問題 6-29 の解答例

Feynman-Hibbs cover
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Problem 6-29
Suppose two perturbations, \(V(x,t)\) and \(U(x,t)\), are acting. (Examples include a combination of DC and AC electronic fields or a combination of electric and magnetic fields.) Suppose further that a certain transition cannot occur with either \(V\) or \(U\) alone, but can occur only when both act together. Under the special assumption that both \(V\) and \(U\) are constant in time, show that the matrix element determining the transition element is given by

\begin{equation}
M_{n\to m}=\sum_{k} \frac{V_{m k}U_{k n}+U_{m k}V_{k n}}{E_{k}-E_{n}}
\tag{6-111}
\end{equation}

Next, suppose both potentials are periodic in time but have different frequencies, \(\omega_{1}\) and \(\omega_{2}\). What then is the matrix element?


(解答) 仮定から行列要素は \(V_I(t)+U_I(t)\) と書ける.従って, 2次の遷移振幅 \(\lambda_{mn}^{(2)}\) に相当する 2次の係数 \(c_m^{(2)}(T)\) は次である:

\begin{align}
&c_m^{(2)}(T)=\left(\frac{-i}{\hbar}\right)^{2}\sum_k \int_0^{T}dt_1 \int_0^{t_1}dt_2\,
e^{i\omega_{mk}t_1}\,\bigl[V_{mk}(t_1)+U_{mk}(t_1)\bigr]\,e^{i\omega_{kn}t_2}\,\bigl[V_{kn}(t_2)+U_{kn}(t_2)
\bigr]\\
&\quad =\left(\frac{-i}{\hbar}\right)^{2}\sum_k \int_0^{T}dt_1\,e^{i\omega_{mk}t_1} \int_0^{t_1}dt_2\,
e^{i\omega_{kn}t_2}\,\bigl[V_{mk}(t_1)+U_{mk}(t_1)\bigr]\,\bigl[V_{kn}(t_2)+U_{kn}(t_2)\bigr]
\tag{1}
\end{align}

ここでポテンシャルは \(V\) 或は \(U\) の一方だけでは遷移は起こらないと仮定し, かつそれらは時間に依存しないとすれば, 上式中の行列要素は次に書ける:
\begin{align}
&\bigl[V_{mk}(t_1)+U_{mk}(t_1)\bigr]\,\bigl[V_{kn}(t_2)+U_{kn}(t_2)\bigr]\notag\\
&\quad =V_{mk}(t_1)V_{kn}(t_2)+V_{mk}(t_1)U_{kn}(t_2)+U_{mk}(t_1)V_{kn}(t_2)+U_{mk}(t_1)U_{kn}(t_2)\\
&\quad =V_{mk}(t_1)U_{kn}(t_2)+U_{mk}(t_1)V_{kn}(t_2)=V_{mk}U_{kn}+U_{mk}V_{kn}
\tag{2}
\end{align}

すると式 (1) は, 式 (6-98) のときと同様にして次のようになる:
\begin{align}
c_m^{(2)}(T)&=\left(\frac{-i}{\hbar}\right)^{2}\sum_k (V_{mk}U_{kn}+U_{mk}V_{kn})\int_0^{T}dt_1
\,e^{i\omega_{mk}t_1} \int_0^{t_1}dt_2\,e^{i\omega_{kn}t_2}\\
&=\sum_k\frac{V_{mk}U_{kn}+U_{mk}V_{kn}}{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{3}
\end{align}

よってこの場合の「遷移の行列要素」(matrix element for the transition) は, 式 (6-99) のときと同様な議論により次に書ける:
\begin{equation}
M_{n\to m}=\sum_k \frac{V_{mk}U_{kn}+U_{mk}V_{kn}}{E_k-E_n}
\tag{4}
\end{equation}

次に, 両方のポテンシャルは周期的に時間変化する場合を考える.この場合, 問題 6-24 に倣ってポテンシャルを次のように表わすことにする:
\begin{align}
V(t)+U(t)&=V(x)(e^{i\omega_1t}+e^{-i\omega_1t})+U(x)(e^{i\omega_2t}+e^{-i\omega_2t})\\
&=2V(x)\cos\omega_1t +2U(x)\cos\omega_2t
\tag{5}
\end{align}

