微扰论(perturbation theory)
2024-9-28
定态微扰
非简并情况
Total Hamiltonian can be splitted into two parts,
$$H=H_0+H^\prime$$
where $H_0$ is the unperturbed Hamiltonian, whose eigenfunctions and eigenvalues can be solved by whatever way, and $H^\prime$ is the perturbation. Introduce a real parameter $\lambda\in\left[0,1\right]$, so that by tuning $\lambda$ we can control if the perturbation is switched on or not, and how strong it is
$$H=H_0+\lambda H^\prime$$
$\lambda$ will finally be set to 1, so that the perturbation is fully switched on. The perturbed eigenfuntions and eigenvalues are the power series expansion of parameter ,$\lambda$
$$\begin{equation} \begin{split} f=f_0 + \lambda f^{(1)} +\lambda f^{(2)} +\cdots \\ \epsilon=\epsilon_0 + \lambda \epsilon^{(1)} +\lambda \epsilon^{(2)} +\cdots \\ \end{split} \end{equation}$$
$f^{(n)}$ 与 $\epsilon^{(n)}$ 分别是第 $n$ 级微扰近似. $f_0$ 与 $\epsilon_0$ 是无微扰情况下的定态薛定谔方程的解 $H_0f_0=\epsilon_0 f_0$. 将上式带入定态薛定谔方程
$$\begin{equation} \left(H_0+H^\prime\right)\left(f^{(0)} + \lambda f^{(1)} +\lambda f^{(2)} +\cdots\right) = \left(\epsilon^{(0)} + \lambda \epsilon^{(1)} +\lambda \epsilon^{(2)} +\cdots\right)\left(f^{(0)} + \lambda f^{(1)} +\lambda f^{(2)} +\cdots\right) \end{equation}$$
注意到左右两边 $\lambda$ 的幂次应相等 (这也是引入 $\lambda$ 的原因), 有
$$\begin{equation} \begin{split} \left(H_0-\epsilon_0\right)f_0 & = 0\:\:\: \text{TISE}\\ \left(H_0-\epsilon_0\right)f^{(1)} & = \left(\epsilon^{(1)}-H^\prime\right)f_0 \\ \left(H_0-\epsilon_0\right)f^{(2)} & = \left(\epsilon^{(1)}-H^\prime\right)f^{(1)} +\epsilon^{(2)}f_0 \\ \left(H_0-\epsilon_0\right)f^{(3)} & = \left(\epsilon^{(1)}-H^\prime\right)f^{(2)} +\epsilon^{(2)}f^{(1)} +\epsilon^{(3)}f_0 \\ \left(H_0-\epsilon_0\right)f^{(4)} & = \left(\epsilon^{(1)}-H^\prime\right)f^{(3)} +\epsilon^{(2)}f^{(2)} +\epsilon^{(3)}f^{(1)} +\epsilon^{(4)} f_0 \\ \vdots \\ \left(H_0-\epsilon_0\right)f^{(n)} & = \left(\epsilon^{(1)}-H^\prime\right)f^{(n-1)} +\sum_{i=2}^n \epsilon^{(i)} f^{(n-i)} \\ \end{split} \end{equation}$$
左乘 $\langle{f_0}\vert$ 可以得到能量的各级修正 (Note $\langle{f_0}\vert{H_0}\vert{f^{(n)}}\rangle$ = $\langle{f^{(n)}}\vert{H_0 f_0}\rangle^\dagger$= $\epsilon_0 \langle{f^{(n)}}\vert{f_0}\rangle$=0. And $f_0$ is orthogonal to $f^{(n)}$)
$$\begin{equation} \begin{split} \langle{f_0}\vert{H_0-\epsilon_0}\vert{f^{(n)}}\rangle & = \langle{f_0}\vert{\epsilon^{(1)}-H^\prime}\vert{f^{(n-1)}}\rangle +\sum_{i=2}^n \langle{f_0}\vert{\epsilon^{(i)}}\vert{f^{(n-i)}}\rangle \\ 0 & = -\langle{f_0}\vert{H^\prime}\vert{f^{(n-1)}}\rangle +\epsilon^{(n)} \\ \Rightarrow\:\:\: & \epsilon^{(n)} =\langle{f_0}\vert{H^\prime}\vert{f^{(n-1)}}\rangle \\ \end{split} \end{equation}$$
