\(\)
Problem 6-27
Derive the perturbation expansion up through the terms of the second order for potentials periodic in time.
(解答) 2次の遷移振幅 \(\lambda_{mn}^{(2)}\) の式 (6-74) に相当する2次の係数 \(c_m^{(2)}(t)\) の式は次である：
\begin{align}
c_m^{\ (2)}(t)&=e^{i(E_mt_2-E_nt_1)/\hbar}\lambda_{mn}^{(2)}\\
&=\left(-\frac{i}{\hbar}\right)^{2}\sum_k\int_{t_1}^{t_2}dt_4\int_{t_1}^{t_4}dt_3\,e^{i\omega_{mk}t_4}
\,V_{mk}(t_4)\,e^{i\omega_{kn}t_3}\,V_{kn}(t_3)
\tag{1}
\end{align}
c_m^{\ (2)}(t)&=e^{i(E_mt_2-E_nt_1)/\hbar}\lambda_{mn}^{(2)}\\
&=\left(-\frac{i}{\hbar}\right)^{2}\sum_k\int_{t_1}^{t_2}dt_4\int_{t_1}^{t_4}dt_3\,e^{i\omega_{mk}t_4}
\,V_{mk}(t_4)\,e^{i\omega_{kn}t_3}\,V_{kn}(t_3)
\tag{1}
\end{align}
ただし \(\omega_{mk}\) と \(\omega_{kn}\) は次で定義された：
\begin{equation}
E_m-E_k=\hbar\omega_{mk},\quad E_k-E_n=\hbar\omega_{kn}
\tag{2}
\end{equation}
E_m-E_k=\hbar\omega_{mk},\quad E_k-E_n=\hbar\omega_{kn}
\tag{2}
\end{equation}
また次式が成り立つことに注意する：
\begin{equation}
\omega_{mk}+\omega_{kn}=\frac{E_m-E_k}{\hbar}+\frac{E_k-E_n}{\hbar}=\frac{E_m-E_n}{\hbar}=\omega_{mn}
\tag{3}
\end{equation}
\omega_{mk}+\omega_{kn}=\frac{E_m-E_k}{\hbar}+\frac{E_k-E_n}{\hbar}=\frac{E_m-E_n}{\hbar}=\omega_{mn}
\tag{3}
\end{equation}
題意より摂動ポテンシャル \(V(t)\) は, 問題 (6-24) と同じ「調和摂動」を用いよう：
\begin{equation}
V(t)=\mathcal{V}e^{i\omega t}+\mathcal{V}^{\dagger}e^{-i\omega t}
\tag{4}
\end{equation}
V(t)=\mathcal{V}e^{i\omega t}+\mathcal{V}^{\dagger}e^{-i\omega t}
\tag{4}
\end{equation}
\(t_1=0,\,t_2=T\) とした式 (1) の積分に於いて, ポテンシャル \(V(t)\) の行列要素にこの式(4)を適用すると,
\begin{align}
&c_m^{\ (2)}(T)=\left(-\frac{i}{\hbar}\right)^{2}\sum_k\int_0^{T}dt_4\int_0^{t_{4}}dt_{3}\,
e^{i\omega_{mk}t_{4}}\,V_{mk}(t_{4})\,e^{i\omega_{kn}t_{3}}\,V_{kn}(t_{3})\\
&=\left(-\frac{i}{\hbar}\right)^{2}\sum_k\int_0^{T}dt_{4}\,e^{i\omega_{mk}t_{4}}\,V_{mk}(t_{4})
\int_0^{t_{4}}dt_{3}\,e^{i\omega_{kn}t_{3}}V_{kn}(t_{3})\\
&=\sum_k\left(-\frac{i}{\hbar}\right)\int_0^{T}dt_{4}\,e^{i\omega_{mk}t_{4}}\,\big(\mathcal{V}_{mk}
\,e^{i\omega t_{4}}+\mathcal{V}_{mk}^{\dagger}e^{-i\omega t_{4}}\big)\times
\left(-\frac{i}{\hbar}\right)\int_0^{t_{4}}dt_{3}\,e^{i\omega_{kn}t_{3}}\,\big(\mathcal{V}_{kn}
\,e^{i\omega t_{3}}+\mathcal{V}_{kn}^{\dagger}e^{-i\omega t_{3}}\big)\\
&=\sum_k\left(-\frac{i}{\hbar}\right)\int_0^{T}dt_{4}\,\big(\mathcal{V}_{mk}
\,e^{i(\omega_{mk}+\omega)t_{4}}+\mathcal{V}_{mk}^{\dagger}e^{i(\omega_{mk}-\omega)t_{4}}\big)\times
\left[\frac{1-e^{i(\omega_{kn}+\omega)t_{4}}}{\hbar\omega_{kn}+\hbar\omega}
\mathcal{V}_{kn}+\frac{1-e^{i(\omega_{kn}-\omega)t_{4}}}{\hbar\omega_{kn}-\hbar\omega}\mathcal{V}_{kn}^{\dagger}\right]\\
&=\sum_k\left(-\frac{i}{\hbar}\right)\int_0^{T}dt_{4}\,
\Bigg[\frac{\mathcal{V}_{mk}\mathcal{V}_{kn}}{\hbar\omega_{kn}+\hbar\omega}
