Page 195 - PRINCIPLES OF QUANTUM MECHANICS as Applied to Chemistry and Chemical Physics
P. 195
186 The hydrogen atom
For k 0, equation (6.70) gives
Z
ÿ1
hr i nl (6:71)
2
n a 0
For k 1, equation (6.70) gives
2 3a 0 a 2 0 ÿ1
hri nl ÿ l(l 1) hr i nl 0
n 2 Z Z 2
or
a 0 2
hri nl [3n ÿ l(l 1)] (6:72)
2Z
For k 2, equation (6.70) gives
3 2 5a 0 a 2
3
hr i nl ÿ hri nl 2[l(l 1) ÿ ] 0 0
n 2 Z 4 Z 2
or
2 2
n a
2 0 2
hr i nl [5n ÿ 3l(l 1) 1] (6:73)
2Z 2
4
3
For higher values of k, equation (6.70) leads to hr i nl , hr i nl , ...
ÿ3 ÿ2
For k ÿ1, equation (6.70) relates hr i nl to hr i nl
Z
ÿ3 ÿ2
hr i nl hr i nl (6:74)
l(l 1)a 0
ÿ4
ÿ5
For k ÿ2, ÿ3, ... , equation (6.70) gives successively hr i nl , hr i nl , ...
ÿ2
expressed in terms of hr i nl .
ÿ2
Although the expectation value hr i nl cannot be obtained from equation
(6.70), it can be evaluated by regarding the azimuthal quantum number l as the
parameter in the Hellmann±Feynman theorem (equation (3.71)). Thus, we
have
^
@E n @ H l
(6:75)
@l @l
^
where the Hamiltonian operator H l is given by equation (6.18) and the energy
^
levels E n by equation (6.57). The derivative @ H l =@l is just
^ 2
@ H l "
(2l 1) (6:76)
@l 2ìr 2
In the derivation of (6.57), the quantum number n is shown to be the value of l
plus a positive integer. Accordingly, we have @n=@l 1 and
2
2 2
2
2
2
Z e9 @ Z e9 @n @ Z "
@E n ÿ2 ÿ2 ÿ3
ÿ n ÿ n n (6:77)
@l 2a 0 @l 2a 0 @l @n ìa 2
0
2
2
2
2
where a ì " =ìe9 has been replaced by a 0 " =m e e9 . Substitution of
equations (6.76) and (6.77) into (6.75) gives the desired result