本文主要讨论Magnus Expansion。

  量子力学中的时间演化算符为： $$ \hat{U}(t,t_0)=\overleftarrow{T}\exp\left[\int_{t_0}^t i\frac{\hat{H}(t’)}{\hbar}\mathrm{d}t’\right] $$

\(\overleftarrow{T}\)为时序算符(time-ordering operator)，

  我们希望将其写成如下形式： $$ \hat{U}(t,t_0)=\exp\left[\sum_{n=1}^\infty\hat{\Omega}_n(t,t_0)\right] $$

  时间演化算符满足 $$ i\hbar\frac{\mathrm{d}\hat{U}}{\mathrm{d}t}=\hat{H}\hat{U} $$

  改写上式，以便以后做微扰 $$ i\hbar\frac{\mathrm{d}\hat{U}}{\mathrm{d}t}=\lambda\hat{H}\hat{U} $$

并假设 $$ \hat{U}=1+\sum_{n=1}^\infty\lambda^n\hat{P}_n $$

于是得到： $$ i\hbar\sum_{n=1}^\infty\lambda^n\frac{\partial\hat{P}_n}{\partial t}=\lambda\hat{H}\hat{U}=\lambda\hat{H}[1+\sum_{n=1}^\infty\lambda^n\hat{P}_n] $$

对\(\lambda\)逐项对比，得到： $$ i\hbar\frac{\partial\hat{P}_1}{\partial t}=\hat{H} $$

$$ i\hbar\frac{\partial\hat{P}_n}{\partial t}=\hat{H}\hat{P}_{n-1} $$

即 $$ \hat{P}_1(t,t_0)=-i\int_{t_0}^t\frac{\hat{H}(t’)}{\hbar}\mathrm{d}t' $$

$$ \hat{P}_n(t,t_0)=(-i)^n\int_{t_0}^t\dots\int_{t_0}^{t_{n-1}}\frac{\hat{H}(t_1)\dots\hat{H}(t_n)}{\hbar^n}\mathrm{d}t_n\dots\mathrm{d}t_1 $$

  假设\(\hat{U}\)能写成： $$ \hat{U}=\exp\left[\sum_{n=1}^\infty\lambda^n\hat{\Omega}_n\right]=\exp\hat{\Omega} $$

于是有： $$\begin{align} \hat{\Omega}=\sum_{n=1}^\infty\lambda^n\hat{\Omega}_n&=\ln\hat{U}\\ &=\ln(1+\sum_{n=1}^\infty\lambda^n\hat{P}_n)\\ &=\sum_n\lambda^n\hat{P}_n-\frac{1}{2}\sum_{m,n}(\lambda^n\hat{P}_n)(\lambda^m\hat{P}_m)+\frac{1}{3}\sum_{m,n,k}(\lambda^n\hat{P}_n)(\lambda^m\hat{P}_m)(\lambda^k\hat{P}_k) \end{align}$$

  最后一步保留三项，于是可以得到： $$ \hat{\Omega}_1=\hat{P}_1=-\frac{i}{\hbar}\int\mathrm{d}t\hat{H}(t) $$ $$ \hat{\Omega}_2=\hat{P}_2-\frac{1}{2}\hat{P}^2_1=-\frac{1}{2\hbar^2}\iint\mathrm{d}t_1\mathrm{d}t_2[\hat{H}(t_1),\hat{H}(t_2)] $$ $$\begin{align} \hat{\Omega}_3&=\hat{P}_3-\frac{1}{2}(\hat{P}_1\hat{P}_2+\hat{P}_2\hat{P}_1)+\frac{1}{3}\hat{P}^3_1\\ &=\frac{i}{6\hbar^3}\iiint\mathrm{d}t_1\mathrm{d}t_2\mathrm{d}t_3\left(\left[\hat{H}(t_1),[\hat{H}(t_2),\hat{H}(t_3)]\right]+\left[[\hat{H}(t_1),\hat{H}(t_2)],\hat{H}(t_3)\right]\right) \end{align}$$

  如果不同时刻的Hamilton量对易，则回到： $$ \hat{U}=\exp\left[-\frac{i}{\hbar}\int\mathrm{d}t\hat{H}(t)\right] $$