POINT
- ベッセル関数の関係式.
積分表示
積分表示
- Besselの積分表示:
$J_n(x)$$\displaystyle =\frac{1}{2\pi}\int_{0}^{2\pi} \cos\bigl(x\sin\theta-n\theta\bigr)\,\mathrm{d}\theta$ - Hansenの積分表示:
$J_n(x)$$\displaystyle =\frac{1}{\pi i^n}\int_{0}^{\pi} e^{ix\cos\theta}\cos(n\theta) \,\mathrm{d}\theta$
ここでは,Besselの積分表示
\begin{aligned}
J_n(x)
&=\frac{1}{2\pi}\int_{0}^{2\pi} \cos\bigl(x\sin\theta-n\theta\bigr) \,\mathrm{d}\theta
\end{aligned}
を出発点とする.ここで$f(\theta)=e^{i(x\sin\theta-n\theta)}$とするとJ_n(x)
&=\frac{1}{2\pi}\int_{0}^{2\pi} \cos\bigl(x\sin\theta-n\theta\bigr) \,\mathrm{d}\theta
\end{aligned}
\begin{aligned}
&=\frac{1}{2\pi}\int_{0}^{2\pi} \frac{f(\theta)+f(-\theta)}{2} \,\mathrm{d}\theta
\end{aligned}
と表せる.ここで,&=\frac{1}{2\pi}\int_{0}^{2\pi} \frac{f(\theta)+f(-\theta)}{2} \,\mathrm{d}\theta
\end{aligned}
\begin{aligned}
\int_{0}^{2\pi} f(-\theta) \,\mathrm{d}\theta
&= - \int_{0}^{-2\pi} f(\theta) \,\mathrm{d}\theta
= \int_{-2\pi}^{0} f(\theta) \,\mathrm{d}\theta \\
&= \int_{0}^{2\pi} f(\theta - 2\pi) \,\mathrm{d}\theta \\
&= \int_{0}^{2\pi} f(\theta) \,\mathrm{d}\theta
\end{aligned}
だから\int_{0}^{2\pi} f(-\theta) \,\mathrm{d}\theta
&= - \int_{0}^{-2\pi} f(\theta) \,\mathrm{d}\theta
= \int_{-2\pi}^{0} f(\theta) \,\mathrm{d}\theta \\
&= \int_{0}^{2\pi} f(\theta - 2\pi) \,\mathrm{d}\theta \\
&= \int_{0}^{2\pi} f(\theta) \,\mathrm{d}\theta
\end{aligned}
\begin{aligned}
J_n(x)
&=\frac{1}{2\pi}\int_{0}^{2\pi} f(\theta) \,\mathrm{d}\theta
\end{aligned}
となる.つまりJ_n(x)
&=\frac{1}{2\pi}\int_{0}^{2\pi} f(\theta) \,\mathrm{d}\theta
\end{aligned}
\begin{aligned}
J_n(x)
&=\frac{1}{2\pi}\int_{0}^{2\pi} e^{i(x\sin\theta-n\theta)}\,\mathrm{d}\theta
\end{aligned}
J_n(x)
&=\frac{1}{2\pi}\int_{0}^{2\pi} e^{i(x\sin\theta-n\theta)}\,\mathrm{d}\theta
\end{aligned}
ここで,$\theta\rightarrow \theta - \pi/2$と変数変換すると
\begin{aligned}
J_n(x)
&=\frac{1}{2\pi}\int_{-\pi/2}^{3\pi/2} e^{i(x\cos\theta-n\theta)} e^{-in\pi/2} \,\mathrm{d}\theta \\
&=(-i)^n \frac{1}{2\pi}\int_{0}^{2\pi} e^{i(x\cos\theta-n\theta)} \,\mathrm{d}\theta
\end{aligned}
なのでJ_n(x)
&=\frac{1}{2\pi}\int_{-\pi/2}^{3\pi/2} e^{i(x\cos\theta-n\theta)} e^{-in\pi/2} \,\mathrm{d}\theta \\
&=(-i)^n \frac{1}{2\pi}\int_{0}^{2\pi} e^{i(x\cos\theta-n\theta)} \,\mathrm{d}\theta
\end{aligned}
\begin{aligned}
i^n J_n(x)
&=\frac{1}{2\pi}\int_{0}^{2\pi} e^{i(x\cos\theta-n\theta)} \,\mathrm{d}\theta
\end{aligned}
i^n J_n(x)
&=\frac{1}{2\pi}\int_{0}^{2\pi} e^{i(x\cos\theta-n\theta)} \,\mathrm{d}\theta
\end{aligned}
さらに,上式の$\displaystyle \int_{\pi}^{2\pi}$の区間の積分を$\theta\rightarrow -\theta+2\pi$と変数変換すると
\begin{aligned}
&\frac{1}{2\pi}\int_{\pi}^{2\pi} e^{i(x\cos\theta-n\theta)} \,\mathrm{d}\theta \\
&=-\frac{1}{2\pi}\int_{\pi}^{0} e^{i(x\cos\theta+n\theta)} e^{-i2n\pi}\,\mathrm{d}\theta \\
&=\frac{1}{2\pi}\int_{0}^{\pi} e^{i(x\cos\theta+n\theta)} \,\mathrm{d}\theta
\end{aligned}
なので&\frac{1}{2\pi}\int_{\pi}^{2\pi} e^{i(x\cos\theta-n\theta)} \,\mathrm{d}\theta \\
