\(\)
Problem 7-12
Show, if \(V\) is any function of position only, that
\begin{equation}
\left\langle \chi \left| \frac{dV}{dt}\right| \psi\right\rangle
= \left\langle \chi \left| \frac{V(x_{k+1})-V(x_k)}{\varepsilon}\right| \psi\right\rangle
= \frac{i}{\hbar}\int_{-\infty}^{\infty} \chi^{*} \big(HV-VH\big)\psi\,dx
\tag{7-89}
\end{equation}
\left\langle \chi \left| \frac{dV}{dt}\right| \psi\right\rangle
= \left\langle \chi \left| \frac{V(x_{k+1})-V(x_k)}{\varepsilon}\right| \psi\right\rangle
= \frac{i}{\hbar}\int_{-\infty}^{\infty} \chi^{*} \big(HV-VH\big)\psi\,dx
\tag{7-89}
\end{equation}
Consider the case that \(V\) is also a function of the time. Show that the transition element of \(dV/dt\) is equivalent to the transition element of the operator \((i/\hbar)\big(HV-VH\big)+\partial V/\partial t\).
(解答) \(F\) として \(dV/dt\) である場合を考えると, 式 (7-72) に相当するものは,
\begin{equation}
F=\frac{d V}{dt}=\dot{V}(x)=\frac{V(x_{k+1})-V(x_k)}{\varepsilon}
\end{equation}
F=\frac{d V}{dt}=\dot{V}(x)=\frac{V(x_{k+1})-V(x_k)}{\varepsilon}
\end{equation}
であるから,
\begin{align}
\def\BraKet#1#2#3{\langle #1 | #2 | #3 \rangle}
\left\langle\,\chi\left|\frac{d V}{dt}\,\right|\psi\right\rangle
&=\left\langle\,\chi\left|\,
\frac{V(x_{k+1})-V(x_k)}{\varepsilon}\,\right|\psi\right\rangle\notag\\
&=\frac{1}{\varepsilon}\Big\{\BraKet{\chi}{V(x_{k+1})}{\psi}-\BraKet{\chi}{V(x_k)}{\psi}\Big\}
\tag{1}
\end{align}
\def\BraKet#1#2#3{\langle #1 | #2 | #3 \rangle}
\left\langle\,\chi\left|\frac{d V}{dt}\,\right|\psi\right\rangle
&=\left\langle\,\chi\left|\,
\frac{V(x_{k+1})-V(x_k)}{\varepsilon}\,\right|\psi\right\rangle\notag\\
&=\frac{1}{\varepsilon}\Big\{\BraKet{\chi}{V(x_{k+1})}{\psi}-\BraKet{\chi}{V(x_k)}{\psi}\Big\}
\tag{1}
\end{align}
本文の式 (7-73) から式 (7-74) までの議論と同様な手順を行なう．まず, 上式 (1) の第1項を \(\varepsilon\) の1次まで展開すると,
\begin{align}
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
\BraKet{\chi}{V(x_{k+1})}{\psi}&=\int dx\,\chi^{*}(x,t+\varepsilon)V(x,t+\varepsilon) \psi(x,t+\varepsilon)\notag\\
&=\int dx\,\chi^{*}(x)\left(1+\frac{i\varepsilon}{\hbar}H\right)
\left(V(x)+\varepsilon\ppdiff{V}{t}\right)\left(1-\frac{i\varepsilon}{\hbar}H\right)\psi(x)\notag\\
&\approx \int dx\,\chi^{*}(x)V(x)\psi(x)+\frac{i\varepsilon}{\hbar}\int dx\,\chi^{*}(x)HV(x)\psi(x)\notag\\
&\qquad +\varepsilon\int dx\,\chi^{*}(x)\ppdiff{V}{t}\psi(x) -\frac{i\varepsilon}{\hbar}
\int dx\,\chi^{*}V(x)H\psi(x)
