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I quote Problem 12.4.7 of the 5th edition of Mathematical Methods for Physicists by Arfken, Weber, and Harris:

A plane wave may be expanded in a series of spherical waves by the Rayleigh equation: $$ e^{ikr\cos\gamma} = \sum_{n=0}^\infty a_n j_n(kr) P_n(\cos \gamma). $$ Show that $a_n = i^n(2n+1)$. Hint: Use the orthogonality of the $P_n$ to solve for $a_n j_n(kr)$. Differentiate $n$ times with respect to $kr$ and set $r=0$ to eliminate the $r$-dependence. The remaining integral is $$ \int_{-1}^1 x^n P_n(x) \mathrm dx = \frac{2^{n+1} (n!)^2}{(2n+1)!}, $$ as shown in a previous exercise.

Using the orthogonality condition for the Legendre polynomials (and substitution $x=\cos\gamma$), I obtained $$ \frac{2}{2n+1} a_n j_n(kr) = \int_{-1}^1e^{ixkr}P_n(x)\mathrm dx, $$ differentiating which $n$ times w.r.t. $kr$ gives $$ \frac{2}{2n+1} a_n j_n^{(n)}(kr) = \int_{-1}^1(ix)^n e^{ixkr}P_n(x)\mathrm dx. $$ My trouble comes from the $j_n^{(n)}(kr)$ bit. How does one differentiate the $n$th spherical Bessel function $n$ times? Or does setting $r=0$ automatically tell us its value?

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  • $\begingroup$ There must be a mistake in your work, because when you obtained your second last equation$$\frac{2}{2n+1}a_n j_n(kr)=\int_{-1}^1 \mathrm e^{\mathrm ixkr}P_n(x)\mathrm dx$$ Can't you just immediately set $r=0$ to get $$a_n=\frac{2n+1}{2}\frac{1}{j_n(0)}\int_{-1}^1 P_n(x)\mathrm dx$$ But that integral is zero.... $\endgroup$
    – K.defaoite
    Commented Dec 22, 2022 at 18:05
  • $\begingroup$ @K.defaoite I think the problem with your argument is that $j_n(0) = 0$ for all $n > 0$, as can be seen here: mathworld.wolfram.com/…. $\endgroup$
    – Chris Yang
    Commented Dec 22, 2022 at 18:16
  • $\begingroup$ Yes my mistake. $\endgroup$
    – K.defaoite
    Commented Dec 22, 2022 at 18:17

2 Answers 2

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Note that the spherical Bessel functions admit the integral representation $$j_n(z)=\frac{(-\mathrm i)^n}{2}\int_0^\pi \mathrm e^{\mathrm iz\cos\theta}P_n(\cos\theta)\sin\theta~\mathrm d\theta$$ Which, by the Leibniz rule, means $$j_n^{(n)}(z)=\frac{1}{2}\int_0^\pi \mathrm e^{\mathrm iz\cos\theta}\cos(\theta)^nP_n(\cos\theta)\sin\theta~\mathrm d\theta \\ j_n^{(n)}(0)=\frac{1}{2}\int_0^\pi \cos(\theta)^n P_n(\cos\theta)\sin\theta~\mathrm d\theta$$ Which, by a change of variable $\cos\theta=x$ yields $$j_n^{(n)}(0)=\frac{1}{2}\int_{-1}^1 x^nP_n(x)\mathrm dx$$

Which you already know!

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  • $\begingroup$ How did you go from $j^{(n)}_n$ to $j^{(0)}_n$? Where did $\sin \theta$ go? $\endgroup$
    – Gary
    Commented Dec 22, 2022 at 4:52
  • $\begingroup$ Thanks for the reply! @Gary I think it was a typo, and the correct expression is $j_n^{(n)}(0)$, just setting $z = 0$. The $\sin\theta$ is also supposed to be there, which allows the change of variable. $\endgroup$
    – Chris Yang
    Commented Dec 22, 2022 at 11:36
  • $\begingroup$ This certainly works, though I don't think the authors intended it to be solved this way, since a few problems later we are asked to derive this integral representation for $j_n(z)$ from the result of this plane wave problem... $\endgroup$
    – Chris Yang
    Commented Dec 22, 2022 at 11:48
  • $\begingroup$ Whoops, typo. Sorry. I'll fix. $\endgroup$
    – K.defaoite
    Commented Dec 22, 2022 at 16:42
  • $\begingroup$ @Gary thanks for catching those mistakes. Fixed now $\endgroup$
    – K.defaoite
    Commented Dec 22, 2022 at 16:45
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To expand on @K.defaoite's answer, we can instead use the first formula in the link they provided: $$ j_n(z) = \frac{z^n}{2^{n+1} n!}\int_0^\pi \cos(z\cos\theta) (\sin\theta)^{2n+1}\mathrm d \theta, $$ which does show up earlier in the text (Problem 11.7.8). Differentiating this $n$ times gives \begin{align} j_n^{(n)}(z) &= \frac{1}{2^{n+1} n!}\sum_{k=0}^n {n \choose k} (z^n)^{(k)}\left[\int_0^\pi \cos(z\cos\theta) (\sin\theta)^{2n+1}\mathrm d \theta\right]^{(n-k)} \\ &= \frac{1}{2^{n+1} n!}\sum_{k=0}^n{n \choose k} \frac{n!}{(n-k)!}z^{n-k} \int_0^\pi (\cos\theta)^{n-k}\cos(z\cos\theta) (\sin\theta)^{2n+1}\mathrm d \theta. \end{align} Setting $z=0$, all of the terms in the sum vanish except for the $k=n$ term, so $$ j_n^{(n)}(0) = \frac{1}{2^{n+1}} \int_0^\pi(\sin\theta)^{2n+1}\mathrm d \theta. $$ This integral can be solved by a reduction formula to obtain $$ j_n^{(n)}(0) = \frac{1}{2^{n+1}} \frac{2^{2n+1}(n!)^2}{(2n+1)!} = \frac{2^{n}(n!)^2}{(2n+1)!}, $$ which is the result of @K.defaoite's answer. This method also allows the formula for $\int_0^\pi x^nP_n(x)\mathrm dx$ to be used, instead of the integrals just cancelling out on both sides.

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  • $\begingroup$ Sure, but, I don't see why this is better. What definition for $j_n$ are you given in the text? $\endgroup$
    – K.defaoite
    Commented Dec 22, 2022 at 18:18
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    $\begingroup$ @K.defaoite In the text, general Bessel functions are introduced first, and spherical Bessel functions are defined as $$ j_n(x) = \sqrt{\frac{\pi}{2x}}J_{n+1/2}(x). $$ The formula I gave is derived in an exercise from the integral representation of $J_\nu$. In addition, the formula you used is derived in Problem 12.4.9 from the result of Problem 12.4.7, which is the problem I asked about. Also the $x^nP_n(x)$ integral cancels out on both sides if we use your formula, so the given formula is redundant. Once again, thanks for the answer! $\endgroup$
    – Chris Yang
    Commented Dec 22, 2022 at 18:23

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