すると, 問題 6-24 と同様な手順により \(c_m^{(2)}(T)\) は次となる:
\begin{align}
&c_m^{(2)}(T)\\
&\quad =\left(\frac{-i}{\hbar}\right)^{2}\sum_k \int_0^{T}dt_1\int_0^{t_1}dt_2\,
e^{i\omega_{mk}t_1}\,\{V_{mk}(t_1)+U_{mk}(t_1)\}\,e^{i\omega_{kn}t_2}\,\{V_{kn}(t_2)+U_{kn}(t_2)\}\\
&\quad =\left(\frac{-i}{\hbar}\right)^{2}\sum_k \int_0^{T}dt_1\,e^{i\omega_{mk}t_1}\,
\{V_{mk}(t_1)+U_{mk}(t_1)\}\int_0^{t_1}dt_2\,e^{i\omega_{kn}t_2}\,\{V_{kn}(t_2)+U_{kn}(t_2)\}
\tag{6}
\end{align}

まず, \(t_2\) についての積分は次となる:
\begin{align}
\left(\frac{-i}{\hbar}\right)&\int_0^{t_1}dt_2\,e^{i\omega_{kn}t_2}\,\{V_{kn}(t_2)+U_{kn}(t_2)\}\\
&=\left(\frac{-i}{\hbar}\right)\int_0^{t_1}dt_2\,e^{i\omega_{kn}t_2}\,\left\{V_{kn}(e^{i\omega_1t_2}
+e^{-i\omega_1t_2})+U_{kn}(e^{i\omega_2t_2}+e^{-i\omega_2t_2})\right\}\\
&=\frac{1}{\hbar}\left[\frac{V_{kn}}{\omega_{kn}+\omega_1}\left\{1-e^{i(\omega_{kn}+\omega_1)t_1}\right\}
+\frac{V_{kn}}{\omega_{kn}-\omega_1}\left\{1-e^{i(\omega_{kn}-\omega_1)t_1}\right\}\right.\\
&\qquad\quad\left. +\frac{U_{kn}}{\omega_{kn}+\omega_2}\left\{1-e^{i(\omega_{kn}+\omega_2)t_1}\right\}
+\frac{U_{kn}}{\omega_{kn}-\omega_2}\left\{1-e^{i(\omega_{kn}-\omega_2)t_1}\right\}\right]
\tag{7}
\end{align}

