双腔耦合
双腔耦合
$$
\begin{aligned}H&=\hbar \omega_1a_1^\dagger a_1+\hbar \omega_2 a^\dagger_2 a_2+\hbar g(a_1^\dagger a_2 +a_1 a_2^\dagger)\\
&+i\hbar \sqrt{\kappa_{1,ex}}s_{in}(a_1^\dagger e^{-i\omega_l t}-a_1 e^{i\omega_l t})
\end{aligned}$$
选择变换$U=\exp(i\omega_l a_1^\dagger a t+i\omega_l a_2^\dagger a_2 t)$
得到新哈密顿量为
$$\begin{aligned}H&=-\hbar \Delta_1 a_1^\dagger a_1-\hbar \Delta_2 a^\dagger_2 a_2+\hbar g(a_1^\dagger a_2+a_1 a_2^\dagger)\\&+i\hbar \sqrt{\kappa_{ex,1}}s_{in}(a_1^\dagger-a_1)\end{aligned}$$
其中$\Delta_1=\omega_l-\omega_1,\Delta_2=\omega_l-\omega_2$，写出运动方程
$$
\begin{aligned}
\frac{da_1}{dt}&=i\Delta_1a_1-\frac{\kappa_1}{2}a_1-iga_2+\sqrt{\kappa_{ex,1}}s_{in}\\
\frac{da_2}{dt}&=i\Delta_2 a_2-\frac{\kappa_2}{2}a_2-iga_1
\end{aligned}
$$
在稳态下即个导数为零的时候求解方程，得到
$$
\begin{aligned}
&\left(\Delta_1+i\frac{\kappa_1}{2}\right)a_1-ga_2=i\sqrt{\kappa_{ex,1}}s_{in}\\
&\left(\Delta_2+i\frac{\kappa_2}{2}\right)a_2-ga_1=0
\end{aligned}$$
解得
$$
a_1=\frac{2is_{in}\sqrt{\kappa_{ex,1}}}{(2\Delta_1+i\kappa_1)-\frac{4g^2}{2\Delta_2+i\kappa_2}}
$$
计算输出谱
$$
\begin{aligned}
t&=\frac{s_{out}}{s_{in}}\\
&=\frac{s_{in}-\sqrt{\kappa_{ex,1}}a_1}{s_{in}}\\
&=1-\frac{2i\kappa_{ex,1}}{(2\Delta_1+i\kappa_1)-\frac{4g^2}{2\Delta_2+i\kappa_2}}\\
&=1-\frac{1}{(\frac{\kappa_1}{2\kappa_{ex,1}}-i\frac{\Delta_1}{\kappa_{ex,1}})+\frac{g^2/\kappa_{ex,1}^2}{\frac{\kappa_2}{2\kappa_{ex,1}}-i\frac{\Delta_2}{\kappa_{ex,1}}}}
\end{aligned}
$$