From Wigner hyperbolic rotation to fractional squeezing transformation

Abstract

Based on the usual Wigner-Weyl transformation theory we find that the Wigner hyperbolic rotation in phase space will map onto fractional squeezing operator in Hilbert space. The merit of Weyl ordering and the coherent state representation of Fresnel operator is used in our derivation.

1. Introduction

In quantum mechanics, the mapping from classical functions onto operators is often called the Weyl transform [1], whereas the inverse mapping, from operators to functions, is called the Wigner transform [2]. This invertible representation change then allows one to express quantum mechanics in phase space, hence the Weyl–Wigner transform is the invertible mapping between functions in phase space formulation and Hilbert space operators in the Schrödinger picture.

From the point of view of mathematical physics, the Weyl-Wigner transform is a well-defined integral transform between the phase-space and operator representations, and yields insight into the further workings of quantum mechanics. Most importantly, the Wigner quasi-probability distribution is the Wigner transform of the quantum density matrix, and, conversely, the density matrix is the Weyl transform of the Wigner function [3], [4].

In this work we focus on a Wigner hyperbolic rotation in phase space. A hyperbolic rotation is what we get when we slide all the points on the hyperbola along by some angle. To rotate a hyperbola by v, for example, we'd map each point on the unit hyperbola $(\cosh u,\sinh u)$ to $(\cosh (u+v),\sinh (u+v))$. This is exactly analogous to a “circular rotation”, in which we slide all the points on a circle around by some number of radians. Based on the usual Weyl-Wigner transform theory we shall show in this paper that the Wigner hyperbolic rotation in phase space will map onto the fractional squeezing operator in Hilbert space, which is just the product of the usual squeezing operator $\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]$ and the coordinate-momentum mutual exchanging operator $e^{i\pi a^{†}a/2}$, where $[a,a^{†}]=1$. As a consequence, we may naturally introduce the fractional squeezing transformation (FrST), which is the non-trivial generalization of the fractional Fourier transformation (FrFT) [5], [6], [7]. By fractionality we mean that two successive integration transformations, the βth transformation ${\Im }_{\beta }$ and then the αth transformation ${\Im }_{\alpha }$, are equal to that of the $(\alpha +\beta )$th transformation, ${\Im }_{\alpha }{\Im }_{\beta }={\Im }_{\alpha +\beta }$. As FrFT is a very powerful tool in optical communication, image manipulation and signal analysis, we expect that FrST may also play role in optical signal analysis and design. The way we derive FrST is using the Weyl ordering of operators, which brings much convenience for our goal.

2. The Weyl-Wigner transform and the introduction of Weyl ordering

An operator $H(X,P)$'s Weyl-Wigner transform can be introduced from consulting

$$\langle x|P|x^{\prime }\rangle =-i\frac{d}{dx}\delta (x-x^{\prime })=\underset{-\infty }{\overset{\infty }{\int }}\frac{dp}{2\pi }{e}^{ip(x-x^{\prime })}p$$ (1)

$$\langle x|X|x^{\prime }\rangle =\frac{x+x^{\prime }}{2}\delta (x-x^{\prime })=\underset{-\infty }{\overset{\infty }{\int }}\frac{dp}{2\pi }{e}^{ip(x-x^{\prime })}\frac{x+x^{\prime }}{2}$$ (2)

where $[X,P]=iħ$, $\langle x|$ is the coordinate eigenstate, later we set $ħ=1$ for writing's convenience. It follows the definition of Wigner transform

$$\langle x|H(X,P)|x^{\prime }\rangle =\underset{-\infty }{\overset{\infty }{\int }}\frac{dp}{2\pi }{e}^{ip(x-x^{\prime })}h(\frac{x+x^{\prime }}{2},p)$$ (3)

