光机械哈密顿量的对角化

光机械哈密顿量

$$H=\omega_ca^{\dagger} a+\omega_{m}b^{\dagger}b-g_{om}a^{\dagger} a(b^{\dagger+}+b)$$

其中$\omega_c$为腔模的频率，$\omega_m$为声子频率，$g_{om}$为耦合系数。$a^{\dagger},a，b^{\dagger},b$分别为光子、声子的产生湮灭算符，满足对易关系$[a,a^{\dagger}]=[b,b^{\dagger}]=1$。

现求该哈密顿量下的本征值$E_n$与本征态$\ket{\psi_n}$。假设$n,k$分别为光子数与声子数，本征态分别为$\ket{n},\ket{k}$，满足

$$\omega_ca^{\dagger}a\ket{n}=n\omega_c\ket{n},n=0,1,2,3.....\\\omega_mb^{\dagger}b\ket{k}=k\omega_m\ket{k},k=0,1,2,3.....$$

其态空间为各自子空间的张量积，选用$\ket{n,\psi}=\ket{n}\otimes\ket{\psi}$作为表象。在该表象下，哈密顿量满足，

$$H\ket{n,\psi}=[n\omega_c+\omega_{m}b^{\dagger}b-g_{om}a^{\dagger} a(b^{\dagger+}+b)]\ket{n,\psi}$$

从上式中可发现，对于给定的光子数，跃迁只发生在声子态所在的子空间，故可单独研究声子态子空间，设

$$H_n=\omega_{m}b^{\dagger}b-ng_{om}(b^{\dagger+}+b)$$

显然，$H_n$满足

$$H_n\ket{n,\psi}=(E_n-n\omega_c)\ket{n,\psi}$$

此外，$H_n$还有如下形式，

$$H_n=\omega_mB_n^{\dagger}B_n-\frac{n^2g_{om}^2}{\omega_m}$$

其中，$B_n=b-\frac{ng_{om}}{\omega_m}$，可发现其同样满足玻色子对应关系$[B_n,B_n^{\dagger}]=[b,b^{\dagger}]=1$。引入的新算符在其本征态构成表象下满足

$$B_n^{\dagger}B_n\ket{\mu}=\mu\ket{\mu},\mu=0,1,2,3.....$$

故

$$\begin{aligned} \mu\omega_m-\frac{n^2g_{om}^2}{\omega_m}=E_n-n\omega_c\\\Rightarrow E_n=\mu\omega_m+n\omega_c-\frac{n^2g_{om}^2}{\omega_m} \end{aligned}$$

根据$B_n$与$b$满足位移关系$B_n=b+\beta$，其中，$\beta=-\frac{ng_{om}}{\omega_m}$，得到关系，

$$\begin{aligned} B_n\ket{0_\mu}&=0=(b+\beta)\ket{0_\mu}\\\Rightarrow &b\ket{0_\mu}=-\beta\ket{0_\mu} \end{aligned}$$

故$\ket{\mu}$为湮灭算符$b$的本征态。将$\ket{\mu}$在表象$\ket{n_k}$下展开，即$\ket{0_\mu}=\sum_n c_{n,k}\ket{n_k}$。

一方面，

$$\bra{n_k}b\ket{0_\mu}=-\beta c_{n,k}$$

另一方面，

$$\bra{n_k}b\ket{\mu}=\sqrt{n+1}\bra{n+1}\ket{\mu}=\sqrt{n+1}c_{n+1,k}$$

故

$$c_{n+1,k}=\frac{-\beta}{\sqrt{n+1}}c_{n,k}$$

迭代求解可得，

$$c_{n,k}=\frac{(-\beta)^n}{\sqrt{n!}}c_{0,k}$$

根据归一化条件$1=\sum_n|c_{n,k}|^2$，

$$1=\sum_n\frac{|\beta|^{2n}}{n!}|c_{0,k}|^2=e^{|\beta|^2}|c_{0,k}|^2$$

故，

$$c_{0,k}=e^{-\frac{|\beta|^2}{2}}$$

综上，

$$\begin{aligned} \ket{0_\mu}&=\sum_ne^{-\frac{|\beta|^2}{2}}\frac{(-\beta)^n}{\sqrt{n!}}\frac{(b^{\dagger)^n}}{\sqrt{n!}}\ket{0_k}\\&=\sum_ne^{-\frac{|\beta|^2}{2}}\frac{[-\beta b^{\dagger}]^n}{n!}\ket{0_k}\\&=e^{-\beta(b^{\dagger}-b)}\ket{0_k} \end{aligned}$$

