实际中，我们能轻易制备的态只有相干态和热态，而 Fock \text{Fock} Fock squeezed state \text{squeezed state} squeezed state
正交变量
在相干态中，我们定义了两个正交变量 X ^ \hat{X} X ^ Y ^ \hat{Y} Y ^ Y Y Y P P P
X ^ : = 1 2 ( a ^ + a ^ † ) Y ^ : = 1 2 i ( a ^ − a ^ † ) \hat{X}:=\frac{1}{2}(\hat{a}+\hat{a}^\dagger)\qquad\hat{Y}:=\frac{1}{2i}(\hat{a}-\hat{a}^\dagger)
X ^ := 2 1 ( a ^ + a ^ † ) Y ^ := 2 i 1 ( a ^ − a ^ † )
他们的平均值具有以下特点：
⟨ α ∣ X ^ ∣ α ⟩ = 1 2 ( α + α ∗ ) = R e α ⟨ α ∣ Y ^ ∣ α ⟩ = I m α \langle\alpha|\hat{X}|\alpha\rangle=\frac{1}{2}(\alpha+\alpha^*)=\mathrm{Re}\alpha\quad\langle\alpha|\hat{Y}|\alpha\rangle=\mathrm{Im}\alpha
⟨ α ∣ X ^ ∣ α ⟩ = 2 1 ( α + α ∗ ) = Re α ⟨ α ∣ Y ^ ∣ α ⟩ = Im α
而且他们的方差均为 1 4 \frac{1}{4} 4 1
⟨ δ X ^ 2 ⟩ = 1 4 ⟨ α ∣ ( a ^ + a ^ † ) ( a ^ + a ^ † ) ∣ α ⟩ − 1 4 ( α + α ∗ ) 2 = 1 4 = ⟨ δ Y ^ ⟩ 2 \begin{aligned}
\langle\delta\hat{X}^{2}\rangle & =\frac{1}{4}\langle\alpha|(\hat{a}+\hat{a}^{\dagger})(\hat{a}+\hat{a}^{\dagger})|\alpha\rangle-\frac{1}{4}(\alpha+\alpha^{*})^{2} =\frac{1}{4}=\langle\delta\hat{Y}\rangle^{2}
\end{aligned} ⟨ δ X ^ 2 ⟩ = 4 1 ⟨ α ∣ ( a ^ + a ^ † ) ( a ^ + a ^ † ) ∣ α ⟩ − 4 1 ( α + α ∗ ) 2 = 4 1 = ⟨ δ Y ^ ⟩ 2
那么得到相干态的图像
严格来讲，上图将相干态画成圆形是不准确的，因为我们还没考虑到任意方向的涨落，所以我们定义任意方向的正交变量：
X ^ ϕ : = 1 2 ( a ^ e − i ϕ + a ^ † e i ϕ ) Y ^ ϕ : = 1 2 i ( a ^ e − i ϕ − a ^ † e i ϕ ) \hat{X}_{\phi}:=\frac{1}{2}(\hat{a}e^{-i\phi}+\hat{a}^{\dagger}e^{i\phi}) \qquad\hat{Y}_{\phi}:=\frac{1}{2i}(\hat{a}e^{-i\phi}-\hat{a}^{\dagger}e^{i\phi})
X ^ ϕ := 2 1 ( a ^ e − i ϕ + a ^ † e i ϕ ) Y ^ ϕ := 2 i 1 ( a ^ e − i ϕ − a ^ † e i ϕ )
将两个正交变量用于复平面中，那么相当于对最原始的变量 X ^ \hat{X} X ^ Y ^ \hat{Y} Y ^
X ^ ϕ + i Y ^ ϕ = a ^ e − i ϕ = ( X ^ + i Y ^ ) e − i ϕ \hat{X}_\phi+i\hat{Y}_\phi=\hat{a}e^{-i\phi}=(\hat{X}+i\hat{Y})e^{-i\phi}
X ^ ϕ + i Y ^ ϕ = a ^ e − i ϕ = ( X ^ + i Y ^ ) e − i ϕ
这样我们就可以计算任意方向的涨落值，可以证明任意方向的涨落值都是相同的。
压缩态的方差
我们定义一个压缩算符：
S ^ ξ : = exp [ 1 2 ξ ∗ a ^ 2 − 1 2 ξ ( a ^ † ) 2 ] \hat{S}_\xi:=\exp\left[\frac{1}{2}\xi^*\hat{a}^2-\frac{1}{2}\xi(\hat{a}^\dagger)^2\right]
S ^ ξ := exp [ 2 1 ξ ∗ a ^ 2 − 2 1 ξ ( a ^ † ) 2 ]
其中 ξ ≡ r e i θ , r > 0 \xi\equiv re^{i\theta},\quad r>0 ξ ≡ r e i θ , r > 0 Fock \text{Fock} Fock
∣ α , ξ ⟩ : = D ^ α S ^ ξ ∣ 0 ⟩ |\alpha,\xi\rangle:=\hat{D}_\alpha\hat{S}_\xi|0\rangle
∣ α , ξ ⟩ := D ^ α S ^ ξ ∣0 ⟩
利用 BCH \text{BCH} BCH
S ^ ξ a ^ S ^ ξ † = a ^ cosh r + a ^ † e i θ sinh r S ^ ξ † a ^ S ^ ξ = a ^ cosh r − a ^ † e i θ sinh r \hat{S}_\xi\hat{a}\hat{S}_\xi^\dagger=\hat{a}\cosh r+\hat{a}^\dagger e^{i\theta}\sinh r \\[1em]
\hat{S}_\xi^\dagger\hat{a}\hat{S}_\xi=\hat{a}\cosh r-\hat{a}^\dagger e^{i\theta}\sinh r S ^ ξ a ^ S ^ ξ † = a ^ cosh r + a ^ † e i θ sinh r S ^ ξ † a ^ S ^ ξ = a ^ cosh r − a ^ † e i θ sinh r
对 α ^ \hat{\alpha} α ^ ∣ α , ξ ⟩ |\alpha,\xi\rangle ∣ α , ξ ⟩
