(a) When l = 0 and -V_{0}<E<0 the radial equation (6.56),
-\frac{\hbar ^{2} }{2M} \frac{d^{2}U_{nl}(r) }{dr^{2} } +\left [V(r)+\frac{l(l+1)\hbar ^{2}}{2Mr^{2}} \right] U_{nl}(r)=E_{n} U_{nl}(r), (6.56)
-\frac{\hbar ^{2} }{2\mu } \frac{d^{2}U_{nl}(r) }{dr^{2} } +\left [\frac{l(l+1)\hbar ^{2}}{2\mu r^{2}} +V(r)\right] U_{nl}(r)=E_{n} U_{nl}(r), (6.310)
can be written inside the well, call it region (1), as
U^{\prime \prime }_{n} (r)_{1} +k^{2}_{1} U_{n}(r)_{1} =0, 0\leq r\leq a, (6.311)
and outside the well, call it region (2), as
U^{\prime \prime }_{n} (r)_{2} +k^{2}_{2} U_{n}(r)_{2} =0, r>a, (6.312)
where U^{\prime \prime }_{n} (r)=d^{2}U_{n}(r)/dr^{2}, U_{n}(r)_{1}=rR_{n}(r)_{1}, U_{n}(r)_{2}=rR_{n}(r)_{2}, k_{1} =\sqrt{2\mu (V_{0}+E)/\hbar ^{2}} and k_{2}= \sqrt {-2\mu E/\hbar ^{2}} . Since U_{n}(r)_{1} must vanish at r = 0, while U_{n}(r)_{2} has to be finite at r\longrightarrow \infty, the respective solutions of (6.311) and (6.312) are given by
U_{n}(r)_{1}=A\sin (k_{1}r), 0\leq r\leq a , (6.313)
U_{n}(r)_{2}=Be^{-k_{2}r}, r>a. (6.314)
The corresponding radial functions are
R_{n}(r)_{1}=A\frac{\sin (k_{1}r)}{r} , R_{n}(r)_{2}=B\frac{e^{-k_{2}r}}{r}. (6.315)
(b) Since the logarithmic derivative of the radial function is continuous at r = a, we can write
\frac{R^{\prime }_{n} (a)_{1} }{R_{n}(a)_{1}} = \frac {R^{\prime }_{n} (a)_{2} }{R_{n}(a)_{2}}. (6.316)
From (6.315) we have
\frac{R^{\prime }_{n} (a)_{1} }{R_{n}(a)_{1}} =k_{1}\cot (k_{1}a)-\frac{1}{a} , \frac{R^{\prime }_{n} (a)_{2} }{R_{n}(a)_{2}}=-k_{2}-\frac{1}{a}. (6.317)
Substituting (6.317) into (6.316) we obtain
-k_{1}\cot (k_{1}a)=k_{2} (6.318)
or
\sqrt{\frac{2\mu }{\hbar ^{2}}(V_{0}+E) } \cot \left[\sqrt {\frac{2\mu }{\hbar ^{2}}(V_{0}+E)} a\right] =-\sqrt{-\frac{2\mu E}{\hbar ^{2}} } , (6.319)
since k_{1}=\sqrt{2\mu (V_{0}+E)/ \hbar ^{2}} and k_{2}=\sqrt{-2\mu E/\hbar ^{2}} .
(c) In the limit E\rightarrow 0, the system has very few bound states; in this limit, equation (6.319) becomes
\sqrt{\frac{2\mu V_{0}}{\hbar ^{2}} } \cot \left(\sqrt{\frac {2\mu V_{0}}{\hbar ^{2}} } a\right) =0, (6.320)
which leads to a\sqrt{2\mu V_{0n}/\hbar ^{2}} =(2n+1)\pi /2; hence
V_{0n}=\frac{\pi ^{2} \hbar ^{2}}{8\mu a^{2} } (2n+1)^{2}, n=0, 1, 2, 3, …. (6.321)
Thus, the minimum values of V_{0} corresponding to one, two, and three bound states are respectively
V_{00}=\frac{\pi ^{2} \hbar ^{2}}{8\mu a^{2} }, V_{01}=\frac{9\pi ^{2} \hbar ^{2}}{8\mu a^{2} }, V_{02}=\frac{25\pi ^{2} \hbar ^{2}}{8\mu a^{2} } . (6.322)
(d) Using the notation \alpha =ak_{1} and \beta =ak_{2} we can, on the one hand, write
\alpha ^{2}+\beta ^{2} =\frac{2\mu a^{2} V_{0}}{\hbar ^{2}} , (6.323)
and, on the other hand, reduce the transcendental equation (6.318) to
-\alpha \cot \alpha =\beta , (6.324)
since k_{1}=\sqrt{2\mu (V_{0}+E)/ \hbar ^{2}} and k_{2}=\sqrt{-2\mu E/\hbar ^{2}} .
As shown in Figure 6.7, when \pi /2<\alpha <3\pi /2, which in the limit of E\rightarrow 0 leads to
\frac{\pi ^{2} \hbar ^{2}}{8\mu a^{2} }<V_{0}<\frac{9\pi ^{2} \hbar ^{2}}{8\mu a^{2} }, (6.325)
there exists only one bound state, since the circle intersects only once with the curve -\alpha \cot \alpha.
Similarly, there are two bound states if 3\pi /2<\alpha <5\pi /2 or
\frac{9\pi ^{2} \hbar ^{2}}{8\mu a^{2} }<V_{0}<\frac{25\pi ^{2} \hbar ^{2}}{8\mu a^{2} }, (6.326)
and three bound states if 5\pi /2<\alpha <7\pi /2:
\frac{25\pi ^{2} \hbar ^{2}}{8\mu a^{2} }<V_{0}<\frac{49\pi ^{2} \hbar ^{2}}{8\mu a^{2} }. (6.327)
(e) Since m_{p} c^{2} \backsimeq 938MeV and m_{n} c^{2} \backsimeq 940MeV, the reduced mass of the deuteron is given by \mu c^{2}=(m_{p} c^{2})(m_{n} c^{2}) /(m_{p} c^{2}+m_{n} c^{2}) \backsimeq 469.5MeV. Since a=2\times 10^{-15} m the minimum value of V_{0} corresponding to one bound state is
V_{0}=\frac{\pi ^{2} \hbar ^{2}}{8\mu a^{2} }=\frac{\pi ^{2}(\hbar c)^{2} }{8(\mu c^{2})a^{2}} =\frac{\pi ^{2}(197MeV fm)^{2}}{8(469.5MeV)(2\times 10^{-15} m)^{2}} \backsimeq 25.5MeV . (6.328)
