贡献者： addis
\begin{equation}
E_n^2 = \sum_{m}^{E_m\ne E_n} \frac{ \left\lvert \left\langle \psi_m^0 \middle| H^1 \middle| \psi_n^0 \right\rangle \right\rvert ^2}{E_n^0-E_m^0}~.
\end{equation}
注意 $\psi_n^0$ 必须取好量子态。
1. 推导
类似于一阶微扰的推导（子节 4 ），若式 6 展开括号后仅保留 $\lambda^2$ 的项，得
\begin{equation}
H^0\psi_n^2 + H^1\psi_n^1 = E_n^0\psi_n^2 + E_n^1\psi_n^1 + E_n^2\psi_n^0~.
\end{equation}
两边左乘任意 $\psi_m^0$ 得
\begin{equation}
\left\langle \psi_m^0 \middle| H^0 \middle| \psi_n^2 \right\rangle + \left\langle \psi_m^0 \middle| H^1 \middle| \psi_n^1 \right\rangle = E_n^0 \left\langle \psi_m^0 \middle| \psi_n^2 \right\rangle + E_n^1 \left\langle \psi_m^0 \middle| \psi_n^1 \right\rangle + \delta_{m,n}E_n^2~.
\end{equation}
其中
\begin{equation}
\left\langle \psi_m^0 \middle| H^0 \middle| \psi_n^2 \right\rangle = \left\langle H^0\psi_m^0 \middle| \psi_n^2 \right\rangle = E_m^0 \left\langle \psi_m^0 \middle| \psi_n^2 \right\rangle ~.
\end{equation}
代入
式 3 得
\begin{equation}
\left\langle \psi_m^0 \middle| H^1 \middle| \psi_n^1 \right\rangle = (E_n^0-E_m^0) \left\langle \psi_m^0 \middle| \psi_n^2 \right\rangle + E_n^1 \left\langle \psi_m^0 \middle| \psi_n^1 \right\rangle + \delta_{m,n}E_n^2~.
\end{equation}
我们要求该式对所有可能的 $m,n$ 都成立。
非简并
考虑对角元（$m=n$），式 5 要求
\begin{equation}
E_n^2 = \left\langle \psi_n^0 \middle| H^1 \middle| \psi_n^1 \right\rangle - E_n^1 \left\langle \psi_n^0 \middle| \psi_n^1 \right\rangle ~.
\end{equation}
把波函数的一阶微扰（
式 8 ）
\begin{equation}
\psi_n^1 = \sum_m^{E_m\ne E_n} \frac{ \left\langle \psi_m^0 \middle| H^1 \middle| \psi_n^0 \right\rangle }{E_n^0 - E_m^0}\psi_m^0~.
\end{equation}
代入
式 6 发现第二项为 0，得二阶能量修正（
式 1 ）
\begin{equation}
E_n^2 = \left\langle \psi_n^0 \middle| H^1 \middle| \psi_n^1 \right\rangle = \sum_{m}^{E_m\ne E_n} \frac{ \left\lvert \left\langle \psi_m^0 \middle| H^1 \middle| \psi_n^0 \right\rangle \right\rvert ^2}{E_n^0-E_m^0}~.
\end{equation}
式 5 要求所有非对角元（$m\ne n$）满足
\begin{equation} \left\langle \psi_m^0 \middle| \psi_n^2 \right\rangle =
\frac{1}{E_n^0-E_m^0}\sum_{m'}^{E_{m'}\ne E_n}\frac{ \left\langle \psi_m^0 \middle| H^1 \middle| \psi_{m'}^0 \right\rangle \left\langle \psi_{m'}^0 \middle| H^1 \middle| \psi_n^0 \right\rangle }{E_n^0 - E_{m'}^0} - \frac{E_n^1 \left\langle \psi_m^0 \middle| H^1 \middle| \psi_n^0 \right\rangle }{(E_n^0 - E_m^0)^2}~.
\end{equation}
同样假设 $m=n$ 时上式为 0，就得到了二阶波函数修正 $\psi_n^2$。
二阶能量修正的意义
根据式 6 有
\begin{equation}
\begin{aligned}
E_n &= E_n^0 + \lambda E_n^1 + \lambda^2 E_n^2 + \dots \\
&= \left\langle \psi_n^0 + \lambda\psi_n^1 + \lambda^2\psi_n^2 + \dots \middle| H^0 + \lambda H^1 \middle| \psi_n^0 + \lambda\psi_n^1 + \lambda^2 \psi_n^2 + \dots \right\rangle ~.
\end{aligned}
\end{equation}
右边展开后如果按 $\lambda$ 的幂合并同类项，是否会和左边的各项对应？容易验证这对 $E_n^0$ 和 $E_n^1$ 来说都是正确的。但对 $\lambda^2$ 有
\begin{equation}
E_n^2 = 2 \left\langle \psi_n^0 \middle| H^1 \middle| \psi_n^1 \right\rangle + \left\langle \psi_n^1 \middle| H^0 \middle| \psi_n^1 \right\rangle + 2 \left\langle \psi_n^0 \middle| H^0 \middle| \psi_n^2 \right\rangle ~.
\end{equation}
如果这成立，那么对比
式 8 ，就意味着
\begin{equation}
E_n^2 + \left\langle \psi_n^1 \middle| H^0 \middle| \psi_n^1 \right\rangle + 2 \left\langle \psi_n^0 \middle| H^0 \middle| \psi_n^2 \right\rangle = 0~.
\end{equation}
这是否可以证明成立？把
式 7 代入可得
\begin{equation}
\begin{aligned}
\left\langle \psi_n^1 \middle| H^0 \middle| \psi_n^1 \right\rangle &= \sum_{m}^{E_m\ne E_n} \frac{ \left\lvert \left\langle \psi_m^0 \middle| H^1 \middle| \psi_n^0 \right\rangle \right\rvert ^2}{E_n^0 - E_m^0} \frac{E_m^0}{E_n^0 - E_m^0} = -E_n^2 + E_n^0 \left\langle \psi_n^2 \middle| \psi_n^2 \right\rangle ~.
\end{aligned}
\end{equation}
最后一项
\begin{equation}
\left\langle \psi_n^0 \middle| H^0 \middle| \psi_n^2 \right\rangle = E_n^0 \left\langle \psi_n^0 \middle| \psi_n^2 \right\rangle = 0~.
\end{equation}
也就是说
式 11 右边多出了一个 $E_n^0 \left\langle \psi_n^2 \middle| \psi_n^2 \right\rangle $ 项，属于 $\lambda^4$ 数量级。