⟨ α , ξ ∣ a ^ ∣ α , ξ ⟩ = ⟨ 0 ∣ S ^ ξ † D ^ α † a ^ D ^ α S ^ ξ ∣ 0 ⟩ = ⟨ 0 ∣ S ^ ξ † ( a ^ + α ) S ^ ξ ∣ 0 ⟩ = α \langle\alpha,\xi|\hat{a}|\alpha,\xi\rangle= \langle0|\hat{S}_\xi^\dagger\hat{D}_\alpha^\dagger\hat{a}\hat{D}_\alpha\hat{S}_\xi|0\rangle =\langle0|\hat{S}_\xi^\dagger(\hat{a}+\alpha)\hat{S}_\xi|0\rangle=\alpha
⟨ α , ξ ∣ a ^ ∣ α , ξ ⟩ = ⟨ 0∣ S ^ ξ † D ^ α † a ^ D ^ α S ^ ξ ∣0 ⟩ = ⟨ 0∣ S ^ ξ † ( a ^ + α ) S ^ ξ ∣0 ⟩ = α
对 a ^ † a ^ ^ \hat{\hat{a}^\dagger \hat{a}} a ^ † a ^ ^ ∣ α , ξ ⟩ |\alpha,\xi\rangle ∣ α , ξ ⟩
⟨ α , ξ ∣ a ^ † a ^ ∣ α , ξ ⟩ = ⟨ 0 ∣ ( α ∗ + a ^ † cosh r − a ^ e − i θ sinh r ) ( α + a ^ cosh r − a ^ † e i θ sinh r ) ∣ 0 ⟩ = ∣ α ∣ 2 + sinh 2 r \begin{aligned}
\langle\alpha,\xi|\hat{a}^{\dagger}\hat{a}|\alpha,\xi\rangle & =\langle0|\left(\alpha^{*}+\hat{a}^{\dagger}\cosh r-\hat{a}e^{-i\theta}\sinh r\right)\left(\alpha+\hat{a}\cosh r-\hat{a}^{\dagger}e^{i\theta}\sinh r\right)|0\rangle \\
& =|\alpha|^{2}+\sinh^{2}r
\end{aligned} ⟨ α , ξ ∣ a ^ † a ^ ∣ α , ξ ⟩ = ⟨ 0∣ ( α ∗ + a ^ † cosh r − a ^ e − i θ sinh r ) ( α + a ^ cosh r − a ^ † e i θ sinh r ) ∣0 ⟩ = ∣ α ∣ 2 + sinh 2 r
求 α ^ 2 \hat{\alpha}^2 α ^ 2 ∣ α , ξ ⟩ |\alpha,\xi\rangle ∣ α , ξ ⟩
⟨ α , ξ ∣ a ^ 2 ∣ α , ξ ⟩ = α 2 − e i θ sinh r cosh r \langle\alpha,\xi|\hat{a}^2|\alpha,\xi\rangle=\alpha^2-e^{i\theta}\sinh r\cosh r
⟨ α , ξ ∣ a ^ 2 ∣ α , ξ ⟩ = α 2 − e i θ sinh r cosh r
利用上三式，可以计算出 X ^ \hat{X} X ^ Y ^ \hat{Y} Y ^
⟨ δ X ^ 2 ⟩ = 1 4 ⟨ ( a ^ + a ^ † ) 2 ⟩ − ⟨ X ^ ⟩ 2 = 1 4 ( cosh 2 r − sinh 2 r cos θ ) ⟨ δ Y ^ 2 ⟩ = 1 4 ⟨ ( a ^ − a ^ † ) 2 ⟩ − ⟨ Y ^ ⟩ 2 = 1 4 ( cosh 2 r + sinh 2 r cos θ ) \begin{aligned}
\langle\delta\hat{X}^{2}\rangle & =\frac{1}{4}\langle(\hat{a}+\hat{a}^{\dagger})^{2}\rangle-\langle\hat{X}\rangle^{2} =\frac{1}{4}(\cosh2r-\sinh2r\cos\theta) \\
\langle\delta\hat{Y}^{2}\rangle & =\frac{1}{4}\langle(\hat{a}-\hat{a}^{\dagger})^{2}\rangle-\langle\hat{Y}\rangle^{2} =\frac{1}{4}(\cosh2r+\sinh2r\cos\theta)
\end{aligned} ⟨ δ X ^ 2 ⟩ ⟨ δ Y ^ 2 ⟩ = 4 1 ⟨( a ^ + a ^ † ) 2 ⟩ − ⟨ X ^ ⟩ 2 = 4 1 ( cosh 2 r − sinh 2 r cos θ ) = 4 1 ⟨( a ^ − a ^ † ) 2 ⟩ − ⟨ Y ^ ⟩ 2 = 4 1 ( cosh 2 r + sinh 2 r cos θ )
可以看到，压缩态在某个方向的涨落小于相干态 1 4 \frac{1}{4} 4 1 1 4 \frac{1}{4} 4 1 θ \theta θ r r r
我们取涨落值的两个极限情况
θ = 0 : ⟨ δ X ^ 2 ⟩ = 1 4 e − 2 r ⟨ δ Y ^ 2 ⟩ = 1 4 e 2 r θ = π : ⟨ δ X ^ 2 ⟩ = 1 4 e 2 r ⟨ δ Y ^ 2 ⟩ = 1 4 e − 2 r \theta=0{:}\quad\langle\delta\hat{X}^2\rangle=\frac{1}{4}e^{-2r}\quad\langle\delta\hat{Y}^2\rangle=\frac{1}{4}e^{2r} \\[1em]
\theta=\pi{:}\quad\langle\delta\hat{X}^2\rangle=\frac{1}{4}e^{2r}\quad\langle\delta\hat{Y}^2\rangle=\frac{1}{4}e^{-2r} θ = 0 : ⟨ δ X ^ 2 ⟩ = 4 1 e − 2 r ⟨ δ Y ^ 2 ⟩ = 4 1 e 2 r θ = π : ⟨ δ X ^ 2 ⟩ = 4 1 e 2 r ⟨ δ Y ^ 2 ⟩ = 4 1 e − 2 r
那么压缩态的图像如下
当 θ \theta θ 0 0 0 π \pi π 90 ∘ 90^\circ 9 0 ∘
