旋转Frame下哈密顿量的表示

在量子光学中，我们会经常对哈密顿量作基矢变换，使得哈密顿量的形式更加简洁，比如Rotating Frame的变换，或这相互作用表象。我将在这个笔记中对这个变换作简单推导，并给出JC系统的例子来说明如何操作。

Rotating Frame可以看做是线性、幺正的基矢变换

Rotating Frame其实就是相当于做了一次线性的、幺正的基矢变换，假设这个变换表示为 ，那么在这个变换下，哈密顿量的变换过程为

$$H \longrightarrow U H U+i \hbar \dot{U} U^{\dagger}=\tilde{H}$$

从薛定谔方程出发，

$$i\hbar \partial_{t}\psi=\hat{H}\psi$$

由于变换的幺正性，因此，因此可以在薛定谔方程两边同时左乘，并在右边添加单位变换

$$i\hbar\hat{U}\partial_{t}\psi=\hat{U}\hat{H}(\hat{U}^{\dagger}\hat{U})\psi$$

根据求导规则

$$\partial_{t}(\hat{U}\psi)=\partial_{t}(\hat{U})\psi+\hat{U}\partial_{t}\psi$$

通过添加单位变换，并综合式子（3）可以得到

$$\begin{aligned} \partial_{t}(\hat{U}\psi)&=\partial_{t}(\hat{U})(\hat{U}^{\dagger}\hat{U})\psi+\hat{U}\partial_{t}\psi\\ &=\partial_{t}(\hat{U})(\hat{U}^{\dagger}\hat{U})\psi-\frac{i}{\hbar}\hat{U}\hat{H}(\hat{U}^{\dagger}\hat{U})\psi\\ &=-\frac{i}{\hbar}\left[\hat{U}\hat{H}\hat{U}^{\dagger}+i\hbar (\partial_{t}\hat{U})\hat{U}^{\dagger})\right]\hat{U}\psi\\ \implies \partial_{t}\tilde{\psi}&=-\frac{i}{\hbar}\hat{\tilde{H}}\tilde{\psi} \end{aligned}$$

这里我们定义了新表象下的哈密顿量和量子态：

$$\hat{\tilde{H}}=\hat{U}\hat{H}\hat{U}^{\dagger}+i\hbar (\partial_{t}\hat{U})\hat{U}^{\dagger}), \quad \tilde{\psi}=\hat{U}\psi$$

相互作用表象是一种特殊的基矢变换，即将哈密顿量分为两部分

$$\hat{H}=\hat{H}_{0}+\hat{V}$$

原来的薛定谔方程在相互作用下表示为

$$\frac{\partial \tilde{\psi }}{\partial t}=-\frac{i }{\hbar }\hat{V}_{I} \tilde{\psi }$$

这里

$$\begin{aligned} \hat{V}_{I}&=e^{i\hat{H}_{0}t}\hat{V}e^{-i\hat{H}_{0}t}\\ \tilde{\psi}&=e^{i\hat{H}_{0}t}\psi \end{aligned}$$

因此通过定义，就可以按照Rotating Frame的方式得到一样的结果

$$\begin{aligned} & \hat{V}_I=\hat{U} H \hat{U}^{\dagger}+i \hbar\left(\partial_t \hat{U}\right) \hat{U}^{\dagger} \\ & =\hat{U}\left(\hat{H}_0+\hat{V}\right) \hat{U}^{\dagger}+i \hbar\left(\partial_t e^{i \hat{H}_0 t / \hbar}\right) \hat{U}^{\dagger} \\ & =\hat{U}\left(\hat{H}_0+\hat{V}\right) \hat{U}^{\dagger}-\hat{H}_0 \\ & =\hat{U} \tilde{V} \hat{U}^{\dagger} \end{aligned}$$

JC系统哈密顿量的Rotating Frame的推导

我们以JC系统的哈密顿量为例来说明：

$$\hat{H}=\hbar \omega_e \hat{\sigma}_{+} \hat{\sigma}_{-}+\hbar \omega_0 \hat{a}^{\dagger} \hat{a}+g\left(\hat{a}^{\dagger} \hat{\sigma}_{-}+\hat{a} \hat{\sigma}_{+}\right)$$

可以分为耦合部分和非耦合部分，

$$\begin{aligned} & \hat{H}=\hat{H}_0+\hat{H}_V \\ & \hat{H}_0=\hbar \omega_e \hat{\sigma}_{+} \hat{\sigma}_{-}+\hbar \omega_0 \hat{a}^{\dagger} \hat{a} \\ & \hat{H}_V=g\left(\hat{a}^{\dagger} \hat{\sigma}_{-}+\hat{a} \hat{\sigma}_{+}\right) \end{aligned}$$

