized, and the phase is annoying of course) hyperbolic rotations of the annihilation $a$ and creation $a^{\dagger}$ operators of some bosonic modes.

I define the squeezing operator as $$ S\left(\zeta\right) = e^{\Sigma} \;\; ; \;\; \Sigma = \frac{\zeta^{\ast} aa -\zeta a^{\dagger}a^{\dagger}}{2}$$ for any complex parameter $\zeta$.

I would like to know the rule for $S\left(\zeta_{1}\right)\cdot S\left(\zeta_{2}\right)$ ?

For instance, we know that $$D\left(\alpha\right)\cdot D\left(\beta\right)=D\left(\alpha + \beta\right) e^{\left(\alpha \beta^{\ast}-\beta\alpha^{\ast} \right)/2}$$ for the coherent / displacement operator $D\left(\alpha\right)=e^{\Delta}$ with $\Delta = \alpha^{\ast} a - \alpha a^{\dagger}$.

A reference for $S\left(\zeta_{1}\right)\cdot S\left(\zeta_{2}\right)$ would be sufficient.


I) OP is asking for the composition formula for so-called single mode squeezing operators, see eq. (8) below. We will not here prove the composition formula (8), but only give partial hints and references.

The key is to realize that one can identify

$$\tag{1} \sigma_{+}~:=~\frac{1}{2} a^{\dagger}a^{\dagger}, \qquad \sigma_{-}~:=~\frac{1}{2} aa, \qquad\sigma_{3}~:=~a^{\dagger}a+\frac{1}{2} , $$

with generators of a $su(1,1)\cong so(2,1; \mathbb{R})\cong sl(2,\mathbb{R})$ Lie algebra

$$\tag{2} [\sigma_{+},\sigma_{-}]~=~\sigma_{3} , \qquad [\sigma_{3},\sigma_{\pm}]~=~\pm2 \sigma_{\pm}. $$

Here the single mode creation and annihilation operators satisfy

$$\tag{3} [a,a^{\dagger}]~=~1.$$

The squeezing operator

$$ \tag{4} S(z)~:=~\exp\left[-z\sigma_{+} + z^{*} \sigma_{-}\right], \qquad z~\in~ \mathbb{C},$$

may be written in normal-ordered form


cf. e.g. Ref. 1 eq. (1.207), or Ref. 2 eqs. (2.16) and (3.4). Here

$$\tag{6} z~=~re^{i\theta}~\in~ \mathbb{C}, \qquad r~\geq~ 0, \qquad\theta~\in~ \mathbb{R},$$


$$\tag{7} t~:=~e^{i\theta}\tanh(r)~\in~ \mathbb{C}.$$

The composition formula reads

$$\tag{8} S(z_1) S(z_2)~=~S(z_3)\exp\left[ \ln\frac{1+t_1 t_2^{*}}{1+t_1^{*}t_2} \frac{\sigma_{3}}{2}\right], \qquad t_3~:=~\frac{t_1+t_2}{1+t_1^{*}t_2}, $$

cf. e.g. Ref. 2 Exercise 3.8.

II) The squeezing operators (4) may be viewed as elements of $SL(2,\mathbb{C})$. We may use the exponential map

$$\tag{9}\exp: sl(2,\mathbb{C}) ~\longrightarrow~ SL(2,\mathbb{C}) $$

to generalized to operators of the form

$$\tag{10} R(\vec{\alpha}) ~:=~\exp\left[ \vec{\alpha} \cdot \vec{\sigma} \right]~=~\exp\left[ \alpha^{+} \sigma_{+}+\alpha^{3} \sigma_{3}+ \alpha^{-} \sigma_{-}\right],$$ $$ \vec{\alpha}~=~(\alpha^{+},\alpha^{3}, \alpha^{-})\in \mathbb{C}^3. $$

The composition formula (8) generalizes to the Baker-Campbell-Hausdorff (BCH) formula

$$\tag{11} \vec{\gamma}~=~ \vec{f}(\vec{\alpha},\vec{\beta}), $$


$$\tag{12} R(\vec{\alpha}) R(\vec{\beta}) ~=~R(\vec{\gamma}). $$

[See also this Phys.SE post for the corresponding BCH formula for $SU(2)$ and $SO(3;\mathbb{R})$.] Note however that the exponential map (9) is not surjective

$$\tag{13}{\rm Im}(\exp) ~=~\left\{M\in SL(2,\mathbb{C}) \mid {\rm Tr}(M)\neq -2\right\} ~\cup~ \left\{-{\bf 1}\right\} \subsetneq SL(2,\mathbb{C}), $$

which means that the BCH map (12) has singularities.


  1. P. Kok and B.W. Lovett, Introduction to Optical Quantum Information Processing, 2010.

  2. G.S. Agarwal, Quantum Optics, 2012. [Note that Ref. 2 has the opposite sign convention $z\to -z$ in eq. (4), see Ref. 2. eqs. (2.14) and (3.2).]

  3. D.R. Truax, Baker-Campbell-Hausdorff relations and unitarity of $SU(2)$ and $SU(1,1)$ squeeze operators, Phys. Rev. D 31 (1985) 1988.

  • $\begingroup$ Thank for your answer. I was precisely asking for your $\gamma$... Is it given in one of the reference you gave ? (Sorry, it's week-end and I've no access to library.) $\endgroup$ – FraSchelle Jun 8 '13 at 6:32
  • $\begingroup$ Moreover, I highly doubt about you last equation $R\left(\alpha\right)\cdot R\left(\beta\right)=R\left(\gamma\right)$ since it is not even true for the displacement operator $D\left(\alpha\right)$, as I discuss in my question. Would you please elaborate a bit more about your answer please. I'm totally unable to find something related to my question in your references 1 and 3. $\endgroup$ – FraSchelle Jun 8 '13 at 13:16
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    $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jun 8 '13 at 16:16
  • $\begingroup$ Thanks a lot ! I've made a mistake in my previous remark. I was thinking that the composition law $S(1)S(2)$ also gives a phase factor. I still don't know how to demonstrate it from the disentangling formula (your eq.(5)) that I know. I was struggling with the phase factor. I'll try to understand it on Monday. Thanks for your answer again. $\endgroup$ – FraSchelle Jun 8 '13 at 16:44

You will not have easily a closed general formula, because, first, the commutator of $aa$ and $a^+a^+$, so $4(a^+a + 1/2)$ does not commute with $aa$ and $a^+a^+$, and secondly,worse, because the same thing happens for all the high-order commutators of the Baker-Campbell-Hausdorff relations (see more precisely Chapter "Zassenhaus formula")

It works with the $D\left(\alpha\right)$ because the commutator of $a$ and $a^+$ is $1$, so of course you have $[a, 1] = [a^+, 1] = 0$, and the infinite series of terms begin a short finite list (with $X \sim~ a, Y ~ \sim a^+$ ): $$e^X e^Y = e^{X + Y}e^{\frac{1}{2}[X,Y]}$$

Maybe, you could calculate first the value of your operator in the ground state, that is :

$$ <0|S\left(\zeta_{1}\right)\cdot S\left(\zeta_{2}\right)|0>$$

  • $\begingroup$ Thanks for your answer. That's precisely because I'm not interested in the ground state expectation value that I was asking this question. Even acting on the ground state, the squeeze state has no tabulated expression: it's just a sum of the even power of the expansion of the exponential argument as far as I know. Thanks again. $\endgroup$ – FraSchelle Jun 8 '13 at 11:41

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