\(\)
Problem 6-7
Suppose the potential energy \(V(\mathbf{r})=-e\phi(\mathbf{r})\) is the result of a charge distribution \(\rho(\mathbf{r})\) so that
\begin{equation}
\def\mb#1{\mathbf{#1}}
\nabla^{2} V(\mb{r}) = 4\pi e \rho(\mb{r})\quad\rightarrow\quad
\nabla^{2}\phi(\mb{r})=-4\pi\rho(\mb{r})
\tag{6-48}
\end{equation}
\def\mb#1{\mathbf{#1}}
\nabla^{2} V(\mb{r}) = 4\pi e \rho(\mb{r})\quad\rightarrow\quad
\nabla^{2}\phi(\mb{r})=-4\pi\rho(\mb{r})
\tag{6-48}
\end{equation}
By assuming that \(\rho(\mb{r})\) goes to \(0\) as \(|\mb{r}|\to \infty\), multiplying Eq. (6-48) by \(\exp(i\mb{q}\cdot\mb{r}/\hbar)\), and integrating twice over \(\mb{r}\), show that \(v(\mb{q})\) can be expressed in terms of \(\rho(\mb{r})\) as
\begin{equation}
v(\mb{q})=-\frac{4\pi \hbar^{2} e}{q^{2}} \int e^{i\mb{q}\cdot\mb{r}/\hbar} \rho(\mb{r})\,d^{3}\mb{r}
\tag{6-49}
\end{equation}
v(\mb{q})=-\frac{4\pi \hbar^{2} e}{q^{2}} \int e^{i\mb{q}\cdot\mb{r}/\hbar} \rho(\mb{r})\,d^{3}\mb{r}
\tag{6-49}
\end{equation}
( 解答 ) まず, 式 (6-48) が成り立つことを Maxwell 方程式から確認しておこう．
\begin{align}
\def\mr#1{\mathrm{#1}}
&\mb{E}=-\nabla \phi(\mb{r}),\quad \text{div}\mb{E}=\nabla\cdot\mb{E}=4\pi\rho,\\
&\rightarrow\ \nabla\cdot(-\nabla\phi)=-\nabla^{2}\phi=4\pi\rho \qquad
\mr{therefore}\quad \nabla^{2}\phi=-4\pi\rho
\tag{1}
\end{align}
\def\mr#1{\mathrm{#1}}
&\mb{E}=-\nabla \phi(\mb{r}),\quad \text{div}\mb{E}=\nabla\cdot\mb{E}=4\pi\rho,\\
&\rightarrow\ \nabla\cdot(-\nabla\phi)=-\nabla^{2}\phi=4\pi\rho \qquad
\mr{therefore}\quad \nabla^{2}\phi=-4\pi\rho
\tag{1}
\end{align}
題意に従ってまず一回目の部分積分を行うと,
\begin{align}
I&=\int_{-\infty}^{\infty}\nabla^{2}\,V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}\\
&=\biggl[\nabla V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\biggr]_{-\infty}^{\infty}
-\frac{i\mb{q}}{\hbar}\int_{-\infty}^{\infty}\nabla V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{2}
\end{align}
I&=\int_{-\infty}^{\infty}\nabla^{2}\,V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}\\
&=\biggl[\nabla V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\biggr]_{-\infty}^{\infty}
-\frac{i\mb{q}}{\hbar}\int_{-\infty}^{\infty}\nabla V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{2}
\end{align}
仮定から, 無限遠では電荷分布 \(\rho=0\) なので, 無限遠においては, ポテンシャルエネルギー \(V(\mb{r})\) も電場
\(\mb{E}=-(1/e)\nabla V(\mb{r})\) も共にゼロと考えられる．従って上式の第1項は無視する．残りの部分に, 更に二回目の部分積分を行うと,
\begin{align}
I&=-\frac{i\mb{q}}{\hbar}\int_{-\infty}^{\infty}\nabla V(\mb{r})e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}\\
&=-\frac{i\mb{q}}{\hbar}\biggl[V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\biggr]_{-\infty}^{\infty}
+\left(\frac{i\mb{q}}{\hbar}\right)^{2}\int_{-\infty}^{\infty}V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{3}
\end{align}
I&=-\frac{i\mb{q}}{\hbar}\int_{-\infty}^{\infty}\nabla V(\mb{r})e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}\\
&=-\frac{i\mb{q}}{\hbar}\biggl[V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\biggr]_{-\infty}^{\infty}
+\left(\frac{i\mb{q}}{\hbar}\right)^{2}\int_{-\infty}^{\infty}V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{3}
\end{align}
同じ理由から第1項を無視すると,
\begin{equation}
I=-\frac{|\mb{q}|^{2}}{\hbar^{2}}\int_{-\infty}^{\infty}V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{4}
\end{equation}
I=-\frac{|\mb{q}|^{2}}{\hbar^{2}}\int_{-\infty}^{\infty}V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{4}
\end{equation}
他方, 式 (6-48) を用いた表現では,
\begin{equation}
I=\int_{-\infty}^{\infty}\nabla^{2}\,V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
=\int_{-\infty}^{\infty}4\pi e\,\rho(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{5}
\end{equation}
I=\int_{-\infty}^{\infty}\nabla^{2}\,V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
=\int_{-\infty}^{\infty}4\pi e\,\rho(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{5}
\end{equation}
よって, 式 (4) と式 (5) の右辺同士を等しいと置くならば，
\begin{align}
&-\frac{q^{2}}{\hbar^{2}}\int_{-\infty}^{\infty}V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}
\,d^{3}\mb{r}=4\pi e\int_{-\infty}^{\infty}\rho(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r},\\
&\rightarrow\quad \int_{-\infty}^{\infty}V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
=-\frac{4\pi\hbar^{2}e}{q^{2}}\int_{-\infty}^{\infty}\rho(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{6}
\end{align}
&-\frac{q^{2}}{\hbar^{2}}\int_{-\infty}^{\infty}V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}
\,d^{3}\mb{r}=4\pi e\int_{-\infty}^{\infty}\rho(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r},\\
&\rightarrow\quad \int_{-\infty}^{\infty}V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
=-\frac{4\pi\hbar^{2}e}{q^{2}}\int_{-\infty}^{\infty}\rho(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{6}
\end{align}
従って, \(v(\mb{q})\) の定義式 (6-39) から
\begin{align}
v(\mb{q})&\equiv \int_{-\infty}^{\infty}V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}\\
&=-\frac{4\pi\hbar^{2}e}{q^{2}}\int_{-\infty}^{\infty}\rho(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{7}
\end{align}
v(\mb{q})&\equiv \int_{-\infty}^{\infty}V(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}\\
&=-\frac{4\pi\hbar^{2}e}{q^{2}}\int_{-\infty}^{\infty}\rho(\mb{r})\,e^{i\mb{q}\cdot\mb{r}/\hbar}\,d^{3}\mb{r}
\tag{7}
\end{align}
