\(V(r)=0\) for \(r<a\); \(\infty\) for \(r>a\). Take \(l=0\).
\(r>a\): \(u=0\). \(r<a\): \(u''+k^2u=0\), \(k=\sqrt{2mE}/\hbar\) \(\Rightarrow u=A\sin kr+B\cos kr\).
\(R=u/r\) finite at \(r=0\) \(\Rightarrow B=0\); \(R(a)=0\Rightarrow ka=n\pi\): \[E_{n00}=\frac{n^2\pi^2\hbar^2}{2ma^2},\qquad R_{n0}(r)=\sqrt{\frac2a}\,\frac{\sin(n\pi r/a)}{r}\;(r<a),\qquad n=1,2,3,\ldots\]
Normalization: \(|A|^2\int_0^a\sin^2\frac{n\pi r}{a}dr=1\Rightarrow A=\sqrt{2/a}\).
\(\psi_{n00}=R_{n0}Y_0^0=\dfrac{1}{\sqrt{2\pi a}}\,\dfrac1r\sin\dfrac{n\pi r}{a}\) for \(r<a\), zero otherwise.
Example 11.1 Ground state of the spherical well
For \(l=0\): (a) write the ground state and its energy; (b) plot \(R(r)\); (c) plot \(\rho(r)\).
Solution
(a) \(\psi_{100}=\dfrac{1}{\sqrt{2\pi a}}\,\dfrac{\sin(\pi r/a)}{r}\), \(E_{100}=\dfrac{\pi^2\hbar^2}{2ma^2}\).
(b) \(R(r)=\sqrt{\frac2a}\dfrac{\sin(\pi r/a)}{r}\): finite (\(=\sqrt{2/a}\,\pi/a\)) at \(r=0\), falling to zero at \(r=a\).
(c) \(\rho(r)=\dfrac2a\sin^2\dfrac{\pi r}{a}\): zero at both ends, peak at \(r=a/2\).
Example 11.2 Expectation values in the spherical well
For \((n,0,0)\): (a) \(P(0<r<a/2)\); (b) \(\left\langle r\right\rangle\); (c) \(\left\langle x\right\rangle,\left\langle y\right\rangle,\left\langle z\right\rangle\).
Solution
(a) \(P=\dfrac2a\displaystyle\int_0^{a/2}\sin^2\frac{n\pi r}{a}dr=\frac12\) (for every \(n\)).
(b) \(\left\langle r\right\rangle=\dfrac2a\displaystyle\int_0^ar\sin^2\frac{n\pi r}{a}dr=\frac a2\).
(c) \(\left\langle x\right\rangle\propto\int_0^{2\pi}\cos\phi\,d\phi=0\); likewise \(\left\langle y\right\rangle=\left\langle z\right\rangle=0\) (spherical symmetry).
