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I was wondering if there is a known closed form solution for the zeros of the spherical Bessel functions. While doing a quantum assignment, I came across them as a solution for the spherical infinite potential well. However, I only read about them as just a sequence of numbers, but no generating function or closed form expression. Any suggestions would be great.


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Have you already had a look at Watson's Treatise (cfr.…)? – Pacciu Feb 3 '12 at 1:01
@Pacciu Thanks, I just picked up a copy, and made a list of some books from that thread ;) – kηives Feb 3 '12 at 14:07
up vote 4 down vote accepted

For $n=-1,0$, finding the roots of the spherical Bessel functions $j_n(x)$ and $y_n(x)$ is somewhat easy, since:

$$\begin{array}{ll} j_{-1}(x)&=&\frac{\cos\,x}{x}&\quad&y_{-1}(x)&=&\mathrm{sinc}(x)\\ j_0(x)&=&\mathrm{sinc}(x)&\quad&y_0(x)&=&-\frac{\cos\,x}{x}\\ \end{array}$$

where $\mathrm{sinc}(x)=\dfrac{\sin\,x}{x}$ is the sine cardinal. Solving for zeros of other orders results in rather complicated transcendental equations, which I doubt have closed-form solutions. However, you will want to see these DLMF entries for some more information that can help you in numerically determining the zeros (e.g., asymptotic expansions); approximations derived from formulae there can then be subsequently polished with Newton-Raphson or some other iterative method of choice.

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Cheat Code: spherical Bessel roots are their respective half-integer Bessel roots

According to Abramowitz, 1964, Ch9, pp 440,"Zeros and Their Asymptotic Expansions"

The zeros of $j_{n}(x)$ and $y_{n}(x)$ [i.e.: Spherical Bessel] are the same as $J_{n+1/2}(x)$ and $N_{n+1/2}(x) $ [i.e.: 'Normal' Bessel]

Mathematica Implementation

I used, for the spherical Bessel of the first kind:

nRoots = 4;
nBessel = 3;
SphBesselRoot[l_, k_] := N[BesselJZero[l + 1/2, k]];
Grid[Table[SphBesselRoot[l, i], {l, 0, nBessel}, {i, 1, nRoots}]]

This prints a table of the first four roots (columns) for the four spherical Bessel functions (rows) :

3.14159 6.28319 9.42478 12.5664
4.49341 7.72525 10.9041 14.0662
5.76346 9.09501 12.3229 15.5146
6.98793 10.4171 13.698  16.9236

Comparing to tables of the roots online, this table is in good agreement.

I found this method easier than root finding or asymptotic expressions.

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...and how exactly did you think were these computed internally by Mathematica, by the swish of a unicorn's tail? – J. M. Dec 10 '15 at 3:11
Nope, I didn't have enough money to add unicorns to my computer! I posted this because I could find lots of tables, but few generating functions / ways to compute the tables easily and quickly. – heisenBug Feb 19 at 23:35

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