可以直接得到能量的一阶修正
$$\begin{equation} \epsilon^{(1)} =\langle{f_0}\vert{H^\prime}\vert{f_0}\rangle \end{equation}$$
无微扰下的哈密顿算符特征函数构成完备集
$$H_0 \phi_n=\epsilon_n\phi_n,\:\:\:n=1,2,\cdots$$
$$\sum_n \vert{\phi_n}\rangle\langle{\phi_n}\vert=1$$
假设无微扰时系统处在第 $k$ 个能级, 即 $f_0=\phi_k$, $\epsilon_0=\epsilon_k$. 能量的一阶修正为 $\epsilon^{(1)}=\langle{\phi_k}\vert{H^\prime}\vert{\phi_k}\rangle=H^\prime_{kk}$. 同时一阶微扰近似波函数可以由其展开
$$f^{(1)}=\sum_n a_n\phi_n$$
带入到 $\left(H_0-\epsilon_0\right)f^{(1)} = \left(\epsilon^{(1)}-H^\prime\right)f_0$ 有
$$\left(H_0-\epsilon_k\right)\sum_n a_n\phi_n = \left(\epsilon^{(1)}-H^\prime\right)\phi_k$$
左乘 $\langle{\phi_m}\vert$ 有
$$\begin{equation} \begin{split} \sum_n\left\{\langle{\phi_m}\vert{H_0}\vert{a_n\phi_n}\rangle -\epsilon_k\langle{\phi_m}\vert{a_n\phi_n}\rangle \right\} & = \epsilon^{(1)}\langle{\phi_m}\vert{\phi_k}\rangle -\langle{\phi_m}\vert{H^\prime}\vert{\phi_k}\rangle \\ \sum_n \left\{a_n\epsilon_n\delta_{mn} -\epsilon_ka_n\delta_{mn}\right\} & = H^\prime_{kk}\delta_{mk} -H^\prime_{mk} \\ a_m\epsilon_m -\epsilon_ka_m & = H^\prime_{kk}\delta_{mk} -H^\prime_{mk} \\ \end{split} \end{equation}$$ when $m \ne k$, we have $$\begin{equation} a_m = \frac{H^\prime_{mk}}{\epsilon_k -\epsilon_m} \\ \end{equation}$$ then the first order wavefunction correction is $$\begin{equation} f^{(1)} = \sum_{n\ne k} \frac{H^\prime_{nk}}{\epsilon_k-\epsilon_n}\phi_n \end{equation}$$ Since the second order correction for energy is $\epsilon^{(2)}=\langle{f_0}\vert{H^\prime}\vert{f^{(1)}}\rangle$, where $f_0=\phi_k$. Plugging formula for first order correction wavefunction $f^{(1)}$ into it furnishes $$\begin{equation} \begin{split} \epsilon^{(2)} & = \langle{\phi_k}\vert{H^\prime}\vert{\sum_{n\ne k}\frac{H^\prime_{nk}}{\epsilon_k -\epsilon_n}\phi_n}\rangle \\ & = \sum_{n\ne k} \frac{H^\prime_{nk}}{\epsilon_k -\epsilon_n} \langle{\phi_k}\vert{H^\prime}\vert{\phi_n}\rangle \\ & = \sum_{n\ne k} \frac{\vert{H^\prime_{nk}}\vert^2}{\epsilon_k -\epsilon_n} \\ \end{split} \end{equation}$$ The third order energy correction is $\epsilon^{(3)}=\langle{f_0}\vert{H^\prime}\vert{f^{(2)}}\rangle$. Making use of the relation above $$\begin{equation} \begin{split} (H_0-\epsilon_0)f^{(1)} & = (\epsilon^{(1)}-H^\prime)f_0 \\ (H_0-\epsilon_0)f^{(2)} & = (\epsilon^{(1)}-H^\prime)f^{(1)} +\epsilon^{(2)}f_0 \\ \text{$\langle{f^{(2)}}\vert$ for first} & \text{ and $\langle{f^{(1)}}\vert$ for second} \\ \langle{f^{(2)}}\vert{H_0-\epsilon_0}\vert{f^{(1)}}\rangle & = -\langle{f^{(2)}}\vert{H^\prime}\vert{f_0}\rangle \\ \langle{f^{(1)}}\vert{H_0-\epsilon_0}\vert{f^{(2)}}\rangle & = \langle{f^{(1)}}\vert{\epsilon^{(1)}-H^\prime}\vert{f^{(1)}}\rangle \\ \Rightarrow\:\:\:\epsilon^{(3)} & = \langle{f^{(1)}}\vert{H^\prime-\epsilon^{(1)}}\vert{f^{(1)}}\rangle \\ \end{split} \end{equation}$$ Which means the third order energy correction can be obtained from first order correction wavefunction, instead of the second order. Then we have $$\begin{equation} \begin{split} \epsilon^{(3)} & = \langle{f^{(1)}}\vert{H^\prime-\epsilon^{(1)}}\vert{f^{(1)}}\rangle \\ & = \sum_{n\ne k}\sum_{m \ne k}\frac{H^\prime_{nk}}{\epsilon_k-\epsilon_n}\frac{H^\prime_{mk}}{\epsilon_k-\epsilon_m}\langle{\phi_n}\vert{H^\prime-\epsilon^{(1)}}\vert{\phi_m}\rangle \\ & = \sum_{\substack{n\ne k\\ m\ne k}} \frac{H^\prime_{nk}H^\prime_{mk}}{(\epsilon_k-\epsilon_n)(\epsilon_k-\epsilon_m)} \left\{\langle{\phi_n}\vert{H^\prime}\vert{\phi_m}\rangle - H^\prime_{kk}\delta_{nm} \right\} \\ & = \sum_{\substack{n \ne k\\ m\ne k}}\frac{H^\prime_{nk}H^\prime_{mk}H^\prime_{nm}}{(\epsilon_k-\epsilon_n)(\epsilon_k-\epsilon_m)} -\sum_{n\ne k}\frac{H^\prime_{kk}\vert{H^\prime_{nk}}\vert^2}{(\epsilon_k -\epsilon_n)^2} \\ \end{split} \end{equation}$$ 做一个小结: 无微扰时系统处于第 $k$ 个能级上, 无微扰时的波函数为 (零阶修正) $f_0=\phi_k$, 能量为 $\epsilon_0=\epsilon_k$. 定态微扰的一阶能量修正, 一阶波函数修正, 二阶能量与三阶能量修正为 $$\begin{equation} \begin{split} & \epsilon^{(1)} = H^\prime_{kk} \\ & \epsilon^{(2)} = \sum_{\substack{n\ne k}}\frac{\vert{H^\prime_{nk}}\vert^2}{\epsilon_n -\epsilon_k} \\ & \epsilon^{(3)} = \sum_{\substack{n \ne k\\ m\ne k}}\frac{H^\prime_{nk}H^\prime_{mk}H^\prime_{nm}}{(\epsilon_k-\epsilon_n)(\epsilon_k-\epsilon_m)} -\sum_{n\ne k}\frac{H^\prime_{kk}\vert{H^\prime_{nk}}\vert^2}{(\epsilon_k -\epsilon_n)^2} \\ & f^{(1)} = \sum_{n\ne k} \frac{H^\prime_{nk}}{\epsilon_k -\epsilon_n}\phi_n \\ \end{split} \end{equation}$$ 又有 $\left(H_0-\epsilon_0\right)f^{(2)} = \left(\epsilon^{(1)}-H^\prime\right)f^{(1)} +\epsilon^{(2)}f_0$, 可以得到波函数的二阶修正 $$\begin{equation} \begin{split} & (H_0-\epsilon_k)f^{(2)} = (H^\prime_{kk}-H^\prime)\sum_{n\ne k}\frac{H^\prime_{nk}}{\epsilon_k -\epsilon_n}\phi_n +\sum_{n\ne k}\frac{\vert{H^\prime_{nk}}\vert^2}{\epsilon_k -\epsilon_n}\phi_k \\ & \sum_l (H_0-\epsilon_k)b_l \phi_l = (H^\prime_{kk}-H^\prime)\sum_{n\ne k}\frac{H^\prime_{nk}}{\epsilon_k -\epsilon_n}\phi_n +\sum_{n\ne k}\frac{\vert{H^\prime_{nk}}\vert^2}{\epsilon_k -\epsilon_n}\phi_k \\ & \sum_l (b_l\epsilon_l\phi_l-b_l\epsilon_k\phi_l) = (H^\prime_{kk}-H^\prime)\sum_{n\ne k}\frac{H^\prime_{nk}}{\epsilon_k -\epsilon_n}\phi_n +\sum_{n\ne k}\frac{\vert{H^\prime_{nk}}\vert^2}{\epsilon_k -\epsilon_n}\phi_k \\ & \sum_l \langle{\phi_m}\vert{b_l\epsilon_l-b_l\epsilon_k}\vert{\phi_l}\rangle = \sum_{n\ne k}\frac{H^\prime_{nk}}{\epsilon_k-\epsilon_n}\langle{\phi_m}\vert{H^\prime_{kk}-H^\prime}\vert{\phi_n}\rangle +\sum_{n\ne k}\frac{\vert{H^\prime_{nk}}\vert^2}{\epsilon_k -\epsilon_n}\delta_{mk} \\ & \sum_l \left(b_l\epsilon_l\delta_{ml} -b_l\epsilon_k\delta_{ml}\right) = \sum_{n\ne k}\frac{H^\prime_{nk}}{\epsilon_k -\epsilon_n}\left(H^\prime_{kk}\delta_{mn} -H^\prime_{mn}\right) +\sum_{n\ne k}\frac{\vert{H^\prime_{nk}}\vert^2}{\epsilon_k -\epsilon_n}\delta_{mk} \\ & b_m(\epsilon_m -\epsilon_k) = \sum_{n\ne k}\frac{H^\prime_{nk}H^\prime_{kk}}{\epsilon_k -\epsilon_n}\delta_{mn} -\sum_{n\ne k}\frac{H^\prime_{nk}H^\prime_{mn}}{\epsilon_k -\epsilon_n} + \sum_{n\ne k}\frac{\vert{H^\prime_{nk}}\vert^2}{\epsilon_k -\epsilon_n}\delta_{mk} \\ \end{split} \end{equation}$$ When $m=k$, we only get trivial result. When $m\ne k$ $$\begin{equation} \begin{split} & b_m(\epsilon_m -\epsilon_k) = \sum_{n\ne k}\frac{H^\prime_{nk}H^\prime_{kk}}{\epsilon_k -\epsilon_n}\delta_{mn} -\sum_{n\ne k}\frac{H^\prime_{nk}H^\prime_{mn}}{\epsilon_k -\epsilon_n} + \sum_{n\ne k}\frac{\vert{H^\prime_{nk}}\vert^2}{\epsilon_k -\epsilon_n}\delta_{mk} \\ & b_m(\epsilon_m -\epsilon_k) = \frac{H^\prime_{mk}H^\prime_{kk}}{\epsilon_k -\epsilon_m} -\sum_{n\ne k}\frac{H^\prime_{nk}H^\prime_{mn}}{\epsilon_k -\epsilon_n} +0 \\ & b_m = \sum_{n\ne k}\frac{H^\prime_{nk}H^\prime_{mn}}{(\epsilon_k -\epsilon_n)(\epsilon_k -\epsilon_m)} -\frac{H^\prime_{mk}H^\prime_{kk}}{(\epsilon_k -\epsilon_m)^2} \\ \end{split} \end{equation}$$ and finally $$f^{(2)} = \sum_{m\ne k}\left\{\sum_{n\ne k}\frac{H^\prime_{nk}H^\prime_{mn}}{(\epsilon_k -\epsilon_n)(\epsilon_k -\epsilon_m)} -\frac{H^\prime_{mk}H^\prime_{kk}}{(\epsilon_k -\epsilon_m)^2}\right\}\phi_m$$ Using similar trick, the third order wavefunction correction is (maybe wrong) $$f^{(3)} = \sum_{p\ne k}\left\{ \sum_{\substack{m\ne k\\n\ne k}}\frac{H^\prime_{nk}H^\prime_{mn}H^\prime_{pm}}{(\epsilon_k-\epsilon_n)(\epsilon_k-\epsilon_m)(\epsilon_k-\epsilon_p)} -\sum_{n\ne k}\frac{H^\prime_{kk}H^\prime_{nk}H^\prime_{pn}}{(\epsilon_k-\epsilon_n)(\epsilon_k-\epsilon_p)^2} -\sum_{n\ne k}\frac{H^\prime_{pk}\vert{H^\prime_{nk}}\vert^2}{(\epsilon_k-\epsilon_n)(\epsilon_k-\epsilon_p)^2}-\sum_{m\ne k}\frac{H^\prime_{kk}H^\prime_{mk}H^\prime_{pm}}{(\epsilon_k-\epsilon_m)^2(\epsilon_k-\epsilon_p)}+\frac{\vert{H^\prime_{kk}}\vert^2H^\prime_{pk}}{(\epsilon_k-\epsilon_p)^3} \right\}\phi_p$$含时微扰
$$H(t)=H_0 +H^\prime(t)$$where $i\hbar\frac{\partial}{\partial t}\Phi_n(t)=H_0\Phi_n(t)$, $\Phi_n(t)=\phi_n(\boldsymbol{r})e^{-i\frac{\epsilon_n}{\hbar}t}$ and $\sum_n \vert{\Phi_n(t)}\rangle\langle{\Phi_n(t)}\vert=I$, $\langle{\phi_n}\vert{\phi_m}\rangle=\delta_{nm}$, $H_0 \phi_n=\epsilon_n\phi_n$.
