Question 6.P.12: Calculate 〈nl | P^4 | nl〉 in a stationary state | nl〉 of ...

To calculate 〈nl|\hat{P}^{4}|nl 〉 we may consider expressing \hat{P}^{4} in terms of the hydrogen’s Hamiltonian. Since \hat{H}=\hat{P}^{2}/(2m_{e} )-e^{2} /r we have \hat{P}^{2}=2m_{e}(\hat{H}+e^{2} /r) : hence

=(2m_{e} )^{2}\left[E^{2}_{n}+E_{n} \left\langle nl|\frac {e^{2}}{r}|nl\right\rangle +\left\langle nl|\frac{e^{2}}{r}|nl\right \rangle E_{n}+ \left\langle nl|\frac{e^{4}}{r^{2}}|nl\right\rangle \right] ,             (6.329)

where we have used the fact that |nl 〉 is an eigenstate of \hat{H} :\hat{H}|nl 〉=E_{n}|nl 〉 with E_{n}=-e^{2}/(2a_{0} n^{2} )=-13.6eV/n^{2}. The expectation values of 1/r and 1/r^{2} are given by (6.182) and (6.183), 〈nl |r^{-1}|nl 〉=1/(n^{2}a_{0}) and 〈nl|r^{-2}|nl 〉=2[n^{3}(2l+1)a^{2}_{0}];

〈nl|r^{-1}|nl 〉=\frac{1}{n^{2}a_{0}} ,                    (6.182)

〈nl|r^{-2}|nl 〉=\frac{2}{n^{3}(2l+1)a^{2}_{0} } ,                (6.183)

we can thus rewrite (6.329) as

=(2m_{e} E_{n})^{2}\left[1-4+\frac{8n}{2l+1} \right] ;                                 (6.330)

in deriving the last relation we have used E_{n}=-e^{2}/(2a_{0} n^{2}) . Now, since a_{0}=\hbar ^{2} /(m_{e}e^{2}) , the energy E_{n} becomes E_{n}=-e^{2} /(2a_{0}n^{2})=-m_{e}e^{4}/(2\hbar ^{2}n^{2}) which, when inserted into (6.330), leads to

〈nl|\hat{P}^{4}|nl 〉=\frac{m^{4}_{e} e^{8}}{\hbar ^{4}n^{4}} \left[\frac{8n}{2l+1} -3\right] .                                      (6.331)