光力系统中哈密顿量的线性化

由于辐射压力的存在，光学微腔的共振频率受到微小的改变，我们把这种改变线性展开到第二项，忽略高阶项得其系统哈密顿量为
$$
H=\hbar\omega_m b^\dagger b+\hbar(\omega_0-Gx)a^\dagger a+i\hbar\sqrt{k_{ex}}(a^\dagger s_{in} e^{-i\omega_lt}-a s_{in}^\dagger e^{i\omega_lt})
$$
首先取酉变换$U=\exp{(i\omega_l a^\dagger at)}$，并由$x=x_{ZPF}(b^\dagger +b)$，得哈密顿量变为
$$H=-\hbar\Delta a^\dagger a-\hbar g_0a^\dagger a(b^\dagger + b)+\hbar \omega_m b^\dagger b+i\hbar \sqrt{\kappa_{ex}}(a^\dagger s_{in}-as_{in}^\dagger)$$
这里设$\Delta=\omega_l-\omega_0,g_0=Gx_{ZPF}$。
下面做线性化处理，我们取$a=\overline{\alpha}+\delta a$，此时相互作用的哈密顿量为
$$H_{int}=-\hbar g_0(\overline{\alpha}+\delta a)^\dagger(\overline{\alpha}+\delta a)(b^\dagger+b)=-\hbar g_0\sqrt{n_{cav}}(\delta a+\delta a^\dagger)(b^\dagger + b)$$
所以总的哈密顿量为
$$H=-\hbar \Delta \delta a^\dagger \delta a-\hbar g(\delta a+\delta a^\dagger)(b^\dagger + b)+\hbar \omega_m b^\dagger b$$
这里$g=g_0\sqrt{n_{cav}}=g_0\overline{\alpha}$。特别注意此时的泵浦项会抵消掉！对于相互作用哈密顿量，在不同的失谐量下可以选取不同部分，当在红失谐是，我们舍去$\delta a b+\delta a^\dagger b^\dagger$，此时哈密顿量为
$$H=-\hbar \Delta \delta a^\dagger \delta a-\hbar g(\delta ab^\dagger + \delta a^\dagger b)+\hbar \omega_m b^\dagger b$$
而当蓝失谐时，有哈密顿量为
$$H=-\hbar \Delta \delta a^\dagger \delta a-\hbar g(\delta ab + \delta a^\dagger b^\dagger)+\hbar \omega_m b^\dagger b$$

临时推导

$$H=\hbar \omega_m b^\dagger b+\hbar \omega_{1}a_1^\dagger a_1+\hbar \omega_{2}a_2^\dagger a_2+\hbar G_1(a_1^\dagger b+a_1b^\dagger)+\hbar G_2(a_2^\dagger b+a_2 b^\dagger)+i\hbar\sqrt{\kappa_{ex1}}A_1(a_1^\dagger e^{-i\omega_{l1}t}-a_1 e^{i\omega_{l1}t})+i\hbar\sqrt{\kappa_{ex1}}A_{in}(a_1^\dagger e^{-i\omega_{in}t}-a_1 e^{i\omega_{in}t})+i\hbar\sqrt{\kappa_{ex2}}A_2(a_2^\dagger e^{-i\omega_{l2}t}-a_1 e^{i\omega_{l2}t})$$
$U=e^{i\theta t(a_1^\dagger a_1+a_2^\dagger a_2+b^\dagger b)}$

