非简并微扰论的一二级修正梳理

原文: https://zhuanlan.zhihu.com/p/51579444
发布日期: 2018-12-07
分类: 量子力学基础与量子信息

非简并微扰论的一二级修正:

已知 $H^{(0)} ψ_{n}^{(0)} ⟩ = E_{n}^{(0)} ψ_{n}^{(0)} ⟩$ 是原始体系的哈密顿本征方程.

下面找不到[a][ii]等等式子就请用ctrl+f搜索.

所谓微扰就是在哈密顿量中加入一个很小的哈密顿量 ${H}'}$ .

本征方程就应该是 $(H^{(0)} + H^{'}) ∣ ψ_{n} ⟩ = E_{n} ∣ ψ_{n} ⟩$ ------[a]

因为 ${H}'}$ 很小所以引起的态与能量的变化也是很小的.

可以将新的态与能量像下面这样进行多级展开表示:

$\left\{ \begin{align} & \left| {\psi }_{n} \right\rangle =\left| \psi _{n}^{\left( 0 \right)} \right\rangle +\left| \psi _{n}^{\left( 1 \right)} \right\rangle +\left| \psi _{n}^{\left( 2 \right)} \right\rangle +\cdot \cdot \cdot \\ & {E}_{n}=E_{n}^{\left( 0 \right)}+E_{n}^{\left( 1 \right)}+E_{n}^{\left( 2 \right)}+\cdot \cdot \cdot \\ \end{align} \right.$ -----[b]

若将引入的 ${H}'}$ 看作无穷小修正, 则态和能量后面的项就是前面的高阶无穷小这种感觉. 即 $∣ ψ_{n} ⟩ \approx ψ_{n}^{(0)} ⟩$ 且 $ψ_{n}^{(0)} ⟩ ≫ ψ_{n}^{(1)} ⟩ ≫ ψ_{n}^{(2)} ⟩ ≫ \cdot \cdot \cdot$ (能量情况类似)

所以将**[b]带入[a]**既是:

$\begin{align} & \ \ \ \ \ \left[ {H}^{\left( 0 \right)}+{H}' \right]\left[ \left| \psi _{n}^{\left( 0 \right)} \right\rangle +\left| \psi _{n}^{\left( 1 \right)} \right\rangle +\left| \psi _{n}^{\left( 2 \right)} \right\rangle +\cdot \cdot \cdot \right] \\ & =\left[ E_{n}^{\left( 0 \right)}+E_{n}^{\left( 1 \right)}+E_{n}^{\left( 2 \right)}+\cdot \cdot \cdot \right]\left[ \left| \psi _{n}^{\left( 0 \right)} \right\rangle +\left| \psi _{n}^{\left( 1 \right)} \right\rangle +\left| \psi _{n}^{\left( 2 \right)} \right\rangle +\cdot \cdot \cdot \right] \\ \end{align}$

将他们展开后把等号右边的移到左边, 然后再将同一数量级的项放到一起:

会写成如下形式: $\left[ \cdot \cdot \cdot \right]+\left[ \cdot \cdot \cdot \right]\varepsilon +\left[ \cdot \cdot \cdot \right]{\varepsilon }^{\text{2}+\cdot \cdot \cdot =0$ ( $\varepsilon$ 象征无穷小) 值得一提的是要将 ${H}'}$ 视作一阶项, 即与 $E_{n}^{(1)}$ 或 $ψ_{n}^{(1)} ⟩$ 同级

实际上写出来是这样的:

