(a) The unperturbed Hamiltonian for a hydrogen atom is
where μ is the reduced mass. H is the extra energy of the nucleus and electron due to external field and is
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 nonvanishing matrix elements are those for unperturbed states that have opposite parities. In particular all diagonal elements of H 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 degenerate, the quantum numbers n, l and m have 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,
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 energy λ = ±e|E < 2, 0, 0|z|2, 1, 0 >
Thus, only one matrix element needs to be evaluated, using the unperturbed eigen functions explicitly. (see Table 3.2)
Thus, the linear Stark effect splits the degenerate m = 0 level into two components, with the shift
ΔE = ±3 ae |E|
The corresponding eigen functions are 1/√2 (ψs(0) ∓ ψp(0))
The two components being mixed in equal proportion (Fig. 3.27).
