$2\pi\int_{-\infty}^{\infty}\left|\psi\left(p\right)\right|^{2}\frac{dp}{\left|p\right|}=2C_{\psi}.$
(461)

Eq. (454) can be transformed to

$\text{L.H.S of
Eq.(\ref{15.16})}=2C_{\psi}\int_{-\infty}^{\infty}dp\left\langle
f_{2}\right.\left|p\right\rangle\left\langle
p\right|\left.f_{1}\right\rangle=\text{R.H.S of Eq.(\ref{15.16}).}$ (462)

where

$C_{\psi}\equiv
2\pi\int_{0}^{\infty}\frac{\left|\psi\left(p\right)\right|^{2}}{p}dp<\infty,\text{
}$ (463)

thus the theorem is proved. Especially, when $\left|f_{1}\right\rangle=$
$\left|f_{2}\right\rangle,$ Eq.(449) becomes

$\int_{-\infty}^{\infty}\frac{d\mu}{\mu^{2}}\int_{-\infty}^{\infty}ds|\left\langle\psi\right|U\left(\mu,s\right)\left|f_{1}\right\rangle|^{2}=2C_{\psi}\left\langle
f_{1}\right.\left|f_{1}\right\rangle,$ (464)

which is named isometry of energy.

###  14.4 Inversion formula of WT

Now we can directly derive the inversion formula of WT, i.e. we take
$\left\langle f_{2}\right|=\left\langle x\right|$ in Eq.(449), then using
Eq.(430) we see that Eq.(449) reduces to

$\int_{-\infty}^{\infty}\frac{d\mu}{\mu^{2}}\int_{-\infty}^{\infty}dsW_{\psi}f_{1}\left(\mu,s\right)\left\langle
x\right|U^{\dagger}\left(\mu,s\right)\left|\psi\right\rangle=2C_{\psi}\left\langle
x\right.\left|f_{1}\right\rangle.$ (465)

Due to Eq.(431) we have

$\left\langle
x\right|U^{\dagger}\left(\mu,s\right)=\frac{1}{\sqrt{\left|\mu\right|}}\left\langle
x\right|\int_{-\infty}^{\infty}dx^{\prime}\left|x^{\prime}\right\rangle\left\langle\frac{x^{\prime}-s}{\mu}\right|=\frac{1}{\sqrt{\left|\mu\right|}}\left\langle\frac{x-s}{\mu}\right|.$
(466)

It then follows

$\int_{-\infty}^{\infty}\frac{d\mu}{\mu^{2}}\int_{-\infty}^{\infty}dsW_{\psi}f_{1}\left(\mu,s\right)\frac{1}{\sqrt{\left|\mu\right|}}\left\langle\frac{x-s}{\mu}\right|\left.\psi\right\rangle=2C_{\psi}\left\langle
x\right.\left|f_{1}\right\rangle,$ (467)

which means

$f_{1}\left(x\right)=\frac{1}{2C_{\psi}}\int_{-\infty}^{\infty}\frac{d\mu}{\mu^{2}\sqrt{\left|\mu\right|}}\int_{-\infty}^{\infty}ds\psi\left(\frac{x-s}{\mu}\right)W_{\psi}f_{1}\left(\mu,s\right),$
(468)

this is the inversion formula of WT.

###  14.5 New orthogonal property of mother wavelet in parameter space

Form the Parserval theorem (448) of WT in quantum mechanics we can derive some
new property of mother wavelet [104]. Taking
$\left|f_{1}\right\rangle=\left|x\right\rangle$,
$\left|f_{2}\right\rangle=\left|x^{\prime}\right\rangle$ in (448) one can see
that

$\int_{-\infty}^{\infty}\frac{d\mu}{\mu^{2}\left|\mu\right|}\int_{-\infty}^{\infty}ds\psi\left(\frac{x-s}{\mu}\right)\psi^{\ast}\left(\frac{x^{\prime}-s}{\mu}\right)=2C_{\psi}\delta\left(x-x^{\prime}\right),$
(469)

which is a new orthogonal property of mother wavelet in parameter space
spanned by $\left(\mu,s\right)$. In a similar way, we take
$\left|f_{1}\right\rangle=\left|f_{2}\right\rangle=\left|n\right\rangle,$ a
number state, since $\left\langle n\right.\left|n\right\rangle=1,$ then we
have

$\int_{-\infty}^{\infty}\frac{d\mu}{\mu^{2}}\int_{-\infty}^{\infty}ds|\left\langle\psi\right|U\left(\mu,s\right)\left|n\right\rangle|^{2}=2C_{\psi},$
(470)

or take
$\left|f_{1}\right\rangle=\left|f_{2}\right\rangle=\left|z\right\rangle,$
$\left|z\right\rangle=\exp\left(-\left|z\right|^{2}/2+za^{\dagger}\right)\left|0\right\rangle$
is the coherent state, then

$\int_{-\infty}^{\infty}\frac{d\mu}{\mu^{2}}\int_{-\infty}^{\infty}ds|\left\langle\psi\right|U\left(\mu,s\right)\left|z\right\rangle|^{2}=2C_{\psi}.$
(471)

This indicates that $C_{\psi}$ is $\left|f_{1}\right\rangle$-independent,
which coincides with the expression in (463). Next, we consider a special
example. When the mother wavelet is the Mexican hat (443), we have

$\psi_{M}\left(p\right)\equiv\left\langle
p\right.\left|\psi_{M}\right\rangle=\frac{1}{2}\left(\left\langle
p\right.\left|0\right\rangle-\sqrt{2}\left\langle
p\right.\left|2\right\rangle\right)=\pi^{-1/4}\allowbreak
p^{2}e^{-\frac{1}{2}p^{2}}.$ (472)

where

$\left\langle
p\right.\left|n\right\rangle=\frac{\left(-i\right)^{n}}{\sqrt{2^{n}n!\sqrt{\pi}}}e^{-p^{2}/2}H_{n}\left(p\right).$
(473)

Here $H_{n}\left(p\right)$ is the single-variable Hermit polynomial [105].
Substituting Eq.(472) into Eq.(463) we have

$C_{\psi}\equiv
2\pi\int_{0}^{\infty}\frac{\left|\psi_{M}\left(p\right)\right|^{2}}{p}dp=\sqrt{\pi}.$
(474)

Thus, for the Mexican hat wavelet (443), we see

$\int_{-\infty}^{\infty}\frac{d\mu}{\mu^{2}\left|\mu\right|}\int_{-\infty}^{\infty}ds\psi_{M}\left(\frac{x-s}{\mu}\right)\psi_{M}^{\ast}\left(\frac{x^{\prime}-s}{\mu}\right)=2\sqrt{\pi}\delta\left(x-x^{\prime}\right).$
(475)

Eq.(475) can be checked as follows. Using Eq.(443) and noticing that
$\psi_{M}\left(x\right)=\psi_{M}\left(-x\right)$, we can put the left hand
side of Eq.(475) into

$\displaystyle\mathtt{L.H.S.of}\text{ }(\text{\ref{15.43}})$
$\displaystyle=2\int_{0}^{\infty}du\int_{-\infty}^{\infty}ds\psi_{M}\left(ux-s\right)\psi_{M}^{\ast}\left(ux^{\prime}-s\right)$
$\displaystyle=\left\\{\begin{array}[c]{cc}0,&x\neq x^{\prime}\\\
\frac{3}{2}\int_{0}^{\infty}du\rightarrow\infty,&x=x^{\prime}\end{array}\right.=\mathtt{R.H.S.of}\text{
}(\ref{15.43}).$ (478)

where we have used the integration formulas

$\int_{-\infty}^{\infty}\left(1-s^{2}\right)^{2}\exp\left(-s^{2}\right)ds=\allowbreak\frac{3}{4}\sqrt{\pi},$
(479)

and

$\displaystyle\int_{-\infty}^{\infty}\left(1-s^{2}\right)\left[1-\left(s-b\right)^{2}\right]e^{-s^{2}/2-\left(b-s\right)^{2}/2}ds$
$\displaystyle=\frac{\sqrt{\pi}}{16}e^{-\frac{b^{2}}{4}}\left[12+b^{2}\left(b^{2}-12\right)\right].$
(480)

Next, we examine if the Morlet wavelet obey the formalism (469). The the
Morlet wavelet is defined as [106, 107, 108]

$\psi_{mor}\left(x\right)=\pi^{-1/4}\left(e^{ifx}-e^{-f^{2}/2}\right)e^{-x^{2}/2}.$
(481)

Substituting (481) into the left hand side of (469) yields

$\displaystyle I$
$\displaystyle\equiv\int_{-\infty}^{\infty}\frac{d\mu}{\mu^{2}\left|\mu\right|}\int_{-\infty}^{\infty}ds\psi_{mor}\left(\frac{x-s}{\mu}\right)\psi_{mor}^{\ast}\left(\frac{x^{\prime}-s}{\mu}\right)$
$\displaystyle=\left\\{\begin{array}[c]{cc}0,&x\neq x^{\prime}\\\
2\left(1+e^{-f^{2}}-2e^{-3f^{2}/4}\right)\int_{0}^{\infty}\frac{d\mu}{\mu^{2}}\rightarrow\infty,&x=x^{\prime}\end{array}\right..$
(484)

Thus the Morlet wavelet satisfies Eq.(469).

###  14.6 WT and Wigner-Husimi Distribution Function

Phase space technique has been proved very useful in various branches of
physics. Distribution functions in phase space have been a major topic in
studying quantum mechanics and quantum statistics. Among various phase space
distributions the Wigner function $F_{w}\left(q,p\right)$ [66, 67] is the most
popularly used, since its two marginal distributions lead to measuring
probability density in coordinate space and momentum space, respectively. But
the Wigner distribution function itself is not a probability distribution due
to being both positive and negative. In spite of its some attractive formal
properties, it needs to be improved. To overcome this inconvenience, the
Husimi distribution function $F_{h}\left(q^{\prime},p^{\prime}\right)$ is
introduced [109], which is defined in a manner that guarantees it to be
nonnegative. Its definition is smoothing out the Wigner function by averaging
over a “coarse graining” function,

$F_{h}\left(q,p,\kappa\right)=\int\int_{-\infty}^{\infty}dq^{\prime}dp^{\prime}F_{w}\left(q^{\prime},p^{\prime}\right)\exp\left[-\kappa\left(q^{\prime}-q\right)^{2}-\frac{\left(p^{\prime}-p\right)^{2}}{\kappa}\right],$
(485)

where $\kappa>0$ is the Gaussian spatial width parameter, which is free to be
chosen and which determines the relative resolution in $p$-space versus
$q$-space.

In the following, we shall employ the optical wavelet transformation to study
the Husimi distribution function, this is to say, we shall show that the
Husimi distribution function of a quantum state $\left|\psi\right\rangle$ can
be obtained by making a WT of the Gaussian function $e^{-x^{2}/2},$ i.e.,

$\left\langle\psi\right|\Delta_{h}\left(q,p,\kappa\right)\left|\psi\right\rangle=\frac{e^{-\frac{p^{2}}{\kappa}}}{\sqrt{\pi\kappa}}\left|\int_{-\infty}^{\infty}dx\psi^{\ast}\left(\frac{x-s}{\mu}\right)e^{-x^{2}/2}\right|^{2},$
(486)

where

$s=\frac{-1}{\sqrt{\kappa}}\left(\kappa q+ip\right),\text{
}\mu=\sqrt{\kappa},$ (487)

and $\left\langle\psi\right|\Delta_{h}\left(q,p\right)\left|\psi\right\rangle$
is the Husimi distribution function as well as
$\Delta_{h}\left(q,p,\kappa\right)$ is the Husimi operator,

$\Delta_{h}\left(q,p,\kappa\right)=\frac{2\sqrt{\kappa}}{1+\kappa}\colon\exp\left\\{\frac{-\kappa\left(q-Q\right)^{2}}{1+\kappa}-\frac{\left(p-P\right)^{2}}{1+\kappa}\right\\}\colon,$
(488)

here $\colon\colon$ denotes normal ordering; $Q$ and $P$ are the coordinate
and the momentum operator.

Proof of Eq.(486). According to Eqs.(429) and (430), when
$\left|f\right\rangle$ is the vacuum state $\left|0\right\rangle$,
$e^{-x^{2}/2}=\pi^{1/4}\left\langle x\right.\left|0\right\rangle$, we see that

$\pi^{-1/4}\int_{-\infty}^{\infty}\frac{dx}{\sqrt{\mu}}\psi^{\ast}\left(\frac{x-s}{\mu}\right)e^{-x^{2}/2}dx=\left\langle\psi\right|U\left(\mu,s\right)\left|0\right\rangle.$
(489)

From Eq.(433) it then follows that

$U\left(\mu,s\right)\left|0\right\rangle=\operatorname{sech}^{1/2}\lambda\exp\left[\frac{-s^{2}}{2\left(1+\mu^{2}\right)}-\frac{a^{\dagger}s}{\sqrt{2}}\operatorname{sech}\lambda-\frac{a^{\dagger
2}}{2}\tanh\lambda\right]\left|0\right\rangle.$ (490)

Substituting Eq.(487) and $\tanh\lambda=\frac{\kappa-1}{\kappa+1},$
$\cosh\lambda=\frac{1+\kappa}{2\sqrt{\kappa}}$ into Eq.(490) yields

$\displaystyle
e^{-\frac{p^{2}}{2\kappa}+\frac{ipq}{\kappa+1}}U\left(\mu=\sqrt{\kappa},s=-\sqrt{\kappa}q-ip/\sqrt{\kappa}\right)\left|0\right\rangle$
$\displaystyle=\left(\frac{2\sqrt{\kappa}}{1+\kappa}\right)^{1/2}\exp\left\\{\frac{-\kappa
q^{2}}{2\left(1+\kappa\right)}-\frac{p^{2}}{2\left(1+\kappa\right)}\right.$
$\displaystyle\left.+\frac{\sqrt{2}a^{\dagger}}{1+\kappa}\left(\kappa
q+ip\right)+\frac{1-\kappa}{2\left(1+\kappa\right)}a^{\dagger
2}\right\\}\left|0\right\rangle\left.\equiv\left|p,q\right\rangle_{\kappa}\right.,$
(491)

then the WT of Eq.(489) can be further expressed as

$e^{-\frac{p^{2}}{2\kappa}+\frac{ipq}{\kappa+1}}\int_{-\infty}^{\infty}\frac{dx}{\left(\kappa\pi\right)^{1/4}}\psi^{\ast}\left(\frac{x-s}{\mu}\right)e^{-x^{2}/2}=\left\langle\psi\right.\left|p,q\right\rangle_{\kappa}.$
(492)

Using normally ordered form of the vacuum state projector
$\left|0\right\rangle\left\langle 0\right|=\colon e^{-a^{\dagger}a}\colon,$and
the IWOP method as well as Eq.(LABEL:15.55) we have

$\left|p,q\right\rangle_{\kappa\kappa}\left\langle
p,q\right|=\frac{2\sqrt{\kappa}}{1+\kappa}\colon\exp\left[\frac{-\kappa\left(q-Q\right)^{2}}{1+\kappa}-\frac{\left(p-P\right)^{2}}{1+\kappa}\right]\colon=\Delta_{h}\left(q,p,\kappa\right).$
(493)

Now we explain why $\Delta_{h}\left(q,p,\kappa\right)$ is the Husimi operator.
Using the formula for converting an operator $A$ into its Weyl ordering form
[110]

$\displaystyle A$
$\displaystyle=2\int\frac{d^{2}\beta}{\pi}\left\langle-\beta\right|A\left|\beta\right\rangle\genfrac{}{}{0.0pt}{}{:}{:}\exp\\{2\left(\beta^{\ast}a-a^{\dagger}\beta+a^{\dagger}a\right)\\}\genfrac{}{}{0.0pt}{}{:}{:},$
(494) $\displaystyle d^{2}\beta$ $\displaystyle=d\beta_{1}d\beta_{2},\text{
}\beta=\beta_{1}+i\beta_{2},$

where the symbol $\genfrac{}{}{0.0pt}{}{:}{:}\genfrac{}{}{0.0pt}{}{:}{:}$
denotes the Weyl ordering, $\left|\beta\right\rangle$ is the usual coherent
state, substituting Eq.(493) into Eq.(494) and performing the integration by
virtue of the technique of integration within a Weyl ordered product of
operators, we obtain

$\left|p,q\right\rangle_{\kappa\kappa}\left\langle
p,q\right|=2\genfrac{}{}{0.0pt}{}{:}{:}\exp\left[-\kappa\left(q-Q\right)^{2}-\frac{\left(p-P\right)^{2}}{\kappa}\right]\genfrac{}{}{0.0pt}{}{:}{:}.$
(495)

This is the Weyl ordering form of
$\left|p,q\right\rangle_{\kappa\kappa}\left\langle p,q\right|.$ Then according
to Weyl quantization scheme [37] we know the classical corresponding function
of a Weyl ordered operator is obtained by just replacing $Q\rightarrow
q^{\prime},P\rightarrow p^{\prime},$

$\genfrac{}{}{0.0pt}{}{:}{:}\exp\left[-\kappa\left(q-Q\right)^{2}-\frac{\left(p-P\right)^{2}}{\kappa}\right]\genfrac{}{}{0.0pt}{}{:}{:}\rightarrow\exp\left[-\kappa\left(q-q^{\prime}\right)^{2}-\frac{\left(p-p^{\prime}\right)^{2}}{\kappa}\right],$
(496)

and in this case the Weyl rule is expressed as

$\displaystyle\left|p,q\right\rangle_{\kappa\kappa}\left\langle p,q\right|$
$\displaystyle=2\int
dq^{\prime}dp^{\prime}\genfrac{}{}{0.0pt}{}{:}{:}\delta\left(q^{\prime}-Q\right)\delta\left(p^{\prime}-P\right)\genfrac{}{}{0.0pt}{}{:}{:}\exp\left[-\kappa\left(q-q^{\prime}\right)^{2}-\frac{\left(p-p^{\prime}\right)^{2}}{\kappa}\right]$
$\displaystyle=2\int
dq^{\prime}dp^{\prime}\Delta_{w}\left(q^{\prime},p^{\prime}\right)\exp\left[-\kappa\left(q^{\prime}-q\right)^{2}-\frac{\left(p^{\prime}-p\right)^{2}}{\kappa}\right],$
(497)

where at the last step we used the Weyl ordering form of the Wigner operator
$\Delta_{w}\left(q,p\right)$ [111]

$\Delta_{w}\left(q,p\right)=\genfrac{}{}{0.0pt}{}{:}{:}\delta\left(q-Q\right)\delta\left(p-P\right)\genfrac{}{}{0.0pt}{}{:}{:}.$
(498)

In reference to Eq.(485) in which the relation between the Husimi function and
the WF is shown, we know that the right-hand side of Eq. (497) should be just
the Husimi operator, i.e.

