• Chinese Physics B
  • Vol. 29, Issue 10, (2020)
Ke Zhang1, Lan-Lan Li1, and Hong-Yi Fan2,†
Author Affiliations
  • 1School of Electronic Engineering, Huainan Normal University, Huainan 232038, China
  • 2University of Science and Technology of China, Hefei 30031, China
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    DOI: 10.1088/1674-1056/ab99b2 Cite this Article
    Ke Zhang, Lan-Lan Li, Hong-Yi Fan. Damping of displaced chaotic light field in amplitude dissipation channel[J]. Chinese Physics B, 2020, 29(10): Copy Citation Text show less

    Abstract

    We explore how a displaced chaotic light (DCL) behaves in an amplitude dissipation channel, and what is its time evolution formula of photon number distribution. With the use of the method of integration within ordered product product of operator (IWOP) and the new binomial theorem involving two-variable Hermite polynomials we obtain the evolution law of DCL in the channel.
    ρc=(1ef)efaa,(1)

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    tr(ρcaa)=(1ef)fTr[efaan=0|nn|]=(1ef)fn=0efn=(1ef)f11ef=(ef1)1,(2)

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    ρd=Ceλaefaaeλa,(3)

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    d2απ|αα|=1,(4)

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    |α=exp[12|α|2+αa]|0,a|α=α|α.(5)

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    1=Trρd=Tr[d2απ|αα|ρd]=Cd2απα|eλaefaaeλa|α=Cd2απeλα+λαα|:exp[(ef1)aa]:|α=Cd2απeλα+λαexp[(1ef)|α|2]=C1efexp[|λ|21ef],(6)

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    C=(1ef)exp[|λ|21ef].(7)

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    ρd=(1ef)exp[|λ|21ef]eλaefaaeλa.(8)

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    efaa=:exp[(ef1)aa]:,(9)

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    (1)

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    Tr(ρdaa)=Tr(ρdaa)1=Tr[ρdad2απ|αα|a]1=(1ef)exp(|λ|21ef)×d2απ|α|2α|eλaefaaeλa|α1=(1ef)exp(|λ|21ef)d2απ|α|2eλα+λα×exp[(ef1)|α|2]1=1ef1+|λ|2(1ef)2,(10)

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    dρdt=k(2aρaaaρρaa),(11)

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    ρ(t)=n=0Tnn!ektaaanρ0anektaa=n=0Mnρ0Mn,(12)

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    Mn=Tnn!ektaaan,(13)

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    eλaaaeλaa=aeλ(14)

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    n=0MnMn+=n=0Tnn!ane2ktaaan=n=0Tnn!e2nkt:anan:e2ktaa=:eTe2ktaa:e2ktaa=:e(e2kt1)aa:e2ktaa=1.(15)

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    Trρ(t)=Trn=0Mnρ0Mn=Trρ0.(16)

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    ρd(t)=n=0MnρdMn=Cn=0Tnn!ektaaaneλaefaaeλaanektaa,(17)

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    aneλa=eλa(a+λ)n=eλal=0(nl)λnlal,(18)

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    aneλaefaaeλaan=eλal=0k=0(nl)(nk)λnlalefaaakλnkeλa.(19)

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    efaa=efexp[(1ef)aa](20)

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    alefaaak=efalexp[(1ef)aa]ak=efd2zπzle(1ef)|z|2|zz|zk=efd2zπzlzk:eef|z|2+za+zaaa:=(ef)(l+k)/2d2zπzlzk:e|z|2+ef/2za+ef/2zaaa:=(ef)(l+k)/2(i)l+k:e(ef1)aa×m=0l!k!(1)mm!(lm)!(km)!(ief/2a)km(ief/2a)lm:=(i)l+k(ef)(l+k)/2:e(ef1)aa×Hk,l(iaef/2,iaef/2):(21)

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    Hm,n(x,y)=n+mtmτnexp(tx+τytτ)|t=τ=0=mtmetxnτnexp[τ(yt)]|t=τ=0=mtm[etx(yt)n]|t=0=l=0(ml)ltl(yt)nmltmletx|t=0=l=0min(m,n)m!n!(1)ll!(ml)!(nl)!xmlynl(22)

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    m,n=0tmτnm!n!Hm,n(x,y)=exp(tx+τytτ).(23)

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    l=0k=0(nl)(nk)λnlalefaaakλnk=l=0k=0(nl)(nk)λnl(i)l+k(ef)(l+k)/2:e(ef1)aaHk,l(iaef/2,iaef/2):λnk=|λ|2n:e(ef1)aal=0nk=0n(nl)(nk)(ief/2λ)l(ief/2λ)kHk,l(iaef/2,iaef/2):.(24)

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    r=0lq=0k(lr)(kq)Hr,q(x,y)frgq=fltkHl,k(x+1f,y+1g).(25)

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    aqar=Hr,q(a,a).(26)

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    etaeta=etaetaett=eta+tatt=r=0q=0(t)r(t)qq!r!Hr,q(a,a),(27)

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    etaeta=r=0q=0tqtrq!r!araq,(28)

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    r=0lq=0k(lr)(kq)Hr,q(a,a)frgq=r=0lq=0k(lr)(kq):aqar:frgq=:(ga+1)k(fa+1)l:,(29)

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    l,k=0sltk:(ga+1)k(fa+1)l:l!k!=et(ga+1)es(fa+1)=esf(a+1/f)etg(a+1/g)esftg=esf(a+1/f)+tg(a+1/g)esftg=l,k(sf)l(gt)kl!k!Hl,k(a+1f,a+1g),(30)