$\begin{align} & \ \ \ \ \left[ {H}^{\left( 0 \right)}\left| \psi _{n}^{\left( 0 \right)} \right\rangle -E_{n}^{\left( 0 \right)}\left| \psi _{n}^{\left( 0 \right)} \right\rangle \right] \\ & +\left[ {H}^{\left( 0 \right)}\left| \psi _{n}^{\left( 1 \right)} \right\rangle +{H}'\left| \psi _{n}^{\left( 0 \right)} \right\rangle -E_{n}^{\left( 0 \right)}\left| \psi _{n}^{\left( 1 \right)} \right\rangle -E_{n}^{\left( 1 \right)}\left| \psi _{n}^{\left( 0 \right)} \right\rangle \right] \\ & +\left[ {H}^{\left( 0 \right)}\left| \psi _{n}^{\left( 2 \right)} \right\rangle +{H}'\left| \psi _{n}^{\left( 1 \right)} \right\rangle -E_{n}^{\left( 0 \right)}\left| \psi _{n}^{\left( 2 \right)} \right\rangle -E_{n}^{\left( 1 \right)}\left| \psi _{n}^{\left( 1 \right)} \right\rangle -E_{n}^{\left( 2 \right)}\left| \psi _{n}^{\left( 0 \right)} \right\rangle \right] \\ & +\cdot \cdot \cdot \\ & =0 \\ \end{align}$

整体等于 0 即意味着各个数量级之和均是0, 故可以得到:

$\left\{ \begin{align} & {H}^{\left( 0 \right)}\left| \psi _{n}^{\left( 0 \right)} \right\rangle -E_{n}^{\left( 0 \right)}\left| \psi _{n}^{\left( 0 \right)} \right\rangle =0 \\ & {H}^{\left( 0 \right)}\left| \psi _{n}^{\left( 1 \right)} \right\rangle +{H}'\left| \psi _{n}^{\left( 0 \right)} \right\rangle -E_{n}^{\left( 0 \right)}\left| \psi _{n}^{\left( 1 \right)} \right\rangle -E_{n}^{\left( 1 \right)}\left| \psi _{n}^{\left( 0 \right)} \right\rangle =0 \\ & {H}^{\left( 0 \right)}\left| \psi _{n}^{\left( 2 \right)} \right\rangle +{H}'\left| \psi _{n}^{\left( 1 \right)} \right\rangle -E_{n}^{\left( 0 \right)}\left| \psi _{n}^{\left( 2 \right)} \right\rangle -E_{n}^{\left( 1 \right)}\left| \psi _{n}^{\left( 1 \right)} \right\rangle -E_{n}^{\left( 2 \right)}\left| \psi _{n}^{\left( 0 \right)} \right\rangle =0 \\ &\cdot\cdot\cdot\\ \end{align} \right.$

上述括号内的式子我们自上而下分别称之为 [i] [ii] [iii] ...

我们的目标就是求得 $E_{n}^{(1)}, E_{n}^{(2)} / ψ_{n}^{(1)} ⟩, ψ_{n}^{(2)} ⟩$ 的解析式, 即能量/状态的一二级修正.

值得一提的是, 我们接下来自然是选择 ${ψ_{n}^{(0)} ⟩}$ 作为基矢量, 因为已知条件里面只有它.

A.求解 $E_{n}^{(1)}$ :

将**[ii]**投影到 $ψ_{n}^{(0)} ⟩$ 上有:

$⟨ ψ_{n}^{(0)} H^{(0)} ψ_{n}^{(1)} ⟩ + ⟨ ψ_{n}^{(0)} H^{'} ψ_{n}^{(0)} ⟩ - ⟨ ψ_{n}^{(0)} E_{n}^{(0)} ψ_{n}^{(1)} ⟩ - ⟨ ψ_{n}^{(0)} E_{n}^{(1)} ψ_{n}^{(0)} ⟩ = 0$

化简得到: $E_{n}^{\left( 1 \right)}={H}'}_{nn}$ 即为所求.