Multiplying the left of $\langle x|H(X,P)|x^{\prime }\rangle$ by ${\int }_{-\infty }^{\infty }dx|x\rangle$ and the right by ${\int }_{-\infty }^{\infty }d{x}^{\prime }\langle x^{\prime }|$, then using the completeness relation of coordinate representation

$$\underset{-\infty }{\overset{\infty }{\int }}dx|x\rangle \langle x|=1$$ (4)

we have

$$\underset{-\infty }{\overset{\infty }{\int }}dx\underset{-\infty }{\overset{\infty }{\int }}d{x}^{\prime }|x\rangle \langle x|H(X,P)|x^{\prime }\rangle \langle x^{\prime }|=H(X,P)=\underset{-\infty }{\overset{\infty }{\int }}dx|x\rangle \int d{x}^{\prime }\underset{-\infty }{\overset{\infty }{\int }}\frac{dp}{2\pi }{e}^{ip(x-x^{\prime })}h(\frac{x+x^{\prime }}{2},p)\langle x^{\prime }|=\underset{-\infty }{\overset{\infty }{\iint }}dpdxh(x,p)\underset{-\infty }{\overset{\infty }{\int }}\frac{du}{2\pi }{e}^{-ipu}|x-\frac{u}{2}\rangle \langle x+\frac{u}{2}|\equiv \underset{-\infty }{\overset{\infty }{\iint }}dpdxh(x,p)\mathrm{\Delta }(x,p)$$ (5)

in this way the Wigner operator [2]

$$\mathrm{\Delta }(x,p)=\underset{-\infty }{\overset{\infty }{\int }}\frac{du}{2\pi }{e}^{-ipu}|x-\frac{u}{2}\rangle \langle x+\frac{u}{2}|$$ (6)

is introduced. Then

$$Tr\left[\mathrm{\Delta }(x,p)\mathrm{\Delta }(x^{\prime },p^{\prime })\right]=\underset{-\infty }{\overset{\infty }{\int }}\frac{du}{2\pi }{e}^{-i{p}^{\prime }u}\langle x^{\prime }+\frac{u}{2}|\mathrm{\Delta }(x,p)|x^{\prime }-\frac{u}{2}\rangle =\underset{-\infty }{\overset{\infty }{\int }}\frac{du}{2\pi }{e}^{-i{p}^{\prime }u}\langle x^{\prime }+\frac{u}{2}|\underset{-\infty }{\overset{\infty }{\int }}\frac{dv}{2\pi }{e}^{-ipv}|x-\frac{v}{2}\rangle \langle x+\frac{v}{2}|x^{\prime }-\frac{u}{2}\rangle =\underset{-\infty }{\overset{\infty }{\int }}\frac{du}{2\pi }{e}^{-i{p}^{\prime }u}\underset{-\infty }{\overset{\infty }{\int }}\frac{dv}{2\pi }{e}^{-ipv}\delta (x^{\prime }-x+\frac{u+v}{2})\delta (x-x^{\prime }+\frac{u+v}{2})=\underset{-\infty }{\overset{\infty }{\int }}\frac{du}{2\pi }{e}^{-i{p}^{\prime }u}{e}^{-ip(2x^{\prime }-2x-u)}\delta (x^{\prime }-x)=\frac{1}{2\pi }\delta (x^{\prime }-x)\delta (p^{\prime }-p)$$ (7)

It follows from (5) and (7) that

$$2\pi Tr[H(X,P)\mathrm{\Delta }(x,p)]=2\pi \iint d{p}^{\prime }d{x}^{\prime }h(x^{\prime },p^{\prime })Tr\left[\mathrm{\Delta }(x^{\prime },p^{\prime })\mathrm{\Delta }(x,p)\right]=\iint d{p}^{\prime }d{x}^{\prime }h(x^{\prime },p^{\prime })\delta (x^{\prime }-x)\delta (p^{\prime }-p)=h(x,p)$$ (8)

or

$$h(x,p)=\underset{-\infty }{\overset{\infty }{\int }}du{e}^{-ipu}\langle x+\frac{u}{2}|H(X,P)|x-\frac{u}{2}\rangle$$ (9)

(9) and (5) are invertible transforms, $h(x,p)$ is the Weyl correspondence of $H(X,P)$, or $h(x,p)$ is quantized as $H(X,P)$ according to Weyl rule.

