盖尔曼–劳定理

设 ${\displaystyle |\Psi _{0}\rangle }$ 是 ${\displaystyle H_{0}}$ 的一个本征态，能量为 ${\displaystyle E_{0}}$ 。定义相互作用的哈密顿量为 ${\displaystyle H=H_{0}+gV}$，其中 ${\displaystyle g}$ 是耦合常数， ${\displaystyle V}$ 是相互作用项，定义带参量的哈密顿 ${\displaystyle H_{\epsilon }=H_{0}+e^{-\epsilon |t|}gV}$ ，可以看到，当 ${\displaystyle |t|\rightarrow \infty }$ 时，${\displaystyle H_{\epsilon }=H_{0}}$。而当${\displaystyle t=0}$时，${\displaystyle H_{\epsilon }=H}$。令 ${\displaystyle U_{\epsilon I}}$ 为对应于${\displaystyle H_{\epsilon }}$的相互作用繪景（用下标I表示）下的时间演化算符。盖尔曼–劳定理说的是，若 ${\displaystyle \epsilon \rightarrow 0^{+}}$ 时，

${\displaystyle |\Psi _{\epsilon }^{(\pm )}\rangle ={\frac {U_{\epsilon I}(0,\pm \infty )|\Psi _{0}\rangle }{\langle \Psi _{0}|U_{\epsilon I}(0,\pm \infty )|\Psi _{0}\rangle }}}$

的极限存在，则 ${\displaystyle |\Psi _{\epsilon }^{(\pm )}\rangle }$ 就是 ${\displaystyle H}$ 的本征态。

注意当${\displaystyle \epsilon \rightarrow 0^{+}}$时，随着时间t的增加，相互作用项实际上是以无限慢的速度引入的，这称为绝热连续过程，此时称${\displaystyle H_{\epsilon }}$构成${\displaystyle H_{0}}$与${\displaystyle H}$之间的一个绝热连接。

盖尔曼–劳定理本身并没有说当${\displaystyle |\Psi _{0}\rangle }$是基态时，${\displaystyle |\Psi _{\epsilon }^{(\pm )}\rangle }$也是基态，也就是说，没有排除能级在绝热连接时发生交叉的可能。不过，在微扰论的条件满足的前提下，一般认为当${\displaystyle |\Psi _{0}\rangle }$为基态时，${\displaystyle |\Psi _{\epsilon }^{(\pm )}\rangle }$也是基态。

原始的论文是通过演化算符的戴森展开式来完成证明的，而Molinari则将其有效性推广到微扰论成立的范围之外。下面介绍Molinari的方法。在 ${\displaystyle H_{\epsilon }}$ 中令 ${\displaystyle g=e^{\epsilon \theta }}$，由时间演化算符满足的薛定谔方程

${\displaystyle i\hbar \partial _{t_{1}}U_{\epsilon }(t_{1},t_{2})=H_{\epsilon }(t_{1})U_{\epsilon }(t_{1},t_{2})}$

及条件 ${\displaystyle U_{\epsilon }(t,t)=1}$，可以写出方程的形式解

${\displaystyle U_{\epsilon }(t_{1},t_{2})=1+{\frac {1}{i\hbar }}\int _{t_{2}}^{t_{1}}dt'(H_{0}+e^{\epsilon (\theta -|t'|)}V)U_{\epsilon }(t',t_{2}).}$

先集中考虑 ${\displaystyle 0\geq t_{2}\geq t_{1}}$ 的情形，换元后得到，

${\displaystyle U_{\epsilon }(t_{1},t_{2})=1+{\frac {1}{i\hbar }}\int _{\theta +t_{2}}^{\theta +t_{1}}dt'(H_{0}+e^{\epsilon t'}V)U_{\epsilon }(t'-\theta ,t_{2}).}$

于是有

${\displaystyle \partial _{\theta }U_{\epsilon }(t_{1},t_{2})=\epsilon g\partial _{g}U_{\epsilon }(t_{1},t_{2})=\partial _{t_{1}}U_{\epsilon }(t_{1},t_{2})+\partial _{t_{2}}U_{\epsilon }(t_{1},t_{2}).}$

将上式与前面提到的薛定谔方程及其伴式

${\displaystyle -i\hbar \partial _{t_{1}}U_{\epsilon }(t_{2},t_{1})=U_{\epsilon }(t_{2},t_{1})H_{\epsilon }(t_{1})}$