B.求解 $ψ_{n}^{(1)} ⟩$ :

将 $ψ_{n}^{(1)} ⟩$ 用 ${ψ_{n}^{(0)} ⟩}$ 展开:

$ψ_{n}^{(1)} ⟩ = α \sum ⟨ ψ_{α}^{(0)} ∣ ψ_{n}^{(1)} ⟩ ψ_{α}^{(0)} ⟩$ ---[c]

可见要求的是展开系数 $⟨ ψ_{α}^{(0)} ∣ ψ_{n}^{(1)} ⟩$

将**[ii]**投影到 $ψ_{α}^{(0)} ⟩$ 上有:

$\left[ E_{\alpha }^{\left( 0 \right)}-E_{n}^{\left( 0 \right)} \right]\left\langle \psi _{\alpha }^{\left( 0 \right)} | \psi _{n}^{\left( 1 \right)} \right\rangle +{H}'}_{\alpha n}-E_{n}^{\left( 1 \right)}\left\langle \psi _{\alpha }^{\left( 0 \right)} | \psi _{n}^{\left( 0 \right)} \right\rangle =0$

不难看出, 想要展开系数的表达式的话, 则要求 $a \neq = n$ 这样第三项就为 0.

整理得到: $\left\langle \psi _{\alpha }^{\left( 0 \right)} | \psi _{n}^{\left( 1 \right)} \right\rangle =\frac{H}'}_{\alpha n}{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)}$

将上述结论代回展开式**[c]**得到的 $ψ_{n}^{(1)} ⟩ = α (\neq = n) \sum \frac{H}{^{'}}_{α n} E_{n}^{(0)} - E_{α}^{(0)} ψ_{α}^{(0)} ⟩$ 即为所求.

注意: $α \neq = n$ 可不能忘, 那么 $α = n$ 那一项怎么办? 这里全文最后会讲到.

C.求解 $E_{n}^{(2)}$ :

将**[iii]**投影到 $ψ_{n}^{(0)} ⟩$ 上有:

$E_{n}^{(0)} ⟨ ψ_{n}^{(0)} ∣ ψ_{n}^{(2)} ⟩ + ⟨ ψ_{n}^{(0)} H^{'} ψ_{n}^{(1)} ⟩ - E_{n}^{(0)} ⟨ ψ_{n}^{(0)} ∣ ψ_{n}^{(2)} ⟩ - E_{n}^{(1)} ⟨ ψ_{n}^{(0)} ∣ ψ_{n}^{(1)} ⟩ - E_{n}^{(2)} = 0$

整理得到: $E_{n}^{(2)} = ⟨ ψ_{n}^{(0)} H^{'} ψ_{n}^{(1)} ⟩$

将 $ψ_{n}^{(1)} ⟩$ 用 ${ψ_{n}^{(0)} ⟩}$ 展开后得到的

$E_{n}^{\left( 2 \right)}=\sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}\left\langle \psi _{n}^{\left( 0 \right)} \right|{H}'\left| \psi _{\alpha }^{\left( 0 \right)} \right\rangle }{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)}=\sum\limits_{\alpha \left( \ne n \right)}{\frac{\left| {H}'}_{\alpha n} \right|}^{2}{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)}$ 即为所求.

D.求解 $ψ_{n}^{(2)} ⟩$ :

如同B那样首先将 $ψ_{n}^{(2)} ⟩$ 展开:

$ψ_{n}^{(2)} ⟩ = β \sum ⟨ ψ_{β}^{(0)} ∣ ψ_{n}^{(2)} ⟩ ψ_{β}^{(0)} ⟩$ --------[d]

将**[iii]**投影到 $ψ_{β}^{(0)} ⟩$ 有:

$[E_{β}^{(0)} - E_{n}^{(0)}] ⟨ ψ_{β}^{(0)} ∣ ψ_{n}^{(2)} ⟩ + ⟨ ψ_{β}^{(0)} H^{'} ψ_{n}^{(1)} ⟩ - E_{n}^{(1)} ⟨ ψ_{β}^{(0)} ∣ ψ_{n}^{(1)} ⟩ - E_{n}^{(2)} ⟨ ψ_{β}^{(0)} ∣ ψ_{n}^{(0)} ⟩ = 0$

目的求得 $⟨ ψ_{β}^{(0)} ∣ ψ_{n}^{(2)} ⟩$ 就要求 $β \neq = n$ , 这样最后一项即为0.