$$\begin{aligned}
H=&\hbar (\omega_m-\theta) b^\dagger b+\hbar (\omega_{1}-\theta)a_1^\dagger a_1+\hbar (\omega_{2}-\theta)a_2^\dagger a_2+\hbar G_1(a_1^\dagger b+a_1b^\dagger)+\hbar G_2(a_2^\dagger b+a_2 b^\dagger)\\
&+i\hbar\sqrt{\kappa_{ex1}}A_1(a_1^\dagger e^{i\theta t}e^{-i\omega_{l1}t}-a_1 e^{-i\theta t}e^{i\omega_{l1}t})\\
&+i\hbar\sqrt{\kappa_{ex1}}A_{in}(a_1^\dagger e^{i\theta t}e^{-i\omega_{in}t}-a_1 e^{-i\theta t}e^{i\omega_{in}t})\\
&+i\hbar\sqrt{\kappa_{ex2}}A_2(a_2^\dagger e^{i\theta t}e^{-i\omega_{l2}t}-a_1 e^{-i\theta t}e^{i\omega_{l2}t})\\
=&\hbar (\omega_m-\theta) \beta^\dagger \beta+\hbar (\omega_{1}-\theta)\alpha_1^\dagger \alpha_1+\hbar (\omega_{2}-\theta)\alpha_2^\dagger \alpha_2+\hbar G_1(\alpha_1^\dagger \beta+\alpha_1\beta^\dagger)+\hbar G_2(\alpha_2^\dagger \beta+\alpha_2 \beta^\dagger)\\
&+i\hbar\sqrt{\kappa_{ex1}}A_1(\alpha_1^\dagger e^{-i\omega_{l1}t}-\alpha_1 e^{i\omega_{l1}t})\\
&+i\hbar\sqrt{\kappa_{ex1}}A_{in}(\alpha_1^\dagger e^{-i\omega_{in}t}-\alpha_1 e^{i\omega_{in}t})\\
&+i\hbar\sqrt{\kappa_{ex2}}A_2(\alpha_2^\dagger e^{-i\omega_{l2}t}-\alpha_1 e^{i\omega_{l2}t})\\
\end{aligned}$$

$U=e^{i\omega_{l1}ta_1^\dagger a_1+i\omega_{l2}a_2^\dagger a_2}$
$$\begin{aligned}
H=&\hbar (\omega_m-\theta) \beta^\dagger \beta+\hbar (\omega_{1}-\theta-\omega_{l1})\alpha_1^\dagger \alpha_1+\hbar (\omega_{2}-\theta-\omega_{l2})\alpha_2^\dagger \alpha_2+\hbar G_1(\alpha_1^\dagger \beta e^{i\omega_{l1}t}+\alpha_1\beta^\dagger e^{-i\omega_{l1}t})+\hbar G_2(\alpha_2^\dagger \beta e^{i\omega_{l2}t}+\alpha_2 \beta^\dagger e^{-i\omega_{l2}t})\\
&+i\hbar\sqrt{\kappa_{ex1}}A_1(\alpha_1^\dagger e^{-i\omega_{l1}t}-\alpha_1 e^{i\omega_{l1}t})\\
&+i\hbar\sqrt{\kappa_{ex1}}A_{in}(\alpha_1^\dagger e^{-i\omega_{in}t}-\alpha_1 e^{i\omega_{in}t})\\
&+i\hbar\sqrt{\kappa_{ex2}}A_2(\alpha_2^\dagger e^{-i\omega_{l2}t}-\alpha_1 e^{i\omega_{l2}t})\\
\end{aligned}$$

$$\begin{aligned}
\dot{\alpha}&=-\frac{\kappa}{2}\alpha+i(\Delta+Gx)\alpha+\sqrt{\kappa_{ex}}\alpha_{in}\\
\dot{x}&=v\\
\dot{v}&=-\Omega_m^2x-\Gamma_m v+\hbar G|\alpha|^2/m_{eff}
\end{aligned}$$

$$
\begin{aligned}
\frac{d \sqrt{\kappa_{ex}}\alpha/\alpha_{in}}{d \kappa_{ex}t}
&=-\frac{\kappa/\kappa_{ex}}{2}\frac{\sqrt{\kappa_{ex}}\alpha}{\alpha_{in}}+i(\frac{\Delta}{\kappa_{ex}}+\frac{Gx}{\kappa_{ex}})\frac{\sqrt{\kappa_{ex}}\alpha}{\alpha_{in}}+1\\
\frac{d Gx/\kappa_{ex}}{d\kappa_{ex}t}&=\frac{Gv}{\kappa_{ex}^2}\\
\frac{dGv/\kappa_{ex}^2}{d\kappa_{ex}t}&=-\frac{\Omega_m^2}{\kappa_{ex}^2} \frac{Gx}{\kappa_{ex}}-\frac{\Gamma_m}{\kappa_{ex}} \frac{Gv}{\kappa_{ex}^2}+\frac{\hbar G^2\alpha_{in}^2}{m_{eff}\kappa_{ex}^4}|\sqrt{\kappa_{ex}}\alpha/\alpha_{in}|^2\\
\end{aligned}
$$