Can I create squeezing with a $P^2$ interaction? One approach to squeeze light is through the one-mode squeezing operator, which can be written as $e^{-i H t}$ with $H \sim (a^2 - (a^\dagger)^2)$. My question is, can I create squeezing with $H \sim P^2 \sim (a-a^\dagger)^2$ ?
Is this described anywhere in the literature? I know it might not be easy to implement, and so of limited relevance. The term $P^2$ has also additional terms in the expansion $\sim a^\dagger a$, which I'm not sure how they contribute.
 A: It is probably messy to implement in Fock space (number states), but can be related to the conventional squeeze operator through the methods of Nieto et al. discussed in this answer, in an elaborate CBH rebraiding.
In Hilbert space, however, it is elementary, if not trivial. Let us discuss the effect of the operator $\hat S'=e^{-3\hat p^2/2}$ on a plain Gaussian state centered at the origin in both coordinate and momentum space! Nondimensionalize $\hbar,\omega,m$ by absorbing them in your units, so they are set equal to one, for simplicity; they can be reinstated at will.
You have, then,
$$
\langle x|\psi\rangle = \psi(x) = \frac{1}{\pi^{1/4}} e^{-x^2/2} ={1\over \sqrt{2\pi}}\int dp e^{ipx}\phi(p)~~\leadsto \\
\phi(p)=\langle p|\psi\rangle =\frac{1}{\pi^{1/4}} e^{-p^2/2} , \Longrightarrow \\
\langle \hat x\rangle = \langle \hat p\rangle=0, ~~~ \langle \hat x^2\rangle = \langle \hat p^2\rangle =1/2 ~~~,
$$
hence it saturates the standard Heisenberg uncertainty relation (Kennard),
$$
\Delta x = 1/\sqrt{2} , ~~~ \Delta p = 1/\sqrt{2} ~~~\leadsto \Delta x\Delta p =1/2.
$$
Now consider the un-normalized state
$$ 
\hat S'|\psi\rangle \equiv | \psi'\rangle \leadsto \\
\phi'(p)\propto \frac{1}{\pi^{1/4}} e^{-3p^2/2}   e^{-p^2/2} =\frac{1}{\pi^{1/4}} e^{-2p^2} ~~~~\leadsto \psi'(x) \propto \frac{1}{\pi^{1/4}} e^{-x^2/8} ~~\Longrightarrow \\
\Delta x =  \sqrt{2} , ~~~ \Delta p = 1/(2\sqrt{2}) ~~~\leadsto \Delta x\Delta p =1/2.
$$
(The primed states need their normalization adjusted, but it washes out in the uncertainty relation!)
The new state is also a Gaussian, still saturating the uncertainty  relation as before, even as the width of the x-distribution has expanded by a factor of two, while the p-distribution has been squeezed to half its original width.
The operator $\hat S '$ has squeezed your (coherent) state. In the language of WP, $e^{\zeta}=1/2$.