$\displaystyle\left|p,q\right\rangle_{\kappa\kappa}\left\langle p,q\right|$
$\displaystyle=2\int
dq^{\prime}dp^{\prime}\Delta_{w}\left(q^{\prime},p^{\prime}\right)\exp\left[-\kappa\left(q^{\prime}-q\right)^{2}-\frac{\left(p^{\prime}-p\right)^{2}}{\kappa}\right]$
$\displaystyle=\Delta_{h}\left(q,p,\kappa\right),$ (499)

thus Eq. (486) is proved by combining Eqs.(499) and (492).

Thus the optical WT can be used to study the Husimi distribution function in
quantum optics phase space theory [112].

##  15 Complex Wavelet transformation in entangled state representations

We now turn to 2-dimensional complex wavelet transform (CWT) [113].

###  15.1 CWT and the condition of Mother Wavelet

Since wavelet family involves squeezing transform, we recall that the two-mode
squeezing operator has a natural representation in the entangled state
representation (ESR),
$\exp[\lambda\left(a_{1}^{\dagger}a_{2}^{\dagger}-a_{1}a_{2}\right)]=\frac{1}{\mu}\int_{-\infty}^{\infty}\left|\frac{\eta}{\mu}\right\rangle\left\langle\eta\right|dx,$
$\mu=e^{\lambda}$, thus we are naturally led to studying 2-dimensional CWT in
ESR. Using ESR we can derive some new results more conveniently than using the
direct-product of two single-particle coordinate eigenstates. To be concrete,
we impose the condition on qualified mother wavelets also in
$\left|\eta\right\rangle$ representation,

$\int_{-\infty}^{\infty}\frac{d^{2}\eta}{2\pi}\psi\left(\eta\right)=0,$ (500)

where $\psi\left(\eta\right)=$
$\left\langle\eta\right|\left.\psi\right\rangle.$ Thus we see

$\int_{-\infty}^{\infty}\frac{d^{2}\eta}{2\pi}\left|\eta\right\rangle=\exp\\{-a_{1}^{\dagger}a_{2}^{\dagger}\\}\left|00\right\rangle=\left|\xi=0\right\rangle,$
(501)

and the condition (500) becomes

$\left\langle\xi=0\right|\left.\psi\right\rangle=0.$ (502)

Without loss of generality, assuming

$\left|\psi\right\rangle=\sum_{n,m=0}^{\infty}K_{n,m}a_{1}^{\dagger
n}a_{2}^{\dagger m}\left|00\right\rangle,$ (503)

then using the two-mode coherent $\left|z_{1}z_{2}\right\rangle$ state we can
write (502) as

$\displaystyle\left\langle\xi=0\right|\left.\psi\right\rangle$
$\displaystyle=\left\langle\xi=0\right|\int\frac{d^{2}z_{1}d^{2}z_{2}}{\pi^{2}}\left|z_{1}z_{2}\right\rangle\left\langle
z_{1}z_{2}\right|\sum_{n,m=0}^{\infty}K_{n,m}a_{1}^{\dagger n}a_{2}^{\dagger
m}\left|00\right\rangle$
$\displaystyle=\sum_{n,m=0}^{\infty}K_{n,m}\int\frac{d^{2}z_{1}d^{2}z_{2}}{\pi^{2}}z_{1}^{\ast
n}z_{2}^{\ast m}\exp\left[-|z_{1}|^{2}-|z_{2}|^{2}-z_{1}z_{2}\right]$
$\displaystyle=\sum_{n,m=0}^{\infty}K_{n,m}\int\frac{d^{2}z_{2}}{\pi}\exp\left[-|z_{2}|^{2}\right]z_{2}^{n}z_{2}^{\ast
m}\left(-1\right)^{n}$
$\displaystyle=\sum_{n=0}^{\infty}n!K_{n,n}\left(-1\right)^{n}=0,$ (504)

this is the constraint on the coefficient $K_{n,n}$ in (504), i.e., the
admissibility condition for $\left|\psi\right\rangle$. Thus Eq. (503) is in
the form:

$\left|\psi\right\rangle=\sum_{n=0}^{\infty}n!K_{n,n}\left|n,n\right\rangle.$
(505)

To derive the qualified mother wavelet $\psi\left(\eta\right)=$
$\left\langle\eta\right|\left.\psi\right\rangle$ from
$\left|\psi\right\rangle$, noticing Eq.(353) and (505) we have

$\displaystyle\psi\left(\eta\right)$
$\displaystyle=e^{-\left|\eta\right|^{2}/2}\sum_{n=0}^{\infty}K_{n,n}H_{n,n}\left(\eta^{\ast},\eta\right)\left(-1\right)^{n}$
$\displaystyle=e^{-\left|\eta\right|^{2}/2}\sum_{n=0}^{\infty}n!K_{n,n}L_{n}\left(\left|\eta\right|^{2}\right),$
(506)

where $L_{n}\left(x\right)$ is the Laguerre polynomial. In this case, we may
name the wavelet in Eq. (506) as the Laguerre–Gaussian mother wavelets,
analogous to the name of Laguerre–Gaussian modes in optical propagation. For
example:

(1) When taking $K_{0,0}=\frac{1}{2},$ $K_{1,1}=\frac{1}{2},$ $K_{n,n}=0$ for
$n\geqslant 2,$ so we see

$\left|\psi\right\rangle_{1}=\frac{1}{2}\left(1+a_{1}^{\dagger}a_{2}^{\dagger}\right)\left|00\right\rangle,$
(507)

which differs from the direct-product state $\left(1-a_{1}^{\dagger
2}\right)\left|0\right\rangle_{1}\otimes\left(1-a_{2}^{\dagger
2}\right)\left|0\right\rangle_{2}$. It then follows from Eq. (506) that

$\psi_{1}\left(\eta\right)\equiv\frac{1}{2}\left\langle\eta\right|\left(\left|00\right\rangle+\left|11\right\rangle\right)=e^{-\frac{1}{2}\left|\eta\right|^{2}}\\{1-\frac{1}{2}\left|\eta\right|^{2}\\},$
(508)

which differs from
$e^{-\left(x^{2}+y^{2}\right)/2}(1-x^{2})\left(1-y^{2}\right)$, the direct
product of two 1D Mexican hat wavelets (see also the difference between Figs.
9 and 10).

Figure 9: The Laguree-Gaussian mother wavelet $\psi_{1}\left(\eta\right)$.
Figure 10: 2D Mexican hat mother wavelet (Hermite Gaussian mother wavelet).

(2) when $K_{0,0}=1$, $K_{1,1}=3$, $K_{2,2}=1$, $K_{n,n}=0$ for $n\geqslant
3,$ we have (see Fig. 11)

$\psi_{2}\left(\eta\right)\equiv\left(6-7\left|\eta\right|^{2}+\left|\eta\right|^{4}\right)e^{-\frac{1}{2}\left|\eta\right|^{2}}.$
(509)

Figure 11: Laguerre-Gaussian mother wavelet $\psi_{2}\left(\eta\right).$

(3) when $K_{0,0}=1$, $K_{1,1}=1$, $K_{2,2}=3$, $K_{3,3}=3$, $K_{n,n}=0$ for
$n\geqslant 4,$ the mother wavelet $\psi_{3}\left(\eta\right)$ (see Fig. 12)
reads

$\psi_{3}\left(\eta\right)=\left(14-31\left|\eta\right|^{2}+12\left|\eta\right|^{4}-\left|\eta\right|^{6}\right)e^{-\frac{1}{2}\left|\eta\right|^{2}}.$
(510)

Figure 12: Laguerre-Gaussian mother wavelet $\psi_{3}\left(\eta\right).$

From the figures we can see that as long as the coefficients $K_{n,n}$ satisfy
condition (506), we can construct arbitrary complex mother wavelet by adding
or reducing the number of coefficients, or by adjusting the value of them. And
since only $K_{n,m}$ ($m=n$) survive in all the coefficients, the mother
wavelets obtained are all circularly symmetric on the complex plane.

Moreover, the CWT of a signal function $F\left(\eta\right)$ by $\Psi$ is
defined by

$W_{\psi}F\left(\mu,\kappa\right)=\frac{1}{\mu}\int\frac{d^{2}\eta}{\pi}F\left(\eta\right)\psi^{\ast}\left(\frac{\eta-\kappa}{\mu}\right).$
(511)

Using the $\left\langle\eta\right|$ representation we can treat it from the
quantum mechanically,

$W_{\psi}F\left(\mu,\kappa\right)=\frac{1}{\mu}\int\frac{d^{2}\eta}{\pi}\left\langle\psi\right|\left.\frac{\eta-\kappa}{\mu}\right\rangle\left\langle\eta\right|\left.F\right\rangle=\left\langle\Psi\right|U_{2}\left(\mu,\kappa\right)\left|F\right\rangle,$
(512)

where

$U_{2}\left(\mu,\kappa\right)\equiv\frac{1}{\mu}\int\frac{d^{2}\eta}{\pi}\left|\frac{\eta-\kappa}{\mu}\right\rangle\left\langle\eta\right|,\;\mu=e^{\lambda},$
(513)

is the two-mode squeezing-displacing operator. Using the IWOP technique we can
calculate its normally ordered form,

$\displaystyle U_{2}\left(\mu,\kappa\right)$
$\displaystyle=\operatorname{sech}\lambda\colon\exp\\{\left(a_{1}^{\dagger}a_{2}^{\dagger}-a_{1}a_{2}\right)\tanh\lambda+\left(\operatorname{sech}\lambda-1\right)\left(a_{1}^{\dagger}a_{1}+a_{2}^{\dagger}a_{2}\right)$
$\displaystyle+\frac{1}{2}\left(\sigma^{\ast}a_{2}^{\dagger}-\sigma
a_{1}^{\dagger}\right)\operatorname{sech}\lambda+\frac{1}{1+\mu^{2}}\left(\kappa^{\ast}a_{1}-\kappa
a_{2}-\frac{1}{2}\left|\kappa\right|^{2}\right)\\}\colon.$ (514)

When $\kappa=0,$ it reduces to the usual normally ordered two-mode squeezing
operator. Once the state vector ${}_{M}\left\langle\Psi\right|$ corresponding
to mother wavelet is known, for any state $\left|F\right\rangle$ the matrix
element
${}_{M}\left\langle\Psi\right|U_{2}\left(\mu,\kappa\right)\left|F\right\rangle$
is just the wavelet transform of $F(\eta)$ with respect to
${}_{M}\left\langle\Psi\right|.$ Therefore, various quantum optical field
states can then be analyzed by their wavelet transforms.

###  15.2 Parseval Theorem in CWT

In order to complete the CWT theory, we must ask if the corresponding Parseval
theorem exists [114]. This is important since the inversion formula of CWT may
appear as a lemma of this theorem. Noting that CWT involves two-mode squeezing
transform, so the corresponding Parseval theorem differs from that of the
direct-product of two 1D wavelet transforms, too.

Next let us prove the Parseval theorem for CWT,

$\int_{0}^{\infty}\frac{d\mu}{\mu^{3}}\int\frac{d^{2}\kappa}{\pi}W_{\psi}g_{1}\left(\mu,\kappa\right)W_{\psi}^{\ast}g_{2}\left(\mu,\kappa\right)=C_{\psi}^{\prime}\int\frac{d^{2}\eta}{\pi}g_{2}^{\ast}\left(\eta\right)g_{1}\left(\eta\right),$
(515)

where $\kappa=\kappa_{1}+i\kappa_{2},$ and

$C_{\psi}^{\prime}=4\int_{0}^{\infty}\frac{d\left|\xi\right|}{\left|\xi\right|}\left|\psi\left(\xi\right)\right|^{2}.$
(516)

$\psi\left(\xi\right)$ is the complex Fourier transform of
$\psi\left(\eta\right),\psi\left(\xi\right)=\left\langle\xi\right|\left.\psi\right\rangle=\int_{-\infty}^{\infty}\frac{d^{2}\eta}{\pi}\left\langle\xi\right|\left.\eta\right\rangle\left\langle\eta\right|\left.\psi\right\rangle$.
According to (512) and (513) the quantum mechanical version of Parseval
theorem should be

$\int_{0}^{\infty}\frac{d\mu}{\mu^{3}}\int\frac{d^{2}\kappa}{\pi}\left\langle\psi\right|U_{2}\left(\mu,\kappa\right)\left|g_{1}\right\rangle\left\langle
g_{2}\right|U_{2}^{\dagger}\left(\mu,\kappa\right)\left|\psi\right\rangle=C_{\psi}^{\prime}\left\langle
g_{2}\right.\left|g_{1}\right\rangle.$ (517)

Eq.(517) indicates that once the state vector $\left\langle\psi\right|$
corresponding to mother wavelet is known, for any two states
$\left|g_{1}\right\rangle$ and $\left|g_{2}\right\rangle$, their overlap up to
the factor $C_{\psi}$ (determined by (516)) is just their corresponding
overlap of CWTs in the ($\mu,\kappa$) parametric space.

Next we prove Eq.(515) or (517). In the same procedure as the proof of
Eq.(449). We start with calculating
$U_{2}^{\dagger}\left(\mu,\kappa\right)\left|\xi\right\rangle.$ Using (42) and
(513), we have

$U_{2}^{\dagger}\left(\mu,\kappa\right)\left|\xi\right\rangle=\frac{1}{\mu}\left|\frac{\xi}{\mu}\right\rangle
e^{\frac{i}{\mu}\left(\xi_{1}\kappa_{2}-\xi_{2}\kappa_{1}\right)},$ (518)

it then follows

$\displaystyle\int\frac{d^{2}\kappa}{\pi}U_{2}^{\dagger}\left(\mu,\kappa\right)\left|\xi^{\prime}\right\rangle\left\langle\xi\right|U_{2}\left(\mu,\kappa\right)$
$\displaystyle=\frac{1}{\mu^{2}}\int\frac{d^{2}\kappa}{\pi}e^{\frac{i}{\mu}\left[\left(\xi_{1}^{\prime}-\xi_{1}\right)\kappa_{2}+\left(\xi_{2}-\xi_{2}^{\prime}\right)\kappa_{1}\right]}\left|\frac{\xi^{\prime}}{\mu}\right\rangle\left\langle\frac{\xi}{\mu}\right|$
$\displaystyle=4\pi\left|\frac{\xi}{\mu}\right\rangle\left\langle\frac{\xi}{\mu}\right|\delta\left(\xi_{1}^{\prime}-\xi_{1}\right)\delta\left(\xi_{2}-\xi_{2}^{\prime}\right).$
(519)

Using the completeness of $\left|\xi\right\rangle$ and (519) the left-hand
side (LHS) of (517) can be reformed as

LHS of Eq.(517)
$\displaystyle=\int_{0}^{\infty}\frac{d\mu}{\mu^{3}}\int\frac{d^{2}\kappa
d^{2}\xi
d^{2}\xi^{\prime}}{\pi^{3}}\left\langle\psi\right|\left.\xi\right\rangle$
$\displaystyle\times\left\langle\xi\right|U_{2}\left(\mu,\kappa\right)\left|g_{1}\right\rangle\left\langle
g_{2}\right|U_{2}^{\dagger}\left(\mu,\kappa\right)\left|\xi^{\prime}\right\rangle\left\langle\xi^{\prime}\right|\left.\psi\right\rangle$
$\displaystyle=4\int_{0}^{\infty}\frac{d\mu}{\mu^{3}}\int\frac{d^{2}\xi}{\pi}\left|\psi\left(\xi\right)\right|^{2}\left\langle
g_{2}\right.\left|\frac{\xi}{\mu}\right\rangle\left\langle\frac{\xi}{\mu}\right.\left|g_{1}\right\rangle$
$\displaystyle=\int\frac{d^{2}\xi}{\pi}\left\\{4\int_{0}^{\infty}\frac{d\mu}{\mu}\left|\psi\left(\mu\xi\right)\right|^{2}\right\\}\left\langle
g_{2}\right.\left|\xi\right\rangle\left\langle\xi\right.\left|g_{1}\right\rangle,$
(520)

where the integration value in $\\{..\\}$ is actually $\xi-$independent.
Noting that the mother wavelet $\psi\left(\eta\right)$ in Eq.(506) is just the
function of $\left|\eta\right|,$ so $\psi\left(\xi\right)$ is also the
function of $\left|\xi\right|.$ In fact, using Eqs.(506) and (42), we have

$\psi\left(\xi\right)=e^{-1/2\left|\xi\right|^{2}}\sum_{n=0}^{\infty}K_{n,n}H_{n,n}\left(\left|\xi\right|,\left|\xi\right|\right),$
(521)

where we have used the integral formula

$\int\frac{d^{2}z}{\pi}e^{\zeta\left|z\right|^{2}+\xi z+\eta
z^{\ast}}=-\frac{1}{\zeta}e^{-\frac{\xi\eta}{\zeta}},\text{Re}\left(\zeta\right)<0.$
(522)

So we can rewrite (520) as

$\text{LHS of
(\ref{16.18})}=C_{\psi}^{\prime}\int\frac{d^{2}\xi}{\pi}\left\langle
g_{2}\right.\left|\xi\right\rangle\left\langle\xi\right.\left|g_{1}\right\rangle=C_{\psi}^{\prime}\left\langle
g_{2}\right.\left|g_{1}\right\rangle,$ (523)

where

$C_{\psi}^{\prime}=4\int_{0}^{\infty}\frac{d\mu}{\mu}\left|\psi\left(\mu\xi\right)\right|^{2}=4\int_{0}^{\infty}\frac{d\left|\xi\right|}{\left|\xi\right|}\left|\psi\left(\xi\right)\right|^{2}.$
(524)

Then we have completed the proof of the Parseval theorem for CWT in (517).
Here, we should emphasize that (517) is not only different from the product of
two 1D WTs, but also different from the usual WT in 2D.