Especially, when we take $H(X,P)=e^{isX+itP}$, then $h(x,p)=e^{isx+itp}$, as shown below: using (9) and the completeness of the momentum representation

$$\underset{-\infty }{\overset{\infty }{\int }}dp|p\rangle \langle p|=1$$ (10)

we have

$$\underset{-\infty }{\overset{\infty }{\int }}du{e}^{-ipu}{e}^{-\frac{1}{2}[isX,itP]}\langle x+\frac{u}{2}|e^{isX}{e}^{itP}\underset{-\infty }{\overset{\infty }{\int }}d{p}^{\prime }|p^{\prime }\rangle \langle p^{\prime }|x-\frac{u}{2}\rangle =\underset{-\infty }{\overset{\infty }{\int }}du{e}^{-ipu+i\frac{st}{2}+is(x+\frac{u}{2})}\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}d{p}^{\prime }{e}^{it{p}^{\prime }}\exp [i{p}^{\prime }(x+\frac{u}{2})-i{p}^{\prime }(x-\frac{u}{2})]=\underset{-\infty }{\overset{\infty }{\int }}du{e}^{-ipu+i\frac{st}{2}+is(x+\frac{u}{2})}\delta (t+u)=e^{isx+itp}$$ (11)

i.e., $e^{iux+ivp}$ being the Weyl correspondence of $e^{iuX+ivP}$,

$$e^{iux+ivp}\to e^{iuX+ivP}$$ (12)

although $e^{iux}{e}^{ivp}=e^{iux+ivp}$, but $e^{iuX+ivP}$ is neither equal to $e^{iuX}{e}^{ivP}$, nor $e^{ivP}{e}^{iuX}$, it is actually a special operator ordering which we name Weyl ordering [8], and we can introduce the symbol $\begin{array}{c}:\\ :\end{array}\begin{array}{c}:\\ :\end{array}$ to denote it

Corollary 1

$$e^{iuX+ivP}=\begin{array}{c}:\\ :\end{array}{e}^{iuX+ivP}\begin{array}{c}:\\ :\end{array}$$ (13)

which is similar in spirit to the definition of operator's normal ordering $a^{\mathrm{†}n}{a}^m=:a^{\mathrm{†}n}{a}^m:$ . It is remarkable that operators within Weyl ordering symbol can be permuted,

$$\begin{array}{c}:\\ :\end{array}{e}^{iuX+ivP}\begin{array}{c}:\\ :\end{array}=\begin{array}{c}:\\ :\end{array}{e}^{ivP}{e}^{iuX}\begin{array}{c}:\\ :\end{array}=\begin{array}{c}:\\ :\end{array}{e}^{iuX}{e}^{ivP}\begin{array}{c}:\\ :\end{array}$$ (14)

Equation (14) means the properties of Weyl ordering and is exemplified by the X and P operators.

3. The Weyl ordered form of the Wigner operator

From

$$e^{iuX+ivP}=\begin{array}{c}:\\ :\end{array}{e}^{iuX}{e}^{ivP}\begin{array}{c}:\\ :\end{array}=\iint dpdx{e}^{iux+ivp}\mathrm{\Delta }(x,p)$$ (15)

we infer that the Weyl ordered form of the Wigner operator is

$$\mathrm{\Delta }(x,p)=\begin{array}{c}:\\ :\end{array}\delta (x-X)\delta (p-P)\begin{array}{c}:\\ :\end{array}$$ (16)

this expression is correct as can be checked by performing the following integration

$$\underset{-\infty }{\overset{\infty }{\int }}dx\mathrm{\Delta }(x,p)=\begin{array}{c}:\\ :\end{array}\delta (p-P)\begin{array}{c}:\\ :\end{array}=\delta (p-P)=|p\rangle \langle p|,\underset{-\infty }{\overset{\infty }{\int }}dp\mathrm{\Delta }(x,p)=\begin{array}{c}:\\ :\end{array}\delta (x-X)\begin{array}{c}:\\ :\end{array}=\delta (x-X)=|x\rangle \langle x|$$ (17)

