Relation between Optical Fresnel transformation and quantum tomography in two-mode entangled case

TL;DR

This paper explores the relationship between two-mode optical Fresnel transformations and quantum tomography, showing how entangled states relate to Radon transforms of the two-mode Wigner function, thus linking optical transformations with quantum state reconstruction.

Contribution

It extends previous work by analyzing the two-mode entangled case, revealing how Fresnel transformations correspond to Radon transforms of the two-mode Wigner operator in quantum optics.

Findings

Fresnel transformation maps entangled states to Radon-transformed Wigner functions.

Probability distributions of Fresnel quadrature phase relate directly to quantum tomography.

Similar results are obtained in the frequency domain.

Abstract

Similar in spirit to the preceding work [Opt. Commun. 282 (2009) 3734] where the relation between optical Fresnel transformation and quantum tomography is revealed, we study this kind of relationship in the two-mode entangled case. We show that under the two-mode Fresnel transformation the bipartite entangled state density |eta><eta| becomes density operator F_2|eta><eta|F_2 ^{dag}=|eta>_{r,s}<eta|, which is just the Radon transform of the two-mode Wigner operator (sigma,gama) in entangled form, where F_2 is an two-mode Fresnel operator in quantum optics, and s,r are the complex-value expression of (A, B, C,D). So the probability distribution for the Fresnel quadrature phase is the {tomography (Radon transform of the two-mode Wigner function), correspondingly, {s,r}_<eta|phi>=<eta|F_2{dag}|phi>. Similarly, we find a simial conclusion in the `frequency` domain.

Equations

$${|\eta ⟩}_{s,rs,r}⟨\eta |=\pi \int d^2\gamma d^2\sigma \delta \left({\eta }_2-D{\sigma }_2+B{\gamma }_1\right)\delta \left({\eta }_1-D{\sigma }_1-B{\gamma }_2\right)\Delta \left(\sigma ,\gamma \right),$$

$$F_2|\xi ⟩⟨\xi |F_2^{†}={|\xi ⟩}_{s,rs,r}⟨\xi |=\pi \int d^2\sigma d^2\gamma \delta \left({\xi }_1-A{\sigma }_1-C{\gamma }_2\right)\delta \left({\xi }_2-A{\sigma }_2+C{\gamma }_1\right)\Delta \left(\sigma ,\gamma \right),$$

$$g\left(x^{\prime }\right)=\frac{1}{\sqrt{2\pi iB}}{\int }_{-\infty }^{\infty }\exp \left[\frac{i}{2B}\left(A{x}^2-2x^{\prime }x+D{x}^{\prime 2}\right)\right]f\left(x\right)dx.$$

$$F_1\left(r,s\right)$$

$$⟨x^{\prime }|F_1\left(r,s\right)|x⟩=\frac{1}{\sqrt{2\pi iB}}\exp \left[\frac{i}{2B}\left(A{x}^2-2x^{\prime }x+D{x}^{\prime 2}\right)\right],$$

$$s=\frac{1}{2}\left[A+D-i\left(B-C\right)\right],r=-\frac{1}{2}\left[A-D+i\left(B+C\right)\right],$$

$$g\left(x^{\prime }\right)={\int }_{-\infty }^{\infty }⟨x^{\prime }|F_1\left(r,s\right)|x⟩⟨x|f⟩dx=⟨x^{\prime }|F_1\left(r,s\right)|f⟩,$$

$$F_1|x⟩⟨x|F_1^{†}={|x⟩}_{s,rs,r}⟨x|={\int }_{-\infty }^{\infty }d{p}^{\prime }d{x}^{\prime }\delta \left[x-\left(D{x}^{\prime }-B{p}^{\prime }\right)\right]\Delta \left(x^{\prime },p^{\prime }\right).$$

$${|x⟩}_{s,r}=\frac{{\pi }^{-1/4}}{\sqrt{D+iB}}\exp \left\{-\frac{A-iC}{D+iB}\frac{x^2}{2}+\frac{\sqrt{2}x}{D+iB}{a}^{†}-\frac{D-iB}{D+iB}\frac{a^{†2}}{2}\right\}|0⟩.$$