展开 $ψ_{n}^{(1)} ⟩$ , $E_{n}^{(1)}$ 整理得:

$\left[ E_{\beta }^{\left( 0 \right)}-E_{n}^{\left( 0 \right)} \right]\left\langle \psi _{\beta }^{\left( 0 \right)} | \psi _{n}^{\left( 2 \right)} \right\rangle =\frac{H}'}_{nn}{H}'}_{\beta n}{E_{n}^{\left( 0 \right)}-E_{\beta }^{\left( 0 \right)}-\sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}{H}'}_{\beta \alpha }{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)}$

$\left\langle \psi _{\beta }^{\left( 0 \right)} | \psi _{n}^{\left( 2 \right)} \right\rangle =\sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}{H}'}_{\beta \alpha }{\left[ E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)} \right]\left[ E_{n}^{\left( 0 \right)}-E_{\beta }^{\left( 0 \right)} \right]}-\frac{H}'}_{nn}{H}'}_{\beta n}{\left[ E_{n}^{\left( 0 \right)}-E_{\beta }^{\left( 0 \right)} \right]}^{2}$

则展开式**[d]**为:

$\left| \psi _{n}^{\left( \text{2} \right)} \right\rangle \text{=}\sum\limits_{\beta \left( \ne n \right)}{\left[ \sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}{H}'}_{\beta \alpha }{\left[ E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)} \right]\left[ E_{n}^{\left( 0 \right)}-E_{\beta }^{\left( 0 \right)} \right]}-\frac{H}'}_{nn}{H}'}_{\beta n}{\left[ E_{n}^{\left( 0 \right)}-E_{\beta }^{\left( 0 \right)} \right]}^{2} \right]\left| \psi _{\beta }^{\left( 0 \right)} \right\rangle }\text{+}\left\langle \psi _{n}^{\left( 0 \right)} | \psi _{n}^{\left( 2 \right)} \right\rangle \left| \psi _{n}^{\left( 0 \right)} \right\rangle$

注意这里同样有 $β \neq = n$ , 但我们不能像一阶修正那样通过修改整体相位[1]直接消除这一项,

不过可以通过归一化关系计算 $β = n$ 项的系数 $⟨ ψ_{n}^{(0)} ∣ ψ_{n}^{(2)} ⟩$ :

$∣ ψ_{n} ⟩ = ψ_{n}^{(0)} ⟩ + ψ_{n}^{(1)} ⟩ + ψ_{n}^{(2)} ⟩ + o (ψ_{n}^{(2)} ⟩)$ 是已知的,

应该注意到下面两点:

❶. $∣ ψ_{n} ⟩$ $ψ_{n}^{(0)} ⟩$ 均是正交归一的态, 因为二者都是确定的 $H$ 的本征方程的解.

❷. $ψ_{n}^{(1)} ⟩$ $ψ_{n}^{(2)} ⟩$ ... $ψ_{n}^{(n)} ⟩$ 都是修正项, 都不是归一的, 甚至不能说是一个态.[2]

综上可得:

