本文主要讨论Magnus Expansion。

  量子力学中的时间演化算符为： $$\hat{U}(t,t_0)=\overleftarrow{T}\exp \left[{\int }_{t_0}^{t}i\frac{\hat{H}(t^{\prime })}{\hslash }\mathrm{d}{t}^{\prime }\right]$$

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

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

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

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

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

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

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

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

即 $${\hat{P}}_1(t,t_0)=-i{\int }_{t_0}^t\frac{\hat{H}(t^{\prime })}{\hslash }\mathrm{d}{t}^{\prime }$$

$${\hat{P}}_n(t,t_0)=(-i{)}^n{\int }_{t_0}^t\dots {\int }_{t_0}^{t_{n-1}}\frac{\hat{H}(t_1)\dots \hat{H}(t_n)}{{\hslash }^n}\mathrm{d}{t}_n\dots \mathrm{d}{t}_1$$

  假设$\hat{U}$能写成： $$\hat{U}=\exp \left[\sum _{n=1}^{\mathrm{\infty }}{\lambda }^n{\hat{\mathrm{\Omega }}}_n\right]=\exp \hat{\mathrm{\Omega }}$$

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

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

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