新定义 b ^ : = a ^ e − i θ 2 \hat{b}:=\hat{a}e^{-i\frac{\theta}{2}} b ^ := a ^ e − i 2 θ
S ^ ξ ′ = exp [ 1 2 r b ^ 2 − 1 2 r b ^ † 2 ] \hat{S}_{\xi^{\prime}}=\exp{[\frac{1}{2}r\hat{b}^2-\frac{1}{2}r\hat{b}^{\dagger2}]}
S ^ ξ ′ = exp [ 2 1 r b ^ 2 − 2 1 r b ^ † 2 ]
正交变量变为
X ^ ′ = 1 2 ( b ^ + b ^ † ) = 1 2 ( a ^ e − i θ 2 + a ^ † e i θ 2 ) Y ^ ′ = 1 2 i ( b ^ − b ^ † ) = 1 2 i ( a ^ e − i θ 2 − a ^ † e i θ 2 ) \begin{aligned}
\hat{X}^{\prime}&=\frac{1}{2}(\hat{b}+\hat{b}^\dagger)=\frac{1}{2}(\hat{a}e^{-i\frac{\theta}{2}}+\hat{a}^\dagger e^{i\frac{\theta}{2}}) \\[1em]
\hat{Y}^{\prime}&=\frac{1}{2i}(\hat{b}-\hat{b}^\dagger)=\frac{1}{2i}(\hat{a}e^{-i\frac{\theta}{2}}-\hat{a}^\dagger e^{i\frac{\theta}{2}})
\end{aligned} X ^ ′ Y ^ ′ = 2 1 ( b ^ + b ^ † ) = 2 1 ( a ^ e − i 2 θ + a ^ † e i 2 θ ) = 2 i 1 ( b ^ − b ^ † ) = 2 i 1 ( a ^ e − i 2 θ − a ^ † e i 2 θ )
得到旋转关系
X ^ ′ + i Y ^ ′ = a ^ ⋅ e − i θ 2 = ( X ^ + i Y ^ ) e − i θ 2 \hat{X}^{\prime}+i\hat{Y}^{\prime}=\hat{a}\cdot e^{-i\frac{\theta}{2}}=(\hat{X}+i\hat{Y})e^{-i\frac{\theta}{2}}
X ^ ′ + i Y ^ ′ = a ^ ⋅ e − i 2 θ = ( X ^ + i Y ^ ) e − i 2 θ
当 θ = π \theta=\pi θ = π ϕ = π 2 \phi=\frac{\pi}{2} ϕ = 2 π ϕ = θ 2 \phi=\frac{\theta}{2} ϕ = 2 θ
X ^ θ = π ′ = 1 2 i ( a ^ − a ^ † ) = X ^ ϕ = π 2 Y ^ θ = π ′ = − 1 2 ( a ^ + a ^ † ) = Y ^ ϕ = π 2 \hat{X}_{\theta=\pi}^{\prime}=\frac{1}{2i}(\hat{a}-\hat{a}^{\dagger})=\hat{X}_{\phi=\frac{\pi}{2}}\\[1em]
\hat{Y}_{\theta=\pi}^{\prime}=-\frac{1}{2}(\hat{a}+\hat{a}^{\dagger})=\hat{Y}_{\phi=\frac{\pi}{2}} X ^ θ = π ′ = 2 i 1 ( a ^ − a ^ † ) = X ^ ϕ = 2 π Y ^ θ = π ′ = − 2 1 ( a ^ + a ^ † ) = Y ^ ϕ = 2 π
根据涨落的表达式，压缩的程度与 r r r
D : = 10 lg ⟨ δ X ^ ϕ 2 ⟩ max ⟨ δ X ^ ϕ 2 ⟩ 0 = 20 r ln 10 [ d B ] ≃ 8.69 r [ d B ] D:=10\lg\frac{\langle\delta\hat{X}_\phi^2\rangle_{\max}}{\langle\delta\hat{X}_\phi^2\rangle_0}=\frac{20r}{\ln10}\left[\mathrm{dB}\right]\quad\simeq8.69r\left[\mathrm{dB}\right]
D := 10 lg ⟨ δ X ^ ϕ 2 ⟩ 0 ⟨ δ X ^ ϕ 2 ⟩ m a x = ln 10 20 r [ dB ] ≃ 8.69 r [ dB ]
其中 ⟨ δ X ^ ϕ 2 ⟩ 0 \langle\delta\hat{X}_\phi^2\rangle_{0} ⟨ δ X ^ ϕ 2 ⟩ 0 1 4 \frac{1}{4} 4 1
压缩态的时间演化
压缩态的时间演化为
e − i H ^ t ∣ α , ξ ⟩ = U t D ^ α U t † ⋅ U t S ^ ξ U t † ⋅ U t ∣ 0 ⟩ = D ^ [ α e − i ω t ] S ^ [ ξ e − 2 i ω t ] ∣ 0 ⟩ = ∣ α e − i ω t , ξ e − 2 i ω t ⟩ \begin{aligned}
e^{-i\hat{H}t}|\alpha,\xi\rangle
& =U_t\hat{D}_\alpha U_t^\dagger\cdot U_t\hat{S}_\xi U_t^\dagger\cdot U_t|0\rangle \\[1em]
&= \hat{D}_{[\alpha e^{-i\omega t}]}\hat{S}_{[\xi e^{-2i\omega t}]} |0\rangle \\[1em]
& =|\alpha e^{-i\omega t},\xi e^{-2i\omega t}\rangle
\end{aligned} e − i H ^ t ∣ α , ξ ⟩ = U t D ^ α U t † ⋅ U t S ^ ξ U t † ⋅ U t ∣0 ⟩ = D ^ [ α e − iω t ] S ^ [ ξ e − 2 iω t ] ∣0 ⟩ = ∣ α e − iω t , ξ e − 2 iω t ⟩
当 θ \theta θ