$${\mathcal{K}}_2\left({\eta }^{\prime },\eta \right)=\frac{1}{2iB\pi }\exp \left[\frac{i}{2B}(A{\left|\eta \right|}^2-\left(\eta {\eta }^{\prime \ast }+{\eta }^{\ast }{\eta }^{\prime }\right)+D{\left|{\eta }^{\prime }\right|}^2)\right],$$

$$F_2|\eta ⟩⟨\eta |F_2^{†}={|\eta ⟩}_{s,rs,r}⟨\eta |=\pi \int d^2\gamma d^2\sigma \delta \left({\eta }_2-D{\sigma }_2+B{\gamma }_1\right)\delta \left({\eta }_1-D{\sigma }_1-B{\gamma }_2\right)\Delta \left(\sigma ,\gamma \right),$$

$$F_2\left(r,s\right)=s\int \frac{d^2{z}_1{d}^2{z}_2}{{\pi }^2}|s{z}_1+r{z}_2^{\ast },r{z}_1^{\ast }+s{z}_2⟩⟨z_1,z_2|,$$

$$|s{z}_1+r{z}_2^{\ast },r{z}_1^{\ast }+s{z}_2⟩\equiv {|s{z}_1+r{z}_2^{\ast }⟩}_1\otimes {|r{z}_1^{\ast }+s{z}_2⟩}_2,$$

$$F_2\left(r,s\right)$$

$$+\left(s{z}_1+r{z}_2^{\ast }\right)a_1^{†}+\left(r{z}_1^{\ast }+s{z}_2\right)a_2^{†}+z_1^{\ast }{a}_1+z_2^{\ast }{a}_2-a_1^{†}{a}_1-a_2^{†}{a}_2]:$$

$$=\frac{1}{s^{\ast }}\exp \left(\frac{r}{s^{\ast }}{a}_1^{†}{a}_2^{†}\right):\exp \left[\left(\frac{1}{s^{\ast }}-1\right)\left(a_1^{†}{a}_1+a_2^{†}{a}_2\right)\right]:\exp \left(-\frac{r^{\ast }}{s^{\ast }}{a}_1{a}_2\right)$$

$$=\exp \left(\frac{r}{s^{\ast }}{a}_1^{†}{a}_2^{†}\right)\exp [\left(a_1^{†}{a}_1+a_2^{†}{a}_2+1\right)\ln {\left(s^{\ast }\right)}^{-1}]\exp \left(-\frac{r^{\ast }}{s^{\ast }}{a}_1{a}_2\right).$$

$$F_2\left(r,s\right)a_1{F}_2^{-1}\left(r,s\right)=s^{\ast }{a}_1-r{a}_2^{†},F_2\left(r,s\right)a_2{F}_2^{-1}\left(r,s\right)=s^{\ast }{a}_2-r{a}_1^{†},$$

$$F_2\left(r,s\right)F_2\left(r^{\prime },s^{\prime }\right)$$

$$=s{s}^{\prime }\int \frac{d^2{z}_1{d}^2{z}_2{d}^2{z}_1^{\prime }{d}^2{z}_2^{\prime }}{{\pi }^4}:\exp \{-|s{|}^2\left(|z_1{|}^2+|z_2{|}^2\right)-r^{\ast }s{z}_1{z}_2$$

$$-r{s}^{\ast }{z}_1^{\ast }{z}_2^{\ast }-\frac{1}{2}[|z_1^{\prime }{|}^2+|z_2^{\prime }{|}^2+|s^{\prime }{z}_1^{\prime }+r^{\prime }{z}_2^{\prime \ast }{|}^2+|r^{\prime }{z}_1^{\prime \ast }+s^{\prime }{z}_2^{\prime }{|}^2]$$

$$+\left(s{z}_1+r{z}_2^{\ast }\right)a_1^{†}+\left(r{z}_1^{\ast }+s{z}_2\right)a_2^{†}+z_1^{\prime \ast }{a}_1+z_2^{\prime \ast }{a}_2$$

$$+z_1^{\ast }\left(s^{\prime }{z}_1^{\prime }+r^{\prime }{z}_2^{\prime \ast }\right)+z_2^{\ast }\left(r^{\prime }{z}_1^{\prime \ast }+s^{\prime }{z}_2^{\prime }\right)-a_1^{†}{a}_1-a_2^{†}{a}_2\}:$$

