OMIT
介绍光机械诱导透明(OMIT)
光机械诱导透明
没有泵浦光时的情况
系统哈密顿量为
$$\begin{aligned}
H&=\frac{p^2}{2m_{eff}}+\frac{1}{2}m_{eff}\Omega_m^2 x^2+\hbar \omega_0a^\dagger a+\hbar g_0xa^\dagger a\\
&+i\hbar \sqrt{\kappa_{ex}}(s_{in}a^\dagger e^{-i\omega_ct}-s_{in}^* ae^{i\omega_ct})
\end{aligned}$$
选择变换$U=\exp(i\omega_ca^{\dagger} at)$，得到新的哈密顿量为
$$\begin{aligned}
H&=\frac{p^2}{2m_{eff}}+\frac{1}{2}m_{eff}\Omega_m^2 x^2-\hbar \Delta a^\dagger a+\hbar g_0xa^\dagger a\\
&+i\hbar \sqrt{\kappa_{ex}}(s_{in}a^\dagger-s_{in}^* a)
\end{aligned}$$
这里$\Delta=\omega_c-\omega_0$。写出运动方程（考虑损耗）
$$\frac{da}{dt}=(i(\Delta-g_0x)-\kappa/2)a+\sqrt{\kappa_{ex}}s_{in}$$
$$m_{eff}\frac{d^2x}{dt^2}=-m_{eff}\Omega_m^2x-\hbar g_0a^\dagger a-\Gamma_m m_{eff}\frac{dx}{dt}$$
求得稳态解（各导数为零）为
$$\bar a=\frac{\sqrt{\kappa_{ex}}s_{in}}{-i(\Delta-g_0\bar x)+\kappa/2}$$
$$\bar x=-\frac{\hbar g_0|\bar a|^2}{m_{eff}\Omega_m^2}$$
由此可研究稳态解个数的问题，会出现双稳态现象。
下面将考虑上面微分方程组在稳态解附近的情况，即假设$a=\bar a+\delta a,x=\bar x+\delta x$
所以可以解出小量$\delta a$和$\delta x$的运动方程
$$\begin{aligned}
\dot{\delta a}&=(i(\Delta -g_0\bar x)-\kappa/2-ig_0\delta x)(\bar a+\delta a)+\sqrt{\kappa_{ex}}s_{in}\\
&=(i(\Delta-g_0\bar x)-\kappa/2)\delta a-ig_0\bar a\delta x\\
&=(i\bar \Delta -\kappa/2)\delta a-ig_0\bar a\delta x
\end{aligned}$$
$$\begin{aligned}
m_{eff}\ddot{\delta x}&=-m_{eff}\Omega_m^2(\bar x+\delta x)-\hbar g_0(\bar a^\dagger +\delta a^\dagger)(\bar a+\delta a)-\Gamma_mm_{eff}\dot{\delta x}\\
&=-m_{eff}\Omega_m^2\delta x-\Gamma_mm_{eff}\dot{\delta x}-\hbar g_0 (\delta a \bar a^\dagger + \bar a\delta a^\dagger)
\end{aligned}$$
这里$\bar \Delta=\Delta -g_0\bar x$。经由傅里叶变换，转换到频谱中分析
$$
-i\omega \delta a[\omega]=(i\bar \Delta-\kappa/2)\delta a[\omega]-ig_0\bar a\delta x[\omega]
$$
$$
(-\omega^2 -i\omega\Gamma_m+\Omega_m^2)\delta x[\omega]=-\hbar g_0(\bar a^\dagger \delta a[\omega]+\bar a\delta a[-\omega]^\dagger)/m_{eff}
$$
从而解出
$$
\begin{aligned}
\delta a[\omega]&=\frac{ig_0\bar a}{i(\bar \Delta -\omega)-\kappa/2}\delta x[\omega]\\
\delta a^\dagger[\omega]&=\frac{-ig_0\bar a}{-i(\bar \Delta +\omega)-\kappa/2}\delta x[\omega]\\
\end{aligned}
$$
在机械振动下产生了斯托克斯分量和反斯托克斯分量
有弱的泵浦光参与时的情况
此时由于有弱泵浦的参与，哈密顿量应该为
$$\begin{aligned}
H&=\frac{p^2}{2m_{eff}}+\frac{1}{2}m_{eff}\Omega_m^2 x^2+\hbar \omega_0a^\dagger a+\hbar g_0xa^\dagger a\\
&+i\hbar \sqrt{\kappa_{ex}}(s_{in} a^\dagger-s_{in}^* a)
\end{aligned}$$
其中$s_{in}=s_{c}e^{-i\omega_c t}+\delta s_p e^{-i\omega_p t}=(s_c+\delta s_p e^{-i(\omega_p-\omega_c)t})e^{-i\omega_c t}$
由于泵浦很弱，我们将$\bar a$和$\bar x$去为刚才没有泵浦是所计算出来的表达式，从而得到新的平衡位置附近的运动方程
$$\begin{aligned}
\dot{\delta a}&=(i(\Delta -g_0\bar x)-\kappa/2-ig_0\delta x)(\bar a+\delta a)+\sqrt{\kappa_{ex}}\bar s_{in}\\
&=(i(\Delta-g_0\bar x)-\kappa/2)\delta a-ig_0\bar a\delta x+\sqrt{\kappa_{ex}}\delta s_p e^{-i(\omega_p-\omega_c)t}\\