Wavefunction $\vert{f(t)}\rangle$ satisfies $i\hbar\frac{\partial}{\partial t}f(t)=(H_0+H^\prime(t))f(t)$. Projecting $\vert{f(t)}\rangle$ on basis $\{\Phi_n(t)\vert n=1,2,\cdots\}$ gives $$\begin{equation} \begin{split} \sum_n i\hbar\frac{\partial}{\partial t}\left(a_n(t)\Phi_n(t)\right) & = \sum_n \left(H_0+H^\prime\right)a_n(t)\Phi_n(t) \\ \sum_n \left\{i\hbar\frac{\partial a_n(t)}{\partial t}\Phi_n(t) +i\hbar a_n(t)\frac{\partial}{\partial t}\Phi_n(t)\right\} & = \sum_n \left\{a_n(t)H_0\Phi_n(t) +a_n(t)H^\prime\Phi_n(t)\right\} \\ \end{split} \end{equation}$$ Note that $i\hbar a_n(t)\frac{\partial}{\partial t}\Phi_n(t)=a_n(t)H_0\Phi_n(0)$, thus $$\begin{equation} \begin{split} \sum_n \left\{i\hbar\frac{\partial a_n(t)}{\partial t}\Phi_n(t)\right\} & = \sum_n a_n(t)H^\prime\Phi_n(t) \\ \sum_n i\hbar\frac{\partial a_n(t)}{\partial t} \langle{\phi_m}\vert{\phi_n}\rangle e^{-i\frac{\epsilon_n}{\hbar}t} & = \sum_n a_n(t)\langle{\phi_m}\vert{H^\prime}\vert{\phi_n}\rangle e^{-i\frac{\epsilon_n}{\hbar}t} \\ i\hbar\frac{\partial a_m(t)}{\partial t}e^{-i\frac{\epsilon_m}{\hbar}t} & = \sum_n a_n(t) H^\prime_{mn} e^{-i\frac{\epsilon_n}{\hbar}t} \\ i\hbar\frac{\partial}{\partial t}a_m(t) & = \sum_n a_n(t)H^\prime_{mn}(t) e^{i\omega_{mn}t} \\ \end{split} \end{equation}$$ where $\omega_{mn}=\frac{\epsilon_m -\epsilon_n}{\hbar}$. 上式是含时薛定谔方程的另一种表现形式, 隐含使用相互作用表象. \par 同定态微扰, 引入参数 $\lambda$ 控制微扰打开的程度 $H^\prime(t)=\lambda H^\prime(t)$, 那么同理在微扰下系数 $a_n(t)$ 可以展开成 $\lambda$ 的幂级数 $$a_n(t) =a^{(0)}_n +\lambda a_n^{(1)}(t) +\lambda^2 a_n^{(2)}(t)+\cdots$$ 带入 $i\hbar\frac{\partial}{\partial t}a_m(t)= \sum_n a_n(t)H^\prime_{mn}(t) e^{i\omega_{mn}t}$ 有 $$i\hbar\frac{\partial}{\partial t}\left[a_m^{(0)}+\lambda a_m^{(1)}(t)+\lambda^2 a_m^{(2)}(t)+\cdots \right] =\sum_n \left[a^{(0)}_n +\lambda a_n^{(1)}(t) +\lambda^2 a_n^{(2)}(t)+\cdots \right]\lambda