# Deriving Squeezing and Higher Order Squeezing operators

I have few doubts regarding the Squeezing operation:

The Squeeze Operator for a single mode of an electromagnetic field is given as:

$$\hat{S}(z)=\exp(\frac{1}{2}(z^{*}\hat{a}^{2}-z{\hat{a}^{\dagger2}}))$$

1. How can we derive this equation?

2. Is there a similar operator for higher order squeezing? How can we derive one?

3. After second order squeezing, one of the quadratures will have its variance squeezed by $e^{-2r}$. How can we derive the similar variance change for $2N$ order squeezing?

I am an absolute beginner in this field. Although I am not expecting a spoon-fed answer, refs and guidelines are deeply appreciated.

The squeeze operator cannot be 'derived' in any meaningful sense: the relationship $$\hat{S}(z)=\exp(\frac{1}{2}(z^{*}\hat{a}^{2}-z{\hat{a}^{\dagger2}}))$$ is simply its definition, and it does not need to be justified. What you do need to do, of course, is to provide a suitable sense in which the operator $\hat S(z)$ as defined above does actually act on the system with a 'squeezing' action as normally understood. To show this, one works from the action on the creation and annihilation operators, \begin{align} {\hat {S}}^{\dagger }(z){\hat {a}}{\hat {S}}(z) & ={\hat {a}}\cosh r-e^{i\theta }{\hat {a}}^{\dagger }\sinh r \\ {\hat {S}}^{\dagger }(z){\hat {a}}^{\dagger }{\hat {S}}(z) & ={\hat {a}}^{\dagger }\cosh r-e^{-i\theta }{\hat {a}}\sinh r \end{align} which reads as \begin{align} {\hat {S}}^{\dagger }(r){\hat {a}}{\hat {S}}(z) & ={\hat {a}}\cosh r-{\hat {a}}^{\dagger }\sinh r \\ {\hat {S}}^{\dagger }(r){\hat {a}}^{\dagger }{\hat {S}}(z) & ={\hat {a}}^{\dagger }\cosh r-{\hat {a}}\sinh r \end{align} once you specify $z=r$ with squeezing direction at $\theta=0$ along the position axis. From there, you take the sum and difference of those equations to get the action of the squeezing operator on the position and momentum operators, $\hat x = \frac12(\hat a + \hat a^\dagger)$ and $\hat p = \frac{1}{2i}(\hat a - \hat a^\dagger)$, as \begin{align} {\hat {S}}^{\dagger }(r) \hat x {\hat {S}}(r) & = e^{r} \hat x \\ {\hat {S}}^{\dagger }(r) \hat p {\hat {S}}(r) & = e^{-r} \hat p. \end{align} (Re-do the algebra yourself - there's a nonzero chance I mucked up the constants.) This is a clear re-scaling operation on phase space, which squeezes the position quadrature at the expense of the momentum one, and this is what justifies the name 'squeezing' for the operator $\hat S(z)$ as initially defined.
with some nontrivial citations, and which defines higher-order squeezing as a squeezing that's manifested not in the variance, $\langle (\Delta \hat x)^2\rangle$, but rather on the higher-order moments of the quadrature, $\langle (\Delta \hat x)^{2n}\rangle$, $n>1$. There are some papers that introduce higher-order squeezing operators (including in particular this one), but I don't get the sense that they're very widely used in the modern literature.