$$=\frac{1}{s^{\prime \prime \ast }}\exp \left(\frac{r^{\prime \prime }}{2s^{\prime \prime \ast }}{a}_1^{†}{a}_2^{†}\right):\exp \left\{\left(\frac{1}{s^{\prime \prime \ast }}-1\right)\left(a_1^{†}{a}_1+a_2^{†}{a}_2\right)\right\}:\exp \left(-\frac{r^{\prime \prime \ast }}{2s^{\prime \prime \ast }}{a}_1{a}_2\right)$$

$$=F_2\left(r^{\prime \prime },s^{\prime \prime }\right),$$

$$\left(\begin{array}[]{cc}s^{\prime\prime}&-r^{\prime\prime}\\ -r^{\ast\prime\prime}&s^{\ast\prime\prime}\end{array}\right)=\left(\begin{array}[]{cc}s&-r\\ -r^{\ast}&s^{\ast}\end{array}\right)\left(\begin{array}[]{cc}s^{\prime}&-r^{\prime}\\ -r^{\prime\ast}&s^{\prime\ast}\end{array}\right).$$

$$|\eta ⟩=\exp \left[-\frac{1}{2}{\left|\eta \right|}^2+\eta a_1^{†}-{\eta }^{\ast }{a}_2^{†}+a_1^{†}{a}_2^{†}\right]|00⟩,$$

$$\left(Q_1-Q_2\right)|\eta ⟩=\sqrt{2}{\eta }_1|\eta ⟩,\left(P_1+P_2\right)|\eta ⟩=\sqrt{2}{\eta }_2|\eta ⟩,$$

$$⟨z_1,z_2|\eta ⟩=\exp \left[-\frac{1}{2}({\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|\eta \right|}^2)+\eta z_1^{\ast }-{\eta }^{\ast }{z}_2^{\ast }+z_1^{\ast }{z}_2^{\ast }\right],$$

$$⟨z_1^{\prime },z_2^{\prime }|F_2\left(r,s\right)|z_1,z_2⟩$$

$$+\frac{r}{s^{\ast }}{z}_1^{\prime \ast }{z}_2^{\prime \ast }-\frac{r^{\ast }}{s^{\ast }}{z}_1{z}_2+\frac{1}{s^{\ast }}\left(z_1^{\prime \ast }{z}_1+z_2^{\prime \ast }{z}_2\right)\},$$

$${\mathcal{K}}_2^{\left(r,s\right)}\left({\eta }^{\prime },\eta \right)$$

$$=\int \frac{d^2{z}_1{d}^2{z}_2{d}^2{z}_1^{\prime }{d}^2{z}_2^{\prime }}{{\pi }^5}⟨{\eta }^{\prime }|z_1^{\prime },z_2^{\prime }⟩⟨z_1^{\prime },z_2^{\prime }|F_2\left(r,s\right)|z_1,z_2⟩⟨z_1,z_2|\eta ⟩$$

$$=\frac{1}{s^{\ast }}\int \frac{d^2{z}_1{d}^2{z}_2{d}^2{z}_1^{\prime }{d}^2{z}_2^{\prime }}{{\pi }^5}\exp \left[-({\left|z_1\right|}^2+{\left|z_2\right|}^2+{\left|z_1^{\prime }\right|}^2+{\left|z_2^{\prime }\right|}^2)-\frac{1}{2}({\left|{\eta }^{\prime }\right|}^2+{\left|\eta \right|}^2)\right]$$

$$\times \exp \left[z_1^{\ast }{z}_2^{\ast }+\eta z_1^{\ast }-\frac{r^{\ast }}{s^{\ast }}{z}_1{z}_2+\frac{z_1^{\prime \ast }{z}_1+z_2^{\prime \ast }{z}_2}{s^{\ast }}+\frac{r}{s^{\ast }}{z}_1^{\prime \ast }{z}_2^{\prime \ast }+z_1^{\prime }{z}_2^{\prime }+{\eta }^{\prime \ast }{z}_1^{\prime }-{\eta }^{\prime }{z}_2^{\prime }-{\eta }^{\ast }{z}_2^{\ast }\right]$$