The merit of $\mathrm{\Delta }(x,p)=\begin{array}{c}:\\ :\end{array}\delta (x-X)\delta (p-P)\begin{array}{c}:\\ :\end{array}$ lies in that any similar transformation U for $\mathrm{\Delta }(x,p)$ can be performed in the manner of

$$U\mathrm{\Delta }(x,p)U^{-1}=U\begin{array}{c}:\\ :\end{array}\delta (x-X)\delta (p-P)\begin{array}{c}:\\ :\end{array}{U}^{-1}=\begin{array}{c}:\\ :\end{array}U\delta (x-X)\delta (p-P)U^{-1}\begin{array}{c}:\\ :\end{array},$$ (18)

which shows the property that U can penetrate through the “fence” $\begin{array}{c}:\\ :\end{array}\begin{array}{c}:\\ :\end{array}$, in another word, the Weyl ordering of Weyl-ordered operators is invariant under similar transformations, this was proved by Fan in Ref. [9], [10]. Through the translational effect of the parity operator in the phase space, the negativity of the Wigner function can be linked to the squeezed operator in Ref. [11].

4. Mapping of Wigner hyperbolic rotation onto operator

For the following hyperbolic rotation of $(x,p)$ in phase space

$$\left(\begin{array}{c}\hfill x\hfill \\ \hfill p\hfill \end{array}\right)\to \left(\begin{array}{cc}\hfill \sinh \alpha \hfill & \hfill \cosh \alpha \hfill \\ \hfill -\cosh \alpha \hfill & \hfill -\sinh \alpha \hfill \end{array}\right)\left(\begin{array}{c}\hfill x\hfill \\ \hfill p\hfill \end{array}\right)$$ (19)

the corresponding Wigner operator, according to (16), can be put in Weyl ordering as

$$\mathrm{\Delta }(x\sinh \alpha +p\cosh \alpha ,-x\cosh \alpha -p\sinh \alpha )=\begin{array}{c}:\\ :\end{array}\delta (x\sinh \alpha +p\cosh \alpha -X)\delta (x\cosh \alpha +p\sinh \alpha +P)\begin{array}{c}:\\ :\end{array}$$ (20)

Rewriting the Delta function as

$$\begin{array}{c}:\\ :\end{array}\delta (x\sinh \alpha +p\cosh \alpha -X)\delta (x\cosh \alpha +p\sinh \alpha +P)\begin{array}{c}:\\ :\end{array}=\begin{array}{c}:\\ :\end{array}\delta [\left(\begin{array}{c}\hfill X\hfill \\ \hfill P\hfill \end{array}\right)-\left(\begin{array}{cc}\hfill \sinh \alpha \hfill & \hfill \cosh \alpha \hfill \\ \hfill -\cosh \alpha \hfill & \hfill -\sinh \alpha \hfill \end{array}\right)\left(\begin{array}{c}\hfill x\hfill \\ \hfill p\hfill \end{array}\right)]\begin{array}{c}:\\ :\end{array}$$ (21)

with

$$\det \left(\begin{array}{cc}\hfill \sinh \alpha \hfill & \hfill \cosh \alpha \hfill \\ \hfill -\cosh \alpha \hfill & \hfill -\sinh \alpha \hfill \end{array}\right)=1$$ (22)

Since the inverse of this matrix is

$$\left(\begin{array}{cc}\hfill -\sinh \alpha \hfill & \hfill -\cosh \alpha \hfill \\ \hfill \cosh \alpha \hfill & \hfill \sinh \alpha \hfill \end{array}\right)$$ (23)

so Eq. (21) is equivalent to

$$\begin{array}{c}:\\ :\end{array}\delta [\left(\begin{array}{c}\hfill x\hfill \\ \hfill p\hfill \end{array}\right)-\left(\begin{array}{cc}\hfill -\sinh \alpha \hfill & \hfill -\cosh \alpha \hfill \\ \hfill \cosh \alpha \hfill & \hfill \sinh \alpha \hfill \end{array}\right)\left(\begin{array}{c}\hfill X\hfill \\ \hfill P\hfill \end{array}\right)]\begin{array}{c}:\\ :\end{array}=\begin{array}{c}:\\ :\end{array}\delta (x+P\cosh \alpha +X\sinh \alpha )\delta (p-X\cosh \alpha -P\sinh \alpha )\begin{array}{c}:\\ :\end{array},$$ (24)