&=(i\bar \Delta -\kappa/2)\delta a-ig_0\bar a\delta x+\sqrt{\kappa_{ex}}\delta s_p e^{-i\Delta_m t}
\end{aligned}$$
$$\begin{aligned}
m_{eff}\ddot{\delta x}&=-m_{eff}\Omega_m^2(\bar x+\delta x)-\hbar g_0(\bar a^\dagger +\delta a^\dagger)(\bar a+\delta a)-\Gamma_mm_{eff}\dot{\delta x}\\
&=-m_{eff}\Omega_m^2\delta x-\Gamma_mm_{eff}\dot{\delta x}-\hbar g_0 (\delta a \bar a^\dagger + \bar a\delta a^\dagger)
\end{aligned}$$
为解此方程组，我们假设解的形式为
$$
\begin{aligned}
&\delta a=A^-e^{-i\Delta_m t}+A^+e^{i\Delta_m t}\\
&\delta a^\dagger =A^{+\dagger}e^{-i\Delta _m t}+A^{-\dagger}e^{i\Delta_m t}\\
&\delta x=X e^{-i\Delta_m t}+X^\dagger e^{i\Delta_m t}
\end{aligned}
$$
由于上式是在控制光中考虑的，而我们要考虑的是泵浦光频率的分量，所以我们只用考虑上式中$A^-$的分量，因为前面方程中的泵浦项的$\delta s_p$的相位为$-i\Delta_m t$。代入运动方程，得到等式
$$
\begin{aligned}
&A^- \left(\kappa/2-i \left(\bar \Delta +\Delta _m\right)\right)=\sqrt{\kappa _{\text{ex}}} \delta s_p-i \bar a g_0 X\\
&A^{+\dagger}\left(\kappa/2+i\left(\bar \Delta -\Delta_m\right)\right)=i \bar a^\dagger g_0 X\\
&X \left(\Omega _m^2-\Delta _m^2-i \Gamma _m \Delta _m\right)=-\frac{\hbar g_0}{m_{\text{eff}}}\left(\bar a^\dagger A^-+\bar a A^{+\dagger}\right)
\end{aligned}$$
从而解出
$$
A^-=\frac{1+if}{\kappa/2-i(\bar \Delta+\Delta_m)+2f\bar \Delta}\sqrt{\kappa_{ex}}\delta s_p
$$
其中
$$
\begin{aligned}
f&=\frac{\hbar g_0^2|\bar a|^2\chi}{\kappa/2+i(\bar \Delta -\Delta_m)}\\
\chi&=\frac{1}{m_{eff}(\Omega_m^2-\Delta_m^2-i\Gamma_m\Delta_m)}
\end{aligned}
$$
求泵浦光的投射谱
首先根据输入输出关系$s_{out}=s_{in}-\sqrt{\kappa_{ex}}a$，我们有
$$
\begin{aligned}
s_{out}&=s_c+\delta s_p e^{-i\Delta_m t}-\sqrt{\kappa_{ex}}\left(\bar a+A^-e^{-i\Delta_m t}+A^+e^{i\Delta_m t}\right)\\
&=(s_c-\sqrt{\kappa_{ex}}\bar a)+(\delta s_p-\sqrt{\kappa_{ex}}A^-)e^{-i\Delta_m t}-\sqrt{\kappa_{ex}}A^+e^{i\Delta_m t}
\end{aligned}
$$
我们只关心泵浦光分量，所以有
$$
\begin{aligned}
t_p&=\frac{\delta s_p-\sqrt{\kappa_{ex}}A^-}{\delta s_p}\\
&=1-\frac{1+if}{\kappa/2-i(\bar \Delta+\Delta_m)+2f\bar \Delta}\kappa_{ex}
\end{aligned}
$$
简化
考虑实际实验，我们做如下简化
- 控制光处在resolved-sideband region，$\kappa\ll\Omega_m$，此时$A^+\approx 0$
- $\Delta’=\Delta_m-\Omega_m\approx 0$，此时$\Omega_m^2-\Delta_m^2-i\Gamma_m\Delta_m\approx -\Omega_m(2\Delta’+i\Gamma_m)$
我们有
$$
\begin{aligned}
&A^- \left(\kappa/2-i \left(\bar \Delta +\Delta’+\Omega_m\right)\right)=\sqrt{\kappa _{\text{ex}}} \delta s_p-i \bar a g_0 X\\
&-\Omega_m X \left(2\Delta ‘+i \Gamma _m\right)=-\frac{\hbar g_0}{m_{\text{eff}}}\bar a^\dagger A^-
\end{aligned}$$
容易求得
$$
A^-=\frac{\sqrt{\kappa_{ex}}\delta s_p}{\kappa/2-i(\bar \Delta +\Delta’+\Omega_m)+\frac{|\Omega_c|^2/4}{-i\Delta’+\Gamma_m/2}}
$$
其中$\Omega_c=2g_0\bar ax_{zpf},x_{zpf}=\sqrt{\frac{\hbar}{2m_{eff}\Omega_m}}$
跟EIT的表达式是一样的。