2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.


As for your other questions, I'm not particularly aware of a broad usage of the concept of "higher-order squeezing" in the literature, but a quick search does produce the paper

Higher-Order Squeezing of a Quantum Field. C. K. Hong and L. Mandel. Phys. Rev. Lett. 54, 323 (1985)

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.

Still, if you're interested in going down that road, that should be a good starting point for a literature search, by taking those two papers and looking for interesting papers that cite them, their other references, and so on.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.