\big\{e^{i(\omega_{mk}+\omega)t_4}-e^{i(\omega_{mn}+2\omega)t_4}\big\}
+\frac{\mathcal{V}_{mk}\mathcal{V}_{kn}^{\dagger}}{\hbar\omega_{kn}-\hbar\omega}
\big\{e^{i(\omega_{mk}+\omega)t_4}-e^{i\omega_{mn}t_4}\big\}\\
&\qquad +\frac{\mathcal{V}_{mk}^{\dagger}\mathcal{V}_{kn}}{\hbar\omega_{kn}+\hbar\omega}
\big\{e^{i(\omega_{mk}-\omega)t_4}-e^{i\omega_{mn}t_4}\big\}
+\frac{\mathcal{V}_{mk}^{\dagger}\mathcal{V}_{kn}^{\dagger}}{\hbar\omega_{kn}-\hbar\omega}
\big\{e^{i(\omega_{mk}-\omega)t_4}-e^{i(\omega_{mn}-2\omega)t_4}\big\}\Bigg]
\tag{5}
\end{align}
&c_m^{\ (2)}(T)=\left(-\frac{i}{\hbar}\right)^{2}\sum_k\int_0^{T}dt_4\int_0^{t_{4}}dt_{3}\,
e^{i\omega_{mk}t_{4}}\,V_{mk}(t_{4})\,e^{i\omega_{kn}t_{3}}\,V_{kn}(t_{3})\\
&=\left(-\frac{i}{\hbar}\right)^{2}\sum_k\int_0^{T}dt_{4}\,e^{i\omega_{mk}t_{4}}\,V_{mk}(t_{4})
\int_0^{t_{4}}dt_{3}\,e^{i\omega_{kn}t_{3}}V_{kn}(t_{3})\\
&=\sum_k\left(-\frac{i}{\hbar}\right)\int_0^{T}dt_{4}\,e^{i\omega_{mk}t_{4}}\,\big(\mathcal{V}_{mk}
\,e^{i\omega t_{4}}+\mathcal{V}_{mk}^{\dagger}e^{-i\omega t_{4}}\big)\times
\left(-\frac{i}{\hbar}\right)\int_0^{t_{4}}dt_{3}\,e^{i\omega_{kn}t_{3}}\,\big(\mathcal{V}_{kn}
\,e^{i\omega t_{3}}+\mathcal{V}_{kn}^{\dagger}e^{-i\omega t_{3}}\big)\\
&=\sum_k\left(-\frac{i}{\hbar}\right)\int_0^{T}dt_{4}\,\big(\mathcal{V}_{mk}
\,e^{i(\omega_{mk}+\omega)t_{4}}+\mathcal{V}_{mk}^{\dagger}e^{i(\omega_{mk}-\omega)t_{4}}\big)\times
\left[\frac{1-e^{i(\omega_{kn}+\omega)t_{4}}}{\hbar\omega_{kn}+\hbar\omega}
\mathcal{V}_{kn}+\frac{1-e^{i(\omega_{kn}-\omega)t_{4}}}{\hbar\omega_{kn}-\hbar\omega}\mathcal{V}_{kn}^{\dagger}\right]\\
&=\sum_k\left(-\frac{i}{\hbar}\right)\int_0^{T}dt_{4}\,
\Bigg[\frac{\mathcal{V}_{mk}\mathcal{V}_{kn}}{\hbar\omega_{kn}+\hbar\omega}
\big\{e^{i(\omega_{mk}+\omega)t_4}-e^{i(\omega_{mn}+2\omega)t_4}\big\}
+\frac{\mathcal{V}_{mk}\mathcal{V}_{kn}^{\dagger}}{\hbar\omega_{kn}-\hbar\omega}
\big\{e^{i(\omega_{mk}+\omega)t_4}-e^{i\omega_{mn}t_4}\big\}\\
&\qquad +\frac{\mathcal{V}_{mk}^{\dagger}\mathcal{V}_{kn}}{\hbar\omega_{kn}+\hbar\omega}
\big\{e^{i(\omega_{mk}-\omega)t_4}-e^{i\omega_{mn}t_4}\big\}
+\frac{\mathcal{V}_{mk}^{\dagger}\mathcal{V}_{kn}^{\dagger}}{\hbar\omega_{kn}-\hbar\omega}
\big\{e^{i(\omega_{mk}-\omega)t_4}-e^{i(\omega_{mn}-2\omega)t_4}\big\}\Bigg]
\tag{5}
\end{align}
\(dt_4\) 積分を, 本文の式 (6-98) と同様な仕方で実行するならば次となる：
\begin{align}
c_m^{\ (2)}(T)=\sum_k &\left[\ \left(\frac{-\mathcal{V}_{mk}\mathcal{V}_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}+2\omega)T}}{E_m-E_n+2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\right.\\
&+\left(\frac{-\mathcal{V}_{mk}^{\dagger}\mathcal{V}_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\\
&+\left(\frac{-\mathcal{V}_{mk}\mathcal{V}_{kn}^{\dagger}}{E_k-E_n-\hbar\omega}\right)\left\{