结合就有，

${\displaystyle i\hbar \epsilon g\partial _{g}U_{\epsilon }(t_{1},t_{2})=H_{\epsilon }(t_{1})U_{\epsilon }(t_{1},t_{2})-U_{\epsilon }(t_{1},t_{2})H_{\epsilon }(t_{2}).}$

${\displaystyle H_{\epsilon I}}$与${\displaystyle U_{\epsilon I}}$ 之间的关系式形式上与上式相同，事实上，将上式两边各左乘 ${\displaystyle e^{iH_{0}t_{1}/\hbar }}$，右乘 ${\displaystyle e^{iH_{0}t_{2}/\hbar }}$ ，并利用关系

${\displaystyle U_{\epsilon I}(t_{1},t_{2})=e^{iH_{0}t_{1}/\hbar }U_{\epsilon }(t_{1},t_{2})e^{-iH_{0}t_{2}/\hbar }.}$

就可以得到${\displaystyle H_{\epsilon I}}$与${\displaystyle U_{\epsilon I}}$ 之间的关系式。

现在，令${\displaystyle t_{1}=0,t_{2}=+\infty }$，等式两边作用在${\displaystyle |\Psi _{0}\rangle }$上，并注意到${\displaystyle |\Psi _{0}\rangle }$是${\displaystyle H_{\epsilon }(+\infty )=H_{0}}$的本征态，就有

${\displaystyle \left(H_{\epsilon ,t=0}-E_{0}+i\hbar \epsilon g\partial _{g}\right)U_{\epsilon I}(0,\infty )|\Psi _{0}\rangle =0.}$

对于时间为负值的情况，证明完全类似，最后就得到，

${\displaystyle \left(H_{\epsilon ,t=0}-E_{0}\pm i\hbar \epsilon g\partial _{g}\right)U_{\epsilon I}(0,\pm \infty )|\Psi _{0}\rangle =0.}$

下面以时间为负值为例继续证明，为清晰起见，先把算符写成简略形式，即将${\displaystyle U_{\epsilon I}(0,-\infty )}$简写作${\displaystyle U}$。

${\displaystyle i\hbar \epsilon g\partial _{g}\left(U|\Psi _{0}\rangle \right)=(H_{\epsilon }-E_{0})U|\Psi _{0}\rangle .}$

下面计算 ${\displaystyle i\hbar \epsilon g\partial _{g}|\Psi _{\epsilon }^{-}\rangle }$，把${\displaystyle |\Psi _{\epsilon }^{-}\rangle }$的定义式代入，并利用上面的关系式，可得，

${\displaystyle {\begin{aligned}i\hbar \epsilon g\partial _{g}|\Psi _{\epsilon }^{-}\rangle &={\frac {1}{\langle \Psi _{0}|U|\Psi _{0}\rangle }}(H_{\epsilon }-E_{0})U|\Psi _{0}\rangle -{\frac {U|\Psi _{0}\rangle }{{\langle \Psi _{0}|U|\Psi _{0}\rangle }^{2}}}\langle \Psi _{0}|H_{\epsilon }-E_{0}|\Psi _{0}\rangle \\&=(H_{\epsilon }-E_{0})|\Psi _{\epsilon }^{-}\rangle -|\Psi _{\epsilon }^{-}\rangle \langle \Psi _{0}|H_{\epsilon }-E_{0}|\Psi _{\epsilon }^{-}\rangle \\&=\left|\Psi _{\epsilon }^{-}\rangle .\end{aligned}}}$

式中 ${\displaystyle E^{-}=E_{0}+\langle \Psi _{0}|H_{\epsilon }-H_{0}|\Psi _{\epsilon }^{-}\rangle }$.

即

${\displaystyle \left|\Psi _{\epsilon }^{-}\rangle =0.}$

类似地可证明 ${\displaystyle |\Psi _{\epsilon }^{+}\rangle }$的关系式，综合起来可写成：

${\displaystyle \left|\Psi _{\epsilon }^{\pm }\rangle =0}$

然后取${\displaystyle \epsilon \rightarrow 0^{+}}$的极限，即可证明${\displaystyle |\Psi _{\epsilon }^{\pm }\rangle }$是${\displaystyle H}$的本征函数，本征值分别为${\displaystyle E^{\pm }}$。

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