By noticing $X=\frac{a+a^{†}}{\sqrt{2}}$, $P=\frac{a-a^{†}}{i\sqrt{2}}$, and due to

$$e^{-i\pi a^{†}a/2}{a}^{†}{e}^{i\pi a^{†}a/2}=-i{a}^{†},e^{-i\pi a^{†}a/2}a{e}^{i\pi a^{†}a/2}=ia$$ (25)

so $e^{-i\pi a^{†}a/2}$ causes X and P mutual transformation

$$e^{-i\pi a^{†}a/2}X{e}^{i\pi a^{†}a/2}=P,e^{-i\pi a^{†}a/2}P{e}^{i\pi a^{†}a/2}=-X,$$ (26)

we see Eq. (21) becomes to

$$\begin{array}{c}:\\ :\end{array}\delta (x\sinh \alpha +p\cosh \alpha -X)\delta (x\cosh \alpha +p\sinh \alpha +P)\begin{array}{c}:\\ :\end{array}=\begin{array}{c}:\\ :\end{array}\delta (x+P\cosh \alpha +X\sinh \alpha )\delta (p-X\cosh \alpha -P\sinh \alpha )\begin{array}{c}:\\ :\end{array}=\begin{array}{c}:\\ :\end{array}{e}^{-i\pi a^{†}a/2}\delta (x-X\cosh \alpha +P\sinh \alpha )\delta (p-P\cosh \alpha +X\sinh \alpha )e^{i\pi a^{†}a/2}\begin{array}{c}:\\ :\end{array}=e^{-i\pi a^{†}a/2}\begin{array}{c}:\\ :\end{array}\delta (x-X\cosh \alpha +P\sinh \alpha )\delta (p-P\cosh \alpha +X\sinh \alpha )\begin{array}{c}:\\ :\end{array}{e}^{i\pi a^{†}a/2}.$$ (27)

On the other hand, using the relations

$$\exp \left[\frac{i\alpha }{2}(a^2+a^{†2})\right]a\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]=a\cosh \alpha -i{a}^{†}\sinh \alpha$$ (28)

and

$$\exp \left[\frac{i\alpha }{2}(a^2+a^{†2})\right]a^{†}\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]=a^{†}\cosh \alpha +ia\sinh \alpha$$ (29)

we see the terms on the right side of Eq. (27) can be reformed as

$$\delta (x-X\cosh \alpha +P\sinh \alpha )\delta (p-P\cosh \alpha +X\sinh \alpha )=\delta (x-\frac{a+a^{†}}{\sqrt{2}}\cosh \alpha +\frac{a-a^{†}}{i\sqrt{2}}\sinh \alpha )\delta (p-\frac{a-a^{†}}{i\sqrt{2}}\cosh \alpha +\frac{a+a^{†}}{\sqrt{2}}\sinh \alpha )=\delta [x-\frac{1}{\sqrt{2}}(a\cosh \alpha -i{a}^{†}\sinh \alpha )-\frac{1}{\sqrt{2}}(a^{†}\cosh \alpha +ia\sinh \alpha )]\times \delta [p-\frac{1}{i\sqrt{2}}(a\cosh \alpha -i{a}^{†}\sinh \alpha )+\frac{1}{i\sqrt{2}}(a^{†}\cosh \alpha +ia\sinh \alpha )]=\exp \left[\frac{i\alpha }{2}(a^2+a^{†2})\right]\delta (x-X)\delta (p-P)\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]$$ (30)

On substituting Eqs. (30) into (27) and consulting (24) as well as (21), we can express (20) as

$$\mathrm{\Delta }(x\sinh \alpha +p\cosh \alpha ,-x\cosh \alpha -p\sinh \alpha )=\begin{array}{c}:\\ :\end{array}{e}^{-i\pi a^{†}a/2}\exp \left[\frac{i\alpha }{2}(a^2+a^{†2})\right]\delta (x-X)\delta (p-P)\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]e^{i\pi a^{†}a/2}\begin{array}{c}:\\ :\end{array}=e^{-i\pi a^{†}a/2}\exp \left[\frac{i\alpha }{2}(a^2+a^{†2})\right]\begin{array}{c}:\\ :\end{array}\delta (x-X)\delta (p-P)\begin{array}{c}:\\ :\end{array}\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]e^{i\pi a^{†}a/2}=e^{-i\pi a^{†}a/2}\exp \left[\frac{i\alpha }{2}(a^2+a^{†2})\right]\mathrm{\Delta }(x,p)\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]e^{i\pi a^{†}a/2}$$ (31)

we see that hyperbolic rotation of the Wigner operator $\mathrm{\Delta }(x,p)$ in phase space maps onto the product operator $e^{i\pi a^{†}a/2}\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]$, which we shall prove as the operator causing fractional squeezing transformation.