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    :(ga+1)k(fa+1)l:=fltkHl,k(a+1f,a+1g).(31)

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    r=0lq=0k(lr)(kq)Hr,q(a,a)frgq=fltkHl,k(a+1f,a+1g).(32)

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    r=0lq=0k(lr)(kq)Hr,q(x,y)frgq=fltkHl,k(x+1f,y+1g).(33)

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    l=0k=0(nl)(nk)λnlalefaaakλnk=l=0k=0(nl)(nk)λnl(i)l+k(ef)(l+k)/2:e(ef1)aaHk,l(iaef/2,iaef/2):λnk=|λ|2n:e(ef1)aal=0nk=0n(nl)(nk)(ief/2λ)l(ief/2λ)kHk,l(iaef/2,iaef/2):=(ef)n:e(ef1)aaHn,n(iaef/2+iλef/2,iaef/2+ief/2λ):.(34)

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    Ln(xy)=(1)nn!Hn,n(x,y),(35)

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    aneλaefaaeλaan=eλal=0k=0(nl)(nk)λnlalefaaakλnkeλa=eλa(ef)n:e(ef1)aaHn,n(iaef/2+iλef/2,iaef/2+iλef/2):eλa=eλa(ef)n:e(ef1)aan!(1)nLn[(aef/2+λef/2)(aef/2+λef/2)]:eλa.(36)

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    W=n=0Tnn!aneλaefaaeλaan,(37)

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    ρd(t)=(1ef)exp[|λ|21ef]eκtaaWeκtaa.(38)

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    (1z)1exp[zxz1]=l=0Ln(x)zn(39)

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    W=n=0Tnn!aneλaefaaeλaan=eλan=0Tnn!(ef)n:e(ef1)aan!(1)nLn[(aef/2+λef/2)(aef/2+λef/2)]:eλa=eλan=0(Tef)n:e(ef1)aaLn[(aef/2+λef/2)(aef/2+λef/2)]:eλa=11Tefeλa:e(ef1)aaexp[Tef(aef/2+λef/2)(aef/2+λef/2)Tef1]:eλa,(40)

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    ektaa=aektaaekt(41)

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    ρd(t)=(1ef)exp[|λ|21ef]eκtaaWeκtaa=1ef1Tefexp[T1(ef1)(Tef1)|λ|2]×exp[(λa1Tef)ekt]×eκtaa:exp[(ef1Tef1)aa]:eκtaa×[exp(λa1Tef)ekt]=1ef1Tefexp[T1(ef1)(Tef1)|λ|2]×exp[(λa1Tef)ekt]×exp[aa(lnef1Tef2kt)]×exp[(λa1Tef)ekt]=1ef1Tefexp[T1(ef1)(Tef1)|λ|2]×exp[(λa1Tef)ekt]×exp[aalnef(1T)1Tef]×exp[(λa1Tef)ekt].(42)

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    trρd(t)=1ef1Tefexp[T1(ef1)(Tef1)|λ|2]d2zπz|exp[(λa1Tef)ekt]×exp[aalnef(1T)1Tef]exp[(λa1Tef)ekt]|z=1ef1Tefexp[T1(ef1)(Tef1)|λ|2]d2zπexp[(1ef(1T)1Tef)|z|2+λzekt1Tef+λzekt1Tef]=exp{[e2kt(1ef)(1Tef)+T1(1ef)(1Tef)]|λ|2}=1.(43)

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    tr[ρ(t)aa]=tr[ρ(t)ad2zπ|zz|a]=1ef1Tefexp[T1(ef1)(Tef1)|λ|2]d2zπ|z|2z|exp[(λa+1Tef)ekt]×exp[aalnef(1T)1Tef]exp[(λa1Tef)ekt]|z=1ef1Tefexp[T1(ef1)(Tef1)|λ|2]d2zπ|z|2exp[1ef1Tef|z|2+λz1Tefekt+λz1Tefekt]=1ef1Tefexp[T1(ef1)(Tef1)|λ|2](1Tef)2e2kt2λλ×d2zπexp[1ef1Tef|z|2+λz1Tefekt+λz1Tefekt]=exp[T1(ef1)(Tef1)|λ|2](1Tef)2e2kt2λλexp[e2kt(ef1)(Tef1)|λ|2],(44)

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    tr[ρ(t)aa]=(Tef1)exp[e2kt(ef1)(Tef1)|λ|2]×2λλexp[1ef1|λ|2]=(Tef1)exp[e2kt(ef1)(Tef1)|λ|2]λ×{λef1exp[1ef1|λ|2]}=(Tef1)exp[e2kt(ef1)(Tef1)|λ|2]×{1ef1+|λ|2(ef1)2}exp[1ef1|λ|2]=(Tef1){1ef1+|λ|2(ef1)2}=(Tef1){1ef1+|λ|2e2kt(ef1)2(Tef1)}=Tef1ef1+|λ|2e2kt(ef1)2.(45)

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    tr[ρ(t)aa]=tr[ρ(t)aa]1=ef(T1)ef1+|λ|2e2kt(ef1)2=[1ef1+|λ|2(ef1)2]e2kt.(46)

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    Ke Zhang, Lan-Lan Li, Hong-Yi Fan. Damping of displaced chaotic light field in amplitude dissipation channel[J]. Chinese Physics B, 2020, 29(10):
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