When $\left|g_{2}\right\rangle=\left|\eta\right\rangle,$ by using (513) we see
$\left\langle\eta\right|U_{2}^{\dagger}\left(\mu,\kappa\right)\left|\psi\right\rangle=\frac{1}{\mu}\psi\left(\frac{\eta-\kappa}{\mu}\right),$
then substituting it into (517) yields

$g_{1}\left(\eta\right)=\frac{1}{C_{\psi}^{\prime}}\int_{0}^{\infty}\frac{d\mu}{\mu^{3}}\int\frac{d^{2}\kappa}{\pi\mu}W_{\psi}g_{1}\left(\mu,\kappa\right)\psi\left(\frac{\eta-\kappa}{\mu}\right),$
(525)

which is just the inverse transform of the CWT. Especially, when
$\left|g_{1}\right\rangle=$ $\left|g_{2}\right\rangle,$ Eq. (517) reduces to

$\displaystyle\int_{0}^{\infty}\frac{d\mu}{\mu^{3}}\int\frac{d^{2}\kappa}{\pi}\left|W_{\psi}g_{1}\left(\mu,\kappa\right)\right|^{2}$
$\displaystyle=C_{\psi}^{\prime}\int\frac{d^{2}\eta}{\pi}\left|g_{1}\left(\eta\right)\right|^{2},$
$\displaystyle\text{or
}\int_{0}^{\infty}\frac{d\mu}{\mu^{3}}\int\frac{d^{2}\kappa}{\pi}\left|\left\langle\psi\right|U_{2}\left(\mu,\kappa\right)\left|g_{1}\right\rangle\right|^{2}$
$\displaystyle=C_{\psi}^{\prime}\left\langle
g_{1}\right.\left|g_{1}\right\rangle,$ (526)

which is named isometry of energy.

###  15.3 Orthogonal property of mother wavelet in parameter space

On the other hand, when $\left|g_{1}\right\rangle=\left|\eta\right\rangle,$
$\left|g_{2}\right\rangle=\left|\eta^{\prime}\right\rangle$, Eq.(517) becomes

$\frac{1}{C_{\psi}^{\prime}}\int_{0}^{\infty}\frac{d\mu}{\mu^{5}}\int\frac{d^{2}\kappa}{\pi}\psi\left(\frac{\eta^{\prime}-\kappa}{\mu}\right)\psi^{\ast}\left(\frac{\eta-\kappa}{\mu}\right)=\pi\delta^{(2)}\left(\eta-\eta^{\prime}\right),$
(527)

which is a new orthogonal property of mother wavelet in parameter space
spanned by $\left(\mu,\kappa\right)$. In a similar way, we take
$\left|g_{1}\right\rangle=\left|g_{2}\right\rangle=\left|m,n\right\rangle,$ a
two-mode number state, since $\left\langle
m,n\right.\left|m,n\right\rangle=1,$ then we have

$\int_{0}^{\infty}\frac{d\mu}{\mu^{3}}\int\frac{d^{2}\kappa}{\pi}\left|\left\langle\psi\right|U_{2}\left(\mu,\kappa\right)\left|m,n\right\rangle\right|^{2}=C_{\psi}^{\prime},$
(528)

or take
$\left|g_{1}\right\rangle=\left|g_{2}\right\rangle=\left|z_{1},z_{2}\right\rangle,$
$\left|z\right\rangle=\exp\left(-\left|z\right|^{2}/2+za^{\dagger}\right)\left|0\right\rangle$
is the coherent state, then

$\int_{0}^{\infty}\frac{d\mu}{\mu^{3}}\int\frac{d^{2}\kappa}{\pi}\left|\left\langle\psi\right|U_{2}\left(\mu,\kappa\right)\left|z_{1},z_{2}\right\rangle\right|^{2}=C_{\psi}^{\prime}.$
(529)

Next we examine a special example. When the mother wavelet is
$\psi_{1}\left(\eta\right)$ in (508), using (42) we have
$\psi\left(\xi\right)=\frac{1}{2}\left|\xi\right|^{2}e^{-\frac{1}{2}\left|\xi\right|^{2}},$
which leads to
$C_{\psi}^{\prime}=\int_{0}^{\infty}\left|\xi\right|^{3}e^{-\left|\xi\right|^{2}}d\left|\xi\right|=\frac{1}{2}.$
Thus for $\psi_{1}\left(\eta\right)$, we see

$2\int_{0}^{\infty}\frac{d\mu}{\mu^{5}}\int\frac{d^{2}\kappa}{\pi}\psi_{1}\left(\frac{\eta^{\prime}-\kappa}{\mu}\right)\psi_{1}^{\ast}\left(\frac{\eta-\kappa}{\mu}\right)=\pi\delta^{(2)}\left(\eta-\eta^{\prime}\right).$
(530)

Eq. (530) can be checked as follows. Using (508) and the integral formula

$\displaystyle\int_{0}^{\infty}u\left(1-\frac{ux^{2}}{2}\right)\left(1-\frac{uy^{2}}{2}\right)e^{-u\frac{x^{2}+y^{2}}{2}}du$
$\displaystyle=-\frac{4(x^{4}-4x^{2}y^{2}+y^{4})}{(x^{2}+y^{2})^{4}},\text{
}\operatorname{Re}\left(x^{2}+y^{2}\right)>0,$ (531)

we can put the left-hand side (LHS) of (530) into

$\text{LHS of
(\ref{16.31})}=-\int\frac{d^{2}\kappa}{\pi}\frac{4(x^{4}-4x^{2}y^{2}+y^{4})}{(x^{2}+y^{2})^{4}},$
(532)

where $x^{2}=\left|\eta^{\prime}-\kappa\right|^{2},$
$y^{2}=\left|\eta-\kappa\right|^{2}.$

When $\eta^{\prime}=\eta,$ $x^{2}=y^{2},$

$\text{LHS of
(\ref{16.31})}=\int\allowbreak\frac{d^{2}\kappa}{2\pi\left|\kappa-\eta\right|^{4}}=\int_{0}^{\infty}\allowbreak\int_{0}^{2\pi}\frac{drd\theta}{2\pi
r^{3}}\rightarrow\infty.$ (533)

On the other hand, when $\eta\neq\eta^{\prime}$ and noticing that

$\displaystyle x^{2}$
$\displaystyle=\left(\eta_{1}^{\prime}-\kappa_{1}\right)^{2}+\left(\eta_{2}^{\prime}-\kappa_{2}\right)^{2},$
$\displaystyle y^{2}$
$\displaystyle=\left(\eta_{1}-\kappa_{1}\right)^{2}+\left(\eta_{2}-\kappa_{2}\right)^{2},$
(534)

which leads to
$\mathtt{d}x^{2}\mathtt{d}y^{2}=4\left|J\right|\mathtt{d}\kappa_{1}\mathtt{d}\kappa_{2}$,
where
$J\left(x,y\right)=\left|\begin{array}[c]{cc}\kappa_{1}-\eta_{1}^{\prime}&\kappa_{2}-\eta_{2}^{\prime}\\\
\kappa_{1}-\eta_{1}&\kappa_{2}-\eta_{2}\end{array}\right|$. As a result of
(534), (532) reduces to

$\text{LHS of
(\ref{16.31})}=-4\int_{-\infty}^{\infty}\frac{dxdy}{\pi}\frac{xy(x^{4}-4x^{2}y^{2}+y^{4})}{\left|J\right|(x^{2}+y^{2})^{4}}=0\text{,}$
(535)

where we have noticed that $J\left(x,y\right)$ is the funtion of
$\left(x^{2},y^{2}\right).$ Thus we have

$\text{LHS of
(\ref{16.31})}=\left\\{\begin{array}[c]{cc}\infty,&\eta=\eta^{\prime},\\\
0,&\eta\neq\eta^{\prime}.\end{array}\right.=\text{RHS of (\ref{16.31}).}$
(536)

###  15.4 CWT and Entangled Husimi distribution

Recalling that in Ref.[115], the so-called entangled Husimi operator
$\Delta_{h}\left(\sigma,\gamma,\kappa\right)$ has been introduced, which is
endowed with definite physical meaning, and it is found that the two-mode
squeezed coherent state $\left|\sigma,\gamma\right\rangle_{\kappa}$
representation of $\Delta_{h}\left(\sigma,\gamma,\kappa\right),$
$\Delta_{h}\left(\sigma,\gamma,\kappa\right)=$
$\left|\sigma,\gamma,\kappa\right\rangle\left\langle\sigma,\gamma,\kappa\right|$.
The entangled Husimi operator $\Delta_{h}\left(\sigma,\gamma,\kappa\right)$
and the entangled Husimi distribution $F_{h}\left(\sigma,\gamma,\kappa\right)$
of quantum state $\left|\psi\right\rangle$ are given by

$\Delta_{h}\left(\sigma,\gamma,\kappa\right)=4\int
d^{2}\sigma^{\prime}d^{2}\gamma^{\prime}\Delta_{w}\left(\sigma^{\prime},\gamma^{\prime}\right)\exp\left\\{-\kappa\left|\sigma^{\prime}-\sigma\right|^{2}-\frac{1}{\kappa}\left|\gamma^{\prime}-\gamma\right|^{2}\right\\},$
(537)

and

$F_{h}\left(\sigma,\gamma,\kappa\right)=4\int
d^{2}\sigma^{\prime}d^{2}\gamma^{\prime}F_{w}\left(\sigma^{\prime},\gamma^{\prime}\right)\exp\left\\{-\kappa\left|\sigma^{\prime}-\sigma\right|^{2}-\frac{1}{\kappa}\left|\gamma^{\prime}-\gamma\right|^{2}\right\\},$
(538)

respectively, where
$F_{w}\left(\sigma^{\prime},\gamma^{\prime}\right)=\left\langle\psi\right|\Delta_{w}\left(\sigma^{\prime},\gamma^{\prime}\right)\left|\psi\right\rangle$
is two-mode Wigner function, with
$\Delta_{w}\left(\sigma^{\prime},\gamma^{\prime}\right)$ being the two-mode
Wigner operator. Thus we are naturally led to studying the entangled Husimi
distribution function from the viewpoint of wavelet transformation. In this
subsection, we shall extend the relation between wavelet transformation and
Wigner-Husimi distribution function to the entangled case, that is to say, we
employ the CWT to investigate the entangled Husimi distribution function
(EHDF) by bridging the relation between CWT and EHDF. This is a convenient
approach for calculating various entangled Husimi distribution functions of
miscellaneous two-mode quantum states.

####  15.4.1 CWT and its quantum mechanical version

In Ref.[113], the CWT has been proposed, i.e., the CWT of a complex signal
function $g\left(\eta\right)$ by $\psi$ is defined by

$W_{\psi}g\left(\mu,z\right)=\frac{1}{\mu}\int\frac{d^{2}\eta}{\pi}g\left(\eta\right)\psi^{\ast}\left(\frac{\eta-z}{\mu}\right),$
(539)

whose admissibility condition for mother wavelets,
$\int\frac{d^{2}\eta}{2\pi}\psi\left(\eta\right)=0,$ is examined in the
entangled state representations $\left\langle\eta\right|$ and a family of new
mother wavelets (named the Laguerre–Gaussian wavelets) are found to match the
CWT [113]. In fact, by introducing the bipartite entangled state
representation $\left\langle\eta=\eta_{1}+\mathtt{i}\eta_{2}\right|,$we can
treat (538) quantum mechanically,

$W_{\psi}g\left(\mu,z\right)=\frac{1}{\mu}\int\frac{d^{2}\eta}{\pi}\left\langle\psi\right|\left.\frac{\eta-z}{\mu}\right\rangle\left\langle\eta\right|\left.g\right\rangle=\left\langle\psi\right|U_{2}\left(\mu,z\right)\left|g\right\rangle,$
(540)

where $z=z_{1}+iz_{2}\in C,$ $0<\mu\in R,$
$g\left(\eta\right)\equiv\left\langle\eta\right|\left.g\right\rangle\ $and
$\psi\left(\eta\right)=\left\langle\eta\right|\left.\psi\right\rangle$ are the
wavefunction of state vector $\left|g\right\rangle$ and the mother wavelet
state vector $\left|\psi\right\rangle$ in $\left\langle\eta\right|$
representation, respectively, and

$U_{2}\left(\mu,z\right)\equiv\frac{1}{\mu}\int\frac{d^{2}\eta}{\pi}\left|\frac{\eta-z}{\mu}\right\rangle\left\langle\eta\right|,\;\mu=e^{\lambda},$
(541)

is the two-mode squeezing-displacing operator. Noticing that the two-mode
squeezing operator has its natural expression in $\left\langle\eta\right|$
representation (36), which is different from the direct product of two single-
mode squeezing (dilation) operators, and the two-mode squeezed state is
simultaneously an entangled state, thus we can put Eq.(541) into the following
form,

$U_{2}\left(\mu,z\right)=S_{2}\left(\mu\right)\mathfrak{D}\left(z\right),$
(542)

where $\mathfrak{D}\left(z\right)$ is a two-mode displacement operator,
$\mathfrak{D}\left(z\right)\left|\eta\right\rangle=\left|\eta-z\right\rangle$
and

$\displaystyle\mathfrak{D}\left(z\right)$
$\displaystyle=\int\frac{d^{2}\eta}{\pi}\left|\eta-z\right\rangle\left\langle\eta\right|$
$\displaystyle=\exp\left[iz_{1}\frac{P_{1}-P_{2}}{\sqrt{2}}-iz_{2}\frac{Q_{1}+Q_{2}}{\sqrt{2}}\right]$
$\displaystyle=D_{1}\left(-z/2\right)D_{2}\left(z^{\ast}/2\right).$ (543)

It the follows the quantum mechanical version of CWT is

$W_{\psi}g\left(\mu,\zeta\right)=\left\langle\psi\right|S_{2}\left(\mu\right)\mathfrak{D}\left(z\right)\left|g\right\rangle=\left\langle\psi\right|S_{2}\left(\mu\right)D_{1}\left(-z/2\right)D_{2}\left(z^{\ast}/2\right)\left|g\right\rangle.$
(544)

Eq.(544) indicates that the CWT can be put into a matrix element in the
$\left\langle\eta\right|$ representation of the two-mode displacing and the
two-mode squeezing operators in Eq.(LABEL:e10) between the mother wavelet
state vector $\left|\psi\right\rangle$ and the state vector
$\left|g\right\rangle$ to be transformed. Once the state vector
$\left\langle\psi\right|$ as mother wavelet is chosen, for any state
$\left|g\right\rangle$ the matrix element
$\left\langle\psi\right|U_{2}\left(\mu,z\right)\left|g\right\rangle$ is just
the wavelet transform of $g(\eta)$ with respect to $\left\langle\psi\right|.$
Therefore, various quantum optical field states can then be analyzed by their
wavelet transforms.

####  15.4.2 Relation between CWT and EHDF

In the following we shall show that the EHDF of a quantum state
$\left|\psi\right\rangle$ can be obtained by making a complex wavelet
transform of the Gaussian function $e^{-\left|\eta\right|^{2}/2},$ i.e.,

$\left\langle\psi\right|\Delta_{h}\left(\sigma,\gamma,\kappa\right)\left|\psi\right\rangle=e^{-\frac{1}{\kappa}\left|\gamma\right|^{2}}\left|\int\frac{d^{2}\eta}{\sqrt{\kappa}\pi}e^{-\left|\eta\right|^{2}/2}\psi^{\ast}\left(\frac{\eta-z}{\sqrt{\kappa}}\right)\right|^{2},$
(545)

where $\mu=e^{\lambda}=\sqrt{\kappa},$ $z=z_{1}+iz_{2},$ and

$\displaystyle z_{1}$
$\displaystyle=\frac{\cosh\lambda}{1+\kappa}\left[\gamma^{\ast}-\gamma-\kappa\left(\sigma^{\ast}+\sigma\right)\right],$
(546) $\displaystyle z_{2}$
$\displaystyle=\frac{i\cosh\lambda}{1+\kappa}\left[\gamma+\gamma^{\ast}+\kappa\left(\sigma-\sigma^{\ast}\right)\right],$
(547)

$\Delta_{h}\left(\sigma,\gamma,\kappa\right)$ is named the entangled Husimi
operator by us,

$\displaystyle\Delta_{h}\left(\sigma,\gamma,\kappa\right)$
$\displaystyle=\frac{4\kappa}{\left(1+\kappa\right)^{2}}\colon\exp\left\\{-\frac{\left(a_{1}+a_{2}^{{\dagger}}-\gamma\right)\left(a_{1}^{{\dagger}}+a_{2}-\gamma^{\ast}\right)}{1+\kappa}\right.$
$\displaystyle-\left.\frac{\kappa\left(a_{1}-a_{2}^{{\dagger}}-\sigma\right)\left(a_{1}^{{\dagger}}-a_{2}-\sigma^{\ast}\right)}{1+\kappa}\right\\}\colon.$
(548)

$\left\langle\psi\right|\Delta_{h}\left(\sigma,\gamma,\kappa\right)\left|\psi\right\rangle$
is the Husimi distribution function.

Proof of Eq.(545).