&=-\frac{1}{2\pi}\int_{\pi}^{0} e^{i(x\cos\theta+n\theta)} e^{-i2n\pi}\,\mathrm{d}\theta \\
&=\frac{1}{2\pi}\int_{0}^{\pi} e^{i(x\cos\theta+n\theta)} \,\mathrm{d}\theta
\end{aligned}
\begin{aligned}
i^n J_n(x)
&=\frac{1}{\pi}\int_{0}^{\pi} e^{ix\cos\theta}\cos(n\theta) \,\mathrm{d}\theta
\end{aligned}
i^n J_n(x)
&=\frac{1}{\pi}\int_{0}^{\pi} e^{ix\cos\theta}\cos(n\theta) \,\mathrm{d}\theta
\end{aligned}
平面波
平面波とベッセル関数
\begin{aligned}
e^{ikr\cos\theta}
&=\sum_{n=-\infty}^\infty i^n J_n(kr) e^{in\theta} \\
&=J_0(kr) + 2 \sum_{n=1}^\infty i^n J_n(kr)\cos(n\theta)
\end{aligned}
e^{ikr\cos\theta}
&=\sum_{n=-\infty}^\infty i^n J_n(kr) e^{in\theta} \\
&=J_0(kr) + 2 \sum_{n=1}^\infty i^n J_n(kr)\cos(n\theta)
\end{aligned}
$e^{ikz}=e^{ikr\cos\theta}$のフーリエ級数展開
\begin{aligned}
e^{ikr\cos\theta}
&=\sum_{n=-\infty}^\infty c_n e^{in\theta}
\end{aligned}
を考える.係数を計算するとe^{ikr\cos\theta}
&=\sum_{n=-\infty}^\infty c_n e^{in\theta}
\end{aligned}
\begin{aligned}
c_n
&=\frac{1}{2\pi} \int_{-\pi}^\pi e^{ikr\cos\theta} e^{-in\theta}\,\mathrm{d}\theta \\
&=i^n J_n(kr)
\end{aligned}
なのでc_n
&=\frac{1}{2\pi} \int_{-\pi}^\pi e^{ikr\cos\theta} e^{-in\theta}\,\mathrm{d}\theta \\
&=i^n J_n(kr)
\end{aligned}
\begin{aligned}
e^{ikr\cos\theta}
&=\sum_{n=-\infty}^\infty i^n J_n(kr) e^{in\theta}
\end{aligned}
となる.e^{ikr\cos\theta}
&=\sum_{n=-\infty}^\infty i^n J_n(kr) e^{in\theta}
\end{aligned}
また,$f(\theta)= e^{ikr\cos\theta}$は偶関数なので
\begin{aligned}
e^{ikr\cos\theta}&=\frac{f(\theta)+f(-\theta)}{2} \\
&=\sum_{n=-\infty}^\infty c_n \frac{e^{in\theta}+e^{-in\theta}}{2} \\
&=c_0 + 2\sum_{n=1}^\infty (c_n+c_{-n}) \cos(n\theta)
\end{aligned}
であり,e^{ikr\cos\theta}&=\frac{f(\theta)+f(-\theta)}{2} \\
&=\sum_{n=-\infty}^\infty c_n \frac{e^{in\theta}+e^{-in\theta}}{2} \\
&=c_0 + 2\sum_{n=1}^\infty (c_n+c_{-n}) \cos(n\theta)
\end{aligned}
\begin{aligned}
c_n+c_{-n}
&=\frac{1}{2\pi} \int_{-\pi}^\pi e^{ikr\cos\theta} \frac{e^{in\theta}+e^{-in\theta}}{2}\,\mathrm{d}\theta \\
&=\frac{1}{2\pi} \int_{-\pi}^\pi e^{ikr\cos\theta} \cos(n\theta)\,\mathrm{d}\theta \\
&=\frac{1}{\pi} \int_{0}^\pi e^{ikr\cos\theta} \cos(n\theta)\,\mathrm{d}\theta \\
&=i^n J_n(kr)
\end{aligned}
となる(Hansenの積分表示を使った).したがって,c_n+c_{-n}
&=\frac{1}{2\pi} \int_{-\pi}^\pi e^{ikr\cos\theta} \frac{e^{in\theta}+e^{-in\theta}}{2}\,\mathrm{d}\theta \\
&=\frac{1}{2\pi} \int_{-\pi}^\pi e^{ikr\cos\theta} \cos(n\theta)\,\mathrm{d}\theta \\
&=\frac{1}{\pi} \int_{0}^\pi e^{ikr\cos\theta} \cos(n\theta)\,\mathrm{d}\theta \\
&=i^n J_n(kr)
\end{aligned}
\begin{aligned}
e^{ikr\cos\theta}
&=J_0(kr) + 2\sum_{n=1}^\infty i^n J_n(kr)\cos(n\theta)
\end{aligned}
となる.e^{ikr\cos\theta}
&=J_0(kr) + 2\sum_{n=1}^\infty i^n J_n(kr)\cos(n\theta)
\end{aligned}
参考文献
- 第9章$\text{\sect} 2$円柱関数にBessel関数関連の公式がまとまっています.
- quantum mechanics - Plane wave expansion in cylindrical coordinates - Physics Stack Exchange