(a) First, to obtain 〈nl|r^{-1}|nl 〉, we need simply to insert k = 0 into Kramers’ recursion rule (6.184):
\frac{k+1}{n^{2} } 〈nl|r^{k}|nl 〉-(2k+1)a_{0} 〈nl|r^{k-1}|nl 〉+\frac{ka^{2}_{0} }{4} \left[(2l+1)^{2}-k^{2} \right] 〈nl|r^{k-2}|nl 〉=0. (6.184)
\frac{1}{n^{2}} \left\langle nl|r^{0}|nl\right\rangle -a_{0}\left \langle nl|r^{-1}|nl\right\rangle=0 (6.230)
hence
\left\langle nl|\frac{1}{r}|nl \right\rangle =\frac{1}{n^{2}a_{0}}. (6.231)
Second, an insertion of k = 1 into (6.184) leads to the relation for 〈nl|r|nl 〉:
\frac{2}{n^{2}} 〈nl|r|nl 〉-3a_{0}\left\langle nl|r^{0}|nl \right\rangle+\frac{a^{2}_{0} }{4} \left[(2l+1)^{2}-1 \right] \left\langle nl|r^{-1}|nl\right\rangle=0, (6.232)
and since \left\langle nl|r^{-1}|nl\right\rangle=1/(n^{2}a_{0}), we have
〈nl|r|nl 〉=\frac{1}{2} \left[3n^{2}-l(l+1) \right]a_{0}. (6.233)
Third, substituting k = 2 into (6.184) we get
\frac{3}{n^{2}} \left\langle nl|r^{2}|nl\right\rangle-5a_{0} \left\langle nl|r|nl\right\rangle+\frac{a^{2}_{0} }{4} \left[(2l+1)^{2} -4 \right] \left\langle nl|r^{0}|nl\right\rangle=0, (6.234)
which when combined with 〈nl|r|nl 〉=\frac{1}{2} \left[3n^{2}-l(l+1) \right]a_{0} yields
\left\langle nl|r^{2}|nl\right\rangle=\frac{1}{2}n^{2}\left [5n^{2}+1-3l(l+1) \right]a^{2}_{0}. (6.235)
We can continue in this way to obtain any positive power of r: 〈nl|r^{k}|nl 〉.
(b) Inserting k = -1 into Kramers’ rule,
0+a_{0}\left\langle nl|r^{-2}|nl\right\rangle-\frac{1}{4} \left[(2l+1)^{2}-1 \right]a_{0}\left\langle nl|r^{-3}|nl\right\rangle, (6.236)
we obtain
\left\langle nl|\frac{1}{r^{3} }|nl\right\rangle=\frac{1}{l(l+1) a_{0}} \left\langle nl|\frac{1}{r^{2} }|nl\right\rangle, (6.237)
where the expression for 〈nl|r^{-2}|nl 〉 is given by (6.225);
\left\langle nl|\frac{1}{r^{2} }|nl \right\rangle =\frac{2}{n^{3}(2l+1)a^{2}_{0} }, (6.225)
thus, we have
\left\langle nl|\frac{1}{r^{3} }|nl\right\rangle=\frac{2}{n^{3}l(l+1)(2l+1)a^{3}_{0}}. (6.238)
(c) To obtain the expression for 〈nl|r^{-4}|nl 〉 we need to substitute k = -2 into Kramers’ rule:
-\frac{1}{n^{2}}\left\langle nl|r^{-2}|nl\right\rangle +3a_{0} \left\langle nl|r^{-3}|nl\right\rangle-\frac{a^{2}_{0}}{2} \left [(2l+1)^{2}-4 \right]\left\langle nl|r^{-4}|nl\right\rangle=0. (6.239)
Inserting (6.225) and (6.238) for 〈nl|r^{-2}|nl 〉 and 〈nl|r^{-3}|nl 〉, we obtain
\left\langle nl|\frac{1}{r^{4} }|nl\right\rangle=\frac {4\left [3n^{2}-l(l+1) \right] }{n^{5}l(l+1)(2l+1)\left[(2l+1)^{2}-4 \right] a^{4}_{0}} . (6.240)
We can continue in this way to obtain any negative power of r: 〈nl|r^{-k}|nl 〉.