When the state to be transformed is
$\left|g\right\rangle=\left|00\right\rangle$ (the two-mode vacuum state), by
noticing that
$\left\langle\eta\right.\left|00\right\rangle=e^{-\left|\eta\right|^{2}/2},$
we can express Eq.(540) as

$\frac{1}{\mu}\int\frac{d^{2}\eta}{\pi}e^{-\left|\eta\right|^{2}/2}\psi^{\ast}\left(\frac{\eta-z}{\mu}\right)=\left\langle\psi\right|U_{2}\left(\mu,z\right)\left|00\right\rangle.$
(549)

To combine the CWTs with transforms of quantum states more tightly and
clearly, using the IWOP technique we can directly perform the integral in
Eq.(541) [116]

$\displaystyle U_{2}\left(\mu,z\right)$
$\displaystyle=\operatorname{sech}\lambda\exp\left[-\frac{1}{2\left(1+\mu^{2}\right)}\left|z\right|^{2}+a_{1}^{\dagger}a_{2}^{\dagger}\tanh\lambda+\frac{1}{2}\left(z^{\ast}a_{2}^{\dagger}-za_{1}^{\dagger}\right)\operatorname{sech}\lambda\right]$
$\displaystyle\times\exp\left[\left(a_{1}^{\dagger}a_{1}+a_{2}^{\dagger}a_{2}\right)\ln\operatorname{sech}\lambda\right]\exp\left(\frac{z^{\ast}a_{1}-za_{2}}{1+\mu^{2}}-a_{1}a_{2}\tanh\lambda\right).$
(550)

where we have set $\mu=e^{\lambda}$,
$\operatorname{sech}\lambda=\frac{2\mu}{1+\mu^{2}}$,
$\tanh\lambda=\frac{\mu^{2}-1}{\mu^{2}+1}$, and we have used the operator
identity
$e^{ga^{\dagger}a}=\colon\exp\left[\left(e^{g}-1\right)a^{\dagger}a\right]\colon$.
In particular, when $z=0,$ $U_{2}\left(\mu,z=0\right)$ becomes to the usual
normally ordered two-mode squeezing operator $S_{2}\left(\mu\right)$. From
Eq.(550) it then follows that

$\displaystyle U_{2}\left(\mu,z\right)\left|00\right\rangle$
$\displaystyle=\operatorname{sech}\lambda\exp\left\\{-\frac{\left(z_{1}-iz_{2}\right)\left(z_{1}+iz_{2}\right)}{2\left(1+\mu^{2}\right)}+a_{1}^{\dagger}a_{2}^{\dagger}\tanh\lambda\right.$
$\displaystyle\left.+\frac{1}{2}\left[\left(z_{1}-iz_{2}\right)a_{2}^{\dagger}-\left(z_{1}+iz_{2}\right)a_{1}^{\dagger}\right]\operatorname{sech}\lambda\right\\}\left|00\right\rangle.$
(551)

Substituting Eqs.(546), (547) and $\tanh\lambda=\frac{\kappa-1}{\kappa+1},$
$\cosh\lambda=\frac{1+\kappa}{2\sqrt{\kappa}}$ into Eq.(551) yields

$\displaystyle
e^{-\frac{1}{2\kappa}\left|\gamma\right|^{2}-\frac{\sigma\gamma^{\ast}-\gamma\sigma^{\ast}}{2\left(\kappa+1\right)}}U_{2}\left(\mu,z_{1},z_{2}\right)\left|00\right\rangle$
$\displaystyle=\frac{2\sqrt{\kappa}}{1+\kappa}\exp\left\\{-\frac{\left|\gamma\right|^{2}+\kappa\left|\sigma\right|^{2}}{2\left(\kappa+1\right)}+\frac{\kappa\sigma+\gamma}{1+\kappa}a_{1}^{\dagger}+\frac{\gamma^{\ast}-\kappa\sigma^{\ast}}{1+\kappa}a_{2}^{\dagger}+a_{1}^{\dagger}a_{2}^{\dagger}\frac{\kappa-1}{\kappa+1}\right\\}\allowbreak\left|00\right\rangle\left.\equiv\right.\left|\sigma,\gamma\right\rangle_{\kappa},$
(552)

then the CWT of Eq.(549) can be further expressed as

$e^{-\frac{1}{2\kappa}\left|\gamma\right|^{2}-\frac{\sigma\gamma^{\ast}-\gamma\sigma^{\ast}}{2\left(\kappa+1\right)}}\int\frac{d^{2}\eta}{\mu\pi}e^{-\left|\eta\right|^{2}/2}\psi^{\ast}\left(\frac{\eta-
z_{1}-iz_{2}}{\mu}\right)=\left\langle\psi\right.\left|\sigma,\gamma\right\rangle_{\kappa}.$
(553)

Using normally ordered form of the vacuum state projector
$\left|00\right\rangle\left\langle 00\right|=\colon
e^{-a_{1}^{\dagger}a_{1}-a_{2}^{\dagger}a_{2}}\colon,$ and the IWOP method as
well as Eq.(552) we have

$\displaystyle\left|\sigma,\gamma\right\rangle_{\kappa\kappa}\left\langle\sigma,\gamma\right|$
$\displaystyle=\frac{4\kappa}{\left(1+\kappa\right)^{2}}\colon\exp\left[-\frac{\left|\gamma\right|^{2}+\kappa\left|\sigma\right|^{2}}{\kappa+1}+\frac{\kappa\sigma+\gamma}{1+\kappa}a_{1}^{\dagger}+\frac{\gamma^{\ast}-\kappa\sigma^{\ast}}{1+\kappa}a_{2}^{\dagger}\right.$
$\displaystyle\left.+\frac{\kappa\sigma^{\ast}+\gamma^{\ast}}{1+\kappa}a_{1}+\frac{\gamma-\kappa\sigma}{1+\kappa}a_{2}+\frac{\kappa-1}{\kappa+1}\left(a_{1}^{\dagger}a_{2}^{\dagger}+a_{1}a_{2}\right)-a_{1}^{\dagger}a_{1}-a_{2}^{\dagger}a_{2}\right]\colon$
$\displaystyle=\frac{4\kappa}{\left(1+\kappa\right)^{2}}\colon\exp\left\\{-\frac{\left(a_{1}+a_{2}^{{\dagger}}-\gamma\right)\left(a_{1}^{{\dagger}}+a_{2}-\gamma^{\ast}\right)}{1+\kappa}\right.$
$\displaystyle-\left.\frac{\kappa\left(a_{1}-a_{2}^{{\dagger}}-\sigma\right)\left(a_{1}^{{\dagger}}-a_{2}-\sigma^{\ast}\right)}{1+\kappa}\right\\}\colon\left.=\right.\Delta_{h}\left(\sigma,\gamma,\kappa\right).$
(554)

Now we explain why $\Delta_{h}\left(\sigma,\gamma,\kappa\right)$ is the
entangled Husimi operator. Using the formula for converting an operator $A$
into its Weyl ordering form [37]

$A=4\int\frac{d^{2}\alpha
d^{2}\beta}{\pi^{2}}\left\langle-\alpha,-\beta\right|A\left|\alpha,\beta\right\rangle\genfrac{}{}{0.0pt}{}{:}{:}\exp\\{2\left(\alpha^{\ast}a_{1}-a_{1}^{\dagger}\alpha+\beta^{\ast}a_{2}-a_{2}^{\dagger}\beta+a_{1}^{\dagger}a_{1}+a_{2}^{\dagger}a_{2}\right)\\}\genfrac{}{}{0.0pt}{}{:}{:},$
(555)

where the symbol $\genfrac{}{}{0.0pt}{}{:}{:}\genfrac{}{}{0.0pt}{}{:}{:}$
denotes the Weyl ordering, $\left|\beta\right\rangle$ is the usual coherent
state, substituting Eq.(554) into Eq.(555) and performing the integration by
virtue of the technique of integration within a Weyl ordered product of
operators, we obtain

$\displaystyle\left|\sigma,\gamma\right\rangle_{\kappa\kappa}\left\langle\sigma,\gamma\right|$
$\displaystyle=\frac{16\kappa}{\left(1+\kappa\right)^{2}}\int\frac{d^{2}\alpha
d^{2}\beta}{\pi^{2}}\left\langle-\alpha,-\beta\right|\colon\exp\left\\{-\frac{\left(a_{1}+a_{2}^{{\dagger}}-\gamma\right)\left(a_{1}^{{\dagger}}+a_{2}-\gamma^{\ast}\right)}{1+\kappa}\right.$
$\displaystyle\left.-\frac{\kappa\left(a_{1}-a_{2}^{{\dagger}}-\sigma\right)\left(a_{1}^{{\dagger}}-a_{2}-\sigma^{\ast}\right)}{1+\kappa}\right\\}\colon\left|\alpha,\beta\right\rangle$
$\displaystyle\times\genfrac{}{}{0.0pt}{}{:}{:}\exp\\{2\left(\alpha^{\ast}a_{1}-a_{1}^{\dagger}\alpha+\beta^{\ast}a_{2}-a_{2}^{\dagger}\beta+a_{1}^{\dagger}a_{1}+a_{2}^{\dagger}a_{2}\right)\\}\genfrac{}{}{0.0pt}{}{:}{:}$
$\displaystyle=4\genfrac{}{}{0.0pt}{}{:}{:}\exp\left\\{-\kappa\left(a_{1}-a_{2}^{{\dagger}}-\sigma\right)\left(a_{1}^{{\dagger}}-a_{2}-\sigma^{\ast}\right)-\frac{1}{\kappa}\left(a_{1}+a_{2}^{{\dagger}}-\gamma\right)\left(a_{1}^{{\dagger}}+a_{2}-\gamma^{\ast}\right)\right\\}\genfrac{}{}{0.0pt}{}{:}{:},$
(556)

where we have used the integral formula

$\int\frac{d^{2}z}{\pi}\exp\left(\zeta\left|z\right|^{2}+\xi z+\eta
z^{\ast}\right)=-\frac{1}{\zeta}e^{-\frac{\xi\eta}{\zeta}},\text{Re}\left(\zeta\right)<0.$
(557)

Eq.(556) is the Weyl ordering form of
$\left|\sigma,\gamma\right\rangle_{\kappa\kappa}\left\langle\sigma,\gamma\right|.$
Then according to Weyl quantization scheme we know the Weyl ordering form of
two-mode Wigner operator is given by

$\Delta_{w}\left(\sigma,\gamma\right)=\genfrac{}{}{0.0pt}{}{:}{:}\delta\left(a_{1}-a_{2}^{{\dagger}}-\sigma\right)\delta\left(a_{1}^{{\dagger}}-a_{2}-\sigma^{\ast}\right)\delta\left(a_{1}+a_{2}^{{\dagger}}-\gamma\right)\delta\left(a_{1}^{{\dagger}}+a_{2}-\gamma^{\ast}\right)\genfrac{}{}{0.0pt}{}{:}{:},$
(558)

thus the classical corresponding function of a Weyl ordered operator is
obtained by just replacing
$a_{1}-a_{2}^{{\dagger}}\rightarrow\sigma^{\prime},a_{1}+a_{2}^{{\dagger}}\rightarrow\gamma^{\prime},$
i.e.,

$\displaystyle
4\genfrac{}{}{0.0pt}{}{:}{:}\exp\left\\{-\kappa\left(a_{1}-a_{2}^{{\dagger}}-\sigma\right)\left(a_{1}^{{\dagger}}-a_{2}-\sigma^{\ast}\right)-\frac{1}{\kappa}\left(a_{1}+a_{2}^{{\dagger}}-\gamma\right)\left(a_{1}^{{\dagger}}+a_{2}-\gamma^{\ast}\right)\right\\}\genfrac{}{}{0.0pt}{}{:}{:}$
$\displaystyle\rightarrow
4\exp\left\\{-\kappa\left|\sigma^{\prime}-\sigma\right|^{2}-\frac{1}{\kappa}\left|\gamma^{\prime}-\gamma\right|^{2}\right\\},$
(559)

and in this case the Weyl rule is expressed as

$\displaystyle\left|\sigma,\gamma\right\rangle_{\kappa\kappa}\left\langle\sigma,\gamma\right|$
$\displaystyle=4\int
d^{2}\sigma^{\prime}d^{2}\gamma^{\prime}\genfrac{}{}{0.0pt}{}{:}{:}\delta\left(a_{1}-a_{2}^{{\dagger}}-\sigma\right)\delta\left(a_{1}^{{\dagger}}-a_{2}-\sigma^{\ast}\right)\delta\left(a_{1}+a_{2}^{{\dagger}}-\gamma\right)$
$\displaystyle\times\delta\left(a_{1}^{{\dagger}}+a_{2}-\gamma^{\ast}\right)\genfrac{}{}{0.0pt}{}{:}{:}\exp\left\\{-\kappa\left|\sigma^{\prime}-\sigma\right|^{2}-\frac{1}{\kappa}\left|\gamma^{\prime}-\gamma\right|^{2}\right\\}$
$\displaystyle=4\int
d^{2}\sigma^{\prime}d^{2}\gamma^{\prime}\Delta_{w}\left(\sigma^{\prime},\gamma^{\prime}\right)\exp\left\\{-\kappa\left|\sigma^{\prime}-\sigma\right|^{2}-\frac{1}{\kappa}\left|\gamma^{\prime}-\gamma\right|^{2}\right\\}.$
(560)

In reference to Eq.(538) in which the relation between the entangled Husimi
function and the two-mode Wigner function is shown, we know that the right-
hand side of Eq. (560) should be just the entangled Husimi operator, i.e.

$\left|\sigma,\gamma\right\rangle_{\kappa\kappa}\left\langle\sigma,\gamma\right|=4\int
d^{2}\sigma^{\prime}d^{2}\gamma^{\prime}\Delta_{w}\left(\sigma^{\prime},\gamma^{\prime}\right)\exp\left\\{-\kappa\left|\sigma^{\prime}-\sigma\right|^{2}-\frac{1}{\kappa}\left|\gamma^{\prime}-\gamma\right|^{2}\right\\}=\Delta_{h}\left(\sigma,\gamma,\kappa\right),$
(561)

thus Eq. (545) is proved by combining Eqs.(561) and (553).

Thus we have further extended the relation between wavelet transformation and
Wigner-Husimi distribution function to the entangled case. That is to say, we
prove that the entangled Husimi distribution function of a two-mode quantum
state $\left|\psi\right\rangle$ is just the modulus square of the complex
wavelet transform of $e^{-\left|\eta\right|^{2}/2}$ with
$\psi\left(\eta\right)$ being the mother wavelet up to a Gaussian function,
i.e.,
$\left\langle\psi\right|\Delta_{h}\left(\sigma,\gamma,\kappa\right)\left|\psi\right\rangle=e^{-\frac{1}{\kappa}\left|\gamma\right|^{2}}\left|\int\frac{d^{2}\eta}{\sqrt{\kappa}\pi}e^{-\left|\eta\right|^{2}/2}\psi^{\ast}\left(\left(\eta-z\right)/\sqrt{\kappa}\right)\right|^{2}$.
Thus is a convenient approach for calculating various entangled Husimi
distribution functions of miscellaneous quantum states.

##  16 Symplectic Wavelet transformation (SWT)

In this section we shall generalize the usual wavelet transform to symplectic
wavelet transformation (SWT) by using the coherent state representation [117].

###  16.1 Single-mode SWT

First we are motivated to generalize the usual wavelet transform, which
concerns about dilation, to optical Fresnel transform (we will explain this in
detail in section below), i.e. we shall use the symplectic-
transformed—translated versions of the mother wavelet

$\psi_{r,s;\kappa}\left(z\right)=\sqrt{s^{\ast}}\psi\left[s\left(z-\kappa\right)-r\left(z^{\ast}-\kappa^{\ast}\right)\right]$
(562)

as a weighting function to synthesize the original complex signal
$f\left(z\right)$,

$\displaystyle W_{\psi}f\left(r,s;\kappa\right)$
$\displaystyle=\int\frac{d^{2}z}{\pi}f\left(z\right)\psi_{r,s;\kappa}^{\ast}\left(z\right),\text{
}$ (563) $\displaystyle d^{2}z$ $\displaystyle=dxdy,\text{ }z=x+iy,$

this is named the symplectic-transformed—translated wavelet transform. One can
see that the mother wavelet $\psi$ generates the other wavelets of the family
$\psi^{\ast}\left[s\left(z-\kappa\right)-r\left(z^{\ast}-\kappa^{\ast}\right)\right]$
through a translating transform followed by a symplectic transform, ($r,s$ are
the symplectic transform parameter, $|s|^{2}-|r|^{2}=1,$ $\kappa$ is a
translation parameter, $s$, $r$ and $\kappa\in\mathrm{C}$), this can be seen
more clearly by writing the second transform in matrix form

$\left(\begin{array}[c]{c}z-\kappa\\\
z^{\ast}-\kappa^{\ast}\end{array}\right)\rightarrow
M\left(\begin{array}[c]{c}z-\kappa\\\
z^{\ast}-\kappa^{\ast}\end{array}\right),\text{
}M\equiv\left(\begin{array}[c]{cc}s&-r\\\
-r^{\ast}&s^{\ast}\end{array}\right),$ (564)

where $M$ is a symplectic matrix satisfies $M^{T}JM=J$,
$J=\left(\begin{array}[c]{cc}0&I\\\ -I&0\end{array}\right)$.

Symplectic matrices in Hamiltonian dynamics correspond to canonical
transformations and keep the Poisson bracket invariant, while in matrix optics
they represent ray transfer matrices of optical instruments, such as lenses
and fibers.