在 t = 0 t=0 t = 0 θ = 0 \theta=0 θ = 0 X X X θ = π \theta=\pi θ = π Y Y Y ∣ α e − i ω t ⟩ \ket{\alpha e^{-i\omega t}} ∣ α e − iω t ⟩ ω \omega ω ∣ ξ e − 2 i ω t ⟩ \ket{\xi e^{-2i\omega t}} ∣ ξ e − 2 iω t ⟩ ω \omega ω
同拍测量和压缩光的产生
同拍测量
同拍测量是将待测光场与一个强相干态光场（本地振荡场）在一个 50:50 \text{50:50} 50:50 a ^ \hat{a} a ^ ∣ β ⟩ |\beta\rangle ∣ β ⟩ c ^ \hat{c} c ^
c ^ = 1 2 ( a ^ + i b ^ ) \hat{c}=\frac{1}{\sqrt{2}}(\hat{a}+i\hat{b})
c ^ = 2 1 ( a ^ + i b ^ )
上式中，i i i
⟨ c ^ † c ^ ⟩ = 1 2 ⟨ ( a ^ † − i β ∗ ) ( a ^ + i β ) ⟩ = ∣ β ∣ 2 + ⟨ a ^ † a ^ ⟩ + 1 2 ⟨ i β a ^ † − i β ∗ a ^ ⟩ \begin{aligned}
\langle\hat{c}^{\dagger}\hat{c}\rangle & =\frac{1}{2}\langle(\hat{a}^\dagger-i\beta^*)(\hat{a}+i\beta)\rangle \\
& =|\beta|^{2}+\langle\hat{a}^{\dagger}\hat{a}\rangle+\frac{1}{2}\langle i\beta\hat{a}^{\dagger}-i\beta^{*}\hat{a}\rangle
\end{aligned}
⟨ c ^ † c ^ ⟩ = 2 1 ⟨( a ^ † − i β ∗ ) ( a ^ + i β )⟩ = ∣ β ∣ 2 + ⟨ a ^ † a ^ ⟩ + 2 1 ⟨ i β a ^ † − i β ∗ a ^ ⟩
将最后一项与正交变量对比
X ^ ϕ : = 1 2 ( a ^ e − i ϕ + a ^ † e i ϕ ) \hat{X}_\phi:=\frac{1}{2}(\hat{a}e^{-i\phi}+\hat{a}^\dagger e^{i\phi})
X ^ ϕ := 2 1 ( a ^ e − i ϕ + a ^ † e i ϕ )
我们可以调节 β \beta β
压缩光的产生
压缩光可以通过非线性光学效应产生。我们都知道，若是拿一根带电小棒靠近水流，水流会发生弯曲，同时水流产生了极化：
P ⃗ ( t ) = ε 0 χ E ⃗ ( t ) \vec{P}(t)=\varepsilon_{0}\chi\vec{E}(t)
P ( t ) = ε 0 χ E ( t )
我们在电磁学中学到，极化 P ⃗ \vec{P} P E ⃗ \vec{E} E P ⃗ \vec{P} P E ⃗ \vec{E} E
P ⃗ = ε 0 [ χ ( 1 ) E ⃗ + χ ( 2 ) E ⃗ 2 + χ ( 3 ) E ⃗ 3 + … ] \vec{P}=\varepsilon_{0}[\chi^{(1)}\vec{E}+\chi^{(2)}\vec{E}^{2}+\chi^{(3)}\vec{E}^{3}+\ldots]
P = ε 0 [ χ ( 1 ) E + χ ( 2 ) E 2 + χ ( 3 ) E 3 + … ]
在一些晶体中，高阶项并不为零，比如双折射就是由非线性项引起的。压缩光可以通过参量过程（入射和出射光总能量相等，但是光子数和频率会发生变化）产生。用到的非线性过程有两种：二阶项 χ ( 2 ) E ↔ \chi^{(2)}\overleftrightarrow E χ ( 2 ) E PDC \text{PDC} PDC χ ( 3 ) E 3 \chi^{(3)} E^3 χ ( 3 ) E 3 FWM \text{FWM} FWM
高斯积分
我们常需要用到以下积分
∫ − ∞ ∞ d x e − a x 2 = π / a ∫ − ∞ ∞ d x e − a x 2 + b x = π a e b 2 / 4 a \int_{-\infty}^\infty dxe^{-ax^2}=\sqrt{\pi/a}\qquad\qquad
\int_{-\infty}^\infty dxe^{-ax^2+bx}=\sqrt{\frac{\pi}{a}}e^{b^2/4a} ∫ − ∞ ∞ d x e − a x 2 = π / a ∫ − ∞ ∞ d x e − a x 2 + b x = a π e b 2 /4 a
对于复变量 z = x + i y z=x+iy z = x + i y d z 2 = d x d y dz^2=dxdy d z 2 = d x d y
∫ d 2 z e − a ∣ z ∣ 2 = π / a ∫ d 2 z e − a ∣ z ∣ 2 + λ ∗ z − λ z ∗ = π a e − ∣ λ ∣ 2 / a \int d^2ze^{-a|z|^2}=\pi/a\qquad\qquad
\int d^2ze^{-a|z|^2+\lambda^*z-\lambda z^*}=\frac{\pi}{a}e^{-|\lambda|^2/a} ∫ d 2 z e − a ∣ z ∣ 2 = π / a ∫ d 2 z e − a ∣ z ∣ 2 + λ ∗ z − λ z ∗ = a π e − ∣ λ ∣ 2 / a
对于多元变量 X = ( x 1 , x 2 , … , x n ) T \bm{X}=(x_1,x_2,\ldots,x_n)^T X = ( x 1 , x 2 , … , x n ) T A A A
∫ d n X exp [ − X T ⋅ A ⋅ X ] = π n det A ∫ d n X exp [ − X T ⋅ A ⋅ X + B T ⋅ X ] = π n det A exp [ 1 4 B T ⋅ A − 1 ⋅ B ] \int d^nX\exp\left[-X^T\cdot\mathbf{A}\cdot X\right]=\sqrt{\frac{\pi^n}{\det\mathbf{A}}} \\[1em]
