Bogolyubov变换

这只是一个小技巧啦，用于将 $\hat a^\dagger\hat b^\dagger,\hat a\hat b$ 形式对角化

单态情形

简单起见先计算单态情况，根据哈密顿量厄米性和$[\hat a,\hat a^\dagger]=1$假设

$$H=p\hat a^\dagger\hat a+q\left(\hat a^\dagger\hat a^\dagger+\hat a\hat a\right)\tag{1.1}$$

1写为矩阵形式

$$H=\left(\hat a,\hat a^\dagger\right)\left(\begin{array}{cc}0&q\\q&p\end{array}\right)\left(\begin{array}{c}\hat a^\dagger\\\hat a\end{array}\right)\equiv a^\dagger ha\tag{1.2}$$

我们希望引入变换$U$使得$H=a^\dagger U^\dagger [(U^\dagger)^{-1}hU^{-1}]Ua$括号内为对角，并且$b=Ua$满足

$$[\hat b,\hat b^\dagger]=[u_{21}\hat a^\dagger+u_{22}\hat a,u_{11}\hat a^\dagger+u_{12}\hat a]=u_{11}u_{22}-u_{21}u_{22}=1\tag{1.3}$$

根据$\hat b,\hat b^\dagger$厄米共轭可知$u_{21}^=u_{12},u_{22}^=u_{11}$。所以可设

$$U=\left(\begin{array}{cc} e^{i\alpha}\cosh\theta & e^{i\beta}\sinh\theta\\ e^{-i\beta}\sinh\theta & e^{-i\alpha}\cosh\theta \end{array}\right)\Rightarrow U^{-1}=\left(\begin{array}{cc} e^{-i\alpha}\cosh\theta & -e^{i\beta}\sinh\theta \\ -e^{-i\beta}\sinh\theta & e^{i\alpha}\cosh\theta \end{array}\right)\tag{1.4}$$

所以代入(1.2)得$\alpha=\beta=0$时

$$H=\frac12b^\dagger\left(\begin{array}{cc} p\cosh2\theta-2q\sinh2\theta-p & 2q\cosh2\theta-p\sinh2\theta \\ 2q\cosh2\theta-p\sinh2\theta & p\cosh2\theta-2q\sinh2\theta+p \end{array}\right)b$$

可选取$\tanh2\theta=\frac{2q}{p}$使得对角，并有

$$\sinh2\theta=\frac{2q}{\sqrt{p^2-4q^2}},\cosh2\theta=\frac{p}{\sqrt{p^2-4q^2}}$$

所以

$$H=\frac12\hat b\left(\sqrt{p^2-4q^2}-p\right)\hat b^\dagger +\frac12\hat b^\dagger\left(\sqrt{p^2-4q^2}+p\right)\hat b\tag{1.5}$$

或者写为

$$H=\sqrt{p^2-4q^2}\hat b^\dagger\hat b+E_0\tag{1.6}$$

需要注意一个问题，$\tanh2\theta\in(-1,1)$，要求$|2q|\le|p|$，其物理意义尚不明确

双态情形

其实往往见到的是两套产生-湮灭，记为$\hat a,\hat b$。由厄米性可知，哈密顿量可写为

$$H=p_1\hat a^\dagger\hat a+p_2\hat b^\dagger\hat b+q(\hat a^\dagger\hat b^\dagger+\hat a\hat b)+s(\hat a^\dagger\hat b+\hat a\hat b^\dagger) \tag{2.1}$$

$\hat a^\dagger\hat a^\dagger+\hat a\hat a,\hat b^\dagger\hat b^\dagger+\hat b\hat b$项可在单态情形中消去，因此不写出。$\hat a^\dagger\hat b+\hat a\hat b^\dagger$项可通过重新定义$a,b$消去，即

$$\left(\hat a^\dagger,\hat b^\dagger\right)\left(\begin{array}{cc}p_1 & s\\s&p_2\end{array}\right)\left(\begin{array}{c}\hat a\\\hat b\end{array}\right)=\lambda_1\hat{\tilde a}^\dagger\hat{\tilde a}+\lambda_2\hat{\tilde b}^\dagger\hat{\tilde b}$$

实对称矩阵可以通过正交矩阵对角化，所以变换后的$a,b$仍然满足标准的对易关系。因此(2.1)可设$s=0$

$$H=\left(\hat a^\dagger,\hat b\right)\left(\begin{array}{cc}p_1 & q\\q&p_2\end{array}\right)\left(\begin{array}{c}\hat a\\\hat b^\dagger\end{array}\right)-p_2\tag{2.2}$$

假设变换关系为

$$\left(\begin{array}c\hat\alpha\\\hat \beta^\dagger\end{array}\right)=\left(\begin{array}{cc}u_{11} & u_{12}\\u_{21}&u_{22}\end{array}\right)\left(\begin{array}{c}\hat a\\\hat b^\dagger\end{array}\right)\tag{2.3}$$

则显式得写出来

$$\left\{\begin{aligned} &\hat\alpha=u_{11}\hat a+u_{12}\hat b^\dagger\\ &\hat\beta^\dagger=u_{21}\hat a+u_{22}\hat b^\dagger\\ &\hat\alpha^\dagger=u_{11}^*\hat a^\dagger+u_{12}^*\hat b\\ &\hat\beta=u_{21}^*\hat a^\dagger+u_{22}^*\hat b\\ \end{aligned}\right.\Rightarrow \left\{\begin{aligned} &\left[\hat\alpha,\hat\alpha^\dagger\right]=|u_{11}|^2-|u_{12}|^2=1\\ &\left[\hat\beta,\hat\beta^\dagger\right]=|u_{22}|^2-|u_{21}|^2=1\\ &\left[\hat\alpha,\hat\beta\right]=u_{11}u_{21}^*-u_{12}u_{22}^*=0 \end{aligned}\right.\tag{2.4}$$

当然还有其他对易关系，但是要么是显然的，要么包括在以上三个限制条件中了。

不妨设

$$\left(\begin{array}c\hat\alpha\\\hat \beta^\dagger\end{array}\right)=\left(\begin{array}{cc}\cosh\theta & \sinh\theta\\\sinh\theta&\cosh\theta\end{array}\right)\left(\begin{array}{c}\hat a\\\hat b^\dagger\end{array}\right)$$

回代(2.2)，和单态结果几乎完全一样嘛，得到

$$\begin{aligned} H&=\frac12\hat\alpha^\dagger\left(\sqrt{(p_1+p_2)^2-(q_1+q_2)^2}+(p_1-p_2)\right)\hat\alpha\\ &+\frac12\hat\beta\left(\sqrt{(p_1+p_2)^2-(q_1+q_2)^2}-(p_1-p_2)\right)\hat\beta^\dagger\\ &-p_2\\ &=\frac12\left(\sqrt{(p_1+p_2)^2-(q_1+q_2)^2}+(p_1-p_2)\right)\hat\alpha^\dagger\hat\alpha\\ &+\frac12\left(\sqrt{(p_1+p_2)^2-(q_1+q_2)^2}-(p_1-p_2)\right)\hat\beta^\dagger\hat\beta\\ &+\left(\sqrt{(p_1+p_2)^2-(q_1+q_2)^2}-p_1\right) \end{aligned}$$