令$\mathcal{D}^{\dagger}(\beta)=e^{-\beta(b^{\dagger}-b)}$，可得到重要关系，

$$\ket{0_\mu}=\mathcal{D}^{\dagger}(\beta)\ket{0_k}\\B_n=\mathcal{D}^{\dagger}(\beta) b \mathcal{D}(\beta)=b+\beta$$

对等式两边同时作用产生算符$\frac{(B_n^{\dagger})^n}{\sqrt{n!}}$，即

$$\frac{(B_n^{\dagger})^n}{\sqrt{n!}}\ket{0_\mu}=\frac{(B_n^{\dagger})^n}{\sqrt{n!}}\mathcal{D}^{\dagger}(\beta)\ket{0_k}$$ $$\begin{align} 左式&=\ket{\mu}\\ 右式&=\mathcal{D}^{\dagger}(\beta) \frac{(b^{\dagger})^n}{\sqrt{n!}} \mathcal{D}(\beta)\mathcal{D}^{\dagger}(\beta)\ket{0_k}\\ &=\mathcal{D}^{\dagger}(\beta) \frac{(b^{\dagger})^n}{\sqrt{n!}} \ket{0_k}\\&=\mathcal{D}^{\dagger}(\beta)\ket{k} \end{align}$$

故

$$\ket{\mu}=\mathcal{D}^{\dagger}(\beta)\ket{k}$$

此外，

$$B_n^{\dagger}B_n\ket{\mu}=B_n^{\dagger}B_n\mathcal{D}^{\dagger}(\beta)\ket{k}\\ \Rightarrow\mu\ket{\mu}=\mathcal{D}^{\dagger}(\beta)b^{\dagger}b\mathcal{D}(\beta)\mathcal{D}^{\dagger}(\beta)\ket{k}\\ \Rightarrow\mu\ket{\mu}=\mathcal{D}^{\dagger}(\beta)b^{\dagger}b\ket{k}=k\mathcal{D}^{\dagger}(\beta)\ket{k}\\ \Rightarrow$$

综上所述，

$$\begin{aligned} E_n=&\mu\omega_m+n\omega_c-\frac{n^2g_{om}^2}{\omega_m} \\ \ket{\mu}=&\mathcal{D}^{\dagger}(\beta)\ket{k} \\ \mu=&0,1,3,4,...,\beta=-\frac{ng_{om}}{\omega_m} \end{aligned}$$

哈密顿量可改写为以下形式

$$\begin{aligned} H&=\omega_m(b^{\dagger}-\frac{a^{\dagger}ag_{om}}{\omega_m})(b-\frac{a^{\dagger}ag_{om}}{\omega_m})+\omega_ca^{\dagger}a -\frac{a^{\dagger}aa^{\dagger}ag_{om}^2}{\omega_m}\\ &=\omega_m(b^{\dagger}-\frac{a^{\dagger}ag_{om}}{\omega_m})(b-\frac{a^{\dagger}ag_{om}}{\omega_m})+\omega_ca^{\dagger}a -\frac{a^{\dagger}(a^{\dagger}a+1)ag_{om}^2}{\omega_m}\\ &=\omega_m(b^{\dagger}-\frac{a^{\dagger}ag_{om}}{\omega_m})(b-\frac{a^{\dagger}ag_{om}}{\omega_m})+\omega_ca^{\dagger}a -\frac{g_{om}^2}{\omega_m}a^{\dagger}a-\frac{g_{om}^2}{\omega_m}a^{\dagger}a^{\dagger}aa \end{aligned}$$

当$g_{om}\ll\omega_m$时，$-\frac{g_{om}^2}{\omega_m}a^{\dagger}a^{\dagger}aa$可忽略，由此得到线性化的哈密顿量

$$H=\omega_mB^{\dagger}B+(\omega_c -\frac{g_{om}^2}{\omega_m})a^{\dagger}a, \\ B=(b-\frac{a^{\dagger}ag_{om}}{\omega_m}),[B,B^{\dagger}]=1$$