$$=\frac{1}{\left(r^{\ast }+s^{\ast }-r-s\right)\pi }\exp \left[\frac{\left(r^{\ast }-s\right){\left|\eta \right|}^2-\left(r+s\right){\left|{\eta }^{\prime }\right|}^2+\eta {\eta }^{\prime \ast }+{\eta }^{\ast }{\eta }^{\prime }}{r^{\ast }+s^{\ast }-r-s}-\frac{{\left|{\eta }^{\prime }\right|}^2+{\left|\eta \right|}^2}{2}\right].$$

$${\mathcal{K}}_2^{\left(r,s\right)}\left({\eta }^{\prime },\eta \right)=\frac{1}{2iB\pi }\exp \left[\frac{i}{2B}\left(A{\left|\eta \right|}^2-\left(\eta {\eta }^{\prime \ast }+{\eta }^{\ast }{\eta }^{\prime }\right)+D{\left|{\eta }^{\prime }\right|}^2\right)\right]\equiv {\mathcal{K}}_2^M\left({\eta }^{\prime },\eta \right),$$

Full text

[TABLE]

where $F_{2}$ is an two-mode Fresnel operator in quantum optics, and $s,r$ are the complex-value expression of ${\small(}A,B,C,D).$ So the probability distribution for the Fresnel quadrature phase is the tomography (Radon transform of the two-mode Wigner function), correspondingly, ${}_{s,r}\left\langle\eta\right|\left.\psi\right\rangle=\left\langle\eta\right|F_{2}^{\dagger}\left|\psi\right\rangle.$ Similarly, we find

[TABLE]

where $\left|\xi\right\rangle$ is the conjugated state to $\left|\eta\right\rangle$.

1 Introduction

In Ref.[1], by using the technique of integration within an ordered product (IWOP) of operators and the coherent state representation [2, 3] we have proved that corresponding to optical Fresnel transformation characteristic of ray transfer matrix elements $(A,B,C,D),$ $AD-BC=1$, connecting the input light field $f\left(x\right)$ and output light field $g\left(x^{\prime}\right)$ by Fresnel integration [4, 5, 6]

[TABLE]

there exists Fresnel operator $F_{1}(r,s)$ in quantum optics [7],

[TABLE]

such that in the coordinate $\left\langle x\right|$ representation[1, 8]

[TABLE]

where $(s,r),$ $\left|s\right|^{2}-\left|r\right|^{2}=1$, are related to a classical ray transfer matrix \left(\begin{array}[]{cc}A&B\\ C&D\end{array}\right) by

[TABLE]

the unimodularity condition $AD-BC=1$ is equivalent to $\left|s\right|^{2}-\left|r\right|^{2}=1.$ If we let $f\left(x\right)=\left\langle x\right|\left.f\right\rangle$, then Eq.(1) is expressed as

[TABLE]

which is just the quantum mechanical version of Fresnel transformation.

In a preceding paper [9], we also found that under the Fresnel transformation the pure position density $\left|x\right\rangle\left\langle x\right|$ becomes the tomographic density $\left|x\right\rangle_{s,rs,r}\left\langle x\right|$, which is just the Radon transform of the Wigner operator $\Delta\left(x,p\right),$ i.e.,

[TABLE]

So the probability distribution for the Fresnel quadrature phase is the tomography (Radon transform of Wigner function [10, 11, 12]), and the tomogram of a state $\left|\psi\right\rangle$ is just the wave function of its Fresnel transformed state $F_{1}^{\dagger}\left|\psi\right\rangle,$ i.e. ${}_{s,r}\left\langle x\right|\left.\psi\right\rangle=\left\langle x\right|F_{1}^{\dagger}\left|\psi\right\rangle,$ and $\left|x\right\rangle_{s,r}$ is expressed as

[TABLE]

In this Communication we want to generalize the above conclusion to two-mode entangled case. Firstly, we extend Eq. (3) to the two-dimensional Fresnel transformation,

[TABLE]

where $\eta$ is a complex number, then we construct the two-mode Fresnel operator $F_{2}\left(r,s\right)$ such that its transformation matrix element in the entangled state $\left|\eta\right\rangle$ representation (see below Eq.(16)) is just the two-dimensional Fresnel transformation, i.e., $\mathcal{K}_{2}^{\left(r,s\right)}\left(\eta^{\prime},\eta\right)=\frac{1}{\pi}\left\langle\eta^{\prime}\right|F_{2}\left(r,s\right)\left|\eta\right\rangle,$ then we shall prove

