Question 6.P.9: Find the number of s bound states for a particle of mass m m...

The l = 0 radial equation can be obtained from (6.57):

-\frac{\hbar ^{2} }{2M} \frac{d^{2} U_{nl} (r)}{dr^{2} } +V_{eff} U_{nl} (r)=E_{n} U_{nl} (r)                       (6.57)

\frac{d^{2} U_{n0} (r)}{dr^{2} }+\left[\frac{2mV_{0}}{\hbar ^{2}} \delta (r-a)-k^{2}\right] U_{n0} (r)=0,                    (6.277)

where U_{nl} (r)=U_{n0} (r)=rR_{n0} (r) and k^{2}=-2mE/\hbar ^{2}, since we are looking here at the bound states only, E < 0. The solutions of this equation are

U_{n0} (r)=\begin{cases} U_{n0_{1} } (r)=Ae^{kr}+Be^{-kr}, & 0<r<a, \\ U_{n0_{2} } (r)=Ce^{-kr}, & r>a.\end{cases}                             (6.278)

The energy eigenvalues can be obtained from the boundary conditions. As the wave function vanishes at r = 0, U_{n0} (0)=0, we have A + B = 0 or B = -A ; hence U_{n0_{1} } (r)=D\sinh kr :

U_{n0} (r)=D\sinh kr,                     0<r<a,                   (6.279)

with D = 2A. The continuity condition at r = a of U_{n0} (r), U_{n0_{1} } (a)=U_{n0_{2} } (a), leads to

D\sinh ka=Ce^{-ka}.                                                     (6.280)

To obtain the discontinuity condition for the first derivative of U_{n0} (r) at r = a, we need to integrate (6.277):

\underset{\varepsilon \rightarrow 0}{\lim } [U^{\prime }_{n 0_{2} } (a+\varepsilon )-U^{\prime }_{n0_{1} }(a-\varepsilon )]+ \frac{2mV_{0}}{\hbar ^{2}} U_{n0_{2} } (a)=0              (6.281)

or

-kCe^{-ka}-kD\cosh ka+\frac{2mV_{0}}{\hbar ^{2}}Ce^{-ka}=0.                  (6.282)

Taking Ce^{-ka}=D\sinh ka, as given by (6.280), and substituting it into (6.282), we get

-k\sinh ka-k\cosh ka+\frac{2mV_{0}}{\hbar ^{2}}\sinh ka=0;                (6.283)

hence

\gamma \coth \gamma =\frac{2mV_{0}}{\hbar ^{2}}a-\gamma ,                 (6.284)

where \gamma =ka.

The energy eigenvalues are given by the intersection of the curves f(\gamma )=\gamma \coth \gamma and g (\gamma )=2mV_{0}a/\hbar ^{2}-\gamma . As shown in Figure 6.6, if a<\hbar ^{2}/(2mV_{0}) then no bound state solution can exist, since the curves of f(\gamma ) and g (\gamma ) do not intersect. But if a>\hbar ^{2}/(2mV_{0})the curves intersect only once; hence there is one bound state. We can summarize these results as follows:

a<\frac{\hbar ^{2}}{2mV_{0}} \Longrightarrow no bound states ,                    (6.285)

a>\frac{\hbar ^{2}}{2mV_{0}} \Longrightarrow only one bound state.               (6.286)

The radial wave function is given by

R_{n0} (r)=\frac{1}{r} U_{n0} (r)=\begin{cases} (D/r)\sinh kr, & 0<r<a, \\ (C/r)e^{-kr}, & r>a.\end{cases}                       (6.287)

The normalization of this function yields

=D^{2}\left[\frac{1}{4k}\sinh 2ka-\frac{a}{2} \right] +\frac {C^{2}}{2k}e^{-2ka} .                                                        (6.288)

From (6.280) we have Ce^{-ka}=D\sinh ka, so we can rewrite this relation as

1=D^{2}\left[\frac{1}{4k}\sinh 2ka-\frac{a}{2} \right] +\frac {D^{2}}{2k} \sinh ^{2} ka=D^{2}\left[\frac{\sinh 2ka+2\sinh ^{2}ka}{4k} -\frac{a}{2} \right];      (6.289)

hence

D=\frac{2\sqrt{k} }{\sqrt{\sinh 2ka+2\sinh ^{2}ka-2ak} } .                  (6.290)

The normalized wave function is thus given by \psi _{nlm} (r)=\psi _{n00} (r)=(1/\sqrt{4\pi } )R _{n0} (r) or

\psi _{n00} (r)=\frac{\sqrt{k} }{\sqrt{\pi \sinh 2ka+2\pi \sinh ^{2}ka-2\pi ak} }\begin{cases} (1/r)\sinh (kr), & 0<r<a, \\ (1/r)\sinh (ka)e^{-k(r-a)}, & r>a.\end{cases}   (6.291)