(a) Check that A r j_{1}(k r) satisfies the radial equation with V(r)=0 and \ell=1 .
(b) Determine graphically the allowed energies for the infinite spherical well, when \ell=1 . Show that for large N, E_{N 1} \approx\left(\hbar^{2} \pi^{2} / 2 m a^{2}\right)(N+1 / 2)^{2} Hint: First show that j_{1}(x)=0 \Rightarrow x=\tan x . Plot x and \tan x on the same graph, and locate the points of intersection.
(a)
u=\operatorname{Arj}_{1}(k r)=A\left[\frac{\sin (k r)}{k^{2} r}-\frac{\cos (k r)}{k}\right]=\frac{A}{k}\left[\frac{\sin (k r)}{(k r)}-\cos (k r)\right] .
\frac{d u}{d r}=\frac{A}{k}\left[\frac{k^{2} r \cos (k r)-k \sin (k r)}{(k r)^{2}}+k \sin (k r)\right]=A\left[\frac{\cos (k r)}{k r}-\frac{\sin (k r)}{(k r)^{2}}+\sin (k r)\right] .
\frac{d^{2} u}{d r^{2}}=A\left[\frac{-k^{2} r \sin (k r)-k \cos (k r)}{(k r)^{2}}-\frac{k^{3} r^{2} \cos (k r)-2 k^{2} r \sin (k r)}{(k r)^{4}}+k \cos (k r)\right] .
=A k\left[-\frac{\sin (k r)}{(k r)}-\frac{\cos (k r)}{(k r)^{2}}-\frac{\cos (k r)}{(k r)^{2}}+2 \frac{\sin (k r)}{(k r)^{3}}+\cos (k r)\right] .
=A k\left[\left(1-\frac{2}{(k r)^{2}}\right) \cos (k r)+\left(\frac{2}{(k r)^{3}}-\frac{1}{(k r)}\right) \sin (k r)\right] .
With V = 0 and \ell=1, \text { Eq. } 4.37 \text { reads: } \frac{d^{2} u}{d r^{2}}-\frac{2}{r^{2}} u=-\frac{2 m E}{\hbar^{2}} u=-k^{2} u . In this case the left side is
-\frac{\hbar^{2}}{2 m} \frac{d^{2} u}{d r^{2}}+\left[V+\frac{\hbar^{2}}{2 m} \frac{\ell(\ell+1)}{r^{2}}\right] u=E u . (4.37).
A k\left[\left(1-\frac{2}{(k r)^{2}}\right) \cos (k r)+\left(\frac{2}{(k r)^{3}}-\frac{1}{(k r)}\right) \sin (k r)-\frac{2}{(k r)^{2}}\left(\frac{\sin (k r)}{(k r)}-\cos (k r)\right)\right] .
=A k\left[\cos (k r)-\frac{\sin (k r)}{k r}\right]=-k^{2} u . So this u does satisfy Eq. 4.37.
-\frac{\hbar^{2}}{2 m} \frac{d^{2} u}{d r^{2}}+\left[V+\frac{\hbar^{2}}{2 m} \frac{\ell(\ell+1)}{r^{2}}\right] u=E u . (4.37).
(b) Equation 4.48 j_{\ell}(k a)=0 (4.48).
\Rightarrow j_{1}(z)=0, \text { where } z=k a . \text { Thus } \frac{\sin z}{z^{2}}-\frac{\cos z}{z}=0 or \tan z = z . For high z (large n, if n = 1, 2, 3, . .. counts the allowed energies in increasing order), the intersections occur slightly below z=\left(n+\frac{1}{2}\right) \pi .
\therefore E=\frac{\hbar^{2} k^{2}}{2 m}=\frac{\hbar^{2} z^{2}}{2 m a^{2}}=\frac{\hbar^{2} \pi^{2}}{2 m a^{2}}\left(n+\frac{1}{2}\right)^{2} . QED
