Page 339 - PRINCIPLES OF QUANTUM MECHANICS as Applied to Chemistry and Chemical Physics
P. 339

330                             Appendix H
                                              …              …            …
                                               1              1            1
                                                  í
                                                                              í
                                                 r uu9 dr ˆÿí   r íÿ1 2      r uu9 dr
                                                                   u dr ÿ
                                               0              0            0
                             Combining the integral on the left-hand side with the last one on the right-hand side,
                             we obtain the desired result
                                                      …
                                                       1            í
                                                         í
                                                        r uu9 dr ˆÿ hr íÿ1 i nl                 (H:5)
                                                       0            2
                               To obtain the recurrence relation (H.2), we multiply equation (H.4) by r k‡1 u9 and
                             integrate over r
                             …                     …                …             2  …
                              1                     1            2Z  1           Z   1
                                                                       k
                                r k‡1 u9u0 dr ˆ l(l ‡ 1)  r  kÿ1 uu9 dr ÿ  r uu9 dr ‡  2 2  r  k‡1 uu9 dr
                              0                     0            a 0  0         n a 0  0
                                              l(l ‡ 1)(k ÿ 1)  kÿ2   kZ  kÿ1    (k ‡ 1)Z 2  k
                                          ˆÿ               hr   i nl ‡  hr  i nl ÿ       hr i nl  (H:6)
                                                                                    2 2
                                                    2                a 0          2n a
                                                                                      0
                             where equation (H.5) was applied to the right-hand side. The integral on the left-hand
                             side of (H.6) may be integrated by parts twice to give
                               …                …
                                1                1   d
                                  r k‡1 u9u0 dr ˆÿ  u9  (r k‡1 u9)dr
                                0                0   dr
                                                      …            …
                                                       1            1
                                                          k
                                            ˆÿ(k ‡ 1)    r u9u9 dr ÿ  r k‡1 u9u0 dr
                                                       0            0
                                                    …                …
                                                     1   d            1
                                                             k
                                            ˆ (k ‡ 1)  u   (r u9)dr ÿ  r k‡1 u9u0 dr
                                                     0  dr            0
                                                      …                   …           …
                                                       1                   1           1
                                                                             k
                                            ˆ k(k ‡ 1)  r kÿ1 uu9 dr ‡ (k ‡ 1)  r uu0 dr ÿ  r  k‡1 u9u0 dr
                                                       0                   0           0
                             The integral on the left-hand side and the last integral on the right-hand side may be
                             combined to give
                                  …                                              …
                                   1                (k ÿ 1)k(k ‡ 1)        (k ‡ 1)  1
                                      k‡1                          kÿ2               k
                                     r   u9u0 dr ˆÿ              hr   i nl ‡        r uu0 dr    (H:7)
                                   0                      4                  2    0
                             where equation (H.5) has been used for the ®rst integral on the right-hand side.
                             Substitution of equation (H.4) for u0 in the last integral on the right-hand side of (H.7)
                             yields
                                 …
                                  1               (k ÿ 1)k(k ‡ 1)        (k ‡ 1)l(l ‡ 1)
                                   r k‡1 u9u0 dr ˆÿ             hr  kÿ2 i nl ‡        hr  kÿ2  i nl
                                  0                     4                      2
                                                  (k ‡ 1)Z  kÿ1    (k ‡ 1)Z 2  k
                                                ÿ         hr  i nl ‡   2 2  hr i nl             (H:8)
                                                     a 0             2n a 0
                             Combining equations (H.6) and (H.8), we obtain the recurrence relation (H.2).
   334   335   336   337   338   339   340   341   342   343   344