The effect of the squeezing operator \begin{equation} S = e^{- r (a^2 + a^{\dagger 2}) / 2} \end{equation} on a Wigner phase space representation or W-function of a system with density matrix $\rho$ \begin{equation} W(x, p) = \frac{1}{\pi \hbar} \int_{\mathbb{R}} dy e^{- 2 i p y / \hbar} \langle x + y | \rho | x - y \rangle \end{equation} is \begin{equation} S \circ W(x, p) = W(x e^r, p e^{-r}) \end{equation} (I am using the convention that $x = 2^{-1/2} (a + a^{\dagger})$ and $p = - i 2^{-1/2} (a - a^{\dagger})$). Is there a simple relation for the effect of the squeezing operator on the Husimi phase space representation or Q-function \begin{equation} Q(x, p) = \frac{2}{\pi} \int_{\mathbb{R}^2} du dv e^{- 2 (x - u)^2 - 2 (p - v)^2} W(u, v) , \end{equation} i.e., \begin{equation} S \circ Q(x, p) = ? \end{equation} More generally, I'd also like to know if there is a simple relation for the effect of squeezing on a generalised phase space representation or R-function (sometimes also called an $s$-parametrised W-function) \begin{equation} R(x, p, \tau) = \frac{1}{\pi \tau} \int_{\mathbb{R}} du dv e^{- \tau^{-1} (x - u)^2 - \tau^{-1} (p - v)^2} P(u, v) \end{equation} where $P(x, p)$ is the Glauber-Sudarshan phase space representation or P-function. $P(x, p)$ satisfies \begin{equation} \rho = \int_{\mathbb{C}} d\alpha P(\alpha) |\alpha\rangle \langle \alpha | \end{equation} where $|\alpha\rangle$ are the coherent states of $a$. The reason I ask is that I have developed some numerical algorithms that calculate a generalised phase space representation for certain states very accurately and I'd now like to see the effect of squeezing, preferably without having to perform numerical integration etc.
Isn't this question a bit ambiguous?
More directly, skipping the superfluous creating and annihilation operators, the symplectic dilation operator on phase-space variables is just $$ S= e^{r(x\partial_x- p \partial_p)} ,$$ so that $$ S W(x,p)= W(x e^r , p e^{-r} ) $$ Further note $S\partial_x= e^{-r} \partial_x$ and $S \partial_p= e^r \partial_p $.
The Husimi function is the Weierstrass transform of the WF, $$Q(x,p)=T W(x,p), $$ where $$ T\equiv e^{\frac{\hbar}{4} (\partial_x^2+\partial_p^2)} , $$ eqn (122) of Ref 1, equivalent to the integral kernel you provide (in which you have implicitly chosen/absorbed $\hbar=1/2$.)
Thus, purely formally as you are asking, $$SQ(x,p)=Q(x e^r,p e^{-r})= e^{\frac{\hbar}{4} (e^{-2r}\partial_x^2+e^{2r}\partial_p^2)} W(xe^{r},pe^{-r}) .$$
I assume, however, you are asking the question because you are interested, instead, in the Husimi function of a squeezed state, (ref. 2 Chapter 12.2 and prob 12.1): that is, low-pass-filtering a squeezed WF, SW, with an unsqueezed Gaussian, that is leaving the Weierstrass transform operator T unsqueezed.
Your choices are dictated by the application you are interested in, for example, whether you are fussing with the characteristic star Husimi-Voros product or not, etc.
References:
Thomas L. Curtright, David B. Fairlie, & Cosmas K. Zachos, A Concise Treatise on Quantum Mechanics in Phase Space, World Scientific, 2014. The PDF file is available here.
Wolfgang P. Schleich, Quantum Optics in Phase Space, Wiley , 2011.
-
$\begingroup$ Is there an analogous relation for wave functions? If $\psi(x) = \langle x|\psi \rangle $ then can I write $\langle x |\phi \rangle =\langle x| S(r) |\psi \rangle $ in terms of $\psi(x)$? For sure one can show directly that $|\phi(x)|^2 = e^r |\psi(x e^r)|^2$ but of course one still does not know how the phase of the squeezed wave function is modified. Equivalently, what is the action of the squeezing operator on quadrature eigenstates? $\endgroup$ Dec 12 '17 at 16:20
-
$\begingroup$ Well, you just computed squeezing on $W(x,p=0)$, of course, by the very definition of the Wigner function. The phase information, and its concomitant squeezing is encoded in the p - dependence. That's the point of phase-spacery: to completely forget about wavefunctions, and practice clean living henceforth.... Standard treatises do Fock and coherent states, etc... $\endgroup$ Dec 12 '17 at 17:34