H^\prime_{mn}(t) e^{i\omega_{mn}t}$$ 两边对应幂次相等, 因此 $$\begin{equation} \left\{\begin{split} & i\hbar\frac{\partial}{\partial t}a_m^{(0)} = 0 \\ & i\hbar\frac{\partial}{\partial t}a_m^{(1)}(t) = \sum_n a_n^{(0)}H^\prime_{mn}(t)e^{i\omega_{mn}t} \\ & i\hbar\frac{\partial}{\partial t}a_m^{(2)}(t) = \sum_n a_n^{(1)}(t)H^\prime_{mn}(t)e^{i\omega_{mn}t} \\ & \vdots \\ & i\hbar\frac{\partial}{\partial t}a_m^{(\alpha)}(t) = \sum_n a_n^{(\alpha-1)}(t)H^\prime_{mn}(t)e^{i\omega_{mn}t} \\ \end{split}\right. \end{equation}$$ 说明无微扰时的展开系数 $a_m^{(0)}$ 不含时. 假设微扰在 $t=0$ 时开启, 此时系统处于 $\vert{\Phi_k(t)}\rangle$, $H_0\Phi_k(t)=i\hbar\partial_t \Phi_k(t)$, 所以有 $a_k^{(0)}(0)=1,\:\:a_n^{(0)}(0)=\delta_{nk}$, 所以有 $$i\hbar\frac{\partial}{\partial t}a_m^{(1)}(t)=a_k^{(0)}H^\prime_{mk}(t)e^{i\omega_{mk}t}=H^\prime_{mk}(t)e^{i\omega_{mk}t}$$ 解之可得一级近似 $$\begin{equation} a_m^{(1)}(t) = \frac{1}{i\hbar}\int_0^t H^\prime_{mk}(t^\prime)e^{i\omega_{mk}t^\prime} dt^\prime \\ \end{equation}$$ 在时刻 $t$ 时系统处于 $\vert{\Phi_m(t)}\rangle$ 的概率为 $\vert{a_m(t)}\vert^2$. 对微扰后的展开系数取一级近似 $a_m(t)\approx a_m^{(0)}+a_m^{(1)}=a_m^{(1)}$. 即在微扰作用下系统由初态 $\vert{\Phi_k(t)}\rangle$ 跃迁到 $\vert{\Phi_m(t)}\rangle$ 的概率为 $$W_{k\rightarrow m} = \vert{a_m(t)}\vert^2 \approx \vert{a_m^{(1)}(t)}\vert^2$$$H^\prime$ 不随时间变化
首先证明一个 $\delta$ 函数表达式
$$\lim_{t\rightarrow \infty} \left(\frac{\sin{xt}}{x}\right)^2\frac{1}{\pi t} = \delta(x)$$When $x\ne 0$, $\lim_{t\rightarrow \infty}\left(\frac{\sin{xt}}{x}\right)^2\frac{1}{\pi t}=0$.
When $x=0$, $\lim_{t\rightarrow \infty}\left(\frac{\sin{xt}}{x}\right)^2\frac{1}{\pi t}=\lim_{t\rightarrow \infty}\frac{t}{\pi}=\infty$.
And the integral $$\int_{-\infty}^{\infty}\left(\frac{\sin{xt}}{x}\right)^2\frac{1}{\pi t}dx = \int_{-\infty}^\infty \left(\frac{\sin{xt}}{xt}\right)^2\frac{1}{\pi}dxt = 1$$So $\lim_{t\rightarrow \infty} \left(\frac{\sin{xt}}{x}\right)^2\frac{1}{\pi t} = \delta(x)$.