5. Fractional squeezing operator and FrST

Now we consider

$$\langle p|\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]e^{i\pi a^{†}a/2}|g\rangle \equiv G\left(p\right)$$ (32)

By virtue of the completeness of the coordinate representation $|x\rangle$ we define

$$G\left(p\right)\equiv \underset{-\infty }{\overset{\infty }{\int }}\langle p|\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]e^{i\pi a^{†}a/2}|x\rangle \langle x|g\rangle dx$$ (33)

In Dirac's ket-bra notation we write $\langle x|g\rangle =g\left(x\right)$, and Eq. (33) becomes

$$G\left(p\right)\equiv \underset{-\infty }{\overset{\infty }{\int }}\Re (p,x)g\left(x\right)dx$$ (34)

where

$$\Re (p,x)=\langle p|\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]e^{i\pi a^{†}a/2}|x\rangle$$ (35)

is the integration transform kernel. Besides, using the relation

$$\begin{gathered} e^{i\pi a^{†}a/2}a{e}^{-i\pi a^{†}a/2}=-ia \\ {e}^{i\pi a^{†}a/2}{a}^{†}{e}^{-i\pi a^{†}a/2}=i{a}^{†} \end{gathered}$$ (36)

we can derive

$$e^{i\pi a^{†}a/2}|x\rangle =|p{|}_{p=x}\rangle$$ (37)

Thus, the kernel in Eq. (35) becomes

$$\Re (p,x)=\langle p|\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]|p{|}_{p=x}\rangle .$$ (38)

The squeezing operator $\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]$ has its coherent state representation [12], [13]

$$\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]=\sqrt{\cosh \alpha }\int \frac{d^{2}z}{\pi }|z\cosh \alpha -i{z}^{⁎}\sinh \alpha \rangle \langle z|$$ (39)

where

$$|z\rangle =\exp [-\frac{|z{|}^2}{2}+z{a}^{†}]|0\rangle ,z=\frac{x+ip}{\sqrt{2}}$$ (40)

Employing

$$\langle p|z\rangle ={\pi }^{-1/4}\exp (-\frac{p^2}{2}-\frac{{\left|z\right|}^2}{2}-\sqrt{2}ipz+\frac{z^2}{2})$$ (41)

we see

$$\langle p|z\cosh \alpha -i{z}^{⁎}\sinh \alpha \rangle ={\pi }^{-1/4}\exp [-\frac{p^2}{2}-\frac{{|z\cosh \alpha -i{z}^{⁎}\sinh \alpha |}^2}{2}-\sqrt{2}ip(z\cosh \alpha -i{z}^{⁎}\sinh \alpha )+\frac{{(z\cosh \alpha -i{z}^{⁎}\sinh \alpha )}^2}{2}]$$ (42)

and

$$\langle z|p{|}_{p=x}\rangle ={\pi }^{-1/4}\exp (-\frac{x^2}{2}-\frac{{\left|z\right|}^2}{2}+\sqrt{2}ix{z}^{⁎}+\frac{z^{⁎2}}{2})$$ (43)

Substituting (42)-(43) into (39) we obtain

$$\Re (p,x)=\langle p|\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]e^{i\pi a^{†}a/2}|x\rangle =\frac{1}{\sqrt{2\pi i\sinh \alpha }}\exp [\frac{i(p^2+x^2)}{2\tanh \alpha }-\frac{ipx}{\sinh \alpha }]$$ (44)