\int d^nX\exp\left[-X^T\cdot\mathbf{A}\cdot X+B^T\cdot X\right]=\sqrt{\frac{\pi^n}{\det\mathbf{A}}}\exp\left[\frac{1}{4}B^T\cdot\mathbf{A}^{-1}\cdot B\right]
∫ d n X exp [ − X T ⋅ A ⋅ X ] = det A π n ∫ d n X exp [ − X T ⋅ A ⋅ X + B T ⋅ X ] = det A π n exp [ 4 1 B T ⋅ A − 1 ⋅ B ]
第二式中的 A \mathbf{A} A
相干态 ∣ α 0 ⟩ \ket{\alpha_0} ∣ α 0 ⟩
χ S ( λ ) = e − 1 2 ∣ λ ∣ 2 + λ α 0 ∗ − λ ∗ α 0 \chi_{\mathrm{S}}(\lambda)=e^{-\frac{1}{2}|\lambda|^{2}+\lambda\alpha_{0}^{*}-\lambda^{*}\alpha_{0}}
χ S ( λ ) = e − 2 1 ∣ λ ∣ 2 + λ α 0 ∗ − λ ∗ α 0
利用高斯积分，可以计算出相干态的 Wigner \text{Wigner} Wigner
W ( β ) = 1 π 2 ∫ d 2 λ χ S ( λ ) e − λ β ∗ + λ ∗ β = 1 π 2 ∫ d 2 λ exp [ − 1 2 ∣ λ ∣ 2 + λ ( α 0 ∗ − β ∗ ) − λ ∗ ( α 0 − β ) ] = 2 π exp ( − 2 ∣ β − α 0 ∣ 2 ) \begin{aligned}
W(\beta) & =\frac{1}{\pi^{2}}\int d^{2}\lambda\chi_{\mathrm{S}}(\lambda)e^{-\lambda\beta^{*}+\lambda^{*}\beta} \\
& =\frac{1}{\pi^2}\int d^2\lambda\exp\left[-\frac{1}{2}|\lambda|^2+\lambda(\alpha_0^*-\beta^*)-\lambda^*(\alpha_0-\beta)\right] \\
& =\frac{2}{\pi}\exp{(-2|\beta-\alpha_0|^2)}
\end{aligned} W ( β ) = π 2 1 ∫ d 2 λ χ S ( λ ) e − λ β ∗ + λ ∗ β = π 2 1 ∫ d 2 λ exp [ − 2 1 ∣ λ ∣ 2 + λ ( α 0 ∗ − β ∗ ) − λ ∗ ( α 0 − β ) ] = π 2 exp ( − 2∣ β − α 0 ∣ 2 )
压缩态的 Wigner 函数
压缩相干态的对称序为
χ S ( λ ) = ⟨ α , ξ ∣ e λ a ^ † − λ ∗ a ^ ∣ α , ξ ⟩ \chi_\mathrm{S}(\lambda)=\langle\alpha,\xi|e^{\lambda\hat{a}^\dagger-\lambda^*\hat{a}}|\alpha,\xi\rangle
χ S ( λ ) = ⟨ α , ξ ∣ e λ a ^ † − λ ∗ a ^ ∣ α , ξ ⟩
展开计算得到
⟨ 0 ∣ S ^ ξ † D ^ α † e λ a ^ † − λ ∗ a ^ D ^ α S ^ ξ ∣ 0 ⟩ = ⟨ 0 ∣ e λ ⌊ a ^ † cosh r − a ^ e − i θ sinh r ⌋ − h . c . ∣ 0 ⟩ e λ α ∗ − λ ∗ α = e − 1 2 ∣ λ cosh r + λ ∗ e i θ sinh r ∣ 2 e λ α ∗ − λ ∗ α \begin{aligned}
\langle0|\hat{S}_{\xi}^{\dagger}\hat{D}_{\alpha}^{\dagger}e^{\lambda\hat{a}^{\dagger}-\lambda^{*}\hat{a}}\hat{D}_{\alpha}\hat{S}_{\xi}|0\rangle & =\langle0|e^{\lambda\left\lfloor\hat{a}^{\dagger}\cosh r-\hat{a}e^{-i\theta}\sinh r\right\rfloor-\mathbf{h}.\mathbf{c}.}|0\rangle e^{\lambda\alpha^{*}-\lambda^{*}\alpha} \\
& =e^{-\frac{1}{2}|\lambda\cosh r+\lambda^{*}e^{i\theta}\sinh r|^{2}}e^{\lambda\alpha^{*}-\lambda^{*}\alpha}
\end{aligned} ⟨ 0∣ S ^ ξ † D ^ α † e λ a ^ † − λ ∗ a ^ D ^ α S ^ ξ ∣0 ⟩ = ⟨ 0∣ e λ ⌊ a ^ † c o s h r − a ^ e − i θ s i n h r ⌋ − h . c . ∣0 ⟩ e λ α ∗ − λ ∗ α = e − 2 1 ∣ λ c o s h r + λ ∗ e i θ s i n h r ∣ 2 e λ α ∗ − λ ∗ α
利用高斯积分，可以计算出压缩相干态的 Wigner \text{Wigner} Wigner α = 0 \alpha=0 α = 0
W ∣ 0 , ξ ⟩ ( β ) = 1 π 2 ∫ d 2 λ χ S ( λ ) e − λ β ∗ + λ ∗ β = 2 π exp [ − 2 ∣ β cosh r + β ∗ e − i θ sinh r ∣ 2 ] \begin{aligned}
W_{|0,\xi\rangle}(\beta) & =\frac{1}{\pi^{2}}\int d^{2}\lambda\chi_{\mathrm{S}}(\lambda)e^{-\lambda\beta^{*}+\lambda^{*}\beta} \\