下面先是省略了 $o (ψ_{n}^{(2)} ⟩)$ 带来的三阶项, 再又删除了所有的大于二阶的项

$\begin{align} & \left\langle {\psi }_{n} | {\psi }_{n} \right\rangle =\left\langle \psi _{n}^{\left( 0 \right)} | \psi _{n}^{\left( 0 \right)} \right\rangle +\left\langle \psi _{n}^{\left( 0 \right)} | \psi _{n}^{\left( 1 \right)} \right\rangle +\left\langle \psi _{n}^{\left( 0 \right)} | \psi _{n}^{\left( 2 \right)} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ +\left\langle \psi _{n}^{\left( 1 \right)} | \psi _{n}^{\left( 0 \right)} \right\rangle +\left\langle \psi _{n}^{\left( 1 \right)} | \psi _{n}^{\left( 1 \right)} \right\rangle +\left\langle \psi _{n}^{\left( 1 \right)} | \psi _{n}^{\left( 2 \right)} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ +\left\langle \psi _{n}^{\left( 2 \right)} | \psi _{n}^{\left( 0 \right)} \right\rangle +\left\langle \psi _{n}^{\left( 2 \right)} | \psi _{n}^{\left( 1 \right)} \right\rangle +\left\langle \psi _{n}^{\left( 2 \right)} | \psi _{n}^{\left( 2 \right)} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ =1+\left\langle \psi _{n}^{\left( 0 \right)} | \psi _{n}^{\left( 2 \right)} \right\rangle +\left\langle \psi _{n}^{\left( 1 \right)} | \psi _{n}^{\left( 1 \right)} \right\rangle +\left\langle \psi _{n}^{\left( 2 \right)} | \psi _{n}^{\left( 0 \right)} \right\rangle =1 \\ \end{align}$

不难推出: $⟨ ψ_{n}^{(0)} ∣ ψ_{n}^{(2)} ⟩ + ⟨ ψ_{n}^{(2)} ∣ ψ_{n}^{(0)} ⟩ = - ⟨ ψ_{n}^{(1)} ∣ ψ_{n}^{(1)} ⟩$

接着我们将右边的一阶修正态展开得到:

$\left\langle \psi _{n}^{\left( 0 \right)} | \psi _{n}^{\left( 2 \right)} \right\rangle +\left\langle \psi _{n}^{\left( 2 \right)} | \psi _{n}^{\left( 0 \right)} \right\rangle =-\sum\limits_{\alpha \left( \ne n \right)}{\frac{\left| {H}'}_{\alpha n} \right|}^{2}{\left( E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)} \right)}^{2}$

可以推出:

$\left\langle \psi _{n}^{\left( 0 \right)} | \psi _{n}^{\left( 2 \right)} \right\rangle =-\frac{1}{2}\sum\limits_{\alpha \left( \ne n \right)}{\frac{\left| {H}'}_{\alpha n} \right|}^{2}{\left( E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)} \right)}^{2}$

这里有点小问题, 事实上 $⟨ ψ_{n}^{(0)} ∣ ψ_{n}^{(2)} ⟩$ 完全可以是个复数, 不过我想虚部可以通过整体调整一个相位取消掉.[[3]](#ref_3)

结合前面的展开式**[d]**最终写为:

$\begin{align} & \left| \psi _{n}^{\left( \text{2} \right)} \right\rangle \text{=}\sum\limits_{\beta \left( \ne n \right)}{\left[ \sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}{H}'}_{\beta \alpha }{\left[ E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)} \right]\left[ E_{n}^{\left( 0 \right)}-E_{\beta }^{\left( 0 \right)} \right]}-\frac{H}'}_{nn}{H}'}_{\beta n}{\left[ E_{n}^{\left( 0 \right)}-E_{\beta }^{\left( 0 \right)} \right]}^{2} \right]\left| \psi _{\beta }^{\left( 0 \right)} \right\rangle } \\ & \ \ \ \ \ \ \ \ \ \ -\frac{1}{2}\sum\limits_{\alpha \left( \ne n \right)}{\frac{\left| {H}'}_{\alpha n} \right|}^{2}{\left( E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)} \right)}^{2}\left| \psi _{n}^{\left( 0 \right)} \right\rangle \\ \end{align}$

有的地方可能不包含最后一项, 其实也行吧, 但这也勉强算这篇文章的特色吧(笑. 数量级上来说, $ψ_{n}^{(0)} ⟩$ 的方向上大略是可以忽略这一项的. 因为零阶项就是 $ψ_{n}^{(0)} ⟩$ 本身, 我们整出来的这一修正项相对来说很可能不痛不痒.