[TABLE]

i.e., we show that $\left|\eta\right\rangle_{s,rs,r}\left\langle\eta\right|$ is just the Radon transform of the entangled Wigner operator $\Delta\left(\sigma,\gamma\right)$

Our paper is arranged as follows. In section 2, we briefly review the two-mode Fresnel operator $F_{2}\left(r,s\right)$ and then derive the 2D Fresnel transformation in entangled state $\left|\eta\right\rangle$ representation and introduce a new representation $\left|\eta\right\rangle_{s,r}(=F_{2}\left(r,s\right)\left|\eta\right\rangle)$ in section 3. Section 4 is devoted to proving Eq.(9), i.e., $\left|\eta\right\rangle_{s,r\text{ }s,r}\left\langle\eta\right|$ as Radon transform of entangled Wigner operator. Similar discussions are moved to the Fresnel transformation in its ‘frequency domain’ in section 5. In the last section, we derive the inverse Radon transformation of entangled Wigner operator.

2 Two-mode Fresnel operator

Similar in spirit to the single-mode case, we introduce the two-mode Fresnel operator $F_{2}\left(r,s\right)$ through the following 2-mode coherent state representation [1, 13], i.e.,

[TABLE]

which indicates that $F_{2}\left(r,s\right)$ is a mapping of classical symplectic transform $\left(z_{1},z_{2}\right)\rightarrow\left(sz_{1}+rz_{2}^{\ast},rz_{1}^{\ast}+sz_{2}\right)$ in phase space, where $\left|z_{1},z_{2}\right\rangle=\exp\left\{-\frac{1}{2}\left|z_{1}\right|^{2}-\frac{1}{2}\left|z_{2}\right|^{2}+z_{1}a_{1}^{{\dagger}}+z_{1}a_{1}^{{\dagger}}\right\}\left|00\right\rangle$ is a usual two-mode coherent state. Concretely, the ket in (10) is

[TABLE]

$s$ and $r$ are complex and satisfy the unimodularity condition $\left|s\right|^{2}-\left|r\right|^{2}=1$. Using the IWOP technique [14, 15] and the normal ordering of the vacuum projector $\left|00\right\rangle\left\langle 00\right|=\colon\exp\left(-a_{1}^{\dagger}a_{1}-a_{2}^{\dagger}a_{2}\right)\colon,$ we perform the integral in (10) and obtain

[TABLE]

Thus $F_{2}\left(r,s\right)$ induces the transforms

[TABLE]

and $F_{2}$ is actually a generalized 2-mode squeezing operator [16, 17].

$F_{2}\left(r,s\right)$ abides by the group multiplication rule. Using the IWOP technique and (10) we obtain

[TABLE]

where $\left(r^{\prime\prime},s^{\prime\prime}\right)$ are given by

[TABLE]

Therefore, Eq.(14) is a loyal representation of the multiplication rule for ray transfer matrices in the sense of Matrix Optics.

3 Two-mode Fresnel transformation in entangled state representations

By introducing the bipartite entangled state $\left|\eta\right\rangle$ [18, 19]

[TABLE]

$\left|\eta=\eta_{1}+i\eta_{2}\right\rangle$ is the common eigenstate of relative coordinate $Q_{1}-Q_{2}$ and the total momentum $P_{1}+P_{2}$, i.e.,

[TABLE]

where $Q_{i}=\left(a_{i}+a_{i}^{\dagger}\right)/\sqrt{2},\ P_{i}=\left(a_{i}-a_{i}^{\dagger}\right)/(\mathtt{i}\sqrt{2}),$ $\left(i=1,2.\right),$ are coordinate and momentum operators, respectively. $\left|\eta\right\rangle$ compose a complete set $\int\frac{d^{2}\eta}{\pi}\left|\eta\right\rangle\left\langle\eta\right|=1,$ then using the over-completeness relation of the coherent state and

[TABLE]

as well as

[TABLE]

we can calculate the integral kernel

[TABLE]

Using the relation between $s,r$ and $\left(A,B,C,D\right)$ in Eq.(4) we see that Eq. (20) becomes

[TABLE]

where the superscript $M$ only means the parameters of $\mathcal{K}_{2}^{M}$ are $\left[A,B;C,D\right]$, and the subscript $2$ implies the two-dimensional kernel.