& =\frac{2}{\pi}\exp\left[-2|\beta\cosh r+\beta^*e^{-i\theta}\sinh r|^2\right]
\end{aligned} W ∣0 , ξ ⟩ ( β ) = π 2 1 ∫ d 2 λ χ S ( λ ) e − λ β ∗ + λ ∗ β = π 2 exp [ − 2∣ β cosh r + β ∗ e − i θ sinh r ∣ 2 ]
可以看到压缩态的 Wigner \text{Wigner} Wigner θ = 0 \theta=0 θ = 0
→ θ = 0 exp [ − β x 2 1 2 e − 2 r − β y 2 1 2 e 2 r ] \xrightarrow{\theta=0}\exp\left[-\frac{\beta_x^2}{\frac{1}{2}e^{-2r}}-\frac{\beta_y^2}{\frac{1}{2}e^{2r}}\right]
θ = 0 exp [ − 2 1 e − 2 r β x 2 − 2 1 e 2 r β y 2 ]
可以看到，压缩态的 Wigner \text{Wigner} Wigner β x \beta_x β x β y \beta_y β y
压缩态的 P 函数
压缩相干态的正规序特征函数为
χ N ( λ ) = ⟨ α , ξ ∣ e λ a ^ † e − λ ∗ a ^ ∣ α , ξ ⟩ = e 1 2 ∣ λ ∣ 2 χ S ( λ ) = exp [ 1 2 ∣ λ ∣ 2 − 1 2 ∣ λ cosh r + λ ∗ e i θ sinh r ∣ 2 ] e λ α ∗ − λ ∗ α = exp { − 1 4 ( λ ∗ , λ ) ⋅ [ cosh 2 r − 1 e i θ sinh 2 r e − i θ sinh 2 r cosh 2 r − 1 ] ⋅ [ λ λ ∗ ] } e λ α ∗ − λ ∗ α ] \begin{aligned}
\chi_\mathrm{N}(\lambda)&=\langle\alpha,\xi|e^{\lambda\hat{a}^\dagger}e^{-\lambda^*\hat{a}}|\alpha,\xi\rangle=e^{\frac{1}{2}|\lambda|^2}\chi_\mathrm{S}(\lambda) \\[1em]
& =\exp\left[\frac{1}{2}|\lambda|^{2}-\frac{1}{2}|\lambda\cosh r+\lambda^{*}e^{i\theta}\sinh r|^{2}\right]e^{\lambda\alpha^{*}-\lambda^{*}\alpha} \\[1em]
& \left.=\exp\left\{-\frac{1}{4}(\lambda^{*},\lambda)\cdot\left[
\begin{array}
{cc}\cosh2r-1 & e^{i\theta}\sinh2r \\[1em]
e^{-i\theta}\sinh2r & \cosh2r-1
\end{array}\right]\cdot\left[
\begin{array}
{c}\lambda \\
\lambda^{*}
\end{array}\right]\right\}e^{\lambda\alpha^{*}-\lambda^{*}\alpha}\right]
\end{aligned} χ N ( λ ) = ⟨ α , ξ ∣ e λ a ^ † e − λ ∗ a ^ ∣ α , ξ ⟩ = e 2 1 ∣ λ ∣ 2 χ S ( λ ) = exp [ 2 1 ∣ λ ∣ 2 − 2 1 ∣ λ cosh r + λ ∗ e i θ sinh r ∣ 2 ] e λ α ∗ − λ ∗ α = exp ⎩ ⎨ ⎧ − 4 1 ( λ ∗ , λ ) ⋅ cosh 2 r − 1 e − i θ sinh 2 r e i θ sinh 2 r cosh 2 r − 1 ⋅ [ λ λ ∗ ] ⎭ ⎬ ⎫ e λ α ∗ − λ ∗ α
计算压缩相干态的 P \text{P} P
P ∣ α , ξ ⟩ ( β ) = 1 π 2 ∫ d 2 λ χ N ( λ ) e − λ β ∗ + λ ∗ β P_{|\alpha,\xi\rangle}(\beta)=\frac{1}{\pi^2}\int d^2\lambda\chi_{\mathrm{N}}(\lambda)e^{-\lambda\beta^*+\lambda^*\beta}
P ∣ α , ξ ⟩ ( β ) = π 2 1 ∫ d 2 λ χ N ( λ ) e − λ β ∗ + λ ∗ β
但是，这个积分无法用高斯积分计算出来，因为矩阵 A \mathbf{A} A P \text{P} P
P ( β ) = exp [ q x 4 ∂ β y 2 ] δ ( β y ) ⋅ exp [ q y 4 ∂ β x 2 ] δ ( β x ) = exp [ q x 4 ∂ β y 2 + q y 4 ∂ β x 2 ] δ 2 ( β ) \begin{aligned}
P(\beta)
& =\exp\left[\frac{q_{x}}{4}\partial_{\beta_{y}}^{2}\right]\delta(\beta_{y})\cdot\exp\left[\frac{q_{y}}{4}\partial_{\beta_{x}}^{2}\right]\delta(\beta_{x})=\exp\left[\frac{q_{x}}{4}\partial_{\beta_{y}}^{2}+\frac{q_{y}}{4}\partial_{\beta_{x}}^{2}\right]\delta^{2}(\beta)
\end{aligned} P ( β ) = exp [ 4 q x ∂ β y 2 ] δ ( β y ) ⋅ exp [ 4 q y ∂ β x 2 ] δ ( β x ) = exp [ 4 q x ∂ β y 2 + 4 q y ∂ β x 2 ] δ 2 ( β )
利用该表达式，可以计算正规序的期望值