按照 Rotating frame 的定义（注意我们这里对于算符的变换作了改变， 和前面 是等价的。

$$\begin{aligned} & \hat{H}_{\mathrm{rot}}=i \hbar \frac{\partial \hat{U}_1^{\dagger}}{\partial t} \hat{U}_1+\hat{U}_1^{\dagger} \hat{H} \hat{U}_1 \\ & =i \hbar \frac{\partial \hat{U}_1^{\dagger}}{\partial t} \hat{U}_1+\hat{U}_1^{\dagger}\left(\hat{H}_0+\hat{H}_V\right) \hat{U}_1 \end{aligned}$$

这里 ，我们需要分别计算上面等式右边的三项

• 第一项为： $$\begin{aligned} & i \hbar \frac{\partial \hat{U}_1^{\dagger}}{\partial t} \hat{U}_1 \\ & =i \hbar\left[i \hat{H}_0 / \hbar\right] \hat{U}_1^{\dagger} \hat{U}_1 \\ & =-\hat{H}_0 \end{aligned}$$
• 第二项为 $$\hat{U}_1^{\dagger} \hat{H}_0 \hat{U}_1=e^{i \hat{H}_0 t} \hat{H}_0 e^{-i \hat{H}_0 t}=\hat{H}_0$$
• 第三项为:

为了计算右边的式子： ，需要用到下面的数学展开：

$$e^{-\alpha \hat{A}} \hat{B} e^{\alpha \hat{A}}=\hat{B}-\alpha[\hat{A}, \hat{B}]+\frac{\alpha^2}{2}[\hat{A},[\hat{A}, \hat{B}]]+\ldots \frac{(-\alpha)^n}{2}[\hat{A},[\ldots,[\hat{A}, \hat{B}]] \ldots .]$$

我们只需要考虑不对易的量，比如 $\hat{\sigma}{+}\hat{\sigma}{-}\hat{a}^{\dagger}\hat{a}e^{i \omega_0 \hat{a}^{\dagger} \hat{a} t} \hat{a} e^{-i \omega_0 \hat{a}^{\dagger} \hat{a} t}$ ，我们先要计算对易关系：

$$\begin{aligned} & {\left[\hat{a}^{\dagger} \hat{a}, \hat{a}\right]=\hat{a}^{\dagger} \hat{a} \hat{a}-\hat{a} \hat{a}^{\dagger} \hat{a}=\hat{a}^{\dagger} \hat{a} \hat{a}-\left(\hat{a}^{\dagger} \hat{a}+1\right) \hat{a}=-\hat{a}} \\ & {\left[\hat{a}^{\dagger} \hat{a},\left[\hat{a}^{\dagger} \hat{a}, \hat{a}\right]\right]=\left[\hat{a}^{\dagger} \hat{a},-\hat{a}\right]=(-1)^2 \hat{a}} \end{aligned}$$

可知第 n 项为：

$$\left[\hat{a}^{\dagger} \hat{a},\left[\ldots,\left[\hat{a}^{\dagger} \hat{a}, \hat{a}\right]\right] \ldots .\right]=(-1)^n \hat{a}$$

因此

$$\begin{aligned} & e^{i \omega_0 \hat{a}^{\dagger} \hat{a} t} \hat{a} e^{-i \omega_0 \hat{a}^{\dagger} \hat{a} t}= \\ & =\hat{a}+i \omega_0 t\left[\hat{a}^{\dagger} \hat{a}, \hat{a}\right]+\frac{\left(i \omega_0 t\right)^2}{2!}\left[\hat{a}^{\dagger} \hat{a},\left[\hat{a}^{\dagger} \hat{a}, \hat{a}\right]\right]+\ldots \\ & =\hat{a}-i \omega_0 t \hat{a}+\frac{\left(-i \omega_0 t\right)^2}{2!} \hat{a}+\frac{\left(-i \omega_0 t\right)^3}{3!} \hat{a} \ldots \frac{\left(-i \omega_0 t\right)^n}{n!} \hat{a} \\ & =\hat{a} e^{-i \omega_0 t} \end{aligned}$$

同理由于

$$\begin{aligned} & {\left[\hat{a}^{\dagger} \hat{a}, \hat{a}^{\dagger}\right]=\hat{a}^{\dagger} \hat{a} \hat{a}^{\dagger}-\hat{a}^{\dagger} \hat{a}^{\dagger} \hat{a}=\hat{a}^{\dagger} \hat{a} \hat{a}^{\dagger}-\hat{a}^{\dagger}\left(\hat{a} \hat{a}^{\dagger}-1\right)=\hat{a}^{\dagger}} \\ & {\left[\hat{a}^{\dagger} \hat{a},\left[\hat{a}^{\dagger} \hat{a}, \hat{a}^{\dagger}\right]\right]=\left[\hat{a}^{\dagger} \hat{a}, \hat{a}^{\dagger}\right]=\hat{a}^{\dagger}} \end{aligned}$$