\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\\
&\left.+\left(\frac{-\mathcal{V}_{mk}^{\dagger}\mathcal{V}_{kn}^{\dagger}}{E_k-E_n-\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}-2\omega)T}}{E_m-E_n-2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\ \right]
\tag{6}
\end{align}
c_m^{\ (2)}(T)=\sum_k &\left[\ \left(\frac{-\mathcal{V}_{mk}\mathcal{V}_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}+2\omega)T}}{E_m-E_n+2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\right.\\
&+\left(\frac{-\mathcal{V}_{mk}^{\dagger}\mathcal{V}_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\\
&+\left(\frac{-\mathcal{V}_{mk}\mathcal{V}_{kn}^{\dagger}}{E_k-E_n-\hbar\omega}\right)\left\{
\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\\
&\left.+\left(\frac{-\mathcal{V}_{mk}^{\dagger}\mathcal{V}_{kn}^{\dagger}}{E_k-E_n-\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}-2\omega)T}}{E_m-E_n-2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\ \right]
\tag{6}
\end{align}
さらに問題 6-24 と同様に, \(\mathcal{V}=\mathcal{V}^{\dagger}=V(x)\) の場合を考えるならば,
\begin{align}
c_m^{\ (2)}(T)=\sum_k &\left[\ \left(\frac{-V_{mk} V_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}+2\omega)T}}{E_m-E_n+2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\right.\\
&+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\\
&+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n-\hbar\omega}\right)\left\{
\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\\
&\left.+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n-\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}-2\omega)T}}{E_m-E_n-2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\ \right]
\tag{7}
\end{align}
c_m^{\ (2)}(T)=\sum_k &\left[\ \left(\frac{-V_{mk} V_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}+2\omega)T}}{E_m-E_n+2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\right.\\
&+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\\
&+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n-\hbar\omega}\right)\left\{
\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\\
&\left.+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n-\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}-2\omega)T}}{E_m-E_n-2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\ \right]
\tag{7}
\end{align}
従って, この場合の2次までの摂動展開は, 問題 6-24 の結果も用いることで次となる：
\begin{align}
c^{(0)}_{m}(T)&=\delta_{mn},\\
c^{(1)}_{m}(T)&=V_{mn}\frac{1-e^{i(\omega_{mn}+\omega)T}}{E_m-E_n+\hbar\omega}
+V_{mn}\frac{1-e^{i(\omega_{mn}-\omega)T}}{E_m-E_n-\hbar\omega},\\
c^{(2)}_{m}(T)&=\sum_k\left[\ \left(\frac{-V_{mk}V_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}+2\omega)T}}{E_m-E_n+2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\right.\\
&\qquad+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\\
&\qquad+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n-\hbar\omega}\right)\left\{
\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\\
&\qquad\left.+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n-\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}-2\omega)T}}{E_m-E_n-2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\ \right]
\tag{8}
\end{align}
c^{(0)}_{m}(T)&=\delta_{mn},\\
c^{(1)}_{m}(T)&=V_{mn}\frac{1-e^{i(\omega_{mn}+\omega)T}}{E_m-E_n+\hbar\omega}
+V_{mn}\frac{1-e^{i(\omega_{mn}-\omega)T}}{E_m-E_n-\hbar\omega},\\
c^{(2)}_{m}(T)&=\sum_k\left[\ \left(\frac{-V_{mk}V_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}+2\omega)T}}{E_m-E_n+2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\right.\\
&\qquad+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n+\hbar\omega}\right)
\left\{\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\\
&\qquad+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n-\hbar\omega}\right)\left\{
\frac{1-e^{i\omega_{mn}T}}{E_m-E_n}-\frac{1-e^{i(\omega_{mk}+\omega)T}}{E_m-E_k+\hbar\omega}\right\}\\
&\qquad\left.+\left(\frac{-V_{mk}V_{kn}}{E_k-E_n-\hbar\omega}\right)
\left\{\frac{1-e^{i(\omega_{mn}-2\omega)T}}{E_m-E_n-2\hbar\omega}
-\frac{1-e^{i(\omega_{mk}-\omega)T}}{E_m-E_k-\hbar\omega}\right\}\ \right]
\tag{8}
\end{align}
【 参考 】 参考のために, 「遷移率」も考えてみよう． ここでは注目している特定の状態 \(n\) と \(m\) に対して \(V_{mn}=0\) であるような問題を考える．すると1次の項まではゼロであり, 遷移振幅の計算に入って来る最低次の項は 2次である．上式 (6) の鍵カッコ内の各項は式 (6-98) の各項に似た形をしている．よって同様な議論が出来るであろう．よって, このときの遷移率 \(w_{n\to m}^{(2)}\) は次のようになると推量される(?)：
\begin{align}
w_{n\to m}^{(2)}&=\frac{2\pi}{\hbar}\left|M_{n\to m}^{(1)}\right|^{2}
\Bigl[\delta(E_m-E_n)+\delta(E_m-E_n+2\hbar\omega)\Bigr]\\
&\quad +\frac{2\pi}{\hbar}\sum_k \left|M_{n\to m}^{(2)}\right|^{2}
\Bigl[\delta(E_m-E_n)+\delta(E_m-E_n-2\hbar\omega)\Bigr]
\tag{9}
\end{align}
w_{n\to m}^{(2)}&=\frac{2\pi}{\hbar}\left|M_{n\to m}^{(1)}\right|^{2}
\Bigl[\delta(E_m-E_n)+\delta(E_m-E_n+2\hbar\omega)\Bigr]\\
&\quad +\frac{2\pi}{\hbar}\sum_k \left|M_{n\to m}^{(2)}\right|^{2}
\Bigl[\delta(E_m-E_n)+\delta(E_m-E_n-2\hbar\omega)\Bigr]
\tag{9}
\end{align}
ただし各々の遷移の行列要素は次とした：
\begin{equation}
M_{n\to m}^{(1)}=\sum_k \frac{-V_{mk}V_{kn}}{E_k-E_n+\hbar\omega-i\varepsilon},\quad
M_{n\to m}^{(2)}=\frac{-V_{mk}V_{kn}}{E_k-E_n-\hbar\omega-i\varepsilon}
\tag{10}
\end{equation}
M_{n\to m}^{(1)}=\sum_k \frac{-V_{mk}V_{kn}}{E_k-E_n+\hbar\omega-i\varepsilon},\quad
M_{n\to m}^{(2)}=\frac{-V_{mk}V_{kn}}{E_k-E_n-\hbar\omega-i\varepsilon}
\tag{10}
\end{equation}
[ 式 (9) と式 (10) はあくまでも類推したに過ぎないので非常に怪しい!．後で検討を要するものである．]
