\(\)
Problem 6-28
Show that when a transition is impossible either directly or through a single intermediate state, but requires the use of two intermediate states, it is determined by the matrix element given by
\begin{equation}
M_{n\to m} = -\sum_k \sum_l \frac{V_{m k}V_{k l}V_{l n}}{(E_m – E_k)(E_m -E_l)}
\tag{6-100}
\end{equation}
M_{n\to m} = -\sum_k \sum_l \frac{V_{m k}V_{k l}V_{l n}}{(E_m – E_k)(E_m -E_l)}
\tag{6-100}
\end{equation}
This corresponds to the third-order term in the perturbation expansion.
(解答) 相互作用表示での時間発展演算子及び遷移振幅と係数について,それらの3次の項は次である：
\begin{align}
\def\bra#1{\langle#1|}
\def\ket#1{|#1\rangle}
&\mathscr{U}_I^{(3)}(T,0)=\left(\frac{-i}{\hbar}\right)^{3}\int_0^{T}dt_1 \int_0^{t_1}dt_2 \int_0^{t_2}
dt_3\,V_I(t_1)\,V_I(t_2)\,V_I(t_3),\\
&V_I(t)=e^{iH_0t/\hbar}\,V\,e^{-iH_0t/\hbar},\quad c_m^{(3)}(T)=\bra{m} \mathscr{U}_I^{(3)}(T,0) \ket{n},\quad
\lambda_{mn}^{(3)}(T)=e^{-iE_m T/\hbar} c_{m}^{(3)}(T)
\tag{1}
\end{align}
\def\bra#1{\langle#1|}
\def\ket#1{|#1\rangle}
&\mathscr{U}_I^{(3)}(T,0)=\left(\frac{-i}{\hbar}\right)^{3}\int_0^{T}dt_1 \int_0^{t_1}dt_2 \int_0^{t_2}
dt_3\,V_I(t_1)\,V_I(t_2)\,V_I(t_3),\\
&V_I(t)=e^{iH_0t/\hbar}\,V\,e^{-iH_0t/\hbar},\quad c_m^{(3)}(T)=\bra{m} \mathscr{U}_I^{(3)}(T,0) \ket{n},\quad
\lambda_{mn}^{(3)}(T)=e^{-iE_m T/\hbar} c_{m}^{(3)}(T)
\tag{1}
\end{align}
そこで \(c_{m}^{(3)}(T)\) を具体的に計算して行こう：
\begin{align}
&c_m^{(3)}(T)=\bra{m} \mathscr{U}_I^{(3)}(T,0) \ket{n}
=\bra{m}\left(\frac{-i}{\hbar}\right)^{3}\int_0^{T}dt_1 \int_0^{t_1}dt_2 \int_0^{t_2}dt_3\,V_I(t_1)\,V_I(t_2)\,V_I(t_3)\ket{n}\\
&=\left(\frac{-i}{\hbar}\right)^{3}\int_0^{T} dt_1 \int_0^{t_1}dt_2 \int_0^{t_2}dt_3\,
\bra{m}{(e^{i H_{0} t_{1}/\hbar}\,V\,e^{-i H_{0} t_{1}/\hbar})}\,\bigg(\sum_k\ket{k}\bra{k}\bigg)\,
(e^{i H_{0} t_{1}/\hbar}\,V\,e^{-i H_{0} t_{1}/\hbar})\\
&\quad\times\,\bigg(\sum_l\ket{l}\bra{l}\bigg)\,(e^{i H_{0} t_{1}/\hbar}\,V\,e^{-i H_{0} t_{1}/\hbar})\ket{n}\\
&=\left(\frac{-i}{\hbar}\right)^{3}\sum_k\sum_l\,\int_0^{T} dt_1 \int_0^{t_1} dt_2 \int_0^{t_2} dt_3\,
e^{i E_{m} t_{1}/\hbar}\bra{m} V \ket{k}\,e^{-iE_{k} t_{1}/\hbar}\,e^{i E_{k} t_{2}/\hbar}\bra{k} V \ket{l}
\,e^{-i E_{l} t_{2}/\hbar}\\
&\quad\times\,e^{i E_{l} t_{3}/\hbar}\bra{l} V \ket{n}\,e^{-i E_{n} t_{3}/\hbar}\\