\tag{2}
\end{align}
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
\BraKet{\chi}{V(x_{k+1})}{\psi}&=\int dx\,\chi^{*}(x,t+\varepsilon)V(x,t+\varepsilon) \psi(x,t+\varepsilon)\notag\\
&=\int dx\,\chi^{*}(x)\left(1+\frac{i\varepsilon}{\hbar}H\right)
\left(V(x)+\varepsilon\ppdiff{V}{t}\right)\left(1-\frac{i\varepsilon}{\hbar}H\right)\psi(x)\notag\\
&\approx \int dx\,\chi^{*}(x)V(x)\psi(x)+\frac{i\varepsilon}{\hbar}\int dx\,\chi^{*}(x)HV(x)\psi(x)\notag\\
&\qquad +\varepsilon\int dx\,\chi^{*}(x)\ppdiff{V}{t}\psi(x) -\frac{i\varepsilon}{\hbar}
\int dx\,\chi^{*}V(x)H\psi(x)
\tag{2}
\end{align}
従って, 式(1)は次となる：
\begin{align}
\left\langle\,\chi\left|\frac{d V}{dt}\,\right|\psi\right\rangle
&=\frac{1}{\varepsilon}\Big\{
\BraKet{\chi}{V(x_{k+1})}{\psi}-\BraKet{\chi}{V(x_k)}{\psi}\Big\}\notag\\
&=\frac{i}{\hbar}\int dx\,\chi^{*}(x)HV(x)\psi(x)+\int dx\,\chi^{*}(x)\ppdiff{V}{t}\psi(x)\notag\\
&\qquad -\frac{i}{\hbar}\int dx\,\chi^{*}V(x)H\psi(x)\notag\\
&=\int dx\,\chi^{*}(x)\left[\frac{i}{\hbar}\Big(HV(x)-V(x)H\Big)+\ppdiff{V}{t}\right]\,\psi(x)
\tag{3}
\end{align}
\left\langle\,\chi\left|\frac{d V}{dt}\,\right|\psi\right\rangle
&=\frac{1}{\varepsilon}\Big\{
\BraKet{\chi}{V(x_{k+1})}{\psi}-\BraKet{\chi}{V(x_k)}{\psi}\Big\}\notag\\
&=\frac{i}{\hbar}\int dx\,\chi^{*}(x)HV(x)\psi(x)+\int dx\,\chi^{*}(x)\ppdiff{V}{t}\psi(x)\notag\\
&\qquad -\frac{i}{\hbar}\int dx\,\chi^{*}V(x)H\psi(x)\notag\\
&=\int dx\,\chi^{*}(x)\left[\frac{i}{\hbar}\Big(HV(x)-V(x)H\Big)+\ppdiff{V}{t}\right]\,\psi(x)
\tag{3}
\end{align}
よって,
\begin{equation}
\left\langle\,\chi\left|\frac{d V}{dt}\,\right|\psi\right\rangle
=\left\langle \frac{i}{\hbar}\Big(HV(x)-V(x)H\Big)+\ppdiff{V}{t}\right\rangle
\tag{4}
\end{equation}
\left\langle\,\chi\left|\frac{d V}{dt}\,\right|\psi\right\rangle
=\left\langle \frac{i}{\hbar}\Big(HV(x)-V(x)H\Big)+\ppdiff{V}{t}\right\rangle
\tag{4}
\end{equation}
ただし \(V\) が時間依存していない, 即ち関数 \(V\) に時間 \(t\) が含まれていないならば \(\displaystyle{\ppdiff{V}{t}=0}\) であるから, その場合には式 (2) に於いて \(\displaystyle{\ppdiff{V}{t}=0}\) とすれば, 問題文の式 (7-89) となる：
\begin{equation}
\left\langle\,\chi\left|\frac{d V}{dt}\,\right|\psi\right\rangle
=\frac{i}{\hbar}\int \chi^{*}(x)\,\bigg(HV(x)-V(x)H\bigg)\,\psi(x)\,dx
\tag{5}
\end{equation}
\left\langle\,\chi\left|\frac{d V}{dt}\,\right|\psi\right\rangle
=\frac{i}{\hbar}\int \chi^{*}(x)\,\bigg(HV(x)-V(x)H\bigg)\,\psi(x)\,dx
\tag{5}
\end{equation}