Operating $F_{2}\left(r,s\right)$ on $\left|\eta\right\rangle$ and using Eqs.(12) and (18) yields

[TABLE]

or

[TABLE]

where we have used the integration formula

[TABLE]

Noticing the completeness relation and the orthogonality of $\left|\eta\right\rangle$ we immediately derive

[TABLE]

a generalized entangled state representation $\left|\eta\right\rangle_{s,r}$ with the completeness relation (25). From (23) we can see that

[TABLE]

so we have the eigen-equations for $\left|\eta\right\rangle_{s,r}$ as follows

[TABLE]

We can also check Eqs.(26)-(29) by another way (see Appendix).

4 $\left|\eta\right\rangle_{s,r\text{ }s,r}\left\langle\eta\right|$ as the Radon transform of

entangled Wigner operator

For two-mode correlated system, it is convenient to express the Wigner operator in the $\left|\eta\right\rangle$ representation as [20, 21, 22, 23]

[TABLE]

When $\sigma=\alpha-\beta^{\ast},\;\gamma=\alpha+\beta^{\ast}$, Eq. (30) is just equal to the direct product of two single-mode Wigner operators, i.e., $\Delta\left(\sigma,\gamma\right)=\Delta\left(\alpha,\alpha^{\ast}\right)\otimes\Delta\left(\beta,\beta^{\ast}\right).$ Then according to the Wely correspondence rule [24]

[TABLE]

where $h\left(\sigma,\gamma\right)$ is the Weyl correspondence of $H\left(a_{1}^{\dagger},a_{2}^{\dagger};a_{1},a_{2}\right),$ and

[TABLE]

the classical Weyl correspondence of the projection operator $\left|\eta\right\rangle_{r,sr,s}\left\langle\eta\right|$ can be calculated,

[TABLE]

Then using Eq.(20) we have

[TABLE]

which means the following Weyl correspondence

[TABLE]

so the projector operator $\left|\eta\right\rangle_{s,rs,r}\left\langle\eta\right|$ is just the Radon transformation of $\Delta\left(\sigma,\gamma\right)$, $D$ and $B$ are the Radon transformation parameter. Combining Eqs. (22)-(35) together we complete the proof (9). Therefore, the quantum tomography in two-mode entangled case is expressed as

[TABLE]

where $\left\langle\psi\right|\Delta\left(\sigma,\gamma\right)\left|\psi\right\rangle$ is the Wigner function. So the probability distribution for the Fresnel quadrature phase **(**see Eq. (A11) in the Appendix) is the tomography (Radon transform of the two-mode Wigner function). This is the main result of the present paper. This new relation between quantum tomography and optical Fresnel transform may provide experimentalists to figure out new approach for generating tomography.

5 In the conjugate representation

Next we turn to the “frequency” domain, that is to say, we shall prove that the $(A,C)$ related Radon transform of entangled Wigner operator $\Delta\left(\sigma,\gamma\right)$ is just the pure state density operator $\left|\xi\right\rangle_{s,rs,r}\left\langle\xi\right|,$ i.e.,

[TABLE]

where

[TABLE]

is an entangled state conjugate to $\left|\eta\right\rangle.$ By analogy with the above procedure, we obtain the 2-dimensional Fresnel transformation in its ‘frequency domain’, i.e.,

[TABLE]

where the superscript $N$ means that this transform kernel corresponds to the parameter matrix $N=\left[D,-C,-B,A\right]$. Thus the 2D Fresnel transformation in its ‘frequency domain’ is given by

[TABLE]

Operating $F_{2}\left(r,s\right)$ on $\left|\xi\right\rangle$ we have (also see Appendix)

[TABLE]

or

[TABLE]

Noticing that the entangled Wigner operator in $\left\langle\xi\right|$ representation is expressed as

[TABLE]

and using the classical correspondence of $\left|\xi\right\rangle_{s,rs,r}\left\langle\xi\right|$ which is calculated by

[TABLE]

we obtain

[TABLE]

so the projector operator $\left|\xi\right\rangle_{s,r\text{ }s,r}\left\langle\xi\right|$ is another Radon transformation of the two-mode Wigner operator, with $A$ and $C$ being the Radon transformation parameter (‘frequency’ domain). Therefore, the quantum tomography in ${}_{s,r}\left\langle\xi\right|$ representation is expressed as the Radon transformation of the Wigner function