&=\left(\frac{-i}{\hbar}\right)^{3}\sum_k\sum_l\,\int_0^{T} dt_1 \int_0^{t_1} dt_2 \int_0^{t_2} dt_3\,
e^{i\omega_{m k}t_{1}}\,V_{m k}(t_1)\,e^{i\omega_{k l}t_{2}}\,V_{k l}(t_2)\,e^{i\omega_{l n}t_{3}}\,V_{l n}(t_3)
\tag{2}
\end{align}
&c_m^{(3)}(T)=\bra{m} \mathscr{U}_I^{(3)}(T,0) \ket{n}
=\bra{m}\left(\frac{-i}{\hbar}\right)^{3}\int_0^{T}dt_1 \int_0^{t_1}dt_2 \int_0^{t_2}dt_3\,V_I(t_1)\,V_I(t_2)\,V_I(t_3)\ket{n}\\
&=\left(\frac{-i}{\hbar}\right)^{3}\int_0^{T} dt_1 \int_0^{t_1}dt_2 \int_0^{t_2}dt_3\,
\bra{m}{(e^{i H_{0} t_{1}/\hbar}\,V\,e^{-i H_{0} t_{1}/\hbar})}\,\bigg(\sum_k\ket{k}\bra{k}\bigg)\,
(e^{i H_{0} t_{1}/\hbar}\,V\,e^{-i H_{0} t_{1}/\hbar})\\
&\quad\times\,\bigg(\sum_l\ket{l}\bra{l}\bigg)\,(e^{i H_{0} t_{1}/\hbar}\,V\,e^{-i H_{0} t_{1}/\hbar})\ket{n}\\
&=\left(\frac{-i}{\hbar}\right)^{3}\sum_k\sum_l\,\int_0^{T} dt_1 \int_0^{t_1} dt_2 \int_0^{t_2} dt_3\,
e^{i E_{m} t_{1}/\hbar}\bra{m} V \ket{k}\,e^{-iE_{k} t_{1}/\hbar}\,e^{i E_{k} t_{2}/\hbar}\bra{k} V \ket{l}
\,e^{-i E_{l} t_{2}/\hbar}\\
&\quad\times\,e^{i E_{l} t_{3}/\hbar}\bra{l} V \ket{n}\,e^{-i E_{n} t_{3}/\hbar}\\
&=\left(\frac{-i}{\hbar}\right)^{3}\sum_k\sum_l\,\int_0^{T} dt_1 \int_0^{t_1} dt_2 \int_0^{t_2} dt_3\,
e^{i\omega_{m k}t_{1}}\,V_{m k}(t_1)\,e^{i\omega_{k l}t_{2}}\,V_{k l}(t_2)\,e^{i\omega_{l n}t_{3}}\,V_{l n}(t_3)
\tag{2}
\end{align}
ただし， \(e^{i(E_m-E_k)t_{1}/\hbar}=e^{i\omega_{m k}t_{1}}\) などである．
議論を簡単にするために, ポテンシャル \(V\) が一定値に保たれ, また, 時間に陽に依存しない場合を考えよう．すると \(V_{mk}\) などは全て積分の外へ出るので次となる：
\begin{align}
c_{m}^{(3)}(T)&=\left(\frac{-i}{\hbar}\right)^{3}\sum_k\sum_l\,V_{mk}\,V_{kl}\,V_{ln}\,\int_0^{T}dt_1
\int_0^{t_1}dt_2 \int_0^{t_2}dt_3\,e^{i\omega_{m k}t_{1}}\,e^{i\omega_{k l}t_{2}}\,e^{i\omega_{l n}t_{3}}\\
&=\left(\frac{-i}{\hbar}\right)^{3}\sum_k\sum_l\,V_{m k}\,V_{k l}\,V_{l n}\,\int_0^{T} dt_1
\,e^{i\omega_{m k}t_{1}}\int_0^{t_{1}} dt_2\,e^{i\omega_{k l}t_{2}}\int_0^{t_{2}}dt_3\,e^{i\omega_{l n}t_{3}}
\tag{3}
\end{align}
c_{m}^{(3)}(T)&=\left(\frac{-i}{\hbar}\right)^{3}\sum_k\sum_l\,V_{mk}\,V_{kl}\,V_{ln}\,\int_0^{T}dt_1
\int_0^{t_1}dt_2 \int_0^{t_2}dt_3\,e^{i\omega_{m k}t_{1}}\,e^{i\omega_{k l}t_{2}}\,e^{i\omega_{l n}t_{3}}\\
&=\left(\frac{-i}{\hbar}\right)^{3}\sum_k\sum_l\,V_{m k}\,V_{k l}\,V_{l n}\,\int_0^{T} dt_1
