• Chinese Optics Letters
  • Vol. 13, Issue 3, 030801 (2015)
Shuang Xu, Liyun Hu*, and Jiehui Huang
Author Affiliations
  • Center for Quantum Science and Technology, Jiangxi Normal University, Nanchang 330022, China
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    DOI: 10.3788/COL201513.030801 Cite this Article Set citation alerts
    Shuang Xu, Liyun Hu, Jiehui Huang, "New fractional entangling transform and its quantum mechanical correspondence," Chin. Opt. Lett. 13, 030801 (2015) Copy Citation Text show less

    Abstract

    In this Letter, a new fractional entangling transformation (FrET) is proposed, which is generated in the entangled state representation by a unitary operator exp{iθ(ab+ab)} where a(b) is the Bosonic annihilate operator. The operator is actually an entangled one in quantum optics and differs evidently from the separable operator, exp{iθ(aa+bb)}, of complex fractional Fourier transformation. The additivity property is proved by employing the entangled state representation and quantum mechanical version of the FrET. As an application, the FrET of a two-mode number state is derived directly by using the quantum version of the FrET, which is related to Hermite polynomials.
    Fθ[f(x)]=Kθ(x,y)f(x)dx,(1)

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    Kθ(x,y)=ei(π2θ)2πsinθexp{ix2+y22tanθ+ixysinθ}.(2)

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    Sθ=dxdy|yKθ(x,y)x|exp{(eiθ1)aa}:=exp{iθaa},(3)

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    Fθ[f(x)]=y|Sθ|xx|fdx=y|Sθ|f,(4)

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    dx|xx|=1.(5)

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    Fθ[f](η)=d2ηπKC(η,η)f(η),KC(η,η)=ei(θπ2)2sinθexp[i(|η|2+|η|2)2tanθi(η*η+η*η)2sinθ].(6)

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    d2ηπ|ηη|=1.(7)

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    |η=exp[12|η|2+ηaη*b+ab]|00.(8)

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    Fθ[f](η)=η|exp[iθ(aa+bb)]|f.(9)

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    KC(η,η)=12sinθexp[i(η2+η2+η*2+η*2)4tanθi(ηη+η*η*)2sinθ],(10)

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    Uθ=d2ηd2ηπ2|ηKC(η,η)η|.(11)

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    Uθ=12sinθd2ηd2ηπ2:exp{12(|η|2+|η|2)+ηaη*b+η*aηb+i(η2+η*2+η2+η*2)4tanθi(ηη+η*η*)2sinθ+ab+abaabb}exp[(cosθ1)(aa+bb)+i(ab+ab)sinθ]:(12)

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    d2zπexp(ζ|z|2+ξz+ηz*+fz2+gz*2)=1ζ24fgexp[ζξη+ξ2g+η2fζ24fg].(13)

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    UθaUθ=τexp[(acosθibsinθ)τ]|τ=0=acosθibsinθ,UθbUθ=bcosθiasinθ,(14)

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    θUθ=i(bUθa+aUθb)cosθ(aUθa+bUθb)sinθ=[i(bUθaUθ+aUθbUθ)cosθ(aUθaUθ+bUθbUθ)sinθ]Uθ=i(ab+ab)Uθ,(15)

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    Uθ=exp{iθ(ab+ab)}.(16)

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    Fθ[f](η)=d2ηπKC(η,η)f(η)=d2ηπη|Uθ|ηη||f=η|exp{iθ(ab+ab)}|f,(17)

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    Fθ+α[f(η)]=η|exp{i(θ+α)(ab+ab)}|f=d2ηπη|exp{iθ(ab+ab)}|η×d2ηπη|exp{iα(ab+ab)}|ηf(η)=d2ηπη|exp{iθ(ab+ab)}|ηFα[f(η)]=FθFα[f(η)].(18)

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    f(η)=e12|η|2m!n!m+nαmβnexp[η*αηβ+αβ]|α,β=0=im+ne12|η|2m!n!Hm,n(iη*,iη),(19)

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    Hm,n(x,y)=m+ntmτnexp[tτ+tx+τy]|t=τ=0.(20)

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    Uθ|m,n=1m!n!(acosθ+ibsinθ)m(bcosθ+iasinθ)n|00=1m!n!m+nαmβnea(αcosθ+iβsinθ)+b(iαsinθ+βcosθ)|00|α,β=0=1m!n!m+nαmβn|α¯,β¯|α,β=0,(21)

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    Fθ[f(η)]=η|Uθ|m,n=e12|η|2m!n!m+nαmβnexp[η*α¯ηβ¯+α¯β¯]|α,β=0=l=0min(m,n)(iisin2θ2)m+n2ln!m!e12|η|2cosl2θ4ll!(nl)!(ml)!×Hml(η¯i2isin2θ)Hnl(η¯*i2isin2θ),(22)

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    Hn(x)=ntnexp(2xtt2)|t=0,ddxlHn(x)=2ln!(nl)!Hnl(x).(23)

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