これを式 (6) に代入し, \(t_1\) について積分すると次となる:
\begin{align}
c_m^{(2)}(T)&=\frac{1}{\hbar^{2}}\sum_k\left[\frac{V_{mk}V_{kn}}{(\omega_{kn}+\omega_1)}
\frac{e^{i(\omega_{mn}+2\omega_1)T}-1}{(\omega_{mn}+2\omega_1)}
-\frac{V_{mk}V_{kn}}{(\omega_{kn}+\omega_1)}\frac{e^{i(\omega_{mk}+\omega_1)T}-1}{(\omega_{mk}+\omega_1)}\right.\\
&\quad+\frac{V_{mk}V_{kn}}{(\omega_{kn}+\omega_1)}\frac{e^{i\omega_{mn}T}-1}{\omega_{mn}}
-\frac{V_{mk}V_{kn}}{(\omega_{kn}+\omega_1)}\frac{e^{i(\omega_{mk}-\omega_1)T}-1}{(\omega_{mk}-\omega_1)}\\
&\quad+\frac{U_{mk}V_{kn}}{(\omega_{kn}+\omega_1)}\frac{e^{i(\omega_{mn}+\omega_1+\omega_2)T}-1}{(\omega_{mn}+\omega_1+\omega_2)}
-\frac{U_{mk}V_{kn}}{(\omega_{kn}+\omega_1)}\frac{e^{i(\omega_{mk}+\omega_2)T}-1}{(\omega_{mk}+\omega_2)}\\
&\quad+\frac{U_{mk}V_{kn}}{(\omega_{kn}+\omega_1)}\frac{e^{i(\omega_{mn}+\omega_1-\omega_2)T}-1}
{(\omega_{mn}+\omega_1-\omega_2)}-\frac{U_{mk}V_{kn}}{(\omega_{kn}+\omega_1)}
\frac{e^{i(\omega_{mk}-\omega_2)T}-1}{(\omega_{mk}-\omega_2)}\\
&\quad+\frac{V_{mk}V_{kn}}{(\omega_{kn}-\omega_1)}\frac{e^{i\omega_{mn}T}-1}{\omega_{mn}}
-\frac{V_{mk}V_{kn}}{(\omega_{kn}-\omega_1)}\frac{e^{i(\omega_{mk}+\omega_1)T}-1}{(\omega_{mk}+\omega_1)}\\
&\quad+\frac{V_{mk}V_{kn}}{(\omega_{kn}-\omega_1)}\frac{e^{i(\omega_{mn}-2\omega_1)T}-1}{(\omega_{mn}-2\omega_1)}
-\frac{V_{mk}V_{kn}}{(\omega_{kn}-\omega_1)}\frac{e^{i(\omega_{mk}-\omega_1)T}-1}{(\omega_{mk}-\omega_1)}\\
&\quad+\frac{U_{mk}V_{kn}}{(\omega_{kn}-\omega_1)}\frac{e^{i(\omega_{mm}-\omega_1+\omega_2)T}-1}
{(\omega_{mn}-\omega_1+\omega_2)}-\frac{U_{mk}V_{kn}}{(\omega_{kn}-\omega_1)}
\frac{e^{i(\omega_{mk}+\omega_2)T}-1}{(\omega_{mk}+\omega_2)}\\
&\quad+\frac{U_{mk}V_{kn}}{(\omega_{kn}-\omega_1)}\frac{e^{i(\omega_{mn}-\omega_1-\omega_2)T}-1}
{(\omega_{mn}-\omega_1-\omega_2)}-\frac{U_{mk}V_{kn}}{(\omega_{kn}-\omega_1)}
\frac{e^{i(\omega_{mk}-\omega_2)T}-1}{(\omega_{mk}-\omega_2)}\\
&\quad+\frac{V_{mk}U_{kn}}{(\omega_{kn}+\omega_2)}\frac{e^{i(\omega_{mn}+\omega_1+\omega_2)T}-1}
{(\omega_{mn}+\omega_1+\omega_2)}-\frac{V_{mk}U_{kn}}{(\omega_{kn}+\omega_2)}
\frac{e^{i(\omega_{mk}+\omega_1)T}-1}{(\omega_{mk}+\omega_1)}\\
&\quad+\frac{V_{mk}U_{kn}}{(\omega_{kn}+\omega_2)}\frac{e^{i(\omega_{mn}-\omega_1+\omega_2)T}-1}
{(\omega_{mn}-\omega_1+\omega_2)}-\frac{V_{mk}U_{kn}}{(\omega_{kn}+\omega_2)}
\frac{e^{i(\omega_{mk}-\omega_1)T}-1}{(\omega_{mk}-\omega_1)}\\
&\quad+\frac{U_{mk}U_{kn}}{(\omega_{kn}+\omega_2)}\frac{e^{i(\omega_{mn}+2\omega_2)T}-1}
{(\omega_{mn}+2\omega_2)}-\frac{U_{mk}U_{kn}}{(\omega_{kn}+\omega_2)}
\frac{e^{i(\omega_{mk}+\omega_2)T}-1}{(\omega_{mk}+\omega_2)}\\
&\quad+\frac{U_{mk}U_{kn}}{(\omega_{kn}+\omega_2)}\frac{e^{i\omega_{mn}T}-1}{\omega_{mn}}
-\frac{U_{mk}U_{kn}}{(\omega_{kn}+\omega_2)}\frac{e^{i(\omega_{mk}-\omega_2)T}-1}{(\omega_{mk}-\omega_2)}\\
&\quad+\frac{V_{mk}U_{kn}}{(\omega_{kn}-\omega_2)}\frac{e^{i(\omega_{mn}+\omega_1-\omega_2)T}-1}
{(\omega_{mn}+\omega_1-\omega_2)}-\frac{V_{mk}U_{kn}}{(\omega_{kn}-\omega_2)}
\frac{e^{i(\omega_{mk}+\omega_1)T}-1}{(\omega_{mk}+\omega_1)}\\
&\quad+\frac{V_{mk}U_{kn}}{(\omega_{kn}-\omega_2)}\frac{e^{i(\omega_{mn}-\omega_1-\omega_2)T}-1}
{(\omega_{mn}-\omega_1-\omega_2)}-\frac{V_{mk}U_{kn}}{(\omega_{kn}-\omega_2)}
\frac{e^{i(\omega_{mk}-\omega_1)T}-1}{(\omega_{mk}-\omega_1)}\\
&\quad+\frac{U_{mk}U_{kn}}{(\omega_{kn}-\omega_2)}\frac{e^{i\omega_{mn}T}-1}{\omega_{mn}}
-\frac{U_{mk}U_{kn}}{(\omega_{kn}-\omega_2)}\frac{e^{i(\omega_{mk}+\omega_2)T}-1}{(\omega_{mk}+\omega_2)}\\
&\quad+\frac{U_{mk}U_{kn}}{(\omega_{kn}-\omega_2)}\frac{e^{i(\omega_{mn}-2\omega_2)T}-1}
{(\omega_{mn}-2\omega_2)}-\frac{U_{mk}U_{kn}}{(\omega_{kn}-\omega_2)}
\frac{e^{i(\omega_{mk}-\omega_2)T}-1}{(\omega_{mk}-\omega_2)}
\tag{8}
\end{align}