####  16.1.1 Properties of symplectic-transformed—translated WT

It is straightforward to evaluate this transform and its reciprocal transform
when $f\left(z\right)$ is the complex Fourier exponentials,
$f\left(z\right)=\exp\left(z\beta^{\ast}-z^{\ast}\beta\right),$ (note that
$z\beta^{\ast}-z^{\ast}\beta$ is pure imaginary):

$\displaystyle W_{\psi}f$
$\displaystyle=\sqrt{s}\int\frac{d^{2}z}{\pi}\exp\left(z\beta^{\ast}-z^{\ast}\beta\right)\psi^{\ast}\left[s\left(z-\kappa\right)-r\left(z^{\ast}-\kappa^{\ast}\right)\right]$
$\displaystyle=\sqrt{s}\int\frac{d^{2}z}{\pi}\exp\left[\left(z+\kappa\right)\beta^{\ast}-\left(z^{\ast}+\kappa^{\ast}\right)\beta\right]\psi^{\ast}\left(sz-
rz^{\ast}\right)$
$\displaystyle=\sqrt{s}\int\frac{d^{2}z^{\prime}}{\pi}\exp\left[\left(s^{\ast}z^{\prime}+rz^{\prime\ast}+\kappa\right)\beta^{\ast}-\left(sz^{\prime\ast}+r^{\ast}z^{\prime}+\kappa^{\ast}\right)\beta\right]\psi^{\ast}\left(z^{\prime}\right)$
$\displaystyle=\exp\left[\kappa\beta^{\ast}-\kappa^{\ast}\beta\right]\sqrt{s}\int\frac{d^{2}z^{\prime}}{\pi}\exp\left[z^{\prime}\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta\right)-z^{\prime\ast}\left(s\beta-r\beta^{\ast}\right)\right]\psi^{\ast}\left(z^{\prime}\right)$
$\displaystyle=\sqrt{s}\exp\left[\kappa\beta^{\ast}-\kappa^{\ast}\beta\right]\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta\right),$
(565)

where $\Phi$ is the complex Fourier transform of $\psi^{\ast}.$ Then we form
the adjoint operation

$\displaystyle W_{\psi}^{\ast}\left(W_{\psi}f\right)\left(z\right)$
$\displaystyle=\sqrt{s^{\ast}}\int\frac{d^{2}\kappa}{\pi}\left(W_{\psi}f\right)\left(r,s;\kappa\right)\psi\left[s\left(z-\kappa\right)-r\left(z^{\ast}-\kappa^{\ast}\right)\right]$
$\displaystyle=\left|s\right|\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta\right)\int\frac{d^{2}\kappa}{\pi}\exp\left[\left(\kappa+z\right)\beta^{\ast}-\left(\kappa^{\ast}+z^{\ast}\right)\beta\right]\psi\left[-s\kappa+r\kappa^{\ast}\right]$
$\displaystyle=\left|s\right|\exp\left(z\beta^{\ast}-z^{\ast}\beta\right)\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta\right)$
$\displaystyle\times\int\frac{d^{2}\kappa^{\prime}}{\pi}\exp\left[-\kappa^{\prime}\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta\right)+\kappa^{\prime\ast}\left(s\beta-r\beta^{\ast}\right)\right]\psi\left(\kappa^{\prime}\right)$
$\displaystyle=\left|s\right|\exp\left(z\beta^{\ast}-z^{\ast}\beta\right)\left|\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta\right)\right|^{2},$
(566)

from which we have

$\int\frac{W_{\psi}^{\ast}\left(W_{\psi}f\right)\left(z\right)d^{2}s}{\left|s\right|^{2}}=\exp\left(z\beta^{\ast}-z^{\ast}\beta\right)\int
d^{2}s\frac{\left|\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta\right)\right|^{2}}{\left|s\right|}$
(567)

so we get the inversion formula

$f\left(z\right)=\exp\left(z\beta^{\ast}-z^{\ast}\beta\right)=\frac{\int
d^{2}sW_{\psi}^{\ast}\left(W_{\psi}f\right)\left(z\right)/\left|s\right|^{2}}{\int
d^{2}s\left|\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta\right)\right|^{2}/\left|s\right|}.$
(568)

Eq.(568) leads us to impose the normalization

$\int
d^{2}s\left|\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta\right)\right|^{2}/\left|s\right|=1,$
(569)

in order to get the wavelet representation

$f\left(z\right)=\int
d^{2}sW_{\psi}^{\ast}\left(W_{\psi}f\right)\left(z\right)/\left|s\right|^{2}.$
(570)

Then we can have a form of Parseval’s theorem for this new wavelet transform:

Preposition:  For any $f$ and $f^{\prime}$ we have

$\int\int
W_{\psi}f\left(r,s;\kappa\right)W_{\psi}f^{\prime\ast}\left(r,s;\kappa\right)\frac{d^{2}\kappa
d^{2}s}{\left|s\right|^{2}}=\int\frac{d^{2}z}{2\pi}f\left(z\right)f^{\prime\ast}\left(z\right).$
(571)

Proof: Let us assume $F(\beta)$ and $F^{\prime}(\beta)$ are the complex
Fourier transform of $f\left(z\right)$ and $f^{\prime}\left(z\right)$
respectively,

$F(\beta)=\int\frac{d^{2}z}{2\pi}f\left(z\right)\exp\left(z\beta^{\ast}-z^{\ast}\beta\right)$
(572)

recall the convolution theorem defined on complex Fourier transform,

$\displaystyle\int
d^{2}zf\left(\alpha-z,\alpha^{\ast}-z^{\ast}\right)f^{\prime}\left(z\right)$
$\displaystyle=\int
d^{2}z\int\frac{d^{2}\beta}{2\pi}F(\beta)e^{\left(\alpha^{\ast}-z^{\ast}\right)\beta-\left(\alpha-z\right)\beta^{\ast}}\int\frac{d^{2}\beta^{\prime}}{2\pi}F^{\prime}(\beta^{\prime})\exp\left(z^{\ast}\beta^{\prime}-z\beta^{\prime\ast}\right)$
$\displaystyle=\int\int d^{2}\beta
d^{2}\beta^{\prime}F^{\prime}(\beta^{\prime})F(\beta)e^{\alpha^{\ast}\beta-\alpha\beta^{\ast}}\delta\left(\beta-\beta^{\prime}\right)\delta\left(\beta^{\ast}-\beta^{\prime\ast}\right)$
$\displaystyle=\int d^{2}\beta
F(\beta)F^{\prime}(\beta)e^{\alpha^{\ast}\beta-\alpha\beta^{\ast}}$ (573)

so from (573) and (562), (563) we see that
$W_{\psi}f\left(r,s;\kappa\right)=\int\frac{d^{2}z}{\pi}f\left(z\right)\psi_{r,s;\kappa}^{\ast}\left(z\right)$
can be considered as a convolution in the form

$\displaystyle\int
d^{2}zf\left(z\right)\psi^{\ast}\left[s\left(z-\kappa\right)-r\left(z^{\ast}-\kappa^{\ast}\right)\right]$
$\displaystyle=\int d^{2}\beta
F(\beta)\Phi^{\ast}\left(s\beta-r\beta^{\ast}\right)\exp\left(\kappa\beta^{\ast}-\kappa^{\ast}\beta\right)$
(574)

It then follows from (573) that

$\displaystyle\int
W_{\psi}f\left(r,s;\kappa\right)W_{\psi}f^{\prime\ast}\left(r,s;\kappa\right)d^{2}\kappa$
$\displaystyle=\left|s\right|\int\frac{d^{2}\beta
d^{2}\beta^{\prime}}{2\pi}F(\beta)\Phi^{\ast}\left(s\beta-r\beta^{\ast}\right)$
$\displaystyle\times
F^{\prime\ast}(\beta^{\prime})\Phi^{\prime}\left(s\beta^{\prime}-r\beta^{\prime\ast}\right)\delta\left(\beta-\beta^{\prime}\right)\delta\left(\beta^{\ast}-\beta^{\prime\ast}\right)$
$\displaystyle=\left|s\right|\int\frac{d^{2}\beta}{2\pi}F(\beta)F^{\prime\ast}(\beta)\left|\Phi\left(s\beta-r\beta^{\ast}\right)\right|^{2},$
(575)

Therefore, using (569) we see that the further integration yields

$\displaystyle\int\frac{d^{2}s}{\left|s\right|^{2}}\int
W_{\psi}f\left(r,s;\kappa\right)W_{\psi}f^{\prime\ast}\left(r,s;\kappa\right)d^{2}\kappa$
$\displaystyle=\int\frac{d^{2}\beta}{2\pi}F(\beta)F^{\prime\ast}(\beta)\int\frac{d^{2}s}{\left|s\right|}\left|\Phi\left(s\beta-r\beta^{\ast}\right)\right|^{2}$
$\displaystyle=\int\frac{d^{2}\beta}{2\pi}F(\beta)F^{\prime\ast}(\beta)=\int\frac{d^{2}z}{2\pi}f\left(z\right)f^{\prime\ast}\left(z\right),$
(576)

which completes the proof.

Theorem: From the Proposition (571) we have

$\int\int
W_{\psi}f\left(r,s;\kappa\right)\psi_{r,s;\kappa}\left(z\right)\frac{d^{2}\kappa
d^{2}s}{\left|s\right|^{2}}=f\left(z\right),$ (577)

that is, there exists an inversion formula for arbitrary function
$f\left(z\right)$. In fact, in Eq. (563) when we take
$f\left(z\right)=\delta\left(z-z^{\prime}\right),$ then

$W_{\psi}f\left(r,s;\kappa\right)=\int\frac{d^{2}z}{\pi}f\left(z\right)\psi_{r,s;\kappa}^{\ast}\left(z\right)=\psi_{r,s;\kappa}^{\ast}\left(z^{\prime}\right).$
(578)

Substituting (578) into (576) we obtain (577).

####  16.1.2 Relation between $W_{\psi}f\left(r,s;\kappa\right)$ and optical
Fresnel transform

Now we explain why the idea of $W_{\psi}f\left(r,s;\kappa\right)$ is
originated from the optical Fresnel transform. We can visualize the
symplectic-transformed—translated wavelet transform in the context of quantum
mechanics, letting $f\left(z\right)\equiv\left\langle
z\right|\left.f\right\rangle$, $\left\langle z\right|$ is the coherent state,
$\left|z\right\rangle=\exp\left[za^{\dagger}-z^{\ast}a\right]\equiv\left|\left(\begin{array}[c]{c}z\\\
z^{\ast}\end{array}\right)\right\rangle$, $|0\rangle$ is the vacuum state in
Fock space, then Eq. (562) can be expressed as

$\displaystyle W_{\psi}f\left(r,s;\kappa\right)$
$\displaystyle=\sqrt{s}\int\frac{d^{2}z}{\pi}\psi^{\ast}\left[s\left(z-\kappa\right)-r\left(z^{\ast}-\kappa^{\ast}\right)\right]f\left(z\right)$
$\displaystyle=\sqrt{s}\int\frac{d^{2}z}{\pi}\left\langle\psi\right|\left.\left(\begin{array}[c]{cc}s&-r\\\
-r^{\ast}&s^{\ast}\end{array}\right)\left(\begin{array}[c]{c}z-\kappa\\\
z^{\ast}-\kappa^{\ast}\end{array}\right)\right\rangle\left\langle
z\right|\left.f\right\rangle$ (583)
$\displaystyle=\left\langle\psi\right|F_{1}\left(r,s,\kappa\right)\left|f\right\rangle,$
(584)

where $F^{\left(r,s,\kappa\right)}$ is defined as

$F_{1}\left(r,s,\kappa\right)=\sqrt{s}\int\frac{d^{2}z}{\pi}\left|sz-
rz^{\ast}\right\rangle\left\langle z+\kappa\right|,\;\;$ (585)

and $\left|sz-
rz^{\ast}\right\rangle\equiv\left|\left(\begin{array}[c]{cc}s&-r\\\
-r^{\ast}&s^{\ast}\end{array}\right)\left(\begin{array}[c]{c}z\\\
z^{\ast}\end{array}\right)\right\rangle.$To know the explicit form of
$F_{1}\left(r,s,\kappa\right)$, we employ the normal ordering of the vacuum
projector $\left|0\right\rangle\left\langle
0\right|=:\exp\left(-a^{\dagger}a\right):$ and the IWOP technique to perform
the integration in (585), which leads to

$\displaystyle F_{1}\left(r,s,\kappa\right)$
$\displaystyle=\exp(\frac{1}{4}\left|\kappa\right|^{2}+\frac{r}{8s}\kappa^{\ast
2}+\frac{r^{\ast}\kappa^{2}}{8s^{\ast}})\exp\left(-\frac{r}{2s^{\ast}}a^{\dagger
2}-\frac{\kappa}{2s^{\ast}}\left(\left|s\right|^{2}+\left|r\right|^{2}\right)a^{\dagger}\right)$
$\displaystyle\exp\left[\left(a^{\dagger}a+\frac{1}{2}\right)\ln\frac{1}{s^{\ast}}\right]\exp\left(\frac{r^{\ast}}{2s^{\ast}}a^{2}-\frac{1}{2s^{\ast}}\left(s^{\ast}\kappa^{\ast}+r^{\ast}\kappa\right)a\right).$
(586)

The transformation matrix element of $F_{1}\left(r,s,\kappa=0\right)$ in the
coordinate representation $\left|x\right\rangle$ is just the kernel of optical
diffraction integration (101) (Fresnel transform), this explains our
motivation to introduce $W_{\psi}f\left(r,s;\kappa\right)$. In particular,
when $r^{\ast}=r\equiv\sinh\lambda$, $s=s^{\ast}\equiv\cosh\lambda,$
$F_{1}\left(r,s,\kappa=0\right)$ reduces to it the well-known single-mode
squeezing operator $\exp[\frac{\lambda}{2}\left(a^{2}-a^{\dagger 2}\right)]$
which corresponds to dilation in the usual WT.

###  16.2 Entangled SWT

In the above subsection, the mother wavelet is gained through a translating
transform followed by a symplectic transform. This motivation arises from the
consideration that symplectic transforms are more general than the dilated
transform, and are useful in Fresnel transform of Fourier optics, e.g. ray
transfer matrices of optical instruments, such as lenses and fibers in matrix
optics, while in quantum optics symplectic transforms correspond to single-
mode Fresnel operator (or generalized SU(1,1) squeezing operator).

Recalling that in section 9 we have introduced the 2-mode entangled Fresnel
operator which is a mapping of classical mixed transformation
$\left(z,z^{\prime}\right)\rightarrow\left(sz+rz^{\prime\ast},sz^{\prime}+rz^{\ast}\right)$
in 2-mode coherent state $\left|z,z^{\prime}\right\rangle$ representation onto
quantum operator $F_{2}\left(r,s\right)$, thus we are naturally led to develop
the SWT in (563) to the so-called entangled SWT (ESWT) [118] for signals
$g\left(z,z^{\prime}\right)$ defined in two complex planes,

$W_{\phi}g\left(r,s;k,k^{\prime}\right)={\displaystyle\iint}\frac{d^{2}zd^{2}z^{\prime}}{\pi^{2}}g\left(z,z^{\prime}\right)\phi_{r,s;k,k^{\prime}}^{\ast}\left(z,z^{\prime}\right),$
(587)

here

$\phi_{r,s;k.k^{\prime}}\left(z,z^{\prime}\right)\equiv
s^{\ast}\phi\left[s\left(z-k\right)+r\left(z^{\prime\ast}-k^{\prime\ast}\right),s\left(z^{\prime}-k^{\prime}\right)+r\left(z^{\ast}-k^{\ast}\right)\right],$
(588)

is used as a weighting function to synthesize the signal
$g\left(z,z^{\prime}\right)$ regarding to two complex planes. One can see that
the mother wavelet $\phi$ generates the family
$\phi^{\ast}\left[s\left(z-k\right)+r\left(z^{\prime\ast}-k^{\prime\ast}\right),s\left(z^{\prime}-k^{\prime}\right)+r\left(z^{\ast}-k^{\ast}\right)\right]$
through a translating transform followed by an entangled symplectic transform.
We emphasize that this transform mixes the two complex planes, which is
different from the tensor product of two independent transforms
$\left(z,z^{\prime}\right)\rightarrow\left[s\left(z-k\right)-r\left(z^{\ast}-k^{\ast}\right),s\left(z^{\prime}-k^{\prime}\right)-r\left(z^{\prime\ast}-k^{\prime\ast}\right)\right]$
given by (584). The new symplectic transform can be seen more clearly by
writing it in matrix form:

$\left(\begin{array}[c]{c}z-k\\\ z^{\ast}-k^{\ast}\\\ z^{\prime}-k^{\prime}\\\
z^{\prime\ast}-k^{{}^{\prime}\ast}\end{array}\right)\longrightarrow\mathcal{M}\left(\begin{array}[c]{c}z-k\\\
z^{\ast}-k^{\ast}\\\ z^{\prime}-k^{\prime}\\\
z^{\prime\ast}-k^{{}^{\prime}\ast}\end{array}\right)\text{,
}\mathcal{M}=\left[\begin{array}[c]{cccc}s&0&0&r\\\ 0&s^{\ast}&r^{\ast}&0\\\
0&r&s&0\\\ r^{\ast}&0&0&s^{\ast}\end{array}\right]$ (589)

where $\mathcal{M}$ is symplectic satisfying
$\mathcal{M}^{T}\mathcal{J}\mathcal{M}=\mathcal{J}$,
$\mathcal{J}=\left[\begin{array}[c]{cc}0&\mathcal{I}\\\
-\mathcal{I}&0\end{array}\right]$ , $\mathcal{I}$ is the $2\times 2$ unit
matrix.