可知第 n 项为：

$$\left[\hat{a}^{\dagger} \hat{a},\left[\ldots,\left[\hat{a}^{\dagger} \hat{a}, \hat{a}^{\dagger}\right]\right] \ldots .\right]=\hat{a}^{\dagger}$$ $$\begin{aligned} & e^{i \omega_0} \hat{a}^{\dagger} \hat{a} t \hat{a}^{\dagger} e^{-i \omega_0} \hat{a}^{\dagger} \hat{a} t= \\ & =\hat{a}^{\dagger}+i \omega_0 t\left[\hat{a}^{\dagger} \hat{a}, \hat{a}^{\dagger}\right]+\frac{\left(i \omega_0 t\right)^2}{2!}\left[\hat{a}^{\dagger} \hat{a},\left[\hat{a}^{\dagger} \hat{a}, \hat{a}^{\dagger}\right]\right]+\ldots \\ & =\hat{a}^{\dagger}+i \omega_0 t \hat{a}^{\dagger}+\frac{\left(i \omega_0 t\right)^2}{2!} \hat{a}^{\dagger}+\frac{\left(i \omega_0 t\right)^3}{3!} \hat{a}^{\dagger} \ldots \frac{\left(i \omega_0 t\right)^n}{n!} \hat{a}^{\dagger} \\ & =\hat{a}^{\dagger} e^{i \omega_0 t} \end{aligned}$$

按照同样的做法，可以得到

$$\begin{aligned} & e^{i \omega_e \hat{\sigma}_{+} \hat{\sigma}_{-} t} \hat{\sigma}_{-} e^{-i \omega_e \hat{\sigma}_{+} \hat{\sigma}_{-} t}=\hat{\sigma}_{-} e^{-i \omega_e t} \\ & e^{i \omega_e \hat{\sigma}_{+} \hat{\sigma}_{-} t} \hat{\sigma}_{+} e^{-i \omega_e \hat{\sigma}_{+} \hat{\sigma}_{-} t}=\hat{\sigma}_{+} e^{i \omega_e t} \end{aligned}$$

这样就得到了：

$$\hat{U}_1^{\dagger} \hat{H}_V \hat{U}_1=g\left(\hat{a}^{\dagger} \hat{\sigma}_{-} e^{-i\left(\omega_e-\omega_0\right) t}+\hat{a} \hat{\sigma}_{+} e^{i\left(\omega_e-\omega_0\right) t}\right)$$

因此最后的哈密顿量为

$$\begin{aligned} & \hat{H}_{\mathrm{rot}}=i \hbar \frac{\partial \hat{U}_1^{\dagger}}{\partial t} \hat{U}_1+\hat{U}_1^{\dagger}\left(\hat{H}_0+\hat{H}_V\right) \hat{U}_1 \\ & =-\hat{H}_0+\hat{H}_0+g\left(\hat{a}^{\dagger} \hat{\sigma}_{-} e^{-i\left(\omega_e-\omega_0\right) t}+\hat{a} \hat{\sigma}_{+} e^{i\left(\omega_e-\omega_0\right) t}\right) \\ & =g\left(\hat{a}^{\dagger} \hat{\sigma}_{-} e^{-i\left(\omega_e-\omega_0\right) t}+\hat{a} \hat{\sigma}_{+} e^{i\left(\omega_e-\omega_0\right) t}\right) \end{aligned}$$

因此，对于Rotating Frame的哈密顿量，我们可以只关注，前两项始终会相互抵消，就像相互作用哈密顿量一样。

实际处理时，我们可能也会重新定义$\hat{U}1=e^{-i\hat{H}{\omega{e}}t/\hbar}\omega{e}$的频率下：

$$\hat{H}_{\omega_{e}}=\hbar \omega_e \hat{\sigma}_{+} \hat{\sigma}_{-}+\hbar \omega_e \hat{a}^{\dagger} \hat{a}$$

此时系统的哈密顿量应该表示为

$$\begin{aligned} & \hat{H}_{\mathrm{rot}}=i \hbar \frac{\partial \hat{U}_1^{\dagger}}{\partial t} \hat{U}_1+\hat{U}_1^{\dagger}\left(\hat{H}_0+\hat{H}_V\right) \hat{U}_1 \\ & =-\hat{H}_0+\hat{H}_0+\hbar(\omega_{0}-\omega_{e})\hat{a}^{\dagger} \hat{a}+g\left(\hat{a}^{\dagger} \hat{\sigma}_{-} e^{-i\left(\omega_e-\omega_0\right) t}+\hat{a} \hat{\sigma}_{+} e^{i\left(\omega_e-\omega_0\right) t}\right) \\ & =\hbar(\omega_{0}-\omega_{e})\hat{a}^{\dagger} \hat{a}+g\left(\hat{a}^{\dagger} \hat{\sigma}_{-} e^{-i\left(\omega_e-\omega_0\right) t}+\hat{a} \hat{\sigma}_{+} e^{i\left(\omega_e-\omega_0\right) t}\right) \end{aligned}$$