Thus Eq. (33) is

$$G\left(p\right)\equiv \frac{1}{\sqrt{2\pi i\sinh \alpha }}\underset{-\infty }{\overset{\infty }{\int }}\exp [\frac{i(p^2+x^2)}{2\tanh \alpha }-\frac{ipx}{\sinh \alpha }]g\left(x\right)dx$$ (45)

By comparison with the αth fractional Fourier transform of a signal $g(x)$ which has the form

$$F_{\alpha }\left[g\right](p)=\sqrt{\frac{\exp [-i(\pi /2-\alpha )]}{2\pi \sin \alpha }}\int \exp [\frac{i(x^2+p^2)}{2\tan \alpha }-\frac{ixp}{\sin \alpha }]g\left(x\right)dx,$$ (46)

we see that by changing the triangular function to the hyperbolic function, $\tan \alpha \to \tanh \alpha$, $\sin \alpha \to \sinh \alpha$ in FrFT, we have introduced a new fractional squeezing transformation (FrST), that is to say, $\Re (p,x)$ is just the integral kernel of fractional squeezing transformation, and $\exp [-\frac{i\alpha }{2}(a^2+a^{†2})]e^{i\pi a^{†}a/2}$ is the fractional squeezing operator.

In sum, by virtue of the usual Wigner-Weyl transformation theory and the merit of Weyl ordering we have found that the Wigner hyperbolic rotation in phase space will map onto fractional squeezing operator in Hilbert space. The coherent state representation of Fresnel operator has been used in our derivation.

CRediT authorship contribution statement

Wei-Feng Wu: Project administration, Methodology. Peng Fu: Formal analysis. Hua-Kui Hu: Methodology.

Declaration of Competing Interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

Appendix A.

Substituting (42)-(43) into (39) we calculate the integral (44)

$$\Re (p,x)=\sqrt{\cosh \alpha }\int \frac{d^{2}z}{\pi }\langle p|z\cosh \alpha -i{z}^{⁎}\sinh \alpha \rangle \langle z|p{|}_{p=x}\rangle =\sqrt{\cosh \alpha }{\pi }^{-1/2}\int \frac{d^{2}z}{\pi }\exp [-\frac{x^2}{2}-\frac{{\left|z\right|}^2}{2}+\sqrt{2}ix{z}^{⁎}+\frac{z^{⁎2}}{2}-\frac{p^2}{2}-\frac{{|z\cosh \alpha -i{z}^{⁎}\sinh \alpha |}^2}{2}-\sqrt{2}ip(z\cosh \alpha -i{z}^{⁎}\sinh \alpha )+\frac{{(z\cosh \alpha -i{z}^{⁎}\sinh \alpha )}^2}{2}]=\sqrt{\cosh \alpha }{\pi }^{-1/2}\exp (-\frac{x^2}{2}-\frac{p^2}{2})\int \frac{d^{2}z}{\pi }\exp [(-\frac{1}{2}-\frac{1}{2}({\sinh }^2\alpha +{\cosh }^2\alpha )-i(\cosh \alpha \sinh \alpha )){\left|z\right|}^2+(-\sqrt{2}ip\cosh \alpha )z+(\sqrt{2}ix-\sqrt{2}p\sinh \alpha )z^{⁎}+(\frac{1}{2}{\cosh }^2\alpha -\frac{1}{2}i\cosh \alpha \sinh \alpha )z^2+(\frac{1}{2}+\frac{1}{2}i\cosh \alpha \sinh \alpha -\frac{1}{2}{\sinh }^2\alpha )z^{⁎}{}^2]$$

With the integral formula

$$\int \frac{d^{2}z}{\pi }\exp (\zeta {\left|z\right|}^2+\xi z+\eta z^{⁎}+f{z}^2+g{z}^{⁎2})=\frac{1}{\sqrt{{\zeta }^2-4fg}}\exp \left[\frac{-\zeta \xi \eta +{\xi }^{2}g+{\eta }^{2}f}{{\zeta }^2-4fg}\right],$$

we obtain Eq. (44)

$$\Re (p,x)=\frac{1}{\sqrt{2\pi i\sinh \alpha }}\exp [\frac{i(p^2+x^2)}{2\tanh \alpha }-\frac{ipx}{\sinh \alpha }]$$