For Eq. (587) being qualified as a new wavelet transform we must prove that it
possesses fundamental properties of the usual wavelet transforms, such as the
admissibility condition, the Parseval’s theorem and the inversion formula.
When $g\left(z,z^{\prime}\right)$ is the complex Fourier exponential,

$g_{1}\left(z,z^{\prime}\right)=\exp\left(z\beta^{\ast}-z^{\ast}\beta+z^{\prime}\gamma^{\ast}-z^{\prime\ast}\gamma\right),$
(590)

according to (587)-(588) we evaluate its ESWT

$\displaystyle W_{\phi}g_{1}$
$\displaystyle={\displaystyle\iint}\frac{d^{2}zd^{2}z^{\prime}}{\pi^{2}}\exp\left(z\beta^{\ast}-z^{\ast}\beta+z^{\prime}\gamma^{\ast}-z^{\prime\ast}\gamma\right)\phi_{r,s;k.k^{\prime}}^{\ast}\left(z,z^{\prime}\right)$
$\displaystyle=s{\displaystyle\iint}\frac{d^{2}zd^{2}z^{\prime}}{\pi^{2}}\phi^{\ast}\left[sz+rz^{\prime\ast},sz^{\prime}+rz^{\ast}\right]$
$\displaystyle\times\exp\left[\left(z+k\right)\beta^{\ast}-\left(z^{\ast}+k^{\ast}\right)\beta+\left(z^{\prime}+k^{\prime}\right)\gamma^{\ast}-\left(z^{\prime\ast}+k^{\prime\ast}\right)\gamma\right].$
(591)

Making the integration variables transform $sz+rz^{\prime\ast}\rightarrow w,$
$sz^{\prime}+rz^{\ast}\rightarrow w^{\prime},$ Eq. (591) becomes

$\displaystyle W_{\phi}g_{1}$
$\displaystyle=s\exp\left(k\beta^{\ast}-k^{\ast}\beta+k^{\prime}\gamma^{\ast}-k^{\prime\ast}\gamma\right){\displaystyle\iint}\frac{d^{2}wd^{2}w^{\prime}}{\pi^{2}}\phi^{\ast}\left(w,w^{\prime}\right)$
$\displaystyle\times\exp\left[w\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma\right)-w^{\ast}\left(s\beta+r\gamma^{\ast}\right)+w^{\prime}\left(s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)-w^{\prime\ast}\left(s\gamma+r\beta^{\ast}\right)\right],$
(592)

the last integration is just the complex Fourier transform (CFT) of
$\phi^{\ast},$ denoting it as $\Phi^{\ast},$ we have

$W_{\phi}g_{1}=s\exp\left(k\beta^{\ast}-k^{\ast}\beta+k^{\prime}\gamma^{\ast}-k^{\prime\ast}\gamma\right)\Phi^{\ast}\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right).$ (593)

Then we form the adjoint operation of (593),

$\displaystyle
W_{\phi}^{\ast}\left(W_{\phi}g_{1}\right)\left(z,z^{\prime}\right)$
$\displaystyle=s^{\ast}{\displaystyle\iint}\frac{d^{2}kd^{2}k^{\prime}}{\pi^{2}}\\{\left(W_{\phi}g\right)\left(r,s;k,k^{\prime}\right)\\}\phi\left[s\left(z-k\right)+r\left(z^{\prime\ast}-k^{\prime\ast}\right),s\left(z^{\prime}-k^{\prime}\right)+r\left(z^{\ast}-k^{\ast}\right)\right]$
$\displaystyle=\left|s\right|^{2}\Phi^{\ast}\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right){\displaystyle\iint}\frac{d^{2}kd^{2}k^{\prime}}{\pi^{2}}\exp\left(k\beta^{\ast}-k^{\ast}\beta+k^{\prime}\gamma^{\ast}-k^{\prime\ast}\gamma\right)$
$\displaystyle\times\phi\left[s\left(z-k\right)+r\left(z^{\prime\ast}-k^{\prime\ast}\right),s\left(z^{\prime}-k^{\prime}\right)+r\left(z^{\ast}-k^{\ast}\right)\right]$
$\displaystyle=\left|s\right|^{2}\Phi^{\ast}\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right){\displaystyle\iint}\frac{d^{2}kd^{2}k^{\prime}}{\pi^{2}}\phi\left[-sk-
rk^{\prime\ast},-sk^{\prime}-rk^{\ast}\right]$
$\displaystyle\times\exp\left[\left(k+z\right)\beta^{\ast}-\left(k^{\ast}+z^{\ast}\right)\beta+\left(k^{\prime}+z^{\prime}\right)\gamma^{\ast}-\left(k^{\prime\ast}+z^{\prime\ast}\right)\gamma\right]$
$\displaystyle=\left|s\right|^{2}\Phi^{\ast}\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)\exp\left(z\beta^{\ast}-z^{\ast}\beta+z^{\prime}\gamma^{\ast}-z^{\prime\ast}\gamma\right){\displaystyle\iint}\frac{d^{2}vd^{2}v^{\prime}}{\pi^{2}}\phi\left(v,v^{\prime}\right)$
$\displaystyle\times\exp\left[-v\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma\right)+v^{\ast}\left(s\beta+r\gamma^{\ast}\right)-v^{\prime}\left(s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)+v^{\prime\ast}\left(s\gamma+r\beta^{\ast}\right)\right],$
(594)

where the integration in the last line is just the CFT of $\phi$ (comparing
with (592)), thus (594) leads to

$W_{\phi}^{\ast}\left(W_{\phi}g_{1}\right)\left(z,z^{\prime}\right)=\left|s\right|^{2}\exp\left(z\beta^{\ast}-z^{\ast}\beta+z^{\prime}\gamma^{\ast}-z^{\prime\ast}\gamma\right)\left|\Phi\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)\right|^{2}.$ (595)

From Eq. (595) we have

$\displaystyle\int
d^{2}sW_{\phi}^{\ast}\left(W_{\phi}g_{1}\right)\left(z,z^{\prime}\right)/\left|s\right|^{4}$
$\displaystyle=\exp\left(z\beta^{\ast}-z^{\ast}\beta+z^{\prime}\gamma^{\ast}-z^{\prime\ast}\gamma\right)$
$\displaystyle\times\int
d^{2}s\left|\Phi\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)\right|^{2}/\left|s\right|^{2},$
(596)

which together with (590) lead to

$g_{1}\left(z,z^{\prime}\right)=\frac{\int
d^{2}sW_{\phi}^{\ast}\left(W_{\phi}g_{1}\right)\left(z,z^{\prime}\right)/\left|s\right|^{4}}{\int
d^{2}s\left|\Phi\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)\right|^{2}/\left|s\right|^{2}}.$
(597)

Eq. (597) implies that we should impose the normalization

$\int d^{2}s\left|\Phi\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)\right|^{2}/\left|s\right|^{2}=1,$
(598)

such that the reproducing process exists

$g_{1}\left(z,z^{\prime}\right)=\int
d^{2}sW_{\phi}^{\ast}\left(W_{\phi}g_{1}\right)\left(z,z^{\prime}\right)/\left|s\right|^{4}.$
(599)

(598) may be named the generalized admissibility condition. Now we can have
the corresponding _Parseval theorem_ : For any $g$ and $g^{\prime}$ we have

${\displaystyle\iiint}W_{\phi}g\left(r,s;k,k^{\prime}\right)W_{\phi}^{\ast}g^{\prime}\left(r,s;k,k^{\prime}\right)\frac{d^{2}kd^{2}k^{\prime}d^{2}s}{\left|s\right|^{4}}={\displaystyle\iint}d^{2}zd^{2}z^{\prime}g\left(z,z^{\prime}\right)g^{\prime\ast}\left(z,z^{\prime}\right).$
(600)

_Proof:_ Assuming $F\left(\beta,\gamma\right)$ and
$F^{\prime}\left(\beta,\gamma\right)$ be CFT of $g\left(z,z^{\prime}\right)$
and $g^{\prime}\left(z,z^{\prime}\right)$, respectively,

$F\left(\beta,\gamma\right)={\displaystyle\iint}\frac{d^{2}zd^{2}z^{\prime}}{\pi^{2}}g\left(z,z^{\prime}\right)\exp\left(z\beta^{\ast}-z^{\ast}\beta+z^{\prime}\gamma^{\ast}-z^{\prime\ast}\gamma\right),$
(601)

recalling the corresponding convolution theorem

$\displaystyle{\displaystyle\iint}d^{2}zd^{2}z^{\prime}g\left(\alpha-z,\alpha^{\ast}-z^{\ast};\alpha^{\prime}-z^{\prime},\alpha^{\prime\ast}-z^{\prime\ast}\right)g^{\prime}\left(z,z^{\prime}\right)$
$\displaystyle={\displaystyle\iint}d^{2}\beta d^{2}\gamma
F\left(\beta,\gamma\right)F^{\prime}\left(\beta,\gamma\right)\exp\left(\alpha^{\ast}\beta-\alpha\beta^{\ast}+\alpha^{\prime\ast}\gamma-\alpha^{\prime}\gamma^{\ast}\right),$
(602)

so from Eqs. (587) and (601)-(602) we see that
$W_{\phi}g\left(r,s;k,k^{\prime}\right)={\displaystyle\iint}\frac{d^{2}zd^{2}z\prime}{\pi^{2}}g\left(z,z^{\prime}\right)\phi_{r,s;k,k^{\prime}}^{\ast}\left(z,z^{\prime}\right)$
can be considered as a convolution in the form (noting that the CFT of
$\phi^{\ast}$ is $\Phi^{\ast},$ see (592)-(593))

$\displaystyle{\displaystyle\iint}d^{2}zd^{2}z^{\prime}g\left(z,z^{\prime}\right)\phi^{\ast}\left[s\left(z-k\right)+r\left(z^{\prime\ast}-k^{\prime\ast}\right),s\left(z^{\prime}-k^{\prime}\right)+r\left(z^{\ast}-k^{\ast}\right)\right]$
$\displaystyle={\displaystyle\iint}d^{2}\beta d^{2}\gamma
F\left(\beta,\gamma\right)\Phi^{\ast}\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)\exp\left(k\beta^{\ast}-k^{\ast}\beta+k^{\prime}\gamma^{\ast}-k^{\prime\ast}\gamma\right).$
(603)

Using Eq. (603) we calculate

$\displaystyle{\displaystyle\iint}W_{\phi}g\left(r,s;k,k^{\prime}\right)W_{\phi}^{\ast}g^{\prime}\left(r,s;k,k^{\prime}\right)d^{2}kd^{2}k^{\prime}$
$\displaystyle=\left|s\right|^{2}{\displaystyle\iiiint}d^{2}\beta d^{2}\gamma
d^{2}\beta^{\prime}d^{2}\gamma^{\prime}F\left(\beta,\gamma\right)\Phi^{\ast}\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)$ $\displaystyle\times
F^{\prime^{\ast}}\left(\beta^{\prime},\gamma^{\prime}\right)\Phi\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)\delta\left(\beta-\beta^{\prime}\right)\delta\left(\beta^{\ast}-\beta^{\prime\ast}\right)\delta\left(\gamma-\gamma^{\prime}\right)\delta\left(\gamma^{\ast}-\gamma^{\prime\ast}\right)$
$\displaystyle=\left|s\right|^{2}{\displaystyle\iint}d^{2}\beta d^{2}\gamma
F\left(\beta,\gamma\right)F^{\prime\ast}\left(\beta,\gamma\right)\left|\Phi\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)\right|^{2}.$ (604)

As a consequence of (598) and (604) the further integration yields

$\displaystyle\int\frac{d^{2}s}{\left|s\right|^{4}}{\displaystyle\iint}W_{\phi}g\left(r,s;k,k^{\prime}\right)W_{\phi}^{\ast}g^{\prime}\left(r,s;k,k^{\prime}\right)d^{2}kd^{2}k^{\prime}$
$\displaystyle={\displaystyle\iint}d^{2}\beta d^{2}\gamma
F\left(\beta,\gamma\right)F^{\prime\ast}\left(\beta,\gamma\right)\int
d^{2}s\left|\Phi\left(s^{\ast}\beta^{\ast}+r^{\ast}\gamma,\text{
}s^{\ast}\gamma^{\ast}+r^{\ast}\beta\right)\right|^{2}/\left|s\right|^{2}$
$\displaystyle={\displaystyle\iint}d^{2}\beta d^{2}\gamma
F\left(\beta,\gamma\right)F^{\prime\ast}\left(\beta,\gamma\right)={\displaystyle\iint}d^{2}zd^{2}z^{\prime}g\left(z,z^{\prime}\right)g^{\prime\ast}\left(z,z^{\prime}\right),$
(605)

which completes the proof.

_Inversion Formula_ :__ From Eq. (600) we have

$g\left(z,z^{\prime}\right)={\displaystyle\iiint}W_{\phi}g\left(r,s;k,k^{\prime}\right)\phi_{r,s;k,k^{\prime}}\left(z,z^{\prime}\right)\frac{d^{2}kd^{2}k^{\prime}d^{2}s}{\pi^{2}\left|s\right|^{4}},$
(606)

that is, there exists an inversion formula for $g\left(z,z^{\prime}\right)$
which represents the original signal $g\left(z,z^{\prime}\right)$ as a
superposition of wavelet functions $\phi_{r,s;k,k^{\prime}},$ with the value
of entangled wavelet transform $W_{\phi}g\left(r,s;k,k^{\prime}\right)$
serving as coefficients. In fact, in Eq. (587) when we take

$g\left(z,z^{\prime}\right)=\delta\left(z-u\right)\delta\left(z^{\ast}-u^{\ast}\right)\delta\left(z^{\prime}-u^{\prime}\right)\delta\left(z^{\prime\ast}-u^{\prime\ast}\right),$
(607)

then

$W_{\phi}g\left(r,s;k,k^{\prime}\right)=\frac{1}{\pi^{2}}\phi_{r,s;k,k^{\prime}}^{\ast}\left(u,u^{\prime}\right).$
(608)

Substituting (607)-(608) into (605), we obtain (606). We can visualize the
ESWT in the context of quantum mechanics, letting
$g\left(z,z^{\prime}\right)=\left\langle
z,z^{\prime}\right|\left.g\right\rangle\ $and using Eqs. (588)-(589), Eq.
(587) is expressed as

$\displaystyle W_{\phi}g\left(r,s;k,k^{\prime}\right)$
$\displaystyle=s{\displaystyle\iint}\frac{d^{2}zd^{2}z^{\prime}}{\pi^{2}}\phi^{\ast}\left[s\left(z-k\right)+r\left(z^{\prime\ast}-k^{\prime\ast}\right),s\left(z^{\prime}-k^{\prime}\right)+r\left(z^{\ast}-k^{\ast}\right)\right]g\left(z,z^{\prime}\right)$
$\displaystyle=s{\displaystyle\iint}\frac{d^{2}zd^{2}z^{\prime}}{\pi^{2}}\left\langle\phi\right.\left|\mathcal{M}\left(\begin{array}[c]{c}z-k\\\
z^{\ast}-k^{\ast}\\\ z^{\prime}-k^{\prime}\\\
z^{\prime\ast}-k^{\prime\ast}\end{array}\right)\right\rangle\left\langle
z,z^{\prime}\right|\left.g\right\rangle=\left\langle\phi\right|F_{2}\left(r,s;k,k^{\prime}\right)\left|g\right\rangle$
(613)

where $F_{2}\left(r,s;k,k^{\prime}\right)$ is defined as

$\begin{array}[c]{c}F_{2}\left(r,s;k,k^{\prime}\right)=s{\displaystyle\iint}\frac{d^{2}zd^{2}z^{\prime}}{\pi^{2}}\left|sz+rz^{\prime\ast},sz^{\prime}+rz^{\ast}\right\rangle\left\langle
z+k,z^{\prime}+k^{\prime}\right|,\\\
\left|sz+rz^{\prime\ast},sz^{\prime}+rz^{\ast}\right\rangle=\left|sz+rz^{\prime\ast}\right\rangle_{1}\otimes\left|sz^{\prime}+rz^{\ast}\right\rangle_{2}.\end{array}$
(614)

When $k=0$ and $k^{\prime}=0$, $F_{2}\left(r,s;k=k^{\prime}=0\right)$ is just
the 2-mode Fresnel operator.

Thus, we have extended the SWT of signals in one complex plane to ESWT of
signals defined in two complex planes, the latter is not the tensor product of
two independent SWTs, this generalization is inevitable, since it resembles
the extending from the single-mode squeezing transform (or Fresnel operator)
to the two-mode squeezing transform (or entangled Fresnel operator) in quantum
optics.

###  16.3 Symplectic-dilation mixed WT

Next we shall introduce a new kind of WT, i.e., symplectic-dilation mixed WT
[119]. Recalling that in Ref. [120] we have constructed a new entangled-
coherent state representation (ECSR) $\left|\alpha,x\right\rangle$,

$\displaystyle\left|\alpha,x\right\rangle$
$\displaystyle=\exp\left[-\frac{1}{2}x^{2}-\frac{1}{4}\left|\alpha\right|^{2}+(x+\frac{\alpha}{2})a_{1}^{\dagger}\right.$
$\displaystyle\left.+(x-\frac{\alpha}{2})a_{2}^{\dagger}-\frac{1}{4}(a_{1}^{\dagger}+a_{2}^{\dagger})^{2}\right]\left|00\right\rangle,$
(615)

which is the common eigenvector of the operator $\left(X_{1}+X_{2}\right)/2$
and $a_{1}-a_{2},$ i.e.,
$\left(a_{1}-a_{2}\right)\left|\alpha,x\right\rangle=\alpha\left|\alpha,x\right\rangle$
and
$\frac{1}{2}(X_{1}+X_{2})\left|\alpha,x\right\rangle=\frac{1}{\sqrt{2}}x\left|\alpha,x\right\rangle,$
where $X_{i}=\frac{1}{\sqrt{2}}(a_{i}+a_{i}^{\dagger})$ is the coordinate
operator, ($i=1,2)$. $\left|\alpha,x\right\rangle$ is complete,

$\int_{-\infty}^{\infty}\frac{\mathtt{d}x}{\sqrt{\pi}}\int\frac{\mathtt{d}^{2}\alpha}{2\pi}\left|\alpha,x\right\rangle\left\langle\alpha,x\right|=1,$
(616)

and exhibits partly non-orthogonal property (for $\alpha)$ and orthonormal
property (for $x),$

$\displaystyle\left\langle\alpha^{\prime},x^{\prime}\right.\left|\alpha,x\right\rangle$
$\displaystyle=\sqrt{\pi}\exp\left[-\frac{1}{4}(\left|\alpha\right|^{2}+\left|\alpha^{\prime}\right|^{2})+\frac{1}{2}\alpha\alpha^{\prime\ast}\right]\delta\left(x^{\prime}-x\right),$
(617)

so $\left|\alpha,x\right\rangle$ possess behavior of both the coherent state
and the entangled state. An interesting question is: Can we introduce a new
kind of continuous WT for which the $\left|\alpha,x\right\rangle$
representation underlies? The answer is affirmative. Our motivation of this
issue comes from the mixed lens-Fresnel transform in classical optics [121]
(see (287) below).