现假设微扰 $H^\prime$ 不含时间, 微扰在 $t=0$ 时打开, $t=0$ 时系统处于状态 $\Phi_k(\vec{\boldsymbol{r}},t)$. 那么从 $\vert \Phi_k\rangle$ 跃迁到 $\vert \Phi_m\rangle$ 的概率为 $W_{k\rightarrow m}(t)=\left\vert{a_m^{(1)}(t)}\right\vert^2=\left\vert \frac{1}{i\hbar}\int_{-\infty}^t H^\prime_{mk}(t^\prime) e^{i\omega_{mk}t^\prime} dt^\prime \right\vert^2$. 由于微扰不含时, 且在 $t<0$ 时未开启, 因此有
$$\begin{equation} \begin{split} W_{k\rightarrow m}(t) & = \frac{\left\vert H^\prime_{mk} \right\vert^2}{\hbar^2}\left\vert \int_0^t e^{i\omega_{mk}t^\prime}dt^\prime \right\vert^2 \\ & = \frac{\left\vert H^\prime_{mk} \right\vert^2}{\hbar^2} \left\vert \frac{-i\left(e^{i\omega_{mk}t}-1\right)}{\omega_{mk}} \right\vert^2 \\ & = \frac{4\left\vert H^\prime_{mk} \right\vert^2}{\hbar^2\omega_{mk}^2}\sin^2{\frac{\omega_{mk}t}{2}} \\ & = \frac{\left\vert H^\prime_{mk} \right\vert^2}{\hbar^2}\left(\frac{\sin{\frac{\omega_{mk}t}{2}}}{\frac{\omega_{mk}}{2}}\right)^2 \\ \lim_{t\rightarrow \infty} & \Rightarrow \frac{\left\vert H^\prime_{mk} \right\vert^2}{\hbar^2} \pi t \delta\left(\frac{\omega_{mk}}{2}\right) \\ & = \frac{2\pi t \left\vert H^\prime_{mk} \right\vert^2}{\hbar}\delta\left(\epsilon_{m} - \epsilon_{k}\right) \\ \end{split} \end{equation}$$ 单位时间的跃迁概率为 (跃迁速率) $$w_{k\rightarrow m}=\frac{2\pi \left\vert H^\prime_{mk} \right\vert^2}{\hbar}\delta\left(\epsilon_{m} - \epsilon_{k}\right)$$$\delta$ 函数只有出现在积分中才有意义, 假设末态 $\vert \Phi_m \rangle$ 是连续分布的, 那么末态 $\vert \Phi_m \rangle$ 的态密度为 $\rho(\epsilon_m)$, 处于能量范围 $\epsilon_m \sim \epsilon_m+d\epsilon$ 内的末态数目为 $\rho(\epsilon_m)d\epsilon_m$. 跃迁速率是从初态 $\vert \Phi_k \rangle$ 到连续分布末态的跃迁速率之和
$$\begin{equation} w_{k\rightarrow m} = \int \frac{2\pi \left\vert H^\prime_{mk} \right\vert^2}{\hbar}\delta\left(\epsilon_{m} - \epsilon_{k}\right)\rho(\epsilon_m)d\epsilon_m = \frac{2\pi \left\vert H^\prime_{mk} \right\vert^2}{\hbar}\rho(\epsilon_k) \end{equation}$$该式称为 Fermi's golden rule.
系统由 $\vert\Phi_k(t) \rangle$ 跃迁到 $\vert\Phi_m(t) \rangle$, 由于 $\delta$ 函数的限制, 要求 $\epsilon_k=\epsilon_m$, 即初末态能量相等, 也就是弹性散射. 对于 $\delta$ 函数, 需要引入态密度并积分, 由于初末态能量相等, 故而初末态的波矢 $\vec{\boldsymbol{k}}$ 的方向不同. 如果末态不构成连续的简并态, 则不能取极限为 $\delta$ 函数.
$H^\prime$ 随时间周期变化
Assume $H^\prime(t)=H^\prime e^{i\omega t}$, then transition probability is
$$\begin{equation} \begin{split} W_{k\rightarrow m}(t) & = \left\vert \frac{1}{i\hbar}\int_0^t H^\prime_{mk} e^{i\omega t^\prime} e^{i\omega_{mk}t^\prime} dt^\prime \right\vert^2 \\ & = \frac{\left\vert H^\prime_{mk} \right\vert^2}{\hbar^2} \frac{1}{\left(\omega+\omega_{mk}\right)^2} \left\vert \cos{\left(\omega+\omega_{mk}\right)t + i \sin{\left(\omega+\omega_{mk}\right)t}} -1 \right\vert^2 \\ & = \frac{4\left\vert H^\prime_{mk} \right\vert^2}{\hbar^2} \frac{1}{\left(\omega+\omega_{mk}\right)^2} \sin^2{\frac{\omega+\omega_{mk}}{2}t} \\ & = \frac{2\left\vert H^\prime_{mk} \right\vert^2}{\hbar^2} \pi t \delta\left(\omega_{mk}+\omega\right) \\ \end{split} \end{equation}$$transition rate is
$$w_{k\rightarrow m} = \frac{2\left\vert H^\prime_{mk} \right\vert^2}{\hbar} \pi \delta\left(\epsilon_m - \epsilon_k + \hbar\omega\right)$$If $H^\prime(t)=H^\prime e^{-i\omega t}$, corresponding transition rate is
$$w_{k\rightarrow m} = \frac{2\left\vert H^\prime_{mk} \right\vert^2}{\hbar} \pi \delta\left(\epsilon_m - \epsilon_k - \hbar\omega\right)$$