Find the l = 0 energy and wave function of a particle of mass m that is subject to the following central potential V(r)=\begin{cases} 0, & a<r<b, \\ \infty , & elsewhere.\end{cases}
Find the l = 0 energy and wave function of a particle of mass m that is subject to the following central potential V(r)=\begin{cases} 0, & a<r<b, \\ \infty , & elsewhere.\end{cases}
This particle moves between two concentric, hard spheres of radii r = a and r = b. The l = 0 radial equation between a < r < b can be obtained from (6.57):
-\frac{\hbar ^{2} }{2M} \frac{d^{2}U_{nl}(r) }{dr^{2} } +V_{eff} (r)U_{nl}(r)=E_{n} U_{nl}(r) (6.57)
\frac{d^{2} U_{n0}(r) }{dr^{2} } +k^{2} U_{n0}(r)=0, (6.252)
where U_{n0}(r)=rR_{n0}(r) and k^{2}=2mE/\hbar ^{2} . Since the solutions of this equation must satisfy the condition U_{n0}(a)=0, we may write
U_{n0}(r)=A\sin [k(r-a)]; (6.253)
the radial wave function is zero elsewhere, i.e.,U_{n0}(r)=0 for 0 < r < a and r > b.
Moreover, since the radial function must vanish at r = b, U_{n0}(b)=0, we have
A\sin [k(b-a)]=0\Longrightarrow k(b-a)=n\pi, n=1, 2, 3, …. (6.254)
Coupled with the fact that k^{2}=2mE/(\hbar ^{2}), this condition leads to the energy
E_{n} =\frac{\hbar ^{2}k^{2}}{2m} =\frac{\pi ^{2} \hbar ^{2}}{2m(a-b)^{2} } , n=1, 2, 3, …. (6.255)
We can normalize the radial function (6.253) to obtain the constant A:
1=∫^{b}_{a} r^{2}R^{2}_{n0} (r)dr=∫^{b}_{a}U^{2}_{n0}(r)dr=A^{2} ∫^{b}_{a}\sin ^{2} [k(r-a)]dr
=\frac{A^{2}}{2} ∫^{b}_{a}\left\{1-\cos [2k(r-a)]\right\} dr=\frac{b-a}{2} A^{2}; (6.256)
hence A=\sqrt{2/(b-a)}. Since k_{n} =n\pi /(b-a) the normalized radial function is given by
R_{n0}(r)=\frac{1}{r} U_{n0}(r)=\begin{cases} \sqrt{\frac{2}{b-a} } \frac{1}{r}\sin \left[\frac{n\pi (r-a)}{b-a} \right] , & a<r<b, \\ 0, & elsewhere.\end{cases} (6.257)
To obtain the total wave function \psi _{nlm} (\vec{r}), we need simply to divide the radial function by a 1/\sqrt{4\pi } factor, because in this case of l = 0 the wave function \psi _{n00} (r) depends on no angular degrees of freedom, it depends only on the radius:
\psi _{n00} (r)=\frac{1}{\sqrt{4\pi } } R_{n0}(r)=\begin{cases} \sqrt{\frac{2}{4\pi (b-a)} } \frac{1}{r}\sin \left[\frac{n\pi (r-a)}{b-a} \right] , & a<r<b, \\ 0, & elsewhere.\end{cases} (6.258)