By synthesizing (429) and (563) and in reference to (616) we propose the mixed
WT for $g\left(\alpha,x\right)$ ($\alpha=\alpha_{1}+\mathtt{i}\alpha_{2}$):

$W_{\psi}g\left(s,r,\kappa;\mathrm{a},b\right)\equiv\int_{-\infty}^{\infty}\frac{\mathtt{d}x}{\sqrt{\pi}}\int\frac{\mathtt{d}^{2}\alpha}{2\pi}g\left(\alpha,x\right)\psi_{s,r,\kappa;\mathrm{a},b}^{\ast}\left(\alpha,x\right).$
(618)

where $\mathtt{d}^{2}\alpha=\mathtt{d}\alpha_{1}\mathtt{d}\alpha_{2},$ the
family of mother wavelet $\psi$ involves both the the symplectic transform of
$\alpha$ and the dilation-transform of $x$,

$\psi_{s,r,\kappa;\mathrm{a},b}\left(\alpha,x\right)=\sqrt{\frac{s^{\ast}}{\left|\mathrm{a}\right|}}\psi\left[s\left(\alpha-\kappa\right)-r\left(\alpha^{\ast}-\kappa^{\ast}\right),\frac{x-b}{\mathrm{a}}\right].$
(619)

Letting $g\left(\alpha,x\right)\equiv$
$\left\langle\alpha,x\right|\left.g\right\rangle,$ then (618) can be expressed
as quantum mechanical version

$W_{\psi}g\left(s,r,\kappa;\mathrm{a},b\right)=\left\langle\psi\right|U\left(s,r,\kappa;\mathrm{a},b\right)\left|g\right\rangle,$
(620)

where $U\left(s,r,\kappa;\mathrm{a},b\right)$ is defined as

$\displaystyle U\left(s,r,\kappa;\mathrm{a},b\right)$
$\displaystyle=\sqrt{\frac{s}{\left|\mathrm{a}\right|}}\int_{-\infty}^{\infty}\frac{\mathtt{d}x}{\sqrt{\pi}}\int\frac{\mathtt{d}^{2}\alpha}{2\pi}$
$\displaystyle\times\left|s\alpha-r\alpha^{\ast},\frac{x-b}{\mathrm{a}}\right\rangle\left\langle\alpha+\kappa,x\right|.$
(621)

$U\left(s,r,\kappa=0;\mathrm{a},b=0\right)$ is just the generalized squeezing
operator, which causes a lens-Fresnel mixed transform.

For Eq. (618) being qualified as a new WT we must prove that it possesses
fundamental properties of the usual WTs, such as the admissibility condition,
the Parseval theorem and the inversion formula. It is straightforward to
evaluate the transform (618) and its reciprocal transform when
$g\left(\alpha,x\right)$ is the exponential
$g_{1}\left(\alpha,x\right)=\exp\left(\alpha^{\ast}\beta-\alpha\beta^{\ast}-\mathtt{i}px\right),$

$\displaystyle W_{\psi}g_{1}$
$\displaystyle=\sqrt{\frac{s}{\left|\mathrm{a}\right|}}e^{\kappa^{\ast}\beta-\kappa\beta^{\ast}-\mathtt{i}pb}\int_{-\infty}^{\infty}\frac{\mathtt{d}x}{\sqrt{\pi}}\int\frac{\mathtt{d}^{2}\alpha}{2\pi}$
$\displaystyle\times\psi^{\ast}(s\alpha-r\alpha^{\ast},\frac{x}{\mathrm{a}})e^{\alpha^{\ast}\beta-\alpha\beta^{\ast}-\mathtt{i}px}.$
(622)

Making the integration variables transform $s\alpha-r\alpha^{\ast}\rightarrow
w,\frac{x}{\mathrm{a}}\rightarrow x^{\prime},$ leading to
$\mathtt{d}^{2}\alpha\rightarrow\mathtt{d}^{2}w$ and
$\int_{-\infty}^{\infty}\mathtt{d}x\rightarrow\left|\mathrm{a}\right|\int_{-\infty}^{\infty}\mathtt{d}x^{\prime}$,
(622) becomes

$W_{\psi}g_{1}=\sqrt{s\left|\mathrm{a}\right|}\Phi^{\ast}\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta,\text{
}\mathrm{a}p\right)e^{\kappa^{\ast}\beta-\kappa\beta^{\ast}-\mathtt{i}pb},$
(623)

where $\Phi^{\ast}$ is just the Fourier transform of $\psi^{\ast},$

$\displaystyle\Phi^{\ast}\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta,\text{
}\mathrm{a}p\right)$
$\displaystyle=\int_{-\infty}^{\infty}\frac{\mathtt{d}x^{\prime}}{\sqrt{\pi}}\int\frac{\mathtt{d}^{2}w}{2\pi}\psi^{\ast}\left(w,x^{\prime}\right)$
$\displaystyle\times
e^{w^{\ast}\left(s\beta-r\beta^{\ast}\right)-w\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta\right)-\mathtt{i}\mathrm{a}px^{\prime}}.$
(624)

Then we perform the adjoint WT of (618), using (619) and (623) we see

$\displaystyle W_{\psi}^{\ast}\left(W_{\psi}g_{1}\right)\left(\alpha,x\right)$
$\displaystyle=\sqrt{\frac{s^{\ast}}{\left|\mathrm{a}\right|}}\int_{-\infty}^{\infty}\frac{\mathtt{d}b}{\sqrt{\pi}}\int\frac{\mathtt{d}^{2}\kappa}{2\pi}W_{\psi}g_{1}$
$\displaystyle\times\psi\left[s\left(\alpha-\kappa\right)-r\left(\alpha^{\ast}-\kappa^{\ast}\right),\frac{x-b}{\mathrm{a}}\right]$
$\displaystyle=\left|s\right|\left|\mathrm{a}\right|g_{1}\left(\alpha,x\right)\Phi^{\ast}\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta,\text{
}\mathrm{a}p\right)\int_{-\infty}^{\infty}\frac{\mathtt{d}b^{\prime}}{\sqrt{\pi}}$
$\displaystyle\times\int\frac{\mathtt{d}^{2}\kappa^{\prime}}{2\pi}e^{\kappa^{\prime}\beta^{\ast}-\kappa^{\prime\ast}\beta+\mathtt{i}\mathrm{a}pb^{\prime}}\psi\left(s\kappa^{\prime}-r\kappa^{\prime\ast},b^{\prime}\right)$
$\displaystyle=\left|s\right|\left|\mathrm{a}\right|g_{1}\left(\alpha,x\right)\left|\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta,\text{
}\mathrm{a}p\right)\right|^{2}.$ (625)

From Eq. (625) we obtain

$\displaystyle\int_{-\infty}^{\infty}\frac{\mathtt{d}\mathrm{a}}{\mathrm{a}^{2}}\int\frac{\mathtt{d}^{2}s}{\left|s\right|^{2}}W_{\psi}^{\ast}\left(W_{\psi}g_{1}\right)\left(\alpha,x\right)$
$\displaystyle=g_{1}\left(\alpha,x\right)\int_{-\infty}^{\infty}\frac{\mathtt{d}\mathrm{a}}{\left|\mathrm{a}\right|}\int\frac{\mathtt{d}^{2}s}{\left|s\right|}\left|\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta,\text{
}\mathrm{a}p\right)\right|^{2},$ (626)

which leads to

$g_{1}\left(\alpha,x\right)=\frac{\int_{-\infty}^{\infty}\frac{\mathtt{d}\mathrm{a}}{\mathrm{a}^{2}}\int\frac{\mathtt{d}^{2}s}{\left|s\right|^{2}}W_{\psi}^{\ast}\left(W_{\psi}g_{1}\right)\left(\alpha,x\right)}{\int_{-\infty}^{\infty}\frac{\mathtt{d}\mathrm{a}}{\left|\mathrm{a}\right|}\int\frac{\mathtt{d}^{2}s}{\left|s\right|}\left|\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta,\text{
}\mathrm{a}p\right)\right|^{2}}.$ (627)

Eq. (627) implies that we should impose the normalization

$\int_{-\infty}^{\infty}\frac{\mathtt{d}\mathrm{a}}{\left|\mathrm{a}\right|}\int\frac{\mathtt{d}^{2}s}{\left|s\right|}\left|\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta,\text{
}\mathrm{a}p\right)\right|^{2}=1,$ (628)

such that the reproducing process exists

$g_{1}\left(\alpha,x\right)=\int_{-\infty}^{\infty}\frac{\mathtt{d}\mathrm{a}}{\mathrm{a}^{2}}\int\frac{\mathtt{d}^{2}s}{\left|s\right|^{2}}W_{\psi}^{\ast}\left(W_{\psi}g_{1}\right)\left(\alpha,x\right).$
(629)

(628) may be named the generalized admissibility condition. Now we can have
the corresponding _Parseval theorem_ : For any $g$ and $g^{\prime}$ we have

$\displaystyle\int_{-\infty}^{\infty}\frac{\mathtt{d}\mathrm{a\mathtt{d}}b}{\mathrm{a}^{2}}\int\frac{\mathtt{d}^{2}\kappa\mathtt{d}^{2}s}{\left|s\right|^{2}}W_{\psi}g\left(s,r,\kappa;\mathrm{a},b\right)W_{\psi}^{\ast}g^{\prime}\left(s,r,\kappa;\mathrm{a},b\right)$
$\displaystyle=\int_{-\infty}^{\infty}\mathtt{d}x\int\mathtt{d}^{2}\alpha
g\left(\alpha,x\right)g^{\prime\ast}\left(\alpha,x\right).$ (630)

_Proof:_ Assuming $F\left(\beta,p\right)$ and $F^{\prime}\left(\beta,p\right)$
be the Fourier transforms of $g\left(\alpha,x\right)$ and
$g^{\prime}\left(\alpha,x\right)$, respectively,

$F\left(\beta,p\right)=\int_{-\infty}^{\infty}\frac{\mathtt{d}x}{\sqrt{2\pi}}\int\frac{\mathtt{d}^{2}\alpha}{\pi}g\left(\alpha,x\right)e^{\alpha\beta^{\ast}-\alpha^{\ast}\beta+\mathtt{i}px},$
(631)

In order to prove (630), we first calculate
$W_{\psi}g\left(s,r,\kappa;\mathrm{a},b\right)$. In similar to deriving
Eq.(221), using (LABEL:14), (217) and the inversion formula of (631) we have

$\displaystyle W_{\psi}g\left(s,r,\kappa;\mathrm{a},b\right)$
$\displaystyle=\sqrt{\frac{s}{\left|\mathrm{a}\right|}}\int_{-\infty}^{\infty}\frac{\mathtt{d}p}{\sqrt{2\pi}}\int\frac{\mathtt{d}^{2}\beta}{\pi}F\left(\beta,p\right)\int_{-\infty}^{\infty}\frac{\mathtt{d}x}{\sqrt{\pi}}\int\frac{\mathtt{d}^{2}\alpha}{2\pi}$
$\displaystyle\times
e^{\alpha^{\ast}\beta-\alpha\beta^{\ast}-\mathtt{i}px}\psi^{\ast}\left[s\left(\alpha-\kappa\right)-r\left(\alpha^{\ast}-\kappa^{\ast}\right),\frac{x-b}{\mathrm{a}}\right]$
$\displaystyle=\sqrt{s\left|\mathrm{a}\right|}\int_{-\infty}^{\infty}\frac{\mathtt{d}p}{\sqrt{2\pi}}\int\frac{\mathtt{d}^{2}\beta}{\pi}F\left(\beta,p\right)$
$\displaystyle\times\Phi^{\ast}\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta,\text{
}\mathrm{a}p\right)e^{\kappa^{\ast}\beta-\kappa\beta^{\ast}-\mathtt{i}pb}.$
(632)

It then follows

$\displaystyle\int_{-\infty}^{\infty}\mathtt{d}b\int\mathtt{d}^{2}\kappa
W_{\psi}g\left(s,r,\kappa;\mathrm{a},b\right)W_{\psi}^{\ast}g^{\prime}\left(s,r,\kappa;\mathrm{a},b\right)$
$\displaystyle=\left|\mathrm{a}s\right|\int_{-\infty}^{\infty}\mathtt{d}p\mathtt{d}p^{\prime}\int\mathtt{d}^{2}\beta\mathtt{d}^{2}\beta^{\prime}F\left(\beta,p\right)F^{\prime\ast}\left(\beta^{\prime},p^{\prime}\right)$
$\displaystyle\times\Phi^{\ast}\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta,\text{
}\mathrm{a}p\right)\Phi\left(s^{\ast}\beta^{\prime\ast}-r^{\ast}\beta^{\prime},\mathrm{a}p^{\prime}\right)$
$\displaystyle\times\int_{-\infty}^{\infty}\frac{\mathtt{d}b}{2\pi}\int\frac{\mathtt{d}^{2}\kappa}{\pi^{2}}e^{\kappa^{\ast}\left(\beta-\beta^{\prime}\right)-\kappa\left(\beta^{\ast}-\beta^{\prime\ast}\right)+\mathtt{i}\left(p^{\prime}-p\right)b}$
$\displaystyle=\left|\mathrm{a}s\right|\int_{-\infty}^{\infty}\mathtt{d}p\int\mathtt{d}^{2}\beta
F\left(\beta,p\right)F^{\prime\ast}\left(\beta,p\right)\left|\Phi\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta,\mathrm{a}p\right)\right|^{2}.$
(633)

Substituting (633) into the left-hand side (LHS) of (630) and using (628) we
see

LHS of (271)
$\displaystyle=\int_{-\infty}^{\infty}\mathtt{d}p\int\mathtt{d}^{2}\beta
F\left(\beta,p\right)F^{\prime\ast}\left(\beta,p\right)$
$\displaystyle\times\int_{-\infty}^{\infty}\frac{\mathtt{d}\mathrm{a}}{\left|\mathrm{a}\right|}\int\frac{\mathtt{d}^{2}s}{\left|s\right|}\left|\Phi^{\ast}\left(s^{\ast}\beta^{\ast}-r^{\ast}\beta,\mathrm{a}p\right)\right|^{2}$
$\displaystyle=\int_{-\infty}^{\infty}\mathtt{d}p\int\mathtt{d}^{2}\beta
F\left(\beta,p\right)F^{\prime\ast}\left(\beta,p\right).$ (634)

Thus we complete the proof of Eq.(630).

_Inversion Formula_ :__ From Eq. (630) we have

$g\left(\alpha,x\right)=\int_{-\infty}^{\infty}\frac{\mathtt{d\mathrm{a}d}b}{\sqrt{\pi}\mathrm{a}^{2}}\int\frac{\mathtt{d}^{2}\kappa\mathtt{d}^{2}s}{2\pi\left|s\right|^{2}}W_{\psi}g\left(s,r,\kappa;\mathrm{a},b\right)\psi_{s,r,\kappa;\mathrm{a},b}\left(\alpha,x\right),$
(635)

that is the inversion formula for the original signal $g\left(\alpha,x\right)$
expressed by a superposition of wavelet functions
$\psi_{s,r,\kappa;\mathrm{a},b}\left(\alpha,x\right),$ with the value of
continuous WT $W_{\psi}g\left(s,r,\kappa;\mathrm{a},b\right)$ serving as
coefficients. In fact, in Eq. (618) when we take
$g\left(\alpha,x\right)=\delta\left(\alpha-\alpha^{\prime}\right)\delta\left(\alpha^{\ast}-\alpha^{\prime\ast}\right)\delta\left(x-x^{\prime}\right),$
then

$W_{\psi}g\left(s,r,\kappa;\mathrm{a},b\right)=\frac{1}{2\pi\sqrt{\pi}}\psi_{s,r,\kappa;\mathrm{a},b}^{\ast}\left(\alpha^{\prime},x^{\prime}\right).$
(636)

Substituting (636) into (634) yields (635).

We can visualize the new WT $W_{\psi}g\left(s,r,\kappa;\mathrm{a},b\right)$ in
the context of quantum optics. Noticing that the generalized squeezing
operator $U\left(s,r,\kappa=0;\mathrm{a},b=0\right)$ in (621) is an image of
the combined mapping of the classical real dilation transform $x\rightarrow$
$x/\mathrm{a}$ ($\mathrm{a}>0$) and the classical complex symplectic transform
$\left(\alpha,\alpha^{\ast}\right)\rightarrow\left(s\alpha-r\alpha^{\ast},s^{\ast}\alpha^{\ast}-r^{\ast}\alpha\right)$
in $\left|\alpha,x\right\rangle$ representation, one can use the technique of
integration within normal product of operators to perform the integration in
(621) to derive its explicit form (see Eq. (15) in Ref. [r12a]). The transform
matrix element of $U\left(s,r,\kappa=0;\mathrm{a},b=0\right)$ in the entangled
state representation $\left|\eta\right\rangle$ is

$\displaystyle\left\langle\eta\right|U\left(s,r,\kappa=0;\mathrm{a},b=0\right)\left|\eta^{\prime}\right\rangle$
$\displaystyle=\sqrt{\frac{s}{\mathrm{a}}}\int_{-\infty}^{\infty}\frac{\mathtt{d}x}{\sqrt{\pi}}\int\frac{\mathtt{d}^{2}\alpha}{2\pi}\left\langle\eta\right.\left|s\alpha-r\alpha^{\ast},\frac{x}{\mathrm{a}}\right\rangle\left\langle\alpha,x\right|\left.\eta^{\prime}\right\rangle.$
(637)

In Fock space $\left|\eta=\eta_{1}+\mathtt{i}\eta_{2}\right\rangle$ is two-
mode EPR entangled state in (29).

Then using (615) and (29), we obtain

$\left\langle\eta\right.\left|\alpha,x\right\rangle=\frac{1}{\sqrt{2}}\exp\left[-\frac{\alpha^{2}+\left|\alpha\right|^{2}}{4}-\frac{1}{2}\eta_{1}^{2}+\eta_{1}\alpha-\mathtt{i}\eta_{2}x\right].$
(638)

Substituting (638) into (637) and using (85), we obtain

$\displaystyle\left\langle\eta\right|U\left(s,r,\kappa=0;\mathrm{a},b=0\right)\left|\eta^{\prime}\right\rangle$
$\displaystyle=\frac{\pi}{\sqrt{\mathrm{a}}}\delta\left(\eta_{2}^{\prime}-\eta_{2}/\mathrm{a}\right)\frac{1}{\sqrt{2\mathtt{i}\pi
B}}$
$\displaystyle\times\exp\left[\frac{\mathtt{i}}{2B}\left(A\eta_{1}^{\prime
2}-2\eta_{1}\eta_{1}^{\prime}+D\eta_{1}^{2}\right)\right].$ (639)

which is just the kernel of a mixed lens$-$Fresnel transform, i.e., the
variable $\eta_{1}$ of the object experiences a generalized Fresnel transform,
while $\eta_{2}$ undergoes a lens transformation. Thus, based on
$\left|\alpha,x\right\rangle$ we have introduced SDWT which involves both the
real variable dilation-transform and complex variable symplectic transform,
corresponding to the lens-Fresnel mixed transform in classical optics.