⟨ a ^ † n a ^ n ⟩ = ∫ d 2 β ∣ β ∣ 2 n P ( β ) = ∫ d 2 β ∣ β ∣ 2 n D ^ δ ( 2 ) ( β ) = D ^ ( β x 2 + β y 2 ) n ∣ β → 0 \langle\hat{a}^{\dagger n}\hat{a}^{n}\rangle=\int d^{2}\beta\left|\beta\right|^{2n}P(\beta)=\int d^{2}\beta\left|\beta\right|^{2n}\hat{\mathcal{D}}\delta^{(2)}(\beta)\quad=\hat{\mathcal{D}}(\beta_{x}^{2}+\beta_{y}^{2})^{n}|_{\beta\to0}
⟨ a ^ † n a ^ n ⟩ = ∫ d 2 β ∣ β ∣ 2 n P ( β ) = ∫ d 2 β ∣ β ∣ 2 n D ^ δ ( 2 ) ( β ) = D ^ ( β x 2 + β y 2 ) n ∣ β → 0
其中 D ^ \hat{\mathcal{D}} D ^ P \text{P} P
⟨ a ^ † a ^ ⟩ = 2 ( q x 4 + q y 4 ) = e 2 r + e − 2 r − 2 4 = sinh 2 r ⟨ a ^ 2 ⟩ = D ^ β 2 ∣ β → 0 = 2 ( q y 4 − q x 4 ) = − sinh 2 r \langle\hat{a}^{\dagger}\hat{a}\rangle=2(\frac{q_{x}}{4}+\frac{q_{y}}{4})=\frac{e^{2r}+e^{-2r}-2}{4}=\sinh^{2}r \\[1em]
\langle\hat{a}^{2}\rangle=\hat{\mathcal{D}}\beta^{2}|_{\beta\to0}=2(\frac{q_{y}}{4}-\frac{q_{x}}{4})=-\sinh2r ⟨ a ^ † a ^ ⟩ = 2 ( 4 q x + 4 q y ) = 4 e 2 r + e − 2 r − 2 = sinh 2 r ⟨ a ^ 2 ⟩ = D ^ β 2 ∣ β → 0 = 2 ( 4 q y − 4 q x ) = − sinh 2 r
这确实和我们之前计算方法得出的结果是一致的。
压缩态光子数统计
将压缩态展开在 Fock \text{Fock} Fock
∣ ξ ⟩ = ∑ n = 0 ∞ C n ∣ n ⟩ |\xi\rangle=\sum_{n=0}^{\infty}C_{n}|n\rangle
∣ ξ ⟩ = n = 0 ∑ ∞ C n ∣ n ⟩
那么有
a ^ ∣ 0 ⟩ = 0 ⇒ S ^ ξ a ^ S ^ ξ † ⋅ S ^ ξ ∣ 0 ⟩ = 0 ⇒ ⟨ n ∣ ( a ^ cosh r + a ^ † e i θ sinh r ) ∣ ξ ⟩ = 0 \hat{a}|0\rangle=0\quad\Rightarrow\quad\hat{S}_\xi\hat{a}\hat{S}_\xi^\dagger\cdot\hat{S}_\xi|0\rangle=0\quad\Rightarrow\quad\langle n|(\hat{a}\cosh r+\hat{a}^\dagger e^{i\theta}\sinh r)|\xi\rangle=0
a ^ ∣0 ⟩ = 0 ⇒ S ^ ξ a ^ S ^ ξ † ⋅ S ^ ξ ∣0 ⟩ = 0 ⇒ ⟨ n ∣ ( a ^ cosh r + a ^ † e i θ sinh r ) ∣ ξ ⟩ = 0
⇒ n + 1 C n + 1 cosh r + n C n − 1 e i θ sinh r = 0 \Rightarrow\quad\sqrt{n+1}C_{n+1}\cosh r+\sqrt{n}C_{n-1}e^{i\theta}\sinh r=0
⇒ n + 1 C n + 1 cosh r + n C n − 1 e i θ sinh r = 0
根据上述关系，得到压缩后的 Fock \text{Fock} Fock
∣ 0 , ξ ⟩ = 1 cosh r ∑ m = 0 ∞ ( 2 m ) ! m ! ( − e i θ 2 tanh r ) m ∣ 2 m ⟩ |0,\xi\rangle=\frac{1}{\sqrt{\cosh r}}\sum_{m=0}^{\infty}\frac{\sqrt{(2m)!}}{m!}(-\frac{e^{i\theta}}{2}\tanh r)^m|2m\rangle
∣0 , ξ ⟩ = cosh r 1 m = 0 ∑ ∞ m ! ( 2 m )! ( − 2 e i θ tanh r ) m ∣2 m ⟩
根据 P n = ∣ ⟨ n ∣ α , ξ ⟩ ∣ 2 P_n=\left|\langle n|\alpha,\xi\rangle\right|^2 P n = ∣ ⟨ n ∣ α , ξ ⟩ ∣ 2
P 2 m = ( 2 m ) ! ( m ! ) 2 ( tanh r ) 2 m 2 2 m cosh r P 2 m + 1 = 0 P_{2m}=\frac{(2m)!}{(m!)^2}\frac{(\tanh r)^{2m}}{2^{2m}\cosh r} \\[1em]
P_{2m+1}=0
P 2 m = ( m ! ) 2 ( 2 m )! 2 2 m cosh r ( tanh r ) 2 m P 2 m + 1 = 0
可以看到，压缩态中只有偶数光子数分布，奇数光子数分布为零。且光子数分布与 θ \theta θ
再计算一下平移压缩态的光子数分布
⟨ n ∣ α , ξ ⟩ = exp [ − 1 2 ∣ α ∣ 2 − 1 2 α ∗ 2 e i θ tanh r ] cosh r ⋅ [ 1 2 e i θ tanh r ] n 2 n ! H n [ α cosh r + α ∗ e i θ sinh r e i θ sinh 2 r ] \langle n|\alpha,\xi\rangle=\frac{\exp[-\frac{1}{2}|\alpha|^2-\frac{1}{2}\alpha^{*2}e^{i\theta}\tanh r]}{\sqrt{\cosh r}}\cdot\frac{[\frac{1}{2}e^{i\theta}\tanh r]^{\frac{n}{2}}}{\sqrt{n!}}H_n[\frac{\alpha\cosh r+\alpha^*e^{i\theta}\sinh r}{\sqrt{e^{i\theta}\sinh2r}}]