总结如下:

$\left\{ \begin{align} & E_{n}^{\left( 1 \right)}={H}'}_{nn} \\ & \left| \psi _{n}^{\left( 1 \right)} \right\rangle \text{=}\sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)}\left| \psi _{\alpha }^{\left( 0 \right)} \right\rangle } \\ & E_{n}^{\left( 2 \right)}=\sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}\langle \psi _{n}^{\left( 0 \right)}|{H}'\left| \psi _{\alpha }^{\left( 0 \right)} \right\rangle }{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)}=\sum\limits_{\alpha \left( \ne n \right)}{\frac{\left| {H}'}_{\alpha n} \right|}^{2}{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)} \\ & \left| \psi _{n}^{\left( \text{2} \right)} \right\rangle \text{=}\sum\limits_{\beta \left( \ne n \right)}{\left[ \sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}{H}'}_{\beta \alpha }{\left[ E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)} \right]\left[ E_{n}^{\left( 0 \right)}-E_{\beta }^{\left( 0 \right)} \right]}-\frac{H}'}_{nn}{H}'}_{\beta n}{\left[ E_{n}^{\left( 0 \right)}-E_{\beta }^{\left( 0 \right)} \right]}^{2} \right]\left| \psi _{\beta }^{\left( 0 \right)} \right\rangle } \\ & \ \ \ \ \ \ \ \ \ \ -\frac{1}{2}\sum\limits_{\alpha \left( \ne n \right)}{\frac{\left| {H}'}_{\alpha n} \right|}^{2}{\left( E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)} \right)}^{2}\left| \psi _{n}^{\left( 0 \right)} \right\rangle \\ \end{align} \right.$

顺带一提,能量的修正似乎有迹可循: $\left\{ \begin{align} & E_{n}^{\left( 1 \right)}=\left\langle \psi _{n}^{\left( 0 \right)} \right|{H}'\left| \psi _{n}^{\left( 0 \right)} \right\rangle \\ & E_{n}^{\left( 2 \right)}=\langle \psi _{n}^{\left( 0 \right)}|{H}'\left| \psi _{n}^{\left( 1 \right)} \right\rangle \\ \end{align} \right.$

附录:

前文中站开式 $ψ_{n}^{(1)} ⟩ = α (\neq = n) \sum \frac{H}{^{'}}_{α n} E_{n}^{(0)} - E_{α}^{(0)} ψ_{α}^{(0)} ⟩$ 里面缺了 $α = n$ 的一项.

这里其实采取了一个近似手段:

$∣ ψ_{n} ⟩ = ψ_{n}^{(0)} ⟩ + ψ_{n}^{(1)} ⟩ + o (ψ_{n}^{(1)} ⟩)$ 是已知的.

应该注意到下面两点: ❶. $∣ ψ_{n} ⟩$ $ψ_{n}^{(0)} ⟩$ 均是正交归一的态, 因为二者都是确定的 $H$ 的本征方程的解. ❷. $ψ_{n}^{(1)} ⟩$ $ψ_{n}^{(2)} ⟩$ ... $ψ_{n}^{(n)} ⟩$ 都是修正项, 都不是归一的, 甚至不能说是一个态.

所以可以得到:

下列式子忽略了高阶段无穷小项 $o (ψ_{n}^{(1)} ⟩)$ :

$⟨ ψ_{n} ∣ ψ_{n} ⟩ = ⟨ ψ_{n}^{(0)} ∣ ψ_{n}^{(0)} ⟩ + ⟨ ψ_{n}^{(0)} ∣ ψ_{n}^{(1)} ⟩ + ⟨ ψ_{n}^{(1)} ∣ ψ_{n}^{(0)} ⟩ + ⟨ ψ_{n}^{(1)} ∣ ψ_{n}^{(1)} ⟩ = 1$

实际上 $⟨ ψ_{n}^{(1)} ∣ ψ_{n}^{(1)} ⟩$ 也是 $⟨ ψ_{n}^{(0)} ∣ ψ_{n}^{(1)} ⟩$ 的高阶无穷小项, 故略去.

