Question 6.P.2: Show how to obtain the expressions of: (a) 〈nl|r^-2|nl 〉 a...

The starting point is the radial equation (6.127),

-\frac{\hbar ^{2} }{2\mu } \frac{d^{2}U(r) }{dr^{2} } +\left[\frac{l(l+1)\hbar ^{2}}{2\mu r^{2}}-\frac{e^{2} }{r} \right] U(r)=EU(r).               (6.127)

-\frac{\hbar ^{2} }{2\mu } \frac{d^{2}U_{nl} (r) }{dr^{2} } +\left[\frac{l(l+1)\hbar ^{2}}{2\mu r^{2}}-\frac{e^{2} }{r} \right] U_{nl}(r)=E_{n}U_{nl}(r),             (6.218)

which can be rewritten as

\frac{U^{\prime \prime }_{nl}(r) }{U_{nl}(r)} =\frac{l(l+1)}{r^{2}} -\frac{2\mu e^{2}}{\hbar ^{2}} \frac{1}{r}+\frac{\mu ^{2} e^{4}}{\hbar ^{4}n^{2}},                  (6.219)

where U_{nl}(r)=rR_{nl} (r), U^{\prime \prime }_{nl}(r) =d^{2} U_{nl}(r)/dr^{2}, and E_{n} =-\mu e^{4} /(2\hbar ^{2}n^{2}).

(a) To find \left\langle r^{-2} \right\rangle _{nl}, let us treat the orbital quantum number l as a continuous variable and take the first l derivative of (6.219):

\frac{∂}{∂l} \left[\frac{U^{\prime \prime }_{nl}(r)}{U_{nl}(r)} \right] =\frac{2l+1}{r^{2}} -\frac{2\mu ^{2}e^{4} }{\hbar ^{4}n^{3}},                 (6.220)

where we have the fact that n depends on l since, as shown in (6.145), n = N + l + 1; thus ∂n / ∂l =1 . Now since ∫^{\infty }_{0} U^{2} _{nl} (r)dr=∫^{\infty }_{0} r^{2} R^{2}_{nl} (r)dr=1, multiplying both sides of (6.220) by U^{2}_{nl} (r) and integrating over r we get

∫^{\infty }_{0}U^{2}_{nl} (r) \frac{∂}{∂l} \left[\frac {U^{\prime \prime }_{nl}(r)}{U_{nl}(r)} \right]dr=(2l+1)∫^{\infty }_{0} U^{2}_{nl} (r)\frac{1}{r^{2}}dr -\frac{2\mu ^{2}e^{4}}{\hbar ^{4}n^{3}} ∫^{\infty }_{0} U^{2}_{nl} (r)dr,           (6.221)

or

∫^{\infty }_{0}U^{2}_{nl} (r) \frac{∂}{∂l} \left[\frac {U^{\prime \prime }_{nl}(r)}{U_{nl}(r)} \right]dr=(2l+1)\left\langle nl|\frac{1}{r^{2}}|nl \right\rangle -\frac{2\mu ^{2}e^{4}}{\hbar ^{4}n^{3}}.                  (6.222)

The left-hand side of this relation is equal to zero, since

∫^{\infty }_{0}U^{2}_{nl} (r) \frac{∂}{∂l} \left[\frac {U^{\prime \prime }_{nl}(r)}{U_{nl}(r)} \right]dr=∫^{\infty }_{0} U_{nl} (r)\frac{∂U^{\prime \prime }_{nl}(r)}{∂l} dr-∫^{\infty }_{0} U^{\prime \prime }_{nl} (r)\frac{∂U_{nl}(r)}{∂l} dr=0.  (6.223)

We may therefore rewrite (6.222) as

(2l+1)\left\langle nl|\frac{1}{r^{2} }|nl \right\rangle =\frac {2\mu ^{2}e^{4}}{\hbar ^{4}n^{3}};                          (6.224)

hence

\left\langle nl|\frac{1}{r^{2} }|nl \right\rangle=\frac{2}{n^{3}(2l+1)a^{2}_{0} },                            (6.225)

since a_{0} =\hbar ^{2} /(\mu e^{2} ).

(b) To find \left\langle r^{-1} \right\rangle _{nl} we need now to treat the electron’s charge e as a continuous variable in (6.219). The first e-derivative of (6.219) yields

\frac{∂}{∂l} \left[\frac{U^{\prime \prime }_{nl}(r)}{U_{nl}(r)} \right] =-\frac{4\mu e}{\hbar ^{2}} \frac{1}{r} +\frac{4\mu ^{2}e^{3}}{\hbar ^{4}n^{2}}.                       (6.226)

Again, since ∫^{\infty }_{0} U^{2}_{nl} (r)dr=1, multiplying both sides of (6.226) by U^{2}_{nl} (r) and integrating over r we obtain

∫^{\infty }_{0}U^{2}_{nl} (r) \frac{∂}{∂l} \left[\frac {U^{\prime \prime }_{nl}(r)}{U_{nl}(r)} \right]dr=-\frac{4\mu e}{\hbar ^{2}}∫^{\infty }_{0}U^{2}_{nl} (r)\frac{1}{r}dr +\frac{4\mu ^{2}e^{3}}{\hbar ^{4}n^{2}}∫^{\infty }_{0}U^{2}_{nl} (r)dr,       (6.227)

or

∫^{\infty }_{0}U^{2}_{nl} (r) \frac{∂}{∂l} \left[\frac {U^{\prime \prime }_{nl}(r)}{U_{nl}(r)} \right]dr=-\frac{4\mu e}{\hbar ^{2}}\left\langle nl|\frac{1}{r }|nl \right\rangle+\frac{4\mu ^{2}e^{3}}{\hbar ^{4}n^{2}}.                  (6.228)

As shown in (6.223), the left-hand side of this is equal to zero. Thus, we have

\frac{4\mu e}{\hbar ^{2}}\left\langle nl|\frac{1}{r }|nl \right \rangle=\frac{4\mu ^{2}e^{3}}{\hbar ^{4}n^{2}}\Longrightarrow \left\langle nl|\frac{1}{r }|nl \right\rangle=\frac{1}{n^{2}a_{0} },             (6.229)

since a_{0} =\hbar ^{2} /(\mu e^{2} ).