式 (6-98) から式 (6-99) を求めたときの議論と, 問題6-24の式 (6-94) を求めたときの議論とから, この場合の遷移率は次のようになると類推される?. (ただし, これも類推したに過ぎないので, 後で検討を要するものである):
\begin{align}
w_{n\to m}&=\frac{P(n\to m)}{dt}=\frac{2\pi}{\hbar}\bigl|M_{n\to m}^{(1)}\bigr|^{2}\Bigl[\delta(E_m-E_n)
+\delta(E_m-E_n+2\hbar\omega_1)\Bigr]\\
&\quad+\frac{2\pi}{\hbar}\bigl|M_{n\to m}^{(2)}\bigr|^{2}\Bigl[\delta(E_m-E_n+\hbar\omega_1
+\hbar\omega_2)+\delta(E_m-E_n+\hbar\omega_1-\hbar\omega_2)\Bigr]\\
&\quad+\frac{2\pi}{\hbar}\bigl|M_{n\to m}^{(3)}\bigr|^{2}\Bigl[\delta(E_m-E_n)+\delta(E_m-E_n-2\hbar\omega_1)
\Bigr]\\
&\quad+\frac{2\pi}{\hbar}\bigl|M_{n\to m}^{(4)}\bigr|^{2}\Bigl[\delta(E_m-E_n-\hbar\omega_1
+\hbar\omega_2)+\delta(E_m-E_n-\hbar\omega_1-\hbar\omega_2)\Bigr]\\
&\quad+\frac{2\pi}{\hbar}\bigl|M_{n\to m}^{(5)}\bigr|^{2}\Bigl[\delta(E_m-E_n+\hbar\omega_1
+\hbar\omega_2)+\delta(E_m-E_n-\hbar\omega_1+\hbar\omega_2)\Bigr]\\
&\quad+\frac{2\pi}{\hbar}\bigl|M_{n\to m}^{(6)}\bigr|^{2}\Bigl[\delta(E_m-E_n)
+\delta(E_m-E_n+2\hbar\omega_2)\Bigr]\\
&\quad+\frac{2\pi}{\hbar}\bigl|M_{n\to m}^{(7)}\bigr|^{2}\Bigl[\delta(E_m-E_n+\hbar\omega_1
-\hbar\omega_2)+\delta(E_m-E_n-\hbar\omega_1-\hbar\omega_2)\Bigr]\\
&\quad+\frac{2\pi}{\hbar}\bigl|M_{n\to m}^{(8)}\bigr|^{2}\Bigl[\delta(E_m-E_n)
+\delta(E_m-E_n-2\hbar\omega_2)\Bigr]
\tag{9}
\end{align}

ただし, 各項に含まれる「遷移の行列要素」は次である:
\begin{align}
M_{n\to m}^{(1)}&=\sum_k \frac{V_{mk}V_{kn}}{E_k-E_n+\hbar\omega_1-i\varepsilon},\qquad
M_{n\to m}^{(2)}=\sum_k \frac{U_{mk}V_{kn}}{E_k-E_n+\hbar\omega_1-i\varepsilon},\\
M_{n\to m}^{(3)}&=\sum_k \frac{V_{mk}V_{kn}}{E_k-E_n-\hbar\omega_1-i\varepsilon},\qquad
M_{n\to m}^{(4)}=\sum_k \frac{U_{mk}V_{kn}}{E_k-E_n-\hbar\omega_1-i\varepsilon},\\
M_{n\to m}^{(5)}&=\sum_k \frac{V_{mk}U_{kn}}{E_k-E_n+\hbar\omega_2-i\varepsilon},\qquad
M_{n\to m}^{(6)}=\sum_k \frac{U_{mk}U_{kn}}{E_k-E_n+\hbar\omega_2-i\varepsilon},\\
M_{n\to m}^{(7)}&=\sum_k \frac{V_{mk}U_{kn}}{E_k-E_n-\hbar\omega_2-i\varepsilon},\qquad
M_{n\to m}^{(8)}=\sum_k \frac{U_{mk}U_{kn}}{E_k-E_n-\hbar\omega_2-i\varepsilon}
\tag{10}
\end{align}