[TABLE]

and ${}_{s,r}\left\langle\xi\right|=\left\langle\xi\right|F^{\dagger}.$

6 Inverse Radon transformation

Now we consider the inverse Radon transformation. For instance, using (35) we see the Fourier transformation of $\left|\eta\right\rangle_{s,rs,r}\left\langle\eta\right|$ is

[TABLE]

the right-hand side of (47) can be regarded as a special Fourier transformation of $\Delta\left(\sigma,\gamma\right)$, so by making its inverse Fourier transformation, we get

[TABLE]

where $\cos\theta_{1}=\cos\theta_{2}=\frac{D}{\sqrt{B^{2}+D^{2}}},r_{1}=\zeta_{1}\sqrt{B^{2}+D^{2}},r_{2}=\zeta_{2}\sqrt{B^{2}+D^{2}}$ and

[TABLE]

Eq.(48) is just the inverse Radon transformation of entangled Wigner operator in the entangled state representation. This is different from the two independent Radon transformations’ direct product of the two independent single-mode Wigner operators, because in (23) the $\left|\eta\right\rangle_{s,r}$ is an entangled state. Therefore the Wigner function of quantum state $\left|\psi\right\rangle$ can be reconstructed from the tomographic inversion of a set of measured probability distributions $\left|{}_{s,r}\left\langle\eta\right.\left|\psi\right\rangle\right|^{2}$, i.e.,

[TABLE]

In summary, based on the preceding paper [13], we have further extended the relation connecting optical Fresnel transformation with quantum tomography to the entangled case. The tomography representation ${}_{s,r}\left\langle\eta\right|=\left\langle\eta\right|F_{2}^{\dagger}$ is set up, based on which the tomogram of quantum state $\left|\psi\right\rangle$ is just the squared modulus of the wave function ${}_{s,r}\left\langle\eta\right|\left.\psi\right\rangle.$ i.e. the probability distribution for the Fresnel quadrature phase is the tomogram (Radon transform of the Wigner function).

**Acknowledgement: **Work supported by the National Natural Science Foundation of China (Grant No 10874174) and the President Foundation of Chinese Academy of Science, and the Research Foundation of the Education Department of Jiangxi Province.

APPENDIX

In fact, from (13) we have

[TABLE]

and

[TABLE]

it then follow that

[TABLE]

and

[TABLE]

Noticing that $\left[F_{2}\left(Q_{1}-Q_{2}\right)F_{2}^{\dagger},F_{2}\left(P_{1}+P_{2}\right)F_{2}^{\dagger}\right]=0$ and (17) thus the eigenvector equation of communicative operators $D\left(Q_{1}-Q_{2}\right)-B\left(P_{1}-P_{2}\right)$ and $B\left(Q_{1}+Q_{2}\right)+D\left(P_{1}+P_{2}\right)$ is

[TABLE]

thus

[TABLE]

and we name $D\left(Q_{1}-Q_{2}\right)-B\left(P_{1}-P_{2}\right)$ or $B\left(Q_{1}+Q_{2}\right)+D\left(P_{1}+P_{2}\right)$ the Fresnel quadrature phase.

On the other hand, due to the communicative relation $\left[F_{2}\left(Q_{1}+Q_{2}\right)F_{2}^{\dagger},F_{2}\left(P_{1}-P_{2}\right)F_{2}^{\dagger}\right]=0,$ and $\left|\xi\right\rangle$ (the conjugate state to $\left|\eta\right\rangle$) is the common eigen-equation of $\left(Q_{1}+Q_{2}\right)$ and $\left(P_{1}-P_{2}\right)$, i.e.,

[TABLE]

so the common eigenvector of $F_{2}\left(Q_{1}+Q_{2}\right)F_{2}^{\dagger}$ and $F_{2}\left(P_{1}-P_{2}\right)F_{2}^{\dagger}$ is given by

[TABLE]

where we have used the overlap relation of $\left\langle\eta\right|\left.\xi\right\rangle=\frac{1}{2}\exp\left(\frac{\xi\eta^{\ast}-\xi^{\ast}\eta}{2}\right)$. The corresponding eigen-equations of $\left|\xi\right\rangle_{s,r}$ are

[TABLE]