##  17 Fresnel-Hadamard combinatorial transformation

In the theoretical study of quantum computer, of great importance is the
Hadamard transform. This operation is $n$ Hadamard gates acting in parallel on
$n$ qubits. The Hadamard transform produces an equal superposition of all
computational basis states. From the point of view of Deutsch-Jozsa quantum
algorithm, the Hadamard transform is an example of the $N=2^{n}$ quantum
Fourier transform, which can be expressed as [122]

$\left|j\right\rangle=\frac{1}{\sqrt{2^{n}}}{\displaystyle\sum\limits_{k=0}^{2^{n}-1}}e^{2\pi
ijk/2^{n}}\left|k\right\rangle.$ (640)

Now the continuous Hadamard transform, used to go from the coordinate basis
$\left|x\right\rangle$ to the momentum basis, is defined as[123]

$\mathfrak{F}\left|x\right\rangle=\frac{1}{\sqrt{\pi}\sigma}\int_{-\infty}^{\infty}dy\exp\left(2ixy/\sigma^{2}\right)\left|y\right\rangle,$
(641)

where $\sigma$ is the scale length. $\mathfrak{F}$ is named Hadamard operator.
Using the completeness of
$\int_{-\infty}^{\infty}dx\left|x\right\rangle\left\langle x\right|=1,$ we
have

$\mathfrak{F}=\frac{1}{\sqrt{\pi}\sigma}{\displaystyle\iint_{-\infty}^{\infty}}dxdy\exp\left(2ixy/\sigma^{2}\right)\left|y\right\rangle\left\langle
x\right|.$ (642)

The above two transforms (Fresnel transform and Hadmard transform) are
independent of each other, an interesting question thus naturally arises: can
we combine the two transforms together? To put it in another way, can we
construct a combinatorial operator which play the role of both Fresnel
transform and Hadmard transform for two independent optical modes? The answer
is affirmative, in this section we try to construct so-called Fresnel-Hadmard
combinatorial transform.

###  17.1 The Hadamard-Fresnel combinatorial operator

Based on the coherent-entangled representation $\left|\alpha,x\right\rangle$,
and enlightened by Eq. (96) and (641) we now construct the following ket-bra
integration [124]

$U=\frac{\sqrt{s}}{\sqrt{\pi}\sigma}\int\frac{d^{2}\alpha}{\pi}{\displaystyle\iint}dxdy\exp\left(2ixy/\sigma^{2}\right)\left|s\alpha-r\alpha^{\ast},y\right\rangle\left\langle\alpha,x\right|,$
(643)

we name $U$ the Hadamard-Fresnel combinatorial operator.

Substituting Eq.(641) into Eq.(643), and using the two-mode vacuum projector’s
normally ordered form $\left|00\right\rangle\left\langle
00\right|=\colon\exp\left[-a_{1}^{+}a_{1}-a_{2}^{+}a_{2}\right]\colon$ as well
as the IWOP technique we get

$U=\frac{\sqrt{s}}{\sqrt{\pi}\sigma}\colon\int\frac{d^{2}z}{\pi}A\left(z,z^{\ast}\right){\displaystyle\iint}dxdyB(x,y)e^{C}\colon,$
(644)

where

$C\equiv-\frac{\left(a_{1}^{+}+a_{2}^{+}\right)^{2}+\left(a_{1}+a_{2}\right)^{2}}{4}-a_{1}^{+}a_{1}-a_{2}^{+}a_{2},$
$B(x,y)\equiv\exp\left[-\frac{y^{2}+x^{2}}{2}+y\left(a_{1}^{+}+a_{2}^{+}\right)+x\left(a_{1}+a_{2}\right)+\frac{2ixy}{\sigma^{2}}\right],$

and

$A\left(z,z^{\ast}\right)\equiv\exp\left[-\frac{\left|sz-
rz^{\ast}\right|^{2}+\left|z\right|^{2}}{4}+\frac{sz-
rz^{\ast}}{2}\left(a_{1}^{\dagger}-a_{2}^{\dagger}\right)+\frac{z^{\ast}\left(a_{1}-a_{2}\right)}{2}\right],$

they are all within the normal ordering symbol $::$ . Now performing the
integration over $dxdy$ within $::$ and remembering that all creation
operators are commute with all annihilation operators (the essence of the IWOP
technique) so that they can be considered c-number during the integration, we
can finally obtain

$\displaystyle U$
$\displaystyle=\frac{1}{\sqrt{s^{\ast}}}\frac{4\sqrt{\pi}\sigma}{\sqrt{\sigma^{4}+4}}\colon\exp\left\\{-\frac{1}{2}\frac{r}{s^{\ast}}\left(\frac{a_{1}^{{\dagger}}-a_{2}^{{\dagger}}}{\sqrt{2}}\right)^{2}+\frac{\sigma^{4}-4}{2\left(\sigma^{4}+4\right)}\left(\frac{a_{1}^{{\dagger}}+a_{2}^{{\dagger}}}{\sqrt{2}}\right)^{2}\right.$
$\displaystyle+\left(\frac{1}{s^{\ast}}-1\right)\frac{a_{1}^{{\dagger}}-a_{2}^{{\dagger}}}{\sqrt{2}}\frac{a_{1}-a_{2}}{\sqrt{2}}+\left(\frac{4i\sigma^{2}}{\sigma^{4}+4}-1\right)\frac{a_{1}^{{\dagger}}+a_{2}^{{\dagger}}}{\sqrt{2}}\frac{a_{1}+a_{2}}{\sqrt{2}}$
$\displaystyle+\left.\frac{1}{2}\frac{r^{\ast}}{s^{\ast}}\left(\frac{a_{1}-a_{2}}{\sqrt{2}}\right)^{2}+\frac{\sigma^{4}-4}{2\left(\sigma^{4}+4\right)}\left(\frac{a_{1}+a_{2}}{\sqrt{2}}\right)^{2}\right\\}\colon,$
(645)

which is the normally ordered form of Hadamard-Fresnel combinatorial operator.

###  17.2 The properties of Hadamard-Fresnel operator

Note

$\left[\frac{a_{1}-a_{2}}{\sqrt{2}},\frac{a_{1}^{{\dagger}}+a_{2}^{{\dagger}}}{\sqrt{2}}\right]=0,$
(646)

and

$\left[\frac{a_{1}-a_{2}}{\sqrt{2}},\frac{a_{1}^{{\dagger}}-a_{2}^{{\dagger}}}{\sqrt{2}}\right]=1,\text{
\
}\left[\frac{a_{1}+a_{2}}{\sqrt{2}},\frac{a_{1}^{{\dagger}}+a_{2}^{{\dagger}}}{\sqrt{2}}\right]=1,$
(647)

$\frac{a_{1}-a_{2}}{\sqrt{2}}$ can be considered a mode independent of another
mode $\frac{a_{1}^{{\dagger}}+a_{2}^{{\dagger}}}{\sqrt{2}},$ thus we have the
operator identity

$\text{exp}\left[f\left(a_{1}^{{\dagger}}\pm
a_{2}^{{\dagger}}\right)\left(a_{1}\pm
a_{2}\right)\right]=\colon\exp[\frac{1}{2}\left(e^{2f}-1\right)\left(a_{1}^{{\dagger}}\pm
a_{2}^{{\dagger}}\right)\left(a_{1}\pm a_{2}\right)]\colon.$ (648)

Using (648) we can rewrite Eq.(647) as

$U=U_{2}U_{1}=U_{1}U_{2},$ (649)

where

$\displaystyle U_{1}$
$\displaystyle=\frac{4\sqrt{\pi}\sigma}{\sqrt{\sigma^{4}+4}}\exp\left[\frac{\sigma^{4}-4}{2\left(\sigma^{4}+4\right)}\left(\frac{a_{1}^{{\dagger}}+a_{2}^{{\dagger}}}{\sqrt{2}}\right)^{2}\right]$
$\displaystyle\exp\left[\frac{a_{1}^{{\dagger}}+a_{2}^{{\dagger}}}{\sqrt{2}}\frac{a_{1}+a_{2}}{\sqrt{2}}\ln\frac{4i\sigma^{2}}{\left(\sigma^{4}+4\right)}\right]\exp\left[\frac{\sigma^{4}-4}{2\left(\sigma^{4}+4\right)}\left(\frac{a_{1}+a_{2}}{\sqrt{2}}\right)^{2}\right]$
(650)

and

$\displaystyle U_{2}$
$\displaystyle=\exp\left[-\frac{r}{2s^{\ast}}\left(\frac{a_{1}^{{\dagger}}-a_{2}^{{\dagger}}}{\sqrt{2}}\right)^{2}\right]\exp\left[\left(\frac{a_{1}^{{\dagger}}-a_{2}^{{\dagger}}}{\sqrt{2}}\frac{a_{1}-a_{2}}{\sqrt{2}}+\frac{1}{2}\right)\ln\frac{1}{s^{\ast}}\right]$
$\displaystyle\exp\left[\frac{r^{\ast}}{2s^{\ast}}\left(\frac{a_{1}-a_{2}}{\sqrt{2}}\right)^{2}\right],$
(651)

while $U_{2}$ is the Fresnel operator for mode $\frac{a_{1}-a_{2}}{\sqrt{2}},$
$U_{1}$ is named the Hadamard operator for mode
$\frac{a_{1}+a_{2}}{\sqrt{2}}.$

It then follows

$\displaystyle U\frac{a_{1}-a_{2}}{\sqrt{2}}U^{-1}$
$\displaystyle=U_{2}\frac{a_{1}-a_{2}}{\sqrt{2}}U_{2}^{-1}=s^{\ast}\frac{a_{1}-a_{2}}{\sqrt{2}}+r\frac{a_{1}^{{\dagger}}-a_{2}^{{\dagger}}}{\sqrt{2}},$
$\displaystyle U\frac{a_{1}^{{\dagger}}-a_{2}^{{\dagger}}}{\sqrt{2}}U^{-1}$
$\displaystyle=U_{2}\frac{a_{1}^{{\dagger}}-a_{2}^{{\dagger}}}{\sqrt{2}}U_{2}^{-1}=r^{\ast}\frac{a_{1}-a_{2}}{\sqrt{2}}+s\frac{a_{1}^{{\dagger}}-a_{2}^{{\dagger}}}{\sqrt{2}},$
(652)

from which we see the Hadamard-Fresnel combinatorial operator can play the
role of Fresnel transformation for $\frac{a_{1}-a_{2}}{\sqrt{2}}.$ Physically,
$\frac{a_{1}-a_{2}}{\sqrt{2}}$ and $\frac{a_{1}+a_{2}}{\sqrt{2}}$ can be two
output fields of a beamsplitter.

In a similar way, we have

$\displaystyle U\frac{a_{1}+a_{2}}{\sqrt{2}}U^{-1}$
$\displaystyle=U_{1}\frac{a_{1}+a_{2}}{\sqrt{2}}U_{1}^{-1}=\frac{1}{4i\sigma^{2}}\left[\left(\sigma^{4}+4\right)\frac{a_{1}+a_{2}}{\sqrt{2}}-\left(\sigma^{4}-4\right)\frac{a_{1}^{{\dagger}}+a_{2}^{{\dagger}}}{\sqrt{2}}\right],$
$\displaystyle U\frac{a_{1}^{{\dagger}}+a_{2}^{{\dagger}}}{\sqrt{2}}U^{-1}$
$\displaystyle=U_{1}\frac{a_{1}^{{\dagger}}+a_{2}^{{\dagger}}}{\sqrt{2}}U_{1}^{-1}=\frac{1}{4i\sigma^{2}}\left[-\left(\sigma^{4}+4\right)\frac{a_{1}^{{\dagger}}+a_{2}^{{\dagger}}}{\sqrt{2}}+\left(\sigma^{4}-4\right)\frac{a_{1}+a_{2}}{\sqrt{2}}\right]$
(653)

which for the quadrature
$X_{i}=\left(a_{i}+a_{i}^{{\dagger}}\right)/\sqrt{2},$
$P_{i}=\left(a_{i}-a_{i}^{{\dagger}}\right)/\sqrt{2}i,$ $\left(i=1,2\right),$
leads to

$U\frac{X_{1}+X_{2}}{2}U^{-1}=\frac{\sigma^{2}}{4}\left(P_{1}+P_{2}\right),\text{
\ \
}U\left(P_{1}+P_{2}\right)U^{-1}=-\frac{4}{\sigma^{2}}\frac{X_{1}+X_{2}}{2},$
(654)

from which we see that the Hadamard-Fresnel combinatorial operator also plays
the role of exchanging the total momentum—average position followed by a
squeezing transform, with the squeezing parameter being
$\frac{\sigma^{2}}{4}.$

The mutual transform in (654) can be realized by

$\displaystyle
e^{i\frac{\pi}{2}\left(a_{1}^{{\dagger}}a_{1}+a_{2}^{{\dagger}}a_{2}\right)}\left(X_{1}+X_{2}\right)e^{-i\frac{\pi}{2}\left(a_{1}^{{\dagger}}a_{1}+a_{2}^{{\dagger}}a_{2}\right)}$
$\displaystyle=P_{1}+P_{2},\text{ }$ (655) $\displaystyle\text{\
}e^{i\frac{\pi}{2}\left(a_{1}^{{\dagger}}a_{1}+a_{2}^{{\dagger}}a_{2}\right)}\left(P_{1}+P_{2}\right)e^{-i\frac{\pi}{2}\left(a_{1}^{{\dagger}}a_{1}+a_{2}^{{\dagger}}a_{2}\right)}$
$\displaystyle=-\left(X_{1}+X_{2}\right)\text{\ }$ (656)

while the two-mode squeezing operator is
$S_{2}=\exp\left[\ln\frac{2}{\sigma^{2}}\left(a_{1}^{{\dagger}}a_{2}^{{\dagger}}-a_{1}a_{2}\right)\right],$
therefore

$U_{1}=S_{2}^{-1}e^{i\frac{\pi}{2}\left(a_{1}^{{\dagger}}a_{1}+a_{2}^{{\dagger}}a_{2}\right)}.$
(657)

From Eq.(649) and Eq.(657), we see that the Hadamard-Fresnel combinatorial
operator can be decomposed as

$U=U_{2}S_{2}^{-1}e^{i\frac{\pi}{2}\left(a_{1}^{{\dagger}}a_{1}+a_{2}^{{\dagger}}a_{2}\right)}=S_{2}^{-1}e^{i\frac{\pi}{2}\left(a_{1}^{{\dagger}}a_{1}+a_{2}^{{\dagger}}a_{2}\right)}U_{2}.$
(658)

It can be also seen that $U$ is unitary, $U^{+}U=UU^{+}=1$.

In this section, we have introduced the Fresnel-Hadamard combinatorial
operator by virtue of the IWOP technique. This unitary operator plays the role
of both Fresnel transformation for mode $\frac{a_{1}-a_{2}}{\sqrt{2}}$ and
Hadamard transformation for mode $\frac{a_{1}+a_{2}}{\sqrt{2}},$ respectively,
and the two transformations are combinatorial. We have shown that the two
transformations are concisely expressed in the coherent-entangled state
representation as a projective operator in integration form. We also found
that the Fresnel-Hadamard operator can be decomposed as
$U_{2}S_{2}^{-1}e^{i\frac{\pi}{2}\left(a_{1}^{{\dagger}}a_{1}+a_{2}^{{\dagger}}a_{2}\right)},$
a Fresnel operator $U_{2}$, a two-mode squeezing operator $S_{2}^{-1}$ and the
total momentum-average position exchanging operator. Physically,
$\frac{a_{1}-a_{2}}{\sqrt{2}}$ and $\frac{a_{1}+a_{2}}{\sqrt{2}}$ can be two
output fields of a beamsplitter. If an optical device can be designed for
Fresnel-Hadamard combinatorial transform, then it can be directly applied to
these two output fields of the beamsplitter.

In summary, although quantum optics and classical optics are so different, no
matter in the mathematical tools they employed or in a conceptual view
(quantum optics concerning the wave-particle duality of optical field with an
emphasis on its nonclassical properties, whereas classical optics works on the
distribution aqnd propagation of the light waves), that it is a new
exploration to link them systematically. However, In this review, via the
route of developing Dirac’s symbolic method we have revealed some links
between them by mapping classical symplectic transformation in the coherent
state representation onto quantum unitary operators (GFO), throughout our
discussion the IWOP technique is indispensable for the derivation. We have
resorted to the quantum optical interpretation of various classical optical
transformations by adopting quantum optics concepts such as the coherent
states, squeezed states, and entangled states, etc. Remarkably, we have
endowed complex fractional Fourier transform, Hankel transform with quantum
optical representation-transform interpretation. Our formalism, starting from
quantum optics theory, not only provides quantum mechanical account of various
classical optical transformations, but also have found their way back to some
new classical transformations, e.g. entangled Fresnel transform, Fresnel-
wavelet transform, etc, which may have realistic optical interpretation in the
future. As Dirac predicted, functions that have been applied in classical
optical problems may be translated in an operator language in quantum
mechanics, and vice-versa. We expect that the content of this work may play
some role in quantum states engineering, i.e., optical field states’
preparation and design.

Once the correspondence in this respect between the two distinct fields is
established, the power of Dirac’s symbolic method can be fully displayed to
solve some new problems in classical optics, e.g., to find new eigen-modes of
some optical transforms; to extend the research region of classical optics
theoretically by introducing new transforms (for example, the entangled
Fresnel transforms), which may bring attention of experimentalists, who may
get new ideas to implement these new classical optical transformations.

Acknowledgement: This work supported by the National Natural Science
Foundation of China, Grant No. 10775097 and 10874174, and a grant from the Key
Programs Foundation of Ministry of Education of China (No. 210115), and the
Research Foundation of the Education Department of Jiangxi Province of China
(grant no. GJJ10097).

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