3.3 Solutions 231
∂E
∂Z=0; this yieldsZ=27
16
E
(
27
16
)
=−
(
2 e^2
2 a 0)(
27
16
) 2
=−(2)(13.6)
(
27
16
) 2
=− 77 .45 eVThe estimated value of ionization energy 77.4 eV may be compared to the
74.4 eV derived by perturbation theory and an experimental value of 78.6 eV.
In general the variation method gives better accuracy then the perturbation
theory.3.101 (a) The unperturbed Hamiltonian for a hydrogen atom is
H 0 =−
^2
2 μ∇^2 −
e^2
r(1)
whereμis the reduced mass.
H′is the extra energy of the nucleus and electron due to external field
and isH′=−eEz=eErcosθ (2)where the polar axis is in the direction of positive z.
Now, the perturbation in (2) is an odd operator since it changes sign
when the coordinates are reflected through the origin. Thus, the only non-
vanishing matrix elements are those for unperturbed states that have oppo-
site parities. In particular all diagonal elements ofH′of hydrogen atom
wave functions are zero. This shows that a non-degenerate state, like the
ground state (n=1) has no first-order Stark effect.
(b) The first excited state (n =2) of hydrogen atom is fourfold degener-
ate, the quantum numbersn, landmhave the values (2,0,0), (2,1,1),
(2,1,0), (2,1,−1). The first one (2S) has even parity while the remaining
three(2P) states have odd parity of the degenerate states only| 2 , 0 , 0 >
and| 2 , 1 , 0 >are mixed by the perturbation, but| 2 , 1 , 1 >and (2, 1 ,− 1 >
are not and do not exhibit the Stark effect. It remains to solve the secular
equation,
∣
∣
∣
∣e|E|< 2 , 0 , 0 |z| 2 , 0 , 0 >−λ e|E|< 2 , 0 , 0 |z| 2 , 1 , 0 >
e|E|< 2 , 1 , 0 |z| 2 , 0 , 0 > e|E|< 2 , 1 , 0 |z| 2 , 1 , 0 >−λ∣
∣
∣
∣=^0
Because the conservation of parity the diagonal elements< 2 , 0 , 0 |z| 2 ,
0 , 0 >and< 2 , 1 , 0 |z| 2 , 1 , 0 >vanish. Hence the first-order change in
energyλ=±e|E< 2 , 0 , 0 |z| 2 , 1 , 0 >
Thus, only one matrix element needs to be evaluated, using the unper-
turbed eigen functions explicitly. (see Table 3.2)