\,e^{i\omega_{m k}t_{1}}\int_0^{t_{1}} dt_2\,e^{i\omega_{k l}t_{2}}\int_0^{t_{2}}dt_3\,e^{i\omega_{l n}t_{3}}
\tag{3}
\end{align}
ここでは次の積分を行なう必要がある：
\begin{equation}
\left(\frac{-i}{\hbar}\right)\int_0^{t} e^{i\omega t^{‘}}\,dt^{‘}=\left(\frac{-i}{\hbar}\right)
\Bigg[\frac{e^{i\omega t}}{i\,\omega}\Bigg]_{0}^{t}=\frac{-i}{\hbar}\frac{1}{i\,\omega}\big(e^{i\omega t}-1\big)
=\frac{-1}{\hbar\omega}\big(e^{i\omega t}-1\big)
\tag{4}
\end{equation}
\left(\frac{-i}{\hbar}\right)\int_0^{t} e^{i\omega t^{‘}}\,dt^{‘}=\left(\frac{-i}{\hbar}\right)
\Bigg[\frac{e^{i\omega t}}{i\,\omega}\Bigg]_{0}^{t}=\frac{-i}{\hbar}\frac{1}{i\,\omega}\big(e^{i\omega t}-1\big)
=\frac{-1}{\hbar\omega}\big(e^{i\omega t}-1\big)
\tag{4}
\end{equation}
この式 (4) と \(\hbar\omega_{ln}=E_l-E_n\) を用いて, まず \(t_3\) についての積分を実行するならば次となる：
\begin{equation}
\int_0^{t_2}dt_3\,e^{i\omega_{l n}t_{3}}=\frac{-1}{\hbar\omega_{l n}}\big(e^{i\omega_{l n}t_{2}}-1\big)
=\frac{-1}{E_l-E_n}\big(e^{i\omega_{l n}t_2}-1\big)
\tag{5}
\end{equation}
\int_0^{t_2}dt_3\,e^{i\omega_{l n}t_{3}}=\frac{-1}{\hbar\omega_{l n}}\big(e^{i\omega_{l n}t_{2}}-1\big)
=\frac{-1}{E_l-E_n}\big(e^{i\omega_{l n}t_2}-1\big)
\tag{5}
\end{equation}
同様に \(t_2\) の積分に進むと, 問題 6-27 の式 (3) と同じく \(\omega_{kl}+\omega_{ln}=\omega_{kn}\) が成り立つことを用いることで次となる：
\begin{align}
&\left(\frac{-i}{\hbar}\right)^{2}\int_0^{t_1}dt_2\,e^{i\omega_{kl}t_2}\int_0^{t_2}dt_3\,e^{i\omega_{l n}t_{3}}
=\frac{-1}{E_{l}-E_{n}}\left(\frac{-i}{\hbar}\right)\int_0^{t_1}dt_2\,e^{i\omega_{k l}t_{2}}\big(e^{i\omega_{l n}t_{2}}-1\big)\\
&\quad=\frac{-1}{E_{l}-E_{n}}\left(\frac{-i}{\hbar}\right)\int_0^{t_{1}}dt_2\,
\big(e^{i\omega_{k n}t_{2}}-e^{i\omega_{k l}t_{2}}\big)\\
&\quad =\frac{-1}{E_{l}-E_{n}}\left\{\frac{-1}{\hbar\omega_{k n}}\big( e^{i\omega_{k n}t_{1}}-1 \big)
-\frac{-1}{\hbar\omega_{k l}}\big(e^{i\omega_{kl}t_1}-1\big)\right\}\\
&\quad\simeq \frac{e^{i\omega_{k n}t_{1}}-1}{(E_{l}-E_{n})(E_{k}-E_{n})}
\tag{6}
\end{align}
&\left(\frac{-i}{\hbar}\right)^{2}\int_0^{t_1}dt_2\,e^{i\omega_{kl}t_2}\int_0^{t_2}dt_3\,e^{i\omega_{l n}t_{3}}
=\frac{-1}{E_{l}-E_{n}}\left(\frac{-i}{\hbar}\right)\int_0^{t_1}dt_2\,e^{i\omega_{k l}t_{2}}\big(e^{i\omega_{l n}t_{2}}-1\big)\\
&\quad=\frac{-1}{E_{l}-E_{n}}\left(\frac{-i}{\hbar}\right)\int_0^{t_{1}}dt_2\,
\big(e^{i\omega_{k n}t_{2}}-e^{i\omega_{k l}t_{2}}\big)\\
&\quad =\frac{-1}{E_{l}-E_{n}}\left\{\frac{-1}{\hbar\omega_{k n}}\big( e^{i\omega_{k n}t_{1}}-1 \big)
-\frac{-1}{\hbar\omega_{k l}}\big(e^{i\omega_{kl}t_1}-1\big)\right\}\\
&\quad\simeq \frac{e^{i\omega_{k n}t_{1}}-1}{(E_{l}-E_{n})(E_{k}-E_{n})}
\tag{6}
\end{align}
ただし, 式 (6-98) での議論と同じ理由から, 最終段で第2項目は無視した．
さらに \(t_1\) 積分を行なうならば, 同様な議論により次となる：
\begin{align}
\left(\frac{-i}{\hbar}\right)^{3}&\int_0^{T}dt_1\,e^{i\omega_{mk}t_1}\int_0^{t_1}dt_2\,e^{i\omega_{kl}t_2}
\int_0^{t_2}dt_3\,e^{i\omega_{ln}t_3}\\
&=\frac{1}{(E_l-E_n)(E_k-E_n)}\left(\frac{-i}{\hbar}\right)\int_0^{T}dt_1\,e^{i\omega_{mk}t_1}
\big(e^{i\omega_{kn}t_1}-1\big)\\
&=\frac{1}{(E_l-E_n)(E_k-E_n)}\left(\frac{-i}{\hbar}\right)
\int_0^{T}dt_1\,\big(e^{i\omega_{mn}t_1}-e^{i\omega_{mk}t_1}\big)\\
&=\frac{-1}{(E_l-E_n)(E_k-E_n)}\left\{\frac{1}{\hbar\omega_{mn}}\big(e^{i\omega_{mn}T}-1\big)
-\frac{1}{\hbar\omega_{mk}}\big(e^{i\omega_{mk}T}-1\big)\right\}\\
\tag{7}
\end{align}
\left(\frac{-i}{\hbar}\right)^{3}&\int_0^{T}dt_1\,e^{i\omega_{mk}t_1}\int_0^{t_1}dt_2\,e^{i\omega_{kl}t_2}
\int_0^{t_2}dt_3\,e^{i\omega_{ln}t_3}\\
&=\frac{1}{(E_l-E_n)(E_k-E_n)}\left(\frac{-i}{\hbar}\right)\int_0^{T}dt_1\,e^{i\omega_{mk}t_1}
\big(e^{i\omega_{kn}t_1}-1\big)\\
&=\frac{1}{(E_l-E_n)(E_k-E_n)}\left(\frac{-i}{\hbar}\right)
\int_0^{T}dt_1\,\big(e^{i\omega_{mn}t_1}-e^{i\omega_{mk}t_1}\big)\\
&=\frac{-1}{(E_l-E_n)(E_k-E_n)}\left\{\frac{1}{\hbar\omega_{mn}}\big(e^{i\omega_{mn}T}-1\big)
-\frac{1}{\hbar\omega_{mk}}\big(e^{i\omega_{mk}T}-1\big)\right\}\\
\tag{7}
\end{align}
以上の結果から, 式 (3) の \(c_m^{(3)}(T)\) は近似的に次のような形になる：
\begin{equation}
c_m^{(3)}(T)=\sum_k\sum_l \frac{-V_{mk}V_{kl}V_{ln}}{(E_l-E_n)(E_k-E_n)}
\left(\frac{e^{i\omega_{mn}T}-1}{E_m-E_n}-\frac{e^{i\omega_{mk}T}-1}{E_m-E_k}\right)
\tag{8}
\end{equation}
c_m^{(3)}(T)=\sum_k\sum_l \frac{-V_{mk}V_{kl}V_{ln}}{(E_l-E_n)(E_k-E_n)}
\left(\frac{e^{i\omega_{mn}T}-1}{E_m-E_n}-\frac{e^{i\omega_{mk}T}-1}{E_m-E_k}\right)
\tag{8}
\end{equation}
すると式 (6-99) を求めたときと同じ議論から, このときの \(M_{n\to m}\) は次であると言える：
\begin{equation}
M_{n\to m}=-\sum_k\sum_l \frac{V_{mk}V_{kl}V_{ln}}{(E_l-E_n)(E_k-E_n)}
\tag{9}
\end{equation}
M_{n\to m}=-\sum_k\sum_l \frac{V_{mk}V_{kl}V_{ln}}{(E_l-E_n)(E_k-E_n)}
\tag{9}
\end{equation}