⟨ n ∣ α , ξ ⟩ = cosh r exp [ − 2 1 ∣ α ∣ 2 − 2 1 α ∗ 2 e i θ tanh r ] ⋅ n ! [ 2 1 e i θ tanh r ] 2 n H n [ e i θ sinh 2 r α cosh r + α ∗ e i θ sinh r ]
此时，光子数分布与 θ \theta θ ⟨ n ^ ⟩ = ∣ α ∣ 2 + sinh 2 r \langle\hat{n}\rangle=|\alpha|^2+\sinh^2r ⟨ n ^ ⟩ = ∣ α ∣ 2 + sinh 2 r α \alpha α
当 ∣ α ∣ ∼ sinh r |\alpha| \sim \sinh r ∣ α ∣ ∼ sinh r
当 ∣ α ∣ ≫ sinh r |\alpha| \gg \sinh r ∣ α ∣ ≫ sinh r
我们分别对幅度压缩和相位压缩计算光子数分布的方差
幅度压缩: ⟨ δ n ^ 2 ⟩ = ∣ α ∣ 2 e − 2 r + 1 2 sinh 2 ( 2 r ) ≃ ⟨ n ^ ⟩ e − 2 r 相位压缩: ⟨ δ n ^ 2 ⟩ = ∣ α ∣ 2 e 2 r + 1 2 sinh 2 ( 2 r ) ≃ ⟨ n ^ ⟩ e 2 r \begin{aligned}
\text{幅度压缩:}\quad\langle\delta\hat{n}^2\rangle & =|\alpha|^2e^{-2r}+\frac{1}{2}\sinh^2(2r)\simeq\langle\hat{n}\rangle e^{-2r} \\[1em]
\text{相位压缩:}\quad\langle\delta\hat{n}^2\rangle & =|\alpha|^2e^{2r}+\frac{1}{2}\sinh^2(2r)\simeq\langle\hat{n}\rangle e^{2r}
\end{aligned} 幅度压缩 : ⟨ δ n ^ 2 ⟩ 相位压缩 : ⟨ δ n ^ 2 ⟩ = ∣ α ∣ 2 e − 2 r + 2 1 sinh 2 ( 2 r ) ≃ ⟨ n ^ ⟩ e − 2 r = ∣ α ∣ 2 e 2 r + 2 1 sinh 2 ( 2 r ) ≃ ⟨ n ^ ⟩ e 2 r
这样我们就可以分辨出第二幅图中的三条曲线
泊松: ⟨ δ n 2 ⟩ = ⟨ n ⟩ 亚泊松: ⟨ δ n 2 ⟩ < ⟨ n ⟩ 超泊松: ⟨ δ n 2 ⟩ > ⟨ n ⟩ \begin{aligned}
\text{泊松: }\langle \delta n^{2} \rangle = \langle n \rangle \\[1em]
\text{亚泊松: }\langle \delta n^{2} \rangle < \langle n \rangle \\[1em]
\text{超泊松: }\langle \delta n^{2} \rangle > \langle n \rangle \\[1em]
\end{aligned} 泊松 : ⟨ δ n 2 ⟩ = ⟨ n ⟩ 亚泊松 : ⟨ δ n 2 ⟩ < ⟨ n ⟩ 超泊松 : ⟨ δ n 2 ⟩ > ⟨ n ⟩
我们通过 P P P δ \delta δ P \text{P} P P \text{P} P P \text{P} P
⟨ δ n ^ 2 ⟩ = ⟨ n ^ 2 ⟩ − ⟨ n ^ ⟩ 2 = ⟨ n ^ ⟩ + [ ⟨ a ^ † 2 a ^ 2 ⟩ − ⟨ a ^ † a ^ ⟩ 2 ] = ⟨ n ^ ⟩ + ∫ d 2 α P ( α ) ∣ α ∣ 4 − 2 ⟨ a ^ † a ^ ⟩ ⟨ a ^ † a ^ ⟩ + ⟨ a ^ † a ^ ⟩ 2 = ⟨ n ^ ⟩ + ∫ d 2 α P ( α ) [ ∣ α ∣ 2 − ⟨ a ^ † a ^ ⟩ ] 2 < ⟨ n ^ ⟩ \begin{aligned}
\langle\delta\hat{n}^{2}\rangle & =\langle\hat{n}^{2}\rangle-\langle\hat{n}\rangle^{2}=\langle\hat{n}\rangle+\left[\langle\hat{a}^{\dagger2}\hat{a}^{2}\rangle-\langle\hat{a}^{\dagger}\hat{a}\rangle^{2}\right] \\
& =\langle\hat{n}\rangle+\int d^2\alpha P(\alpha)|\alpha|^4-2\langle\hat{a}^\dagger\hat{a}\rangle\langle\hat{a}^\dagger\hat{a}\rangle+\langle\hat{a}^\dagger\hat{a}\rangle^2 \\
& =\langle\hat{n}\rangle+\int d^2\alpha P(\alpha)\left[|\alpha|^2-\langle\hat{a}^\dagger\hat{a}\rangle\right]^2 <\braket{\hat{n}}
\end{aligned} ⟨ δ n ^ 2 ⟩ = ⟨ n ^ 2 ⟩ − ⟨ n ^ ⟩ 2 = ⟨ n ^ ⟩ + [ ⟨ a ^ † 2 a ^ 2 ⟩ − ⟨ a ^ † a ^ ⟩ 2 ] = ⟨ n ^ ⟩ + ∫ d 2 α P ( α ) ∣ α ∣ 4 − 2 ⟨ a ^ † a ^ ⟩ ⟨ a ^ † a ^ ⟩ + ⟨ a ^ † a ^ ⟩ 2 = ⟨ n ^ ⟩ + ∫ d 2 α P ( α ) [ ∣ α ∣ 2 − ⟨ a ^ † a ^ ⟩ ] 2 < ⟨ n ^ ⟩
所以，若方差小于平均值，则说明 P \text{P} P
相干态和热态都是经典态，Fock \text{Fock} Fock
除相干态以外的所有纯态
W ( α ) W\left(\alpha\right) W ( α ) Q ( α ) Q\left(\alpha\right) Q ( α )