上式可化为 $⟨ ψ_{n}^{(0)} ∣ ψ_{n}^{(1)} ⟩ = - ⟨ ψ_{n}^{(1)} ∣ ψ_{n}^{(0)} ⟩$

即说明 $α = n$ 项的系数是个纯虚数.

设 $⟨ ψ_{n}^{(0)} ∣ ψ_{n}^{(1)} ⟩ = iγ$ 其中 $γ$ 是个实数且远小于1.

这样我们就可以如下述推导那样, 通过改变整体相位取消 $α = n$ 这一项:

$\begin{align} & \left| {\psi }_{n} \right\rangle =\left| \psi _{n}^{\left( 0 \right)} \right\rangle +\left| \psi _{n}^{\left( 1 \right)} \right\rangle +o\left( \left| \psi _{n}^{\left( \text{1} \right)} \right\rangle \right) \\ & \ \ \ \ \ \ =\left| \psi _{n}^{\left( 0 \right)} \right\rangle +\left\langle \psi _{n}^{\left( 0 \right)}|\psi _{n}^{\left( 1 \right)} \right\rangle \left| \psi _{n}^{\left( 0 \right)} \right\rangle +\sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)}\left| \psi _{\alpha }^{\left( 0 \right)} \right\rangle }+o\left( \left| \psi _{n}^{\left( \text{1} \right)} \right\rangle \right) \\ & \ \ \ \ \ \ =\left[ 1+\left\langle \psi _{n}^{\left( 0 \right)}|\psi _{n}^{\left( 1 \right)} \right\rangle \right]\left| \psi _{n}^{\left( 0 \right)} \right\rangle +\sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)}\left| \psi _{\alpha }^{\left( 0 \right)} \right\rangle }+o\left( \left| \psi _{n}^{\left( \text{1} \right)} \right\rangle \right) \\ & \ \ \ \ \ \ =\left[ 1+i\gamma \right]\left| \psi _{n}^{\left( 0 \right)} \right\rangle +\sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)}\left| \psi _{\alpha }^{\left( 0 \right)} \right\rangle }+o\left( \left| \psi _{n}^{\left( \text{1} \right)} \right\rangle \right) \\ & \ \ \ \ \ \ \approx {e}^{i\gamma }\left| \psi _{n}^{\left( 0 \right)} \right\rangle +\sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)}\left| \psi _{\alpha }^{\left( 0 \right)} \right\rangle }+o\left( \left| \psi _{n}^{\left( \text{1} \right)} \right\rangle \right) \\ & \ \ \ \ \ \ ={e}^{i\gamma }\left[ \left| \psi _{n}^{\left( 0 \right)} \right\rangle +\sum\limits_{\alpha \left( \ne n \right)}{\frac{H}'}_{\alpha n}{E_{n}^{\left( 0 \right)}-E_{\alpha }^{\left( 0 \right)}\left| \psi _{\alpha }^{\left( 0 \right)} \right\rangle }+o\left( \left| \psi _{n}^{\left( \text{1} \right)} \right\rangle \right) \right] \\ \end{align}$

上面约等号处使用的近似为: ${e}^{i\gamma }=\sum\limits_{n=0}^{\infty }{\frac{\left( i\gamma \right)}^{n}{n!}\approx 1+i\gamma$ .

参考

^还没讲这个问题, 算是剧透吧.
^这里有点尴尬的, 它们确实是矢量, 实际上可以看作是一个态, 但这样去想根本就没意义.
^这里也有点尴尬的, 反正, 反正这就是个近似展开, 有点误差就有点误差吧欸.

非简并微扰论的一二级修正梳理 ​

**A.**求解 En(1)​ : ​

**B.**求解 ​ψn(1)​⟩ : ​

**C.**求解 En(2)​ : ​

**D.**求解 ​ψn(2)​⟩ : ​